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top of the water. If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne? a. −8.60 m b. 8.60 m c. −51.4 m d. 51.4 m 15. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of above the pool. and her takeoff point is How long are her feet in the air? a. b. c. d. e. 1.28 s hose with a level area beyond (a) How would you use the garden hose, stopwatch, marble, measuring tape, and slope to measure displacement and elapsed time? Hint—The marble is the accelerating object, and the length of the hose is total displacement. (b) How would you use the displacement and time data to calculate velocity, average velocity, and acceleration? Which kinematic equations would you use? (c) How would you use the materials, the measured and calculated data, and the flat area below the slope to determine the negative acceleration? What would you measure, and which kinematic equation would you use? 112 Chapter 3 • Test Prep TEST PREP Multiple Choice 3.1 Acceleration 17. Which variable represents displacement? a. a b. d c. t d. v 18. If a velocity increases from 0 to 20 m/s in 10 s, what is the average acceleration? a. 0.5 m/s2 b. 2 m/s2 10 m/s2 c. 30 m/s2 d. 3.2 Representing Acceleration with Equations and Graphs 19. For the motion of a falling object, which graphs are Short Answer 3.1 Acceleration 21. True or False—The vector for a negative acceleration points in the opposite direction when compared to the vector for a positive acceleration. a. True b. False 22. If a car decelerates from to in , what is a. b. c. d. ? -5 m/s -1 m/s 1 m/s 5 m/s 23. How is the vector arrow representing an acceleration of magnitude 3 m/s2 different from the vector arrow representing a negative acceleration of magnitude 3 m/ s2? a. They point in the same direction. b. They are perpendicular, forming a 90° angle between each other. c. They point in opposite directions. d. They are perpendicular, forming a 270° angle between each other. 24. How long does it take to accelerate from 8.0 m/s to 20.0 m/s at a rate of acceleration of 3.0 m/s2? a. 0.25 s b. 4.0 s c. 9.33 s Access for free at openstax.org. straight lines? a. Acceleration versus time only b. Displacement versus time only c. Displacement versus time and acceleration versus time d. Velocity versus time and acceleration versus time 20. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.30×105 m/ s2 for 8.10×10−4 s. What is the bullet’s final velocity when it leaves the barrel, commonly known as the muzzle velocity? a. b. c. d. 7.79 m/s 51.0 m/s 510 m/s 1020 m/s d. 36 s 3.2 Representing Acceleration with Equations and Graphs 25. If a plot of displacement versus time is linear, what can be said about the acceleration? a. Acceleration is 0. b. Acceleration is a non-zero constant. c. Acceleration is positive. d. Acceleration is negative. 26. True or False: —The image shows a velocity vs. time graph for a jet car. If you take the slope at any point on the graph, the jet car’s acceleration will be 5.0 m/s2. a. True b. False 27. When plotted on the blank plots, which answer choice would show the motion of an object that has uniformly accelerated from 2 m/s to 8 m/s in 3 s? Chapter 3 • Test Prep 113 the plot on the right shows a line from (0,2) to (3,2). d. The plot on the left shows a line from (0,8) to (3,2) while the plot on the right shows a line from (0,3) to (3,3). 28. When is a plot of velocity versus time a straight line and a. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,2) to (3,2). b. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,3) to (3,3). c. The plot on the left shows a line from (0,8) to (3,2) while b. c. d. when is it a curved line? a. It is a straight line when acceleration is changing and is a curved line when acceleration is constant. It is a straight line when acceleration is constant and is a curved line when acceleration is changing. It is a straight line when velocity is constant and is a curved line when velocity is changing. It is a straight line when velocity is changing and is a curved line when velocity is constant. Extended Response 3.1 Acceleration 29. A test car carrying a crash test dummy accelerates from and then crashes into a brick wall. Describe to the direction of the initial acceleration vector and compare the initial acceleration vector’s magnitude with respect to the acceleration magnitude at the moment of the crash. a. The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash. b. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be less than the vector of the crash. c. The direction of the initial acceleration vector will point towards the wall, and its magnitude will be more than the acceleration vector of the crash. d. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be more than the acceleration vector of the crash. 30. A car accelerates from rest at a stop sign at a rate of 3.0 m/s2 to a speed of 21.0 m/s, and then immediately begins to decelerate to a stop at the next stop sign at a rate of 4.0 m/s2. How long did it take the car to travel from the first stop sign to the second stop sign? Show your work. a. b. c. d. 1.7 seconds 5.3 seconds 7.0 seconds 12 seconds 3.2 Representing Acceleration with Equations and Graphs 31. True or False: Consider an object moving with constant acceleration. The plot of displacement versus time for such motion is a curved line while the plot of displacement versus time squared is a straight line. a. True b. False 32. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down? a. 0.574 s b. 0.956 s 1.53 s c. 1.91 s d. 114 Chapter 3 • Test Prep Access for free at openstax.org. CHAPTER 4 Forces and Newton’s Laws of Motion Figure 4.1 Newton’s laws of motion describe the motion of the dolphin’s path. (Credit: Jin Jang) Chapter Outline 4.1 Force 4.2 Newton's First Law of Motion: Inertia 4.3 Newton's Second Law of Motion 4.4 Newton's Third Law of Motion Isaac Newton (1642–1727) was a natural philosopher; a great thinker who combined science and philosophy to INTRODUCTION try to explain the workings of nature on Earth and in the universe. His laws of motion were just one part of the monumental work that has made him legendary. The development of Newton’s laws marks the transition from the Renaissance period of history to the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. Drawing upon earlier work by scientists Galileo Galilei and Johannes Kepler, Newton’s laws of motion allowed motion on Earth and in space to be predicted mathematically. In this chapter you will learn about force as well as Newton’s first, second, and third laws of motion. 116 Chapter 4 • Forces and Newton’s Laws of Motion 4.1 Force Section Learning Objectives By the end of this section, you will be able to do the following: • Differentiate between force, net force, and dynamics • Draw a free-body diagram Section Key Terms dynamics external force force free-body diagram net external force net force Defining Force and Dynamics Force is the cause of motion, and motion draws our attention. Motion itself can be beautiful, such as a dolphin jumping out of the water, the flight of a bird, or the orbit of a satellite. The study of motion is called kinematics, but kinematics describes only the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the foundation of dynamics. These laws describe the way objects speed up, slow down, stay in motion, and interact with other objects. They are also universal laws: they apply everywhere on Earth as well as in space. A force pushes or pulls an object. The object being moved by a force could be an inanimate object, a table, or an animate object, a person. The pushing or pulling may be done by a person, or even the gravitational pull of Earth. Forces have different magnitudes and directions; this means that some forces are stronger than others and can act in different directions. For example, a cannon exerts a strong force on the cannonball that is launched into the air. In contrast, a mosquito landing on your arm exerts only a small force on your arm. When multiple forces act on an object, the forces combine. Adding together all of the forces acting on an object gives the total force, or net force. An external force is a force that acts on an object within the system from outsidethe system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force. We discuss further details about net force, external force, and net external force in the coming sections. In mathematical terms, two forces acting in opposite directions have opposite signs(positive or negative). By convention, the negative sign is assigned to any movement to the left or downward. If two forces pushing in opposite directions are added together, the larger force will be somewhat canceled out by the smaller force pushing in the opposite direction. It is important to be consistent with your chosen coordinate system within a problem; for example, if negative values are assigned to the downward direction for velocity, then distance, force, and a
cceleration should also be designated as being negative in the downward direction. Free-Body Diagrams and Examples of Forces For our first example of force, consider an object hanging from a rope. This example gives us the opportunity to introduce a useful tool known as a free-body diagram. A free-body diagram represents the object being acted upon—that is, the free body—as a single point. Only the forces acting onthe body (that is, external forces) are shown and are represented by vectors (which are drawn as arrows). These forces are the only ones shown because only external forces acting on the body affect its motion. We can ignore any internal forces within the body because they cancel each other out, as explained in the section on Newton’s third law of motion. Free-body diagrams are very useful for analyzing forces acting on an object. Access for free at openstax.org. 4.1 • Force 117 Figure 4.2 An object of mass, m, is held up by the force of tension. Figure 4.2 shows the force of tension in the rope acting in the upward direction, opposite the force of gravity. The forces are indicated in the free-body diagram by an arrow pointing up, representing tension, and another arrow pointing down, representing gravity. In a free-body diagram, the lengths of the arrows show the relative magnitude (or strength) of the forces. Because forces are vectors, they add just like other vectors. Notice that the two arrows have equal lengths in Figure 4.2, which means that the forces of tension and weight are of equal magnitude. Because these forces of equal magnitude act in opposite directions, they are perfectly balanced, so they add together to give a net force of zero. Not all forces are as noticeable as when you push or pull on an object. Some forces act without physical contact, such as the pull of a magnet (in the case of magnetic force) or the gravitational pull of Earth (in the case of gravitational force). In the next three sections discussing Newton’s laws of motion, we will learn about three specific types of forces: friction, the normal force, and the gravitational force. To analyze situations involving forces, we will create free-body diagrams to organize the framework of the mathematics for each individual situation. TIPS FOR SUCCESS Correctly drawing and labeling a free-body diagram is an important first step for solving a problem. It will help you visualize the problem and correctly apply the mathematics to solve the problem. Check Your Understanding 1. What is kinematics? a. Kinematics is the study of motion. b. Kinematics is the study of the cause of motion. c. Kinematics is the study of dimensions. d. Kinematics is the study of atomic structures. 2. Do two bodies have to be in physical contact to exert a force upon one another? 118 Chapter 4 • Forces and Newton’s Laws of Motion a. No, the gravitational force is a field force and does not require physical contact to exert a force. b. No, the gravitational force is a contact force and does not require physical contact to exert a force. c. Yes, the gravitational force is a field force and requires physical contact to exert a force. d. Yes, the gravitational force is a contact force and requires physical contact to exert a force. 3. What kind of physical quantity is force? a. Force is a scalar quantity. b. Force is a vector quantity. c. Force is both a vector quantity and a scalar quantity. d. Force is neither a vector nor a scalar quantity. 4. Which forces can be represented in a free-body diagram? a. Internal forces b. External forces c. Both internal and external forces d. A body that is not influenced by any force 4.2 Newton's First Law of Motion: Inertia Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s first law and friction, and • Discuss the relationship between mass and inertia. Section Key Terms friction inertia law of inertia mass Newton’s first law of motion system Newton’s First Law and Friction Newton’s first law of motion states the following: 1. A body at rest tends to remain at rest. 2. A body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. (Recall that constant velocitymeans that the body moves in a straight line and at a constant speed.) At first glance, this law may seem to contradict your everyday experience. You have probably noticed that a moving object will usually slow down and stop unless some effort is made to keep it moving. The key to understanding why, for example, a sliding box slows down (seemingly on its own) is to first understand that a net external force acts on the box to make the box slow down. Without this net external force, the box would continue to slide at a constant velocity (as stated in Newton’s first law of motion). What force acts on the box to slow it down? This force is called friction. Friction is an external force that acts opposite to the direction of motion (see Figure 4.3). Think of friction as a resistance to motion that slows things down. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it lifts the puck slightly, so the puck experiences very little friction as it moves over the surface. With friction almost eliminated, the puck glides along with very little change in speed. On a frictionless surface, the puck would experience no net external force (ignoring air resistance, which is also a form of friction). Additionally, if we know enough about friction, we can accurately predict how quickly objects will slow down. Now let’s think about another example. A man pushes a box across a floor at constant velocity by applying a force of +50 N. (The positive sign indicates that, by convention, the direction of motion is to the right.) What is the force of friction that opposes the motion? The force of friction must be −50 N. Why? According to Newton’s first law of motion, any object moving at constant velocity has no net external force acting upon it, which means that the sum of the forces acting on the object must be zero. The mathematical way to say that no net external force acts on an object is N of force, then the force of friction must be −50 N for the two forces to add up to zero (that is, for the two forces to canceleach So if the man applies +50 or Access for free at openstax.org. other). Whenever you encounter the phrase at constant velocity, Newton’s first law tells you that the net external force is zero. 4.2 • Newton's First Law of Motion: Inertia 119 Figure 4.3 For a box sliding across a floor, friction acts in the direction opposite to the velocity. The force of friction depends on two factors: the coefficient of friction and the normal force. For any two surfaces that are in contact with one another, the coefficient of friction is a constant that depends on the nature of the surfaces. The normal force is the force exerted by a surface that pushes on an object in response to gravity pulling the object down. In equation form, the force of friction is 4.1 where μis the coefficient of friction and N is the normal force. (The coefficient of friction is discussed in more detail in another chapter, and the normal force is discussed in more detail in the section Newton's Third Law of Motion.) Recall from the section on Force that a net external force acts from outside on the object of interest. A more precise definition is that it acts on the system of interest. A system is one or more objects that you choose to study. It is important to define the system at the beginning of a problem to figure out which forces are external and need to be considered, and which are internal and can be ignored. For example, in Figure 4.4 (a), two children push a third child in a wagon at a constant velocity. The system of interest is the wagon plus the small child, as shown in part (b) of the figure. The two children behind the wagon exert external forces on this system (F1, F2). Friction facting at the axles of the wheels and at the surface where the wheels touch the ground two other external forces acting on the system. Two more external forces act on the system: the weight W of the system pulling down and the normal force N of the ground pushing up. Notice that the wagon is not accelerating vertically, so Newton’s first law tells us that the normal force balances the weight. Because the wagon is moving forward at a constant velocity, the force of friction must have the same strength as the sum of the forces applied by the two children. 120 Chapter 4 • Forces and Newton’s Laws of Motion Figure 4.4 (a) The wagon and rider form a systemthat is acted on by external forces. (b) The two children pushing the wagon and child provide two external forces. Friction acting at the wheel axles and on the surface of the tires where they touch the ground provide an external force that act against the direction of motion. The weight W and the normal force N from the ground are two more external forces acting on the system. All external forces are represented in the figure by arrows. All of the external forces acting on the system add together, but because the wagon moves at a constant velocity, all of the forces must add up to zero. Mass and Inertia Inertia is the tendency for an object at rest to remain at rest, or for a moving object to remain in motion in a straight line with constant speed. This key property of objects was first described by Galileo. Later, Newton incorporated the concept of inertia into his first law, which is often referred to as the law of inertia. As we know from experience, some objects have more inertia than others. For example, changing the motion of a large truck is more difficult than changing the motion of a toy truck. In fact, the inertia of an object is proportional to the mass of the object. Mass is a measure of the amount of matter (or stuff) in an object. The quantity
or amount of matter in an object is determined by the number and types of atoms the object contains. Unlike weight (which changes if the gravitational force changes), mass does not depend on gravity. The mass of an object is the same on Earth, in orbit, or on the surface of the moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so mass is usually not determined this way. Instead, the mass of an object is determined by comparing it with the standard kilogram. Mass is therefore expressed in kilograms. TIPS FOR SUCCESS In everyday language, people often use the terms weightand massinterchangeably—but this is not correct. Weight is actually a force. (We cover this topic in more detail in the section Newton's Second Law of Motion.) WATCH PHYSICS Newton’s First Law of Motion This video contrasts the way we thought about motion and force in the time before Galileo’s concept of inertia and Newton’s first law of motion with the way we understand force and motion now. Click to view content (https://www.khanacademy.org/embed_video?v=5-ZFOhHQS68) Access for free at openstax.org. 4.2 • Newton's First Law of Motion: Inertia 121 GRASP CHECK Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects had a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force? a. Gravitational force b. Electrostatic force c. Nuclear force d. Frictional force Virtual Physics Forces and Motion—Basics In this simulation, you will first explore net force by placing blue people on the left side of a tug-of-war rope and red people on the right side of the rope (by clicking people and dragging them with your mouse). Experiment with changing the number and size of people on each side to see how it affects the outcome of the match and the net force. Hit the "Go!" button to start the match, and the “reset all” button to start over. Next, click on the Friction tab. Try selecting different objects for the person to push. Slide the applied forcebutton to the right to apply force to the right, and to the left to apply force to the left. The force will continue to be applied as long as you hold down the button. See the arrow representing friction change in magnitude and direction, depending on how much force you apply. Try increasing or decreasing the friction force to see how this change affects the motion. Click to view content (https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motionbasics_en.html) GRASP CHECK Click on the tab for the Acceleration Lab and check the Sum of Forcesoption. Push the box to the right and then release. Notice which direction the sum of forces arrow points after the person stops pushing the box and lets it continue moving to the right on its own. At this point, in which direction is the net force, the sum of forces, pointing? Why? a. The net force acts to the right because the applied external force acted to the right. b. The net force acts to the left because the applied external force acted to the left. c. The net force acts to the right because the frictional force acts to the right. d. The net force acts to the left because the frictional force acts to the left. Check Your Understanding 5. What does Newton’s first law state? a. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant acceleration unless acted on by a net external force. b. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. c. The rate of change of momentum of a body is directly proportional to the external force applied to the body. d. The rate of change of momentum of a body is inversely proportional to the external force applied to the body. 6. According to Newton’s first law, a body in motion tends to remain in motion at a constant velocity. However, when you slide an object across a surface, the object eventually slows down and stops. Why? a. The object experiences a frictional force exerted by the surface, which opposes its motion. b. The object experiences the gravitational force exerted by Earth, which opposes its motion c. The object experiences an internal force exerted by the body itself, which opposes its motion. d. The object experiences a pseudo-force from the body in motion, which opposes its motion. 122 Chapter 4 • Forces and Newton’s Laws of Motion 7. What is inertia? a. b. c. d. Inertia is an object’s tendency to maintain its mass. Inertia is an object’s tendency to remain at rest. Inertia is an object’s tendency to remain in motion Inertia is an object’s tendency to remain at rest or, if moving, to remain in motion. 8. What is mass? What does it depend on? a. Mass is the weight of an object, and it depends on the gravitational force acting on the object. b. Mass is the weight of an object, and it depends on the number and types of atoms in the object. c. Mass is the quantity of matter contained in an object, and it depends on the gravitational force acting on the object. d. Mass is the quantity of matter contained in an object, and it depends on the number and types of atoms in the object. 4.3 Newton's Second Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s second law, both verbally and mathematically • Use Newton’s second law to solve problems Section Key Terms freefall Newton’s second law of motion weight Describing Newton’s Second Law of Motion Newton’s first law considered bodies at rest or bodies in motion at a constant velocity. The other state of motion to consider is when an object is moving with a changing velocity, which means a change in the speed and/or the direction of motion. This type of motion is addressed by Newton’s second law of motion, which states how force causes changes in motion. Newton’s second law of motion is used to calculate what happens in situations involving forces and motion, and it shows the mathematical relationship between force, mass, and acceleration. Mathematically, the second law is most often written as 4.2 where Fnet (or ∑F) is the net external force, mis the mass of the system, and a is the acceleration. Note that Fnet and ∑F are the same because the net external force is the sum of all the external forces acting on the system. First, what do we mean by a change in motion? A change in motion is simply a change in velocity: the speed of an object can become slower or faster, the direction in which the object is moving can change, or both of these variables may change. A change in velocity means, by definition, that an acceleration has occurred. Newton’s first law says that only a nonzero net external force can cause a change in motion, so a net external force must cause an acceleration. Note that acceleration can refer to slowing down or to speeding up. Acceleration can also refer to a change in the direction of motion with no change in speed, because acceleration is the change in velocity divided by the time it takes for that change to occur, andvelocity is defined by speed and direction. From the equation we see that force is directly proportional to both mass and acceleration, which makes sense. To accelerate two objects from rest to the same velocity, you would expect more force to be required to accelerate the more massive object. Likewise, for two objects of the same mass, applying a greater force to one would accelerate it to a greater velocity. Now, let’s rearrange Newton’s second law to solve for acceleration. We get In this form, we can see that acceleration is directly proportional to force, which we write as 4.3 4.4 where the symbol means proportional to. This proportionality mathematically states what we just said in words: acceleration is directly proportional to the net external Access for free at openstax.org. 4.3 • Newton's Second Law of Motion 123 force. When two variables are directly proportional to each other, then if one variable doubles, the other variable must double. Likewise, if one variable is reduced by half, the other variable must also be reduced by half. In general, when one variable is multiplied by a number, the other variable is also multiplied by the same number. It seems reasonable that the acceleration of a system should be directly proportional to and in the same direction as the net external force acting on the system. An object experiences greater acceleration when acted on by a greater force. It is also clear from the equation that acceleration is inversely proportional to mass, which we write as 4.5 Inversely proportionalmeans that if one variable is multiplied by a number, the other variable must be dividedby the same number. Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. This relationship is illustrated in Figure 4.5, which shows that a given net external force applied to a basketball produces a much greater acceleration than when applied to a car. Figure 4.5 The same force exerted on systems of different masses produces different accelerations. (a) A boy pushes a basketball to make a pass. The effect of gravity on the ball is ignored. (b) The same boy pushing with identical force on a stalled car produces a far smaller acceleration (friction is negligible). Note that the free-body diagrams for the ball and for the car are identical, which allows us to compare the two situations. Applying Newton’s Second Law Before putting Newton’s second law into action, it is important to consider units. The equation units of force in terms of the three basic units of mass, length, and time (recall that acceleration has units of length divided by tim
e squared). The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s2. That is, because is used to define the we have One of the most important applications of Newton’s second law is to calculate weight (also known as the gravitational force), which is usually represented mathematically as W. When people talk about gravity, they don’t always realize that it is an acceleration. When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that the net external force acting on an object is responsible for the acceleration of the object. If air resistance is negligible, the net external force on a falling object is only the gravitational force (i.e., the weight of the object). Weight can be represented by a vector because it has a direction. Down is defined as the direction in which gravity pulls, so weight is normally considered a downward force. By using Newton’s second law, we can figure out the equation for weight. 4.6 Consider an object with mass mfalling toward Earth. It experiences only the force of gravity (i.e., the gravitational force or weight), which is represented by W. Newton’s second law states that the gravitational force, we have Substituting these two expressions into Newton’s second law gives We know that the acceleration of an object due to gravity is g, so we have Because the only force acting on the object is 124 Chapter 4 • Forces and Newton’s Laws of Motion This is the equation for weight—the gravitational force on a mass m. On Earth, now the direction of the weight) of a 1.0-kg object on Earth is 4.7 so the weight (disregarding for 4.8 Although most of the world uses newtons as the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Recall that although gravity acts downward, it can be assigned a positive or negative value, depending on what the positive direction is in your chosen coordinate system. Be sure to take this into consideration when solving problems with weight. When the downward direction is taken to be negative, as is often the case, acceleration due to gravity becomes g = −9.8 m/s2. When the net external force on an object is its weight, we say that it is in freefall. In this case, the only force acting on the object is the force of gravity. On the surface of Earth, when objects fall downward toward Earth, they are never truly in freefall because there is always some upward force due to air resistance that acts on the object (and there is also the buoyancy force of air, which is similar to the buoyancy force in water that keeps boats afloat). Gravity varies slightly over the surface of Earth, so the weight of an object depends very slightly on its location on Earth. Weight varies dramatically away from Earth’s surface. On the moon, for example, the acceleration due to gravity is only 1.67 m/s2. Because weight depends on the force of gravity, a 1.0-kg mass weighs 9.8 N on Earth and only about 1.7 N on the moon. It is important to remember that weight and mass are very different, although they are closely related. Mass is the quantity of matter (how much stuff) in an object and does not vary, but weight is the gravitational force on an object and is proportional to the force of gravity. It is easy to confuse the two, because our experience is confined to Earth, and the weight of an object is essentially the same no matter where you are on Earth. Adding to the confusion, the terms mass and weight are often used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct unit of newtons. Snap Lab Mass and Weight In this activity, you will use a scale to investigate mass and weight. • • 1 bathroom scale 1 table 1. What do bathroom scales measure? 2. When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. 3. The springs provide a measure of your weight (provided you are not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is now divided by 9.80 to give a reading in kilograms, which is a of mass. The scale detects weight but is calibrated to display mass. If you went to the moon and stood on your scale, would it detect the same massas it did on Earth? 4. GRASP CHECK While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? a. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. b. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight. c. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. d. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on Access for free at openstax.org. you that acts in the direction opposite to your weight. 4.3 • Newton's Second Law of Motion 125 TIPS FOR SUCCESS Only net external forceimpacts the acceleration of an object. If more than one force acts on an object and you calculate the acceleration by using only one of these forces, you will not get the correct acceleration for that object. WATCH PHYSICS Newton’s Second Law of Motion This video reviews Newton’s second law of motion and how net external force and acceleration relate to one another and to mass. It also covers units of force, mass, and acceleration, and reviews a worked-out example. Click to view content (https://www.khanacademy.org/embed_video?v=ou9YMWlJgkE) GRASP CHECK True or False—If you want to reduce the acceleration of an object to half its original value, then you would need to reduce the net external force by half. a. True b. False WORKED EXAMPLE What Acceleration Can a Person Produce when Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N parallel to the ground. The mass of the mower is 240 kg. What is its acceleration? Figure 4.6 Strategy Because Fnet and mare given, the acceleration can be calculated directly from Newton’s second law: Fnet = ma. Solution Solving Newton’s second law for the acceleration, we find that the magnitude of the acceleration, a, is given values for net external force and mass gives Entering the 4.9 126 Chapter 4 • Forces and Newton’s Laws of Motion Inserting the units for N yields 4.10 Discussion The acceleration is in the same direction as the net external force, which is parallel to the ground and to the right. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion, because we are given that the net external force is in the direction in which the person pushes. Also, the vertical forces must cancel if there is no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is reasonable for a person pushing a mower; the mower’s speed must increase by 0.21 m/s every second, which is possible. The time during which the mower accelerates would not be very long because the person’s top speed would soon be reached. At this point, the person could push a little less hard, because he only has to overcome friction. WORKED EXAMPLE What Rocket Thrust Accelerates This Sled? Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on humans at high accelerations. Rocket sleds consisted of a platform mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust, T, for the four-rocket propulsion system shown below. The sled’s initial acceleration is the mass of the system is 2,100 kg, and the force of friction opposing the motion is 650 N. Figure 4.7 Strategy The system of interest is the rocket sled. Although forces act vertically on the system, they must cancel because the system does not accelerate vertically. This leaves us with only horizontal forces to consider. We’ll assign the direction to the right as the positive direction. See the free-body diagram in Figure 4.8. Solution We start with Newton’s second law and look for ways to find the thrust T of the engines. Because all forces and acceleration are along a line, we need only consider the magnitudes of these quantities in the calculations. We begin with 4.11 is the net external force in the horizontal direction. We can see from Figure 4.8 that the engine thrusts are in the where same direction (which we call the positive direction), whereas friction opposes the thrust. In equation form, the net external force is Access for free at openstax.org. Newton’s second law tells us that Fnet= ma, so we get After a little algebra, we solve for the total thrust 4T: which means that the individual thrust is Inserting the known values yields 4.3 • Newton's Second Law of Motion 127 4.12 4.13 4.14 4.15 4.16 Discussion The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and to test the apparatus designed to protect fighter pilots in emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g. (Recall that g, the acceleration due to gravity, is of 45 g is ) Living subjects are no longer used, and land speeds of 10,000 km/h have now been obtained with rocket sleds. In this example, as in th
e preceding example, the system of interest is clear. We will see in later examples that choosing the system of interest is crucial—and that the choice is not always obvious. which is approximately An acceleration Practice Problems 9. If 1 N is equal to 0.225 lb, how many pounds is 5 N of force? a. 0.045 lb b. 1.125 lb c. 2.025 lb 5.000 lb d. 10. How much force needs to be applied to a 5-kg object for it to accelerate at 20 m/s2? a. b. c. d. 1 N 10 N 100 N 1,000 N Check Your Understanding 11. What is the mathematical statement for Newton’s second law of motion? a. F = ma b. F = 2ma c. d. F = ma2 12. Newton’s second law describes the relationship between which quantities? a. Force, mass, and time b. Force, mass, and displacement c. Force, mass, and velocity d. Force, mass, and acceleration 13. What is acceleration? a. Acceleration is the rate at which displacement changes. b. Acceleration is the rate at which force changes. c. Acceleration is the rate at which velocity changes. 128 Chapter 4 • Forces and Newton’s Laws of Motion d. Acceleration is the rate at which mass changes. 4.4 Newton's Third Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s third law, both verbally and mathematically • Use Newton’s third law to solve problems Section Key Terms Newton’s third law of motion normal force tension thrust Describing Newton’s Third Law of Motion If you have ever stubbed your toe, you have noticed that although your toe initiates the impact, the surface that you stub it on exerts a force back on your toe. Although the first thought that crosses your mind is probably “ouch, that hurt” rather than “this is a great example of Newton’s third law,” both statements are true. This is exactly what happens whenever one object exerts a force on another—each object experiences a force that is the same strength as the force acting on the other object but that acts in the opposite direction. Everyday experiences, such as stubbing a toe or throwing a ball, are all perfect examples of Newton’s third law in action. Newton’s third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts. Newton’s third law of motion tells us that forces always occur in pairs, and one object cannot exert a force on another without experiencing the same strength force in return. We sometimes refer to these force pairs as action-reactionpairs, where the force exerted is the action, and the force experienced in return is the reaction (although which is which depends on your point of view). Newton’s third law is useful for figuring out which forces are external to a system. Recall that identifying external forces is important when setting up a problem, because the external forces must be added together to find the net force. We can see Newton’s third law at work by looking at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.8. She pushes against the pool wall with her feet and accelerates in the direction opposite to her push. The wall has thus exerted on the swimmer a force of equal magnitude but in the direction opposite that of her push. You might think that two forces of equal magnitude but that act in opposite directions would cancel, but they do not because they act on different systems. In this case, there are two different systems that we could choose to investigate: the swimmer or the wall. If we choose the swimmer to be the system of interest, as in the figure, then motion. Because acceleration is in the same direction as the net external force, the swimmer moves in the direction of is an external force on the swimmer and affects her Because the swimmer is our system (or object of interest) and not the wall, we do not need to consider the force because it originates fromthe swimmer rather than acting onthe swimmer. Therefore, does not Note that the swimmer pushes in the direction directly affect the motion of the system and does not cancel opposite to the direction in which she wants to move. Access for free at openstax.org. 4.4 • Newton's Third Law of Motion 129 Figure 4.8 When the swimmer exerts a force on the wall, she accelerates in the direction opposite to that of her push. This means that the net external force on her is in the direction opposite to This opposition is the result of Newton’s third law of motion, which dictates that the wall exerts a force on the swimmer that is equal in magnitude but that acts in the direction opposite to the force that the swimmer exerts on the wall. Other examples of Newton’s third law are easy to find. As a teacher paces in front of a whiteboard, he exerts a force backward on the floor. The floor exerts a reaction force in the forward direction on the teacher that causes him to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. Another example is the force of a baseball as it makes contact with the bat. Helicopters create lift by pushing air down, creating an upward reaction force. Birds fly by exerting force on air in the direction opposite that in which they wish to fly. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself forward in the water by ejecting water backward through a funnel in its body, which is similar to how a jet ski is propelled. In these examples, the octopus or jet ski push the water backward, and the water, in turn, pushes the octopus or jet ski forward. Applying Newton’s Third Law Forces are classified and given names based on their source, how they are transmitted, or their effects. In previous sections, we discussed the forces called push, weight, and friction. In this section, applying Newton’s third law of motion will allow us to explore three more forces: the normal force, tension, and thrust. However, because we haven’t yet covered vectors in depth, we’ll only consider one-dimensional situations in this chapter. Another chapter will consider forces acting in two dimensions. The gravitational force (or weight) acts on objects at all times and everywhere on Earth. We know from Newton’s second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? The answer is the normal force. The normal force is the force that a surface applies to an object to support the weight of that object; it acts perpendicular to the surface upon which the object rests. If an object on a flat surface is not accelerating, the net external force is zero, and the normal force has the same magnitude as the weight of the system but acts in the opposite direction. In equation form, we write that Note that this equation is only true for a horizontal surface. The word tensioncomes from the Latin word meaning to stretch. Tension is the force along the length of a flexible connector, such as a string, rope, chain, or cable. Regardless of the type of connector attached to the object of interest, one must remember that the connector can only pull (or exert tension) in the direction parallelto its length. Tension is a pull that acts parallel to the connector, and that acts in opposite directions at the two ends of the connector. This is possible because a flexible connector is simply a long series of action-reaction forces, except at the two ends where outside objects provide one member of the actionreaction forces. Consider a person holding a mass on a rope, as shown in Figure 4.9. 4.17 130 Chapter 4 • Forces and Newton’s Laws of Motion Figure 4.9 When a perfectly flexible connector (one requiring no force to bend it) such as a rope transmits a force T, this force must be parallel to the length of the rope, as shown. The pull that such a flexible connector exerts is a tension. Note that the rope pulls with equal magnitude force but in opposite directions to the hand and to the mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that transmits forces of equal magnitude between the two objects but that act in opposite directions. Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton’s second law. If the 5.00 kg mass in the figure is stationary, then its acceleration is zero, so weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain The only external forces acting on the mass are its 4.18 where T and W are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. By substituting mg for Fnet and rearranging the equation, the tension equals the weight of the supported mass, just as you would expect For a 5.00-kg mass (neglecting the mass of the rope), we see that 4.19 4.20 Another example of Newton’s third law in action is thrust. Rockets move forward by expelling gas backward at a high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas, in turn, exerts a large force forward on the rocket in response. This reaction force is called thrust. TIPS FOR SUCCESS A common misconception is that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can expel exhaust gases more easily. LINKS TO PHYSICS Math: Problem-Solving Strategy for Newton’s Laws of Motion The basics of problem solving, presented earlier in this text, are followed here with specific strategies for applying Newton’s laws of motion. These techniques also reinforce
concepts that are useful in many other areas of physics. First, identify the physical principles involved. If the problem involves forces, then Newton’s laws of motion are involved, and it Access for free at openstax.org. is important to draw a careful sketch of the situation. An example of a sketch is shown in Figure 4.10. Next, as in Figure 4.10, use vectors to represent all forces. Label the forces carefully, and make sure that their lengths are proportional to the magnitude of the forces and that the arrows point in the direction in which the forces act. 4.4 • Newton's Third Law of Motion 131 Figure 4.10 (a) A sketch of Tarzan hanging motionless from a vine. (b) Arrows are used to represent all forces. T is the tension exerted on Tarzan by the vine, is the force exerted on the vine by Tarzan, and W is Tarzan’s weight (i.e., the force exerted on Tarzan by Earth’s gravity). All other forces, such as a nudge of a breeze, are assumed to be negligible. (c) Suppose we are given Tarzan’s mass and asked to find the tension in the vine. We define the system of interest as shown and draw a free-body diagram, as shown in (d). is no longer shown because it does not act on the system of interest; rather, acts on the outside world. (d) The free-body diagram shows only the external forces acting on Tarzan. For these to sum to zero, we must have Next, make a list of knowns and unknowns and assign variable names to the quantities given in the problem. Figure out which variables need to be calculated; these are the unknowns. Now carefully define the system: which objects are of interest for the problem. This decision is important, because Newton’s second law involves only external forces. Once the system is identified, it’s possible to see which forces are external and which are internal (see Figure 4.10). If the system acts on an object outside the system, then you know that the outside object exerts a force of equal magnitude but in the opposite direction on the system. A diagram showing the system of interest and all the external forces acting on it is called a free-body diagram. Only external forces are shown on free-body diagrams, not acceleration or velocity. Figure 4.10 shows a free-body diagram for the system of interest. After drawing a free-body diagram, apply Newton’s second law to solve the problem. This is done in Figure 4.10 for the case of Tarzan hanging from a vine. When external forces are clearly identified in the free-body diagram, translate the forces into equation form and solve for the unknowns. Note that forces acting in opposite directions have opposite signs. By convention, forces acting downward or to the left are usually negative. GRASP CHECK If a problem has more than one system of interest, more than one free-body diagram is required to describe the external forces acting on the different systems. a. True b. False 132 Chapter 4 • Forces and Newton’s Laws of Motion WATCH PHYSICS Newton’s Third Law of Motion This video explains Newton’s third law of motion through examples involving push, normal force, and thrust (the force that propels a rocket or a jet). Click to view content (https://www.openstax.org/l/astronaut) GRASP CHECK If the astronaut in the video wanted to move upward, in which direction should he throw the object? Why? a. He should throw the object upward because according to Newton’s third law, the object will then exert a force on him in the same direction (i.e., upward). b. He should throw the object upward because according to Newton’s third law, the object will then exert a force on him in the opposite direction (i.e., downward). c. He should throw the object downward because according to Newton’s third law, the object will then exert a force on him in the opposite direction (i.e., upward). d. He should throw the object downward because according to Newton’s third law, the object will then exert a force on him in the same direction (i.e., downward). WORKED EXAMPLE An Accelerating Subway Train A physics teacher pushes a cart of demonstration equipment to a classroom, as in Figure 4.11. Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. To push the cart forward, the teacher’s foot applies a force of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N. Figure 4.11 Strategy Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. The teacher pushes backward with a force system. Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. As noted in the figure, the friction f opposes the motion and therefore acts opposite the direction of of 150 N. According to Newton’s third law, the floor exerts a forward force of 150 N on the Access for free at openstax.org. 4.4 • Newton's Third Law of Motion 133 We should not include the forces because these are exerted bythe system, not onthe system. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newton’s second law to find the acceleration. , or , Solution Newton’s second law is The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain 4.21 The mass of the system is the sum of the mass of the teacher, cart, and equipment. Insert these values of net F and minto Newton’s second law to obtain the acceleration of the system. 4.22 4.23 4.24 4.25 Discussion None of the forces between components of the system, such as between the teacher’s hands and the cart, contribute to the net external force because they are internal to the system. Another way to look at this is to note that the forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the teacher on the cart is of equal magnitude but in the opposite direction of the force exerted by the cart on the teacher. In this case, both forces act on the same system, so they cancel. Defining the system was crucial to solving this problem. Practice Problems 14. What is the equation for the normal force for a body with mass mthat is at rest on a horizontal surface? a. N = m b. N = mg c. N = mv d. N = g 15. An object with mass mis at rest on the floor. What is the magnitude and direction of the normal force acting on it? a. N = mvin upward direction b. N = mgin upward direction c. N = mvin downward direction d. N = mgin downward direction Check Your Understanding 16. What is Newton’s third law of motion? a. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude and acts in the direction of the applied force. b. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude and acts in the direction of the applied force. c. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude but acts in the direction opposite the direction of the applied force. d. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude but 134 Chapter 4 • Forces and Newton’s Laws of Motion acts in the direction opposite the direction of the applied force. 17. Considering Newton’s third law, why don’t two equal and opposite forces cancel out each other? a. Because the two forces act in the same direction b. Because the two forces have different magnitudes c. Because the two forces act on different systems d. Because the two forces act in perpendicular directions Access for free at openstax.org. Chapter 4 • Key Terms 135 KEY TERMS dynamics the study of how forces affect the motion of objects and systems external force a force acting on an object or system that originates outside of the object or system force a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force free-body diagram a diagram showing all external forces acting on a body Newton’s second law of motion the net external force, on an object is proportional to and in the same direction as the acceleration of the object, a, and also proportional to the object’s mass, m; defined mathematically as or Newton’s third law of motion when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts freefall a situation in which the only force acting on an normal force the force that a surface applies to an object; object is the force of gravity friction an external force that acts in the direction opposite to the direction of motion inertia the tendency of an object at rest to remain at rest, or for a moving object to remain in motion in a straight line and at a constant speed law of inertia Newton’s first law of motion: a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia mass the quantity of matter in a substance; measured in kilograms net external force the sum of all external forces acting on an object or system net force the sum of all forces acting on an object or system Newton’s first law of motion a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia SECTION SUMMARY 4.1 Force • Dynamics is the study of how forces
affect the motion of objects and systems. • Force is a push or pull that can be defined in terms of various standards. It is a vector and so has both magnitude and direction. • External forces are any forces outside of a body that act on the body. A free-body diagram is a drawing of all external forces acting on a body. 4.2 Newton's First Law of Motion: Inertia • Newton’s first law states that a body at rest remains at rest or, if moving, remains in motion in a straight line at a constant speed, unless acted on by a net external force. This law is also known as the law of inertia. Inertia is the tendency of an object at rest to remain at rest or, if moving, to remain in motion at constant velocity. Inertia is related to an object’s mass. • acts perpendicular and away from the surface with which the object is in contact system one or more objects of interest for which only the forces acting on them from the outside are considered, but not the forces acting between them or inside them tension a pulling force that acts along a connecting medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force exerted on the object by the rope is called tension thrust a force that pushes an object forward in response to the backward ejection of mass by the object; rockets and airplanes are pushed forward by a thrust reaction force in response to ejecting gases backward weight the force of gravity, W, acting on an object of mass m; defined mathematically as W = mg, where g is the magnitude and direction of the acceleration due to gravity • Friction is a force that opposes motion and causes an object or system to slow down. • Mass is the quantity of matter in a substance. 4.3 Newton's Second Law of Motion • Acceleration is a change in velocity, meaning a change in speed, direction, or both. • An external force acts on a system from outside the system, as opposed to internal forces, which act between components within the system. • Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to the system’s mass. In equation form, Newton’s second law of motion is • or or . This is sometimes written as • The weight of an object of mass mis the force of gravity that acts on it. From Newton’s second law, weight is 136 Chapter 4 • Key Equations • given by If the only force acting on an object is its weight, then the object is in freefall. 4.4 Newton's Third Law of Motion • Newton’s third law of motion states that when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. • When an object rests on a surface, the surface applies a force on the object that opposes the weight of the object. KEY EQUATIONS 4.2 Newton's First Law of Motion: Inertia Newton's first law of motion or This force acts perpendicular to the surface and is called the normal force. • The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension. When a rope supports the weight of an object at rest, the tension in the rope is equal to the weight of the object. • Thrust is a force that pushes an object forward in response to the backward ejection of mass by the object. Rockets and airplanes are pushed forward by thrust. Newton’s second law of motion to solve weight 4.4 Newton's Third Law of Motion 4.3 Newton's Second Law of Motion Newton’s second law of motion or normal force for a nonaccelerating horizontal surface tension for an object at rest Newton’s second law of motion to solve acceleration CHAPTER REVIEW Concept Items 4.1 Force 1. What is dynamics? a. Dynamics is the study of internal forces. b. Dynamics is the study of forces and their effect on motion. c. Dynamics describes the motion of points, bodies, and systems without consideration of the cause of motion. d. Dynamics describes the effect of forces on each other. 2. Two forces acting on an object are perpendicular to one another. How would you draw these in a free-body diagram? a. The two force arrows will be drawn at a right angle to one another. b. The two force arrows will be pointing in opposite directions. c. The two force arrows will be at a 45° angle to one another. Access for free at openstax.org. d. The two force arrows will be at a 180° angle to one another. 3. A free-body diagram shows the forces acting on an object. How is that object represented in the diagram? a. A single point b. A square box c. A unit circle d. The object as it is 4.2 Newton's First Law of Motion: Inertia 4. A ball rolls along the ground, moving from north to south. What direction is the frictional force that acts on the ball? a. North to south b. South to north c. West to east d. East to west 5. The tires you choose to drive over icy roads will create more friction with the road than your summer tires. Give another example where more friction is desirable. a. Children’s slide b. Air hockey table Ice-skating rink c. Jogging track d. 6. How do you express, mathematically, that no external force is acting on a body? a. Fnet = −1 b. Fnet = 0 c. Fnet = 1 d. Fnet = ∞ 4.3 Newton's Second Law of Motion 7. What does it mean for two quantities to be inversely proportional to each other? a. When one variable increases, the other variable decreases by a greater amount. b. When one variable increases, the other variable also increases. c. When one variable increases, the other variable decreases by the same factor. d. When one variable increases, the other variable also increases by the same factor. Chapter 4 • Chapter Review 137 8. True or False: Newton's second law can be interpreted based on Newton's first law. a. True b. False 4.4 Newton's Third Law of Motion 9. Which forces cause changes in the motion of a system? a. internal forces b. external forces c. both internal and external forces d. neither internal nor external forces 10. True or False—Newton’s third law applies to the external forces acting on a system of interest. a. True b. False 11. A ball is dropped and hits the floor. What is the direction of the force exerted by the floor on the ball? a. Upward b. Downward c. Right d. Left Critical Thinking Items direction. 4.1 Force 12. Only two forces are acting on an object: force A to the left and force B to the right. If force B is greater than force A, in which direction will the object move? a. To the right b. To the left c. Upward d. The object does not move 13. In a free-body diagram, the arrows representing tension and weight have the same length but point away from one another. What does this indicate? a. They are equal in magnitude and act in the same direction. c. The magnitude is xand points in the downward direction. d. The magnitude is 2xand points in the downward direction. 15. Three forces, A, B, and C, are acting on the same object with magnitudes a, b, and c, respectively. Force A acts to the right, force B acts to the left, and force C acts downward. What is a necessary condition for the object to move straight down? a. The magnitude of force A must be greater than the magnitude of force B, so a > b. b. The magnitude of force A must be equal to the magnitude of force B, so a = b. c. The magnitude of force A must be greater than the b. They are equal in magnitude and act in opposite magnitude of force C, so A > C. directions. c. They are unequal in magnitude and act in the same d. The magnitude of force C must be greater than the magnitude of forces A or B, so A < C > B. direction. d. They are unequal in magnitude and act in opposite 4.2 Newton's First Law of Motion: Inertia directions. 14. An object is at rest. Two forces, X and Y, are acting on it. Force X has a magnitude of xand acts in the downward direction. What is the magnitude and direction of Y? a. The magnitude is xand points in the upward direction. 16. Two people push a cart on a horizontal surface by applying forces F1 and F2 in the same direction. Is the magnitude of the net force acting on the cart, Fnet, equal to, greater than, or less than F1 + F2? Why? a. Fnet < F1 + F2 because the net force will not include the frictional force. b. The magnitude is 2xand points in the upward b. Fnet = F1 + F2 because the net force will not include 138 Chapter 4 • Chapter Review the frictional force in acceleration and mass. c. Fnet < F1 + F2 because the net force will include the component of frictional force 4.4 Newton's Third Law of Motion d. Fnet = F1 + F2 because the net force will include the frictional force 17. True or False: A book placed on a balance scale is balanced by a standard 1-kg iron weight placed on the opposite side of the balance. If these objects are taken to the moon and a similar exercise is performed, the balance is still level because gravity is uniform on the moon’s surface as it is on Earth’s surface. a. True b. False 4.3 Newton's Second Law of Motion 18. From the equation for Newton’s second law, we see that Fnet is directly proportional to a and that the constant of proportionality is m. What does this mean in a practical sense? a. An increase in applied force will cause an increase in acceleration if the mass is constant. b. An increase in applied force will cause a decrease in acceleration if the mass is constant. c. An increase in applied force will cause an increase in acceleration, even if the mass varies. d. An increase in applied force will cause an increase 19. True or False: A person accelerates while walking on the ground by exerting force. The ground in turn exerts force F2 on the person. F1 and F2 are equal in magnitude but act in opposite directions. The person is able to walk because the two forces act on the different systems and the net force acting on the person is nonzero. a. True b. False 20. A helicopter pushes air down, which, in turn, pushes the helicop
ter up. Which force affects the helicopter’s motion? Why? a. Air pushing upward affects the helicopter’s motion because it is an internal force that acts on the helicopter. b. Air pushing upward affects the helicopter’s motion because it is an external force that acts on the helicopter. c. The downward force applied by the blades of the helicopter affects its motion because it is an internal force that acts on the helicopter. d. The downward force applied by the blades of the helicopter affects its motion because it is an external force that acts on the helicopter. Problems 4.3 Newton's Second Law of Motion 55.0 kg c. d. 91.9 kg on Earth. What is its 4.4 Newton's Third Law of Motion 21. An object has a mass of weight on the moon? a. b. c. d. 22. A bathroom scale shows your mass as 55 kg. What will it read on the moon? a. 9.4 kg 10.5 kg b. Performance Task 4.4 Newton's Third Law of Motion 24. A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two passengers in the car, each weighing 55 kg. If the magnitude of the force of friction Access for free at openstax.org. 23. A person pushes an object of mass 5.0 kg along the floor by applying a force. If the object experiences a friction force of 10 N and accelerates at 18 m/s2, what is the magnitude of the force exerted by the person? a. −90 N b. −80 N c. 90 N 100 N d. experienced by the car is 45 N, what is the acceleration of the car? a. 0.244 m/s2 b. 0.265 m/s2 c. 4.00 m/s2 d. 4.10 m/s2 TEST PREP Multiple Choice 4.1 Force 25. Which of the following is a physical quantity that can be described by dynamics but not by kinematics? a. Velocity b. Acceleration c. Force 26. Which of the following is used to represent an object in a free-body diagram? a. A point b. A line c. A vector 4.2 Newton's First Law of Motion: Inertia 27. What kind of force is friction? a. External force b. Internal force c. Net force 28. What is another name for Newton’s first law? a. Law of infinite motion b. Law of inertia c. Law of friction 29. True or False—A rocket is launched into space and escapes Earth’s gravitational pull. It will continue to travel in a straight line until it is acted on by another force. a. True b. False 30. A 2,000-kg car is sitting at rest in a parking lot. A bike and rider with a total mass of 60 kg are traveling along a road at 10 km/h. Which system has more inertia? Why? a. The car has more inertia, as its mass is greater than the mass of the bike. b. The bike has more inertia, as its mass is greater than the mass of the car. c. The car has more inertia, as its mass is less than the mass of the bike. d. The bike has more inertia, as its mass is less than the mass of the car. 4.3 Newton's Second Law of Motion 31. In the equation for Newton’s second law, what does Fnet stand for? Internal force a. b. Net external force c. Frictional force 32. What is the SI unit of force? Chapter 4 • Test Prep 139 a. Kg b. dyn c. N 33. What is the net external force on an object in freefall on Earth if you were to neglect the effect of air? a. The net force is zero. b. The net force is upward with magnitude mg. c. The net force is downward with magnitude mg. d. The net force is downward with magnitude 9.8 N. 34. Two people push a 2,000-kg car to get it started. An acceleration of at least 5.0 m/s2 is required to start the car. Assuming both people apply the same magnitude force, how much force will each need to apply if friction between the car and the road is 300 N? a. 4850 N 5150 N b. c. 97000 N 10300 N d. 4.4 Newton's Third Law of Motion 35. One object exerts a force of magnitude F1 on another object and experiences a force of magnitude F2 in return. What is true for F1 and F2? a. F1 > F2 b. F1 < F2 c. F1 = F2 36. A weight is suspended with a rope and hangs freely. In what direction is the tension on the rope? a. parallel to the rope b. perpendicular to the rope 37. A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates at 2 m/s2, what is the force of friction experienced by the system consisting of the person and the object? 30 N a. b. 50 N c. 270 N d. 290 N 38. A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s2. The friction experienced by the swimmer is 100 N. What is the magnitude of the force that the swimmer applies on the wall? a. −490 N b. −290 N c. 290 N d. 490 N 140 Chapter 4 • Test Prep Short Answer 4.1 Force 39. True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the system. a. True b. False 40. By convention, which sign is assigned to an object moving downward? a. A positive sign ( b. A negative sign ( c. Either a positive or negative sign ( d. No sign is assigned ) ) ) 41. A body is pushed downward by a force of 5 units and upward by a force of 2 units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a point, both pointing up with lengths of 5 units and 2 units b. Two force vectors acting at a point, both pointing down with lengths of 5 units and 2 units c. Two force vectors acting at a point, one pointing up with a length of 5 units and the other pointing down with a length of 2 units d. Two force vectors acting at a point, one pointing down with a length of 5 units and the other pointing up with a length of 2 units 42. A body is pushed eastward by a force of four units and southward by a force of three units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a point, one pointing left with a length of 4 units and the other pointing down with a length of 3 units b. Two force vectors acting at a point, one pointing left with a length of 4 units and the other pointing up with a length of 3 units c. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing down with a length of 3 units d. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing up with a length of 3 units 4.2 Newton's First Law of Motion: Inertia 43. A body with mass m is pushed along a horizontal surface by a force F and is opposed by a frictional force f. How would you draw a free-body diagram to represent this situation? a. A dot with an arrow pointing right, labeled F, and an arrow pointing left, labeled f, that is of equal length or shorter than F Access for free at openstax.org. b. A dot with an arrow pointing right, labeled F, and an arrow pointing right, labeled f, that is of equal length or shorter than F c. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing up, labeled f, that is of equal length or longer than F d. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing down, labeled f, that is of equal length or longer than F 44. Two objects rest on a uniform surface. A person pushes both with equal force. If the first object starts to move faster than the second, what can be said about their masses? a. The mass of the first object is less than that of the second object. b. The mass of the first object is equal to the mass of the second object. c. The mass of the first object is greater than that of the second object. d. No inference can be made because mass and force are not related to each other. 45. Two similar boxes rest on a table. One is empty and the other is filled with pebbles. Without opening or lifting either, how can you tell which box is full? Why? a. By applying an internal force; whichever box accelerates faster is lighter and so must be empty b. By applying an internal force; whichever box accelerates faster is heavier and so the other box must be empty c. By applying an external force; whichever box accelerates faster is lighter and so must be empty d. By applying an external force; whichever box accelerates faster is heavier and so the other box must be empty 46. True or False—An external force is required to set a stationary object in motion in outer space away from all gravitational influences and atmospheric friction. a. True b. False 4.3 Newton's Second Law of Motion 47. A steadily rolling ball is pushed in the direction from east to west, which causes the ball to move faster in the same direction. What is the direction of the acceleration? a. North to south b. South to north c. East to west d. West to east 48. A ball travels from north to south at 60 km/h. After being hit by a bat, it travels from west to east at 60 km/ Chapter 4 • Test Prep 141 h. Is there a change in velocity? a. Yes, because velocity is a scalar. b. Yes, because velocity is a vector. c. No, because velocity is a scalar. d. No, because velocity is a vector 49. What is the weight of a 5-kg object on Earth and on the moon? a. On Earth the weight is 1.67 N, and on the moon the weight is 1.67 N. b. On Earth the weight is 5 N, and on the moon the weight is 5 N. b. F c. 2F 30F d. 52. A fish pushes water backward with its fins. How does this propel the fish forward? a. The water exerts an internal force on the fish in the opposite direction, pushing the fish forward. b. The water exerts an external force on the fish in the opposite direction, pushing the fish forward. c. The water exerts an internal force on the fish in the same direction, pushing the fish forward. c. On Earth the weight is 49 N, and on the moon the d. The water exerts an external force on the fish in the weight is 8.35 N. same direction, pushing the fish forward. d. On Earth the weight is 8.35 N, and on the moon the weight is 49 N. 50. An object weighs 294 N on Earth. What is its weight on the moon? 50.1 N a. b. 30.0 N c. 249 N 1461 N d. 4.4 Newton's Third Law of Motion 51. A large truck with mass 30 m crashes into a small sedan with mass m. If the truck exerts a force F on the sedan, what force will the sedan exert on the truck? a. Extended Response 4.1 Force 55. True or False—When two unequal fo
rces act on a body, the body will not move in the direction of the weaker force. a. True b. False 56. In the figure given, what is Frestore? What is its magnitude? a. Frestore is the force exerted by the hand on the spring, and it pulls to the right. b. Frestore is the force exerted by the spring on the hand, and it pulls to the left. 53. True or False—Tension is the result of opposite forces in a connector, such as a string, rope, chain or cable, that pulls each point of the connector apart in the direction parallel to the length of the connector. At the ends of the connector, the tension pulls toward the center of the connector. a. True b. False 54. True or False—Normal reaction is the force that opposes the force of gravity and acts in the direction of the force of gravity. a. True b. False c. Frestore is the force exerted by the hand on the spring, and it pulls to the left. d. Frestore is the force exerted by the spring on the hand, and it pulls to the right. 4.2 Newton's First Law of Motion: Inertia 57. Two people apply the same force to throw two identical balls in the air. Will the balls necessarily travel the same distance? Why or why not? a. No, the balls will not necessarily travel the same distance because the gravitational force acting on them is different. b. No, the balls will not necessarily travel the same distance because the angle at which they are thrown may differ. c. Yes, the balls will travel the same distance because the gravitational force acting on them is the same. d. Yes, the balls will travel the same distance because the angle at which they are thrown may differ. 58. A person pushes a box from left to right and then lets the box slide freely across the floor. The box slows down as it slides across the floor. When the box is sliding 142 Chapter 4 • Test Prep freely, what is the direction of the net external force? a. The net external force acts from left to right. b. The net external force acts from right to left. c. The net external force acts upward. d. The net external force acts downward. 4.3 Newton's Second Law of Motion 59. A 55-kg lady stands on a bathroom scale inside an elevator. The scale reads 70 kg. What do you know about the motion of the elevator? a. The elevator must be accelerating upward. b. The elevator must be accelerating downward. c. The elevator must be moving upward with a constant velocity. d. The elevator must be moving downward with a constant velocity. 60. True or False—A skydiver initially accelerates in his jump. Later, he achieves a state of constant velocity called terminal velocity. Does this mean the skydiver becomes weightless? a. Yes b. No 4.4 Newton's Third Law of Motion 61. How do rockets propel themselves in space? a. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an internal force, pushes the rockets forward. b. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. c. Rockets expel gas in the backward direction at high velocity, and the gas, which is an internal force, pushes the rockets forward. d. Rockets expel gas in the backward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. 62. Are rockets more efficient in Earth’s atmosphere or in outer space? Why? a. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust for the rocket, and Earth has more air friction than outer space. b. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust to the rocket, and Earth has less air friction than the outer space. c. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create more air friction than in outer space. d. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create less air friction than in outer space. Access for free at openstax.org. CHAPTER 5 Motion in Two Dimensions Figure 5.1 Billiard balls on a pool table are in motion after being hit with a cue stick. (Popperipopp, Wikimedia Commons) Chapter Outline 5.1 Vector Addition and Subtraction: Graphical Methods 5.2 Vector Addition and Subtraction: Analytical Methods 5.3 Projectile Motion 5.4 Inclined Planes 5.5 Simple Harmonic Motion In Chapter 2, we learned to distinguish between vectors and scalars; the difference being that a vector has INTRODUCTION magnitude and direction, whereas a scalar has only magnitude. We learned how to deal with vectors in physics by working straightforward one-dimensional vector problems, which may be treated mathematically in the same as scalars. In this chapter, we’ll use vectors to expand our understanding of forces and motion into two dimensions. Most real-world physics problems (such as with the game of pool pictured here) are, after all, either two- or three-dimensional problems and physics is most useful when applied to real physical scenarios. We start by learning the practical skills of graphically adding and subtracting vectors (by using drawings) and analytically (with math). Once we’re able to work with two-dimensional vectors, we apply these skills to problems of projectile motion, inclined planes, and harmonic motion. 144 Chapter 5 • Motion in Two Dimensions 5.1 Vector Addition and Subtraction: Graphical Methods Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the graphical method of vector addition and subtraction • Use the graphical method of vector addition and subtraction to solve physics problems Section Key Terms graphical method head (of a vector) head-to-tail method resultant resultant vector tail vector addition vector subtraction The Graphical Method of Vector Addition and Subtraction Recall that a vector is a quantity that has magnitude and direction. For example, displacement, velocity, acceleration, and force are all vectors. In one-dimensional or straight-line motion, the direction of a vector can be given simply by a plus or minus sign. Motion that is forward, to the right, or upward is usually considered to be positive(+); and motion that is backward, to the left, or downward is usually considered to be negative(−). In two dimensions, a vector describes motion in two perpendicular directions, such as vertical and horizontal. For vertical and horizontal motion, each vector is made up of vertical and horizontal components. In a one-dimensional problem, one of the components simply has a value of zero. For two-dimensional vectors, we work with vectors by using a frame of reference such as a coordinate system. Just as with one-dimensional vectors, we graphically represent vectors with an arrow having a length proportional to the vector’s magnitude and pointing in the direction that the vector points. Figure 5.2 shows a graphical representation of a vector; the total displacement for a person walking in a city. The person first walks nine blocks east and then five blocks north. Her total displacement does not match her path to her final destination. The displacement simply connects her starting point with her ending point using a straight line, which is the shortest distance. We use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction is given by an angle represented by the symbol Note that her displacement would be the same if she had begun by first walking five blocks north and then walking nine blocks east. TIPS FOR SUCCESS In this text, we represent a vector with a boldface variable. For example, we represent a force with the vector F, which has both magnitude and direction. The magnitude of the vector is represented by the variable in italics, F, and the direction of the variable is given by the angle Figure 5.2 A person walks nine blocks east and five blocks north. The displacement is 10.3 blocks at an angle north of east. The head-to-tail method is a graphical way to add vectors. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the pointed end of the arrow. The following steps describe how to use the head-to-tail method for graphical vector addition. Access for free at openstax.org. 1. Let the x-axis represent the east-west direction. Using a ruler and protractor, draw an arrow to represent the first vector (nine blocks to the east), as shown in Figure 5.3(a). 5.1 • Vector Addition and Subtraction: Graphical Methods 145 Figure 5.3 The diagram shows a vector with a magnitude of nine units and a direction of 0°. 2. Let the y-axis represent the north-south direction. Draw an arrow to represent the second vector (five blocks to the north). Place the tail of the second vector at the head of the first vector, as shown in Figure 5.4(b). Figure 5.4 A vertical vector is added. 3. If there are more than two vectors, continue to add the vectors head-to-tail as described in step 2. In this example, we have only two vectors, so we have finished placing arrows tip to tail. 4. Draw an arrow from the tail of the first vector to the head of the last vector, as shown in Figure 5.5(c). This is the resultant, or the sum, of the vectors. 146 Chapter 5 • Motion in Two Dimensions Figure 5.5 The diagram shows the resultant vector, a ruler, and protractor. 5. To find the magnitude of the resultant, measure its length with a ruler. When we deal with vectors analytically in the next section, the magnitude will be calculated by using the Pythagorean theorem. 6. To find the direction of the resultant, use a protractor to measure the angle it makes with the reference direction (in this case, the x-axis). When we deal with vectors analytically in the next section, the direction
will be calculated by using trigonometry to find the angle. WATCH PHYSICS Visualizing Vector Addition Examples This video shows four graphical representations of vector addition and matches them to the correct vector addition formula. Click to view content (https://openstax.org/l/02addvector) GRASP CHECK There are two vectors and . The head of vector touches the tail of vector . The addition of vectors and gives a resultant vector . Can the addition of these two vectors can be represented by the following two equations? ; a. Yes, if we add the same two vectors in a different order it will still give the same resultant vector. b. No, the resultant vector will change if we add the same vectors in a different order. Vector subtraction is done in the same way as vector addition with one small change. We add the first vector to the negative of the vector that needs to be subtracted. A negative vector has the same magnitude as the original vector, but points in the opposite direction (as shown in Figure 5.6). Subtracting the vector B from the vector A, which is written as A − B, is the same as A + (−B). Since it does not matter in what order vectors are added, A − B is also equal to (−B) + A. This is true for scalars as well as vectors. For example, 5 – 2 = 5 + (−2) = (−2) + 5. Access for free at openstax.org. 5.1 • Vector Addition and Subtraction: Graphical Methods 147 Figure 5.6 The diagram shows a vector, B, and the negative of this vector, –B. Global angles are calculated in the counterclockwise direction. The clockwise direction is considered negative. For example, an from the positive x-axis. angle of south of west is the same as the global angle which can also be expressed as Using the Graphical Method of Vector Addition and Subtraction to Solve Physics Problems Now that we have the skills to work with vectors in two dimensions, we can apply vector addition to graphically determine the resultant vector, which represents the total force. Consider an example of force involving two ice skaters pushing a third as seen in Figure 5.7. Figure 5.7 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. In problems where variables such as force are already known, the forces can be represented by making the length of the vectors proportional to the magnitudes of the forces. For this, you need to create a scale. For example, each centimeter of vector length could represent 50 N worth of force. Once you have the initial vectors drawn to scale, you can then use the head-to-tail method to draw the resultant vector. The length of the resultant can then be measured and converted back to the original units using the scale you created. You can tell by looking at the vectors in the free-body diagram in Figure 5.7 that the two skaters are pushing on the third skater with equal-magnitude forces, since the length of their force vectors are the same. Note, however, that the forces are not equal because they act in different directions. If, for example, each force had a magnitude of 400 N, then we would find the magnitude of the total external force acting on the third skater by finding the magnitude of the resultant vector. Since the forces act at a right angle to one another, we can use the Pythagorean theorem. For a triangle with sides a, b, and c, the Pythagorean theorem tells us that Applying this theorem to the triangle made by F1, F2, and Ftot in Figure 5.7, we get 148 Chapter 5 • Motion in Two Dimensions or Note that, if the vectors were not at a right angle to each other Pythagorean theorem to find the magnitude of the resultant vector. Another scenario where adding two-dimensional vectors is necessary is for velocity, where the direction may not be purely east-west or north-south, but some combination of these two directions. In the next section, we cover how to solve this type of problem analytically. For now let’s consider the problem graphically. to one another), we would not be able to use the WORKED EXAMPLE Adding Vectors Graphically by Using the Head-to-Tail Method: A Woman Takes a Walk Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths north of (displacements) on a flat field. First, he walks 25 m in a direction east. Finally, he turns and walks 32 m in a direction Strategy Graphically represent each displacement vector with an arrow, labeling the first A, the second B, and the third C. Make the lengths proportional to the distance of the given displacement and orient the arrows as specified relative to an east-west line. Use the head-to-tail method outlined above to determine the magnitude and direction of the resultant displacement, which we’ll call R. north of east. Then, he walks 23 m heading south of east. Solution (1) Draw the three displacement vectors, creating a convenient scale (such as 1 cm of vector length on paper equals 1 m in the problem), as shown in Figure 5.8. (2) Place the vectors head to tail, making sure not to change their magnitude or direction, as shown in Figure 5.9. Figure 5.8 The three displacement vectors are drawn first. (3) Draw the resultant vector R from the tail of the first vector to the head of the last vector, as shown in Figure 5.10. Figure 5.9 Next, the vectors are placed head to tail. Access for free at openstax.org. 5.1 • Vector Addition and Subtraction: Graphical Methods 149 Figure 5.10 The resultant vector is drawn . (4) Use a ruler to measure the magnitude of R, remembering to convert back to the units of meters using the scale. Use a protractor to measure the direction of R. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since R is south of the eastward pointing axis (the x-axis), we flip the protractor upside down and measure the angle between the eastward axis and the vector, as illustrated in Figure 5.11. Figure 5.11 A ruler is used to measure the magnitude of R, and a protractor is used to measure the direction of R. In this case, the total displacement R has a magnitude of 50 m and points this vector can be expressed as south of east. Using its magnitude and direction, and 5.1 5.2 Discussion The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that it does not matter in what order the vectors are added. Changing the order does not change the resultant. For example, we could add the vectors as shown in Figure 5.12, and we would still get the same solution. 150 Chapter 5 • Motion in Two Dimensions Figure 5.12 Vectors can be added in any order to get the same result. WORKED EXAMPLE Subtracting Vectors Graphically: A Woman Sailing a Boat A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction west of north). If the north of east from her current location, and then travel 30.0 m in a direction woman makes a mistake and travels in the oppositedirection for the second leg of the trip, where will she end up? The two legs of the woman’s trip are illustrated in Figure 5.13. north of east (or Figure 5.13 In the diagram, the first leg of the trip is represented by vector A and the second leg is represented by vector B. Strategy We can represent the first leg of the trip with a vector A, and the second leg of the trip that she was supposed totake with a vector B. Since the woman mistakenly travels in the opposite direction for the second leg of the journey, the vector for second leg of the trip she actuallytakes is −B. Therefore, she will end up at a location A + (−B), or A − B. Note that −B has the same south of east, as illustrated in Figure 5.14. magnitude as B (30.0 m), but is in the opposite direction, Figure 5.14 Vector –B represents traveling in the opposite direction of vector B. We use graphical vector addition to find where the woman arrives A + (−B). Access for free at openstax.org. 5.1 • Vector Addition and Subtraction: Graphical Methods 151 Solution (1) To determine the location at which the woman arrives by accident, draw vectors A and −B. (2) Place the vectors head to tail. (3) Draw the resultant vector R. (4) Use a ruler and protractor to measure the magnitude and direction of R. These steps are demonstrated in Figure 5.15. Figure 5.15 The vectors are placed head to tail. In this case and 5.3 5.4 Discussion Because subtraction of a vector is the same as addition of the same vector with the opposite direction, the graphical method for subtracting vectors works the same as for adding vectors. WORKED EXAMPLE Adding Velocities: A Boat on a River A boat attempts to travel straight across a river at a speed of 3.8 m/s. The river current flows at a speed vriver of 6.1 m/s to the right. What is the total velocity and direction of the boat? You can represent each meter per second of velocity as one centimeter of vector length in your drawing. Strategy We start by choosing a coordinate system with its x-axis parallel to the velocity of the river. Because the boat is directed straight toward the other shore, its velocity is perpendicular to the velocity of the river. We draw the two vectors, vboat and vriver, as shown in Figure 5.16. Using the head-to-tail method, we draw the resulting total velocity vector from the tail of vboat to the head of vriver. 152 Chapter 5 • Motion in Two Dimensions Figure 5.16 A boat attempts to travel across a river. What is the total velocity and direction of the boat? Solution By using a ruler, we find that the length of the resultant vector is 7.2 cm, which means that the magnitude of the total velocity is By using a protractor to measure the angle, we find Discussion If the velocity of the boat and river were e
qual, then the direction of the total velocity would have been 45°. However, since the velocity of the river is greater than that of the boat, the direction is less than 45° with respect to the shore, or xaxis. 5.5 Practice Problems 1. Vector , having magnitude , pointing south of east and vector having magnitude , pointing north of east are added. What is the magnitude of the resultant vector? a. b. c. d. 2. A person walks north of west for and east of south for . What is the magnitude of his displacement? a. b. c. d. Virtual Physics Vector Addition In this simulation (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/) , you will experiment with adding vectors graphically. Click and drag the red vectors from the Grab One basket onto the graph in the middle of the screen. These red vectors can be rotated, stretched, or repositioned by clicking and dragging with your mouse. Check the Show Sum box to display the resultant vector (in green), which is the sum of all of the red vectors placed on the Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 153 graph. To remove a red vector, drag it to the trash or click the Clear All button if you wish to start over. Notice that, if you click on any of the vectors, the component, and Ryis its vertical component. You can check the resultant by lining up the vectors so that the head of the first vector touches the tail of the second. Continue until all of the vectors are aligned together head-to-tail. You will see that the resultant magnitude and angle is the same as the arrow drawn from the tail of the first vector to the head of the last vector. Rearrange the vectors in any order head-to-tail and compare. The resultant will always be the same. is its direction with respect to the positive x-axis, Rx is its horizontal is its magnitude, Click to view content (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/) GRASP CHECK True or False—The more long, red vectors you put on the graph, rotated in any direction, the greater the magnitude of the resultant green vector. a. True b. False Check Your Understanding 3. While there is no single correct choice for the sign of axes, which of the following are conventionally considered positive? a. backward and to the left b. backward and to the right forward and to the right c. forward and to the left d. 4. True or False—A person walks 2 blocks east and 5 blocks north. Another person walks 5 blocks north and then two blocks east. The displacement of the first person will be more than the displacement of the second person. a. True b. False 5.2 Vector Addition and Subtraction: Analytical Methods Section Learning Objectives By the end of this section, you will be able to do the following: • Define components of vectors • Describe the analytical method of vector addition and subtraction • Use the analytical method of vector addition and subtraction to solve problems Section Key Terms analytical method component (of a two-dimensional vector) Components of Vectors For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17. 154 Chapter 5 • Motion in Two Dimensions Figure 5.17 For a right triangle, the sine, cosine, and tangent of θare defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, xis the adjacent side, yis the opposite side, and his the hypotenuse. Since, by definition, length of yby using , we can find the length xif we know hand by using . Similarly, we can find the . These trigonometric relationships are useful for adding vectors. When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components. For example, given a vector like produce it. In this example, common situation in physics and happens to be the least complicated situation trigonometrically. , add to form a right triangle, meaning that the angle between them is 90 degrees. This is a in Figure 5.18, we may want to find what two perpendicular vectors, and and Figure 5.18 The vector , with its tail at the origin of an x- y-coordinate system, is shown together with its x- and y-components, and These vectors form a right triangle. and triangle. are defined to be the components of along the x- and y-axes. The three vectors, , , and , form a right If the vector and y-components, we use the following relationships for a right triangle: is known, then its magnitude (its length) and its angle (its direction) are known. To find and , its x- and is the magnitude of A in the x-direction, where resultant with respect to the x-axis, as shown in Figure 5.19. is the magnitude of A in the y-direction, and is the angle of the Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 155 Figure 5.19 The magnitudes of the vector components and can be related to the resultant vector and the angle with trigonometric identities. Here we see that and Suppose, for example, that Figure 5.20. is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20 We can use the relationships and to determine the magnitude of the horizontal and vertical component vectors in this example. Then A = 10.3 blocks and , so that This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly, 5.6 5.7 indicating that the walker has traveled 5 blocks to the north—a 5-block northward displacement. 156 Chapter 5 • Motion in Two Dimensions Analytical Method of Vector Addition and Subtraction Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components definition, are known, then we can find analytically. How do we do this? Since, by of a vector and we solve for to find the direction of the resultant. Since this is a right triangle, the Pythagorean theorem (x2 + y2 = h2) for finding the hypotenuse applies. In this case, it becomes Solving for A gives In summary, to find the magnitude in Figure 5.21, we use the following relationships: and direction of a vector from its perpendicular components and , as illustrated Figure 5.21 The magnitude and direction of the resultant vector can be determined once the horizontal components and have been determined. Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors are added to produce the resultant as illustrated in Figure 5.22. and Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 157 Figure 5.22 Vectors and are two legs of a walk, and is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of . represent two legs of a walk (two displacements), then If and up at the tip of x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, . There are many ways to arrive at the same point. The person could have walked straight ahead first in the is the total displacement. The person taking the walk ends and If we know and , we can find and using the equations and . 1. Draw in the xand ycomponents of each vector (including the resultant) with a dashed line. Use the equations and angle of by vector A). to find the components. In Figure 5.23, these components are , and with its own x-axis (which is slightly above the x-axis used Vector makes an , , with the x-axis, and vector makes and angle of Figure 5.23 To add vectors and first determine the horizontal and vertical components of each vector. These are the dotted vectors shown in the image. 2. Find the xcomponent of the resultant by adding the xcomponent of the vectors and find the ycomponent of the resultant (as illustrated in Figure 5.24) by adding the ycomponent of the vectors. 158 Chapter 5 • Motion in Two Dimensions Figure 5.24 The vectors and add to give the magnitude of the resultant vector in the horizontal direction, Similarly, the vectors and add to give the magnitude of the resultant vector in the vertical direction, Now that we know the components of we can find its magnitude and direction. 3. To get the magnitude of the resultant R, use the Pythagorean theorem. 4. To get the direction of the resultant WATCH PHYSICS Classifying Vectors and Quantities Example This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions. Click to view content (https://www.youtube.com/embed/Yp0EhcVBxNU) GRASP CHECK . Vector points in the northwest. If the vectors , , and were added points to the northeast. Vector points to the , and , have the same magnitude of . Vector Three vectors, , southwest exactly opposite to vector together, what would be the magnitude of the resultant vector? Why? a. b. c. d. . All of them will cancel each other out. . Two of them will cancel each other out. units. All of them will add together to give the resultant. . Two of them will add together to give the resultant. TIPS FOR SUCCESS In the video, the vectors were represented with an arrow above them rather than in bold. This is a common notation in math classes. Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 159
Using the Analytical Method of Vector Addition and Subtraction to Solve Problems Figure 5.25 uses the analytical method to add vectors. WORKED EXAMPLE An Accelerating Subway Train Add the vector the y-axis is along the north–south directions. A person first walks vector The person then walks in a direction to the vector shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and north of east, represented by in a direction north of east, represented by vector Figure 5.25 You can use analytical models to add vectors. Strategy The components of We will solve for these components and then add them in the x-direction and y-direction to find the resultant. along the x- and y-axes represent walking due east and due north to get to the same ending point. and Solution First, we find the components of and along the x- and y-axes. From the problem, we know that = , and . We find the x-components by using , which gives and Similarly, the y-components are found using and The x- and y-components of the resultant are and 160 Chapter 5 • Motion in Two Dimensions Now we can find the magnitude of the resultant by using the Pythagorean theorem 5.8 so that Finally, we find the direction of the resultant This is Discussion This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, components of . Therefore, the x- and y-components of the resultant . The components of – are the negatives of the are and and the rest of the method outlined above is identical to that for addition. Practice Problems 5. What is the magnitude of a vector whose x-component is 4 cm and whose y-component is 3 cm? 1 cm a. 5 cm b. c. 7 cm d. 25 cm 6. What is the magnitude of a vector that makes an angle of 30° to the horizontal and whose x-component is 3 units? a. 2.61 units 3.00 units b. c. 3.46 units d. 6.00 units Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 161 LINKS TO PHYSICS Atmospheric Science Figure 5.26 This picture shows Bert Foord during a television Weather Forecast from the Meteorological Office in 1963. (BBC TV) Atmospheric science is a physical science, meaning that it is a science based heavily on physics. Atmospheric science includes meteorology (the study of weather) and climatology (the study of climate). Climate is basically the average weather over a longer time scale. Weather changes quickly over time, whereas the climate changes more gradually. The movement of air, water and heat is vitally important to climatology and meteorology. Since motion is such a major factor in weather and climate, this field uses vectors for much of its math. Vectors are used to represent currents in the ocean, wind velocity and forces acting on a parcel of air. You have probably seen a weather map using vectors to show the strength (magnitude) and direction of the wind. Vectors used in atmospheric science are often three-dimensional. We won’t cover three-dimensional motion in this text, but to go from two-dimensions to three-dimensions, you simply add a third vector component. Three-dimensional motion is represented as a combination of x-, y- and zcomponents, where zis the altitude. Vector calculus combines vector math with calculus, and is often used to find the rates of change in temperature, pressure or wind speed over time or distance. This is useful information, since atmospheric motion is driven by changes in pressure or temperature. The greater the variation in pressure over a given distance, the stronger the wind to try to correct that imbalance. Cold air tends to be more dense and therefore has higher pressure than warm air. Higher pressure air rushes into a region of lower pressure and gets deflected by the spinning of the Earth, and friction slows the wind at Earth’s surface. Finding how wind changes over distance and multiplying vectors lets meteorologists, like the one shown in Figure 5.26, figure out how much rotation (spin) there is in the atmosphere at any given time and location. This is an important tool for tornado prediction. Conditions with greater rotation are more likely to produce tornadoes. GRASP CHECK Why are vectors used so frequently in atmospheric science? a. Vectors have magnitude as well as direction and can be quickly solved through scalar algebraic operations. b. Vectors have magnitude but no direction, so it becomes easy to express physical quantities involved in the atmospheric science. c. Vectors can be solved very accurately through geometry, which helps to make better predictions in atmospheric science. d. Vectors have magnitude as well as direction and are used in equations that describe the three dimensional motion of the atmosphere. Check Your Understanding 7. Between the analytical and graphical methods of vector additions, which is more accurate? Why? a. The analytical method is less accurate than the graphical method, because the former involves geometry and 162 Chapter 5 • Motion in Two Dimensions trigonometry. b. The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations. c. The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale. d. The analytical method is more accurate than the graphical method, because the latter is limited by the precision of the drawing. 8. What is a component of a two dimensional vector? a. A component is a piece of a vector that points in either the xor ydirection. b. A component is a piece of a vector that has half of the magnitude of the original vector. c. A component is a piece of a vector that points in the direction opposite to the original vector. d. A component is a piece of a vector that points in the same direction as original vector but with double of its magnitude. 9. How can we determine the global angle (measured counter-clockwise from positive ) if we know and ? a. b. c. d. 10. How can we determine the magnitude of a vector if we know the magnitudes of its components? a. b. c. d. 5.3 Projectile Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the properties of projectile motion • Apply kinematic equations and vectors to solve problems involving projectile motion Section Key Terms air resistance maximum height (of a projectile) projectile projectile motion range trajectory Properties of Projectile Motion Projectile motion is the motion of an object thrown (projected) into the air. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory. As an object travels through the air, it encounters a frictional force that slows its motion called air resistance. Air resistance does significantly alter trajectory motion, but due to the difficulty in calculation, it is ignored in introductory physics. The most important concept in projectile motion is that horizontal and vertical motions areindependent, meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly. Access for free at openstax.org. Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical. 5.3 • Projectile Motion 163 Figure 5.27 The diagram shows the projectile motion of a cannonball shot at a horizontal angle versus one dropped with no horizontal velocity. Note that both cannonballs have the same vertical position over time. We’ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are xand y, as illustrated in Figure 5.28. Figure 5.28 A boy kicks a ball at angle θ, and it is displaced a distance of s along its trajectory. As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple ay = –g = –9.80 m/s2. Note that this definition defines the upwards direction as positive. Because gravity is vertical, ax = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1. (when ) (when ) Table 5.1 Summary of Kinematic Equations (constant a) Where x is position, x0 is initial position, v is velocity, vavg is average velocity, tis time and a is acceleration. 164 Chapter 5 • Motion in Two Dimensions Solve Problems Involving Projectile Motion The following steps are used to analyze projectile motion: 1. Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so and are used. The magnitudes of the displacement along x- and y-axes are called and The magnitudes of the components of the velocity are and , where is the magnitude of the velocity and is its direction. Initial values are denoted with a subscript 0. 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Vertical motion (assuming positive is up ) 3. Solve for the unknowns in the two separate motions (one horizontal and
one vertical). Note that the only common variable between the motions is time . The problem solving procedures here are the same as for one-dimensional kinematics. . We can use the analytical method of vector 4. Recombine the two motions to find the total displacement and velocity addition, which uses displacement and velocity. and to find the magnitude and direction of the total is the direction of the displacement , and is the direction of the velocity . (See Figure 5.29 Access for free at openstax.org. 5.3 • Projectile Motion 165 Figure 5.29 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x- and y-motions are recombined to give the total velocity at any given point on the trajectory. TIPS FOR SUCCESS For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for and . Usually, it is simplest to set the initial position of the object so that and . 166 Chapter 5 • Motion in Two Dimensions WATCH PHYSICS Projectile at an Angle This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle. Click to view content (https://www.khanacademy.org/embed_video?v=ZZ39o1rAZWY) GRASP CHECK Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component is unchanged, what is the effect on the time of flight? a. The time to reach the ground would remain the same since the vertical component is unchanged. b. The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled. c. The time to reach the ground would be halved since the horizontal component of the velocity is doubled. d. The time to reach the ground would be doubled since the horizontal component of the velocity is doubled. WORKED EXAMPLE A Fireworks Projectile Explodes High and Away During a fireworks display like the one illustrated in Figure 5.30, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Figure 5.30 The diagram shows the trajectory of a fireworks shell. Strategy The motion can be broken into horizontal and vertical motions in which be zero and solve for the maximum height. and . We can then define and to Solution for (a) By height we mean the altitude or vertical position above the starting point. The highest point in any trajectory, the maximum height, is reached when ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of vy when the firework reaches its maximum height, we use the following equation to find Access for free at openstax.org. Because and are both zero, the equation simplifies to Solving for gives Now we must find initial velocity of 70.0 m/s, and is the initial angle. Thus, , the component of the initial velocity in the y-direction. It is given by 5.3 • Projectile Motion 167 , where is the and is so that Discussion for (a) Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. Solution for (b) There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use is zero, this equation reduces to . Because Note that the final vertical velocity, , at the highest point is zero. Therefore, Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using equation for . , and solving the quadratic Solution for (c) Because air resistance is negligible, velocity multiplied by time as given by and the horizontal velocity is constant. The horizontal displacement is horizontal , where is equal to zero where is the x-component of the velocity, which is given by Now, The time for both motions is the same, and so is Discussion for (c) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the 168 Chapter 5 • Motion in Two Dimensions forces of the explosion. The expression we found for while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height ; then, This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity. WORKED EXAMPLE Calculating Projectile Motion: Hot Rock Projectile Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31, with a speed of the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. and at an angle above Figure 5.31 The diagram shows the projectile motion of a large rock from a volcano. Strategy Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion. Solution While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using If we take the initial position vertical component of the initial velocity, found from to be zero, then the final position is Now the initial vertical velocity is the 5.9 Substituting known values yields Rearranging terms gives a quadratic equation in This expression is a quadratic equation of the form –20.0. Its solutions are given by the quadratic formula , where the constants are a= 4.90, b= –14.3, and c= Access for free at openstax.org. 5.3 • Projectile Motion 169 This equation yields two solutions t= 3.96 and t= –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore, Discussion The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Practice Problems 11. If an object is thrown horizontally, travels with an average x-component of its velocity equal to , and does not hit the ground, what will be the x-component of the displacement after a. b. c. d. ? 12. If a ball is thrown straight up with an initial velocity of upward, what is the maximum height it will reach? a. b. c. d. The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32. Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons. 170 Chapter 5 • Motion in Two Dimensions Figure 5.32 Trajectories of projectiles on level ground. (a) The greater the initial speed , the greater the range for a given initial angle. (b) The effect of initial angle on the range of a projectile with a given initial speed. Note that any combination of trajectories that add to 90 degrees will have the same range in the absence of air resistance, although the maximum heights of those paths are different. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed shown in the figure above. The initial angle range of a projectile on level groundis also has a dramatic effect on the range. When air resistance is negligible, the , the greater the range, as is the initial speed and where apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally. is the initial angle relative to the horizontal. It is important to note that the range doesn’t Virtual Physics Projectile Motion In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target. Click to view content (https://archive.cnx.org/specials/317dbd00-8e61-4065-b3eb-f2b80db9b7ed/projectile-motion/) Access for free at openstax.org. 5.4 • Inclined Planes 171 GRASP CHECK If a projectile is launched on level ground, what launch angle maximizes the range of the projec
tile? a. b. c. d. Check Your Understanding 13. What is projectile motion? a. Projectile motion is the motion of an object projected into the air, which moves under the influence of gravity. b. Projectile motion is the motion of an object projected into the air which moves independently of gravity. c. Projectile motion is the motion of an object projected vertically upward into the air which moves under the influence of gravity. d. Projectile motion is the motion of an object projected horizontally into the air which moves independently of gravity. 14. What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance? a. The nuclear force b. The gravitational force c. The electromagnetic force d. The contact force 5.4 Inclined Planes Section Learning Objectives By the end of this section, you will be able to do the following: • Distinguish between static friction and kinetic friction • Solve problems involving inclined planes Section Key Terms kinetic friction static friction Static Friction and Kinetic Friction Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice. There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects. Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going. Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. 172 Chapter 5 • Motion in Two Dimensions Figure 5.33 Frictional forces, such as f, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. The magnitude of the frictional force has two forms: one for static friction, the other for kinetic friction. When there is no motion between the objects, the magnitude of static friction fs is is the coefficient of static friction and N is the magnitude of the normal force. Recall that the normal force opposes the where force of gravity and acts perpendicular to the surface in this example, but not always. Since the symbol means less than or equal to, this equation says that static friction can have a maximum value of That is, Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Once an object is moving, the magnitude of kinetic friction fk is given by where is the coefficient of kinetic friction. Friction varies from surface to surface because different substances are rougher than others. Table 5.2 compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact. System Static Friction Kinetic Friction Rubber on dry concrete Rubber on wet concrete Wood on wood 1.0 0.7 0.5 Waxed wood on wet snow 0.14 Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel 0.5 0.6 0.05 0.04 Bone lubricated by synovial fluid 0.016 Shoes on wood 0.9 Table 5.2 Coefficients of Static and Kinetic Friction 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 0.015 0.7 Access for free at openstax.org. 5.4 • Inclined Planes 173 System Static Friction Kinetic Friction Shoes on ice Ice on ice Steel on ice 0.1 0.1 0.4 0.05 0.03 0.02 Table 5.2 Coefficients of Static and Kinetic Friction Since the direction of friction is always opposite to the direction of motion, friction runs parallel to the surface between objects and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N would keep it moving at a constant speed. If the floor were lubricated, both coefficients would be much smaller than they would be without lubrication. The coefficient of friction is unitless and is a number usually between 0 and 1.0. Working with Inclined Planes We discussed previously that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Up until now, we dealt only with normal force in one dimension, with gravity and normal force acting perpendicular to the surface in opposing directions (gravity downward, and normal force upward). Now that you have the skills to work with forces in two dimensions, we can explore what happens to weight and the normal force on a tilted surface such as an inclined plane. For inclined plane problems, it is easier breaking down the forces into their components if we rotate the coordinate system, as illustrated in Figure 5.34. The first step when setting up the problem is to break down the force of weight into components. Figure 5.34 The diagram shows perpendicular and horizontal components of weight on an inclined plane. When an object rests on an incline that makes an angle with the horizontal, the force of gravity acting on the object is divided into two components: A force acting perpendicular to the plane, perpendicular force of weight, acting parallel to the plane, the object, so it acts upward along the plane. , is typically equal in magnitude and opposite in direction to the normal force, The force , opposes the motion of , causes the object to accelerate down the incline. The force of friction, , and a force acting parallel to the plane, . The 174 Chapter 5 • Motion in Two Dimensions It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle to the horizontal, then the magnitudes of the weight components are Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between and . Knowing this property, you can use trigonometry to determine the magnitude of the weight components WATCH PHYSICS Inclined Plane Force Components This video (https://www.khanacademy.org/embed_video?v=TC23wD34C7k) shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. GRASP CHECK Click to view content (https://www.youtube.com/embed/TC23wD34C7k) This video shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. When the surface is flat, you could say that one of the components of the gravitational force is zero; Which one? As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle shrinks while the sine of the angle increases. b. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle increases while the sine of the angle shrinks. c. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle shrinks while the sine of the angle increases. d. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle increases while the sine of the angle shrinks. TIPS FOR SUCCESS Normal force is represented by the variable This should not be confused with the symbol for the newton, which is also represented by the letter N. It is important to tell apart these symbols, especially since the units for normal force ( to be newtons (N). For
example, the normal force, difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! , that the floor exerts on a chair might be One important ) happen To review, the process for solving inclined plane problems is as follows: Access for free at openstax.org. 5.4 • Inclined Planes 175 1. Draw a sketch of the problem. 2. 3. Draw a free-body diagram (which is a sketch showing all of the forces acting on an object) with the coordinate system Identify known and unknown quantities, and identify the system of interest. rotated at the same angle as the inclined plane. Resolve the vectors into horizontal and vertical components and draw them on the free-body diagram. 4. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then Fnet x= 0. If the object does accelerate in that direction, Fnet x= ma. 5. Check your answer. Is the answer reasonable? Are the units correct? WORKED EXAMPLE Finding the Coefficient of Kinetic Friction on an Inclined Plane A skier, illustrated in Figure 5.35(a), with a mass of 62 kg is sliding down a snowy slope at an angle of 25 degrees. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Figure 5.35 Use the diagram to help find the coefficient of kinetic friction for the skier. Strategy . Therefore, The magnitude of kinetic friction was given as 45.0 N. Kinetic friction is related to the normal force N as we can find the coefficient of kinetic friction by first finding the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope. That is, Substituting this into our expression for kinetic friction, we get which can now be solved for the coefficient of kinetic friction μk. Solution Solving for gives Substituting known values on the right-hand side of the equation, Discussion This result is a little smaller than the coefficient listed in Table 5.2 for waxed wood on snow, but it is still reasonable since values 176 Chapter 5 • Motion in Two Dimensions of the coefficients of friction can vary greatly. In situations like this, where an object of mass mslides down a slope that makes an angle θwith the horizontal, friction is given by WORKED EXAMPLE Weight on an Incline, a Two-Dimensional Problem The skier’s mass, including equipment, is 60.0 kg. (See Figure 5.36(b).) (a) What is her acceleration if friction is negligible? (b) What is her acceleration if the frictional force is 45.0 N? Figure 5.36 Now use the diagram to help find the skier's acceleration if friction is negligible and if the frictional force is 45.0 N. Strategy The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Remember that motions along perpendicular axes are independent. We use the symbol perpendicular, and to mean parallel. to mean , and in the free-body diagram. The only external forces acting on the system are the skier’s weight, friction, and the normal force exerted by the ski slope, labeled , direction of either axis, so we must break it down into components along the chosen axes. We define weight parallel to the slope and two separate problems of forces parallel to the slope and forces perpendicular to the slope. to be the component of the component of weight perpendicular to the slope. Once this is done, we can consider the is always perpendicular to the slope and is parallel to it. But is not in the Solution The magnitude of the component of the weight parallel to the slope is the component of the weight perpendicular to the slope is , and the magnitude of (a) Neglecting friction: Since the acceleration is parallel to the slope, we only need to consider forces parallel to the slope. Forces perpendicular to the slope add to zero, since there is no acceleration in that direction. The forces parallel to the slope are the amount of the skier’s weight parallel to the slope acceleration parallel to the slope is and friction . Assuming no friction, by Newton’s second law the Where the net force parallel to the slope , so that is the acceleration. (b) Including friction: Here we now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now Access for free at openstax.org. 5.4 • Inclined Planes 177 and substituting this into Newton’s second law, gives We substitute known values to get or which is the acceleration parallel to the incline when there is 45 N opposing friction. Discussion Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is not. Practice Problems 15. When an object sits on an inclined plane that makes an angle θwith the horizontal, what is the expression for the component of the objects weight force that is parallel to the incline? a. b. c. d. rests on a plane inclined from horizontal. What is the component of the weight force that 16. An object with a mass of is parallel to the incline? a. b. c. d. Snap Lab Friction at an Angle: Sliding a Coin An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in the first Worked Example, the kinetic friction on a slope opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out , and the component of the weight down the slope is equal to . These forces act in Solving for , since we find that • • • 1 coin 1 book 1 protractor 1. Put a coin flat on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. 5.10 178 Chapter 5 • Motion in Two Dimensions 2. Measure the angle of tilt relative to the horizontal and find . GRASP CHECK True or False—If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False Check Your Understanding 17. What is friction? a. Friction is an internal force that opposes the relative motion of an object. b. Friction is an internal force that accelerates an object’s relative motion. c. Friction is an external force that opposes the relative motion of an object. d. Friction is an external force that increases the velocity of the relative motion of an object. 18. What are the two varieties of friction? What does each one act upon? a. Kinetic and static friction both act on an object in motion. b. Kinetic friction acts on an object in motion, while static friction acts on an object at rest. c. Kinetic friction acts on an object at rest, while static friction acts on an object in motion. d. Kinetic and static friction both act on an object at rest. 19. Between static and kinetic friction between two surfaces, which has a greater value? Why? a. The kinetic friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. b. The static friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. c. The kinetic friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. d. The static friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. 5.5 Simple Harmonic Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Hooke’s law and Simple Harmonic Motion • Describe periodic motion, oscillations, amplitude, frequency, and period • Solve problems in simple harmonic motion involving springs and pendulums Section Key Terms amplitude deformation equilibrium position frequency Hooke’s law oscillate period periodic motion restoring force simple harmonic motion simple pendulum Hooke’s Law and Simple Harmonic Motion Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This Access for free at openstax.org. 5.5 • Simple Harmonic Motion 179 second property is known as Hooke’s law. In equation form, Hooke’s law is where x is the amount of deformation (the change in length, for example) produced by the restoring force F, and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke’s law is solving problems involving spring
s and pendulums, which we will cover at the end of this section. Oscillations and Periodic Motion What do an ocean buoy, a child in a swing, a guitar, and the beating of hearts all have in common? They all oscillate. That is, they move back and forth between two points, like the ruler illustrated in Figure 5.37. All oscillations involve force. For example, you push a child in a swing to get the motion started. Figure 5.37 A ruler is displaced from its equilibrium position. Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 5.38. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until it gradually loses all of its energy. The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke’s law describes this situation with the equation F = −kx. Therefore, Hooke’s law describes and applies to the simplest case of oscillation, known as simple harmonic motion. Figure 5.38 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. 180 Chapter 5 • Motion in Two Dimensions When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each vibration of the string takes the same time as the previous one. Periodic motion is a motion that repeats itself at regular time intervals, such as with an object bobbing up and down on a spring or a pendulum swinging back and forth. The time to complete one oscillation (a complete cycle of motion) remains constant and is called the period T. Its units are usually seconds. Frequency fis the number of oscillations per unit time. The SI unit for frequency is the hertz (Hz), defined as the number of oscillations per second. The relationship between frequency and period is As you can see from the equation, frequency and period are different ways of expressing the same concept. For example, if you get a paycheck twice a month, you could say that the frequency of payment is two per month, or that the period between checks is half a month. If there is no friction to slow it down, then an object in simple motion will oscillate forever with equal displacement on either side of the equilibrium position. The equilibrium position is where the object would naturally rest in the absence of force. The maximum displacement from equilibrium is called the amplitude X. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, shown in Figure 5.39, the units of amplitude and displacement are meters. Figure 5.39 An object attached to a spring sliding on a frictionless surface is a simple harmonic oscillator. When displaced from equilibrium, the object performs simple harmonic motion that has an amplitude X and a period T. The object’s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T. The mass mand the force constant k are the onlyfactors that affect the period and frequency of simple harmonic motion. The period of a simple harmonic oscillator is given by and, because f= 1/T, the frequency of a simple harmonic oscillator is Access for free at openstax.org. 5.5 • Simple Harmonic Motion 181 WATCH PHYSICS Introduction to Harmonic Motion This video shows how to graph the displacement of a spring in the x-direction over time, based on the period. Watch the first 10 minutes of the video (you can stop when the narrator begins to cover calculus). Click to view content (https://www.khanacademy.org/embed_video?v=Nk2q-_jkJVs) GRASP CHECK If the amplitude of the displacement of a spring were larger, how would this affect the graph of displacement over time? What would happen to the graph if the period was longer? a. Larger amplitude would result in taller peaks and troughs and a longer period would result in greater separation in time between peaks. b. Larger amplitude would result in smaller peaks and troughs and a longer period would result in greater distance between peaks. c. Larger amplitude would result in taller peaks and troughs and a longer period would result in shorter distance between peaks. d. Larger amplitude would result in smaller peaks and troughs and a longer period would result in shorter distance between peaks. Solving Spring and Pendulum Problems with Simple Harmonic Motion Before solving problems with springs and pendulums, it is important to first get an understanding of how a pendulum works. Figure 5.40 provides a useful illustration of a simple pendulum. Figure 5.40 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of −mgsinθ toward the equilibrium position—that is, a restoring force. Everyday examples of pendulums include old-fashioned clocks, a child’s swing, or the sinker on a fishing line. For small displacements of less than 15 degrees, a pendulum experiences simple harmonic oscillation, meaning that its restoring force is directly proportional to its displacement. A pendulum in simple harmonic motion is called a simple pendulum. A pendulum has an object with a small mass, also known as the pendulum bob, which hangs from a light wire or string. The equilibrium position for a pendulum is where the angle is zero (that is, when the pendulum is hanging straight down). It makes sense that without any force applied, this is where the pendulum bob would rest. The displacement of the pendulum bob is the arc length s. The weight mg has components mg cos along the string and mg sin tangent to the arc. Tension in the string exactly cancels the component mg cos parallel to the string. This leaves a net restoring force back toward the equilibrium position that runs tangent to the arc and equals −mg sin . 182 Chapter 5 • Motion in Two Dimensions For a simple pendulum, The period is The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass or amplitude. However, note that Tdoes depend on g. This means that if we know the length of a pendulum, we can actually use it to measure gravity! This will come in useful in Figure 5.40. TIPS FOR SUCCESS Tension is represented by the variable T, and period is represented by the variable T. It is important not to confuse the two, since tension is a force and period is a length of time. WORKED EXAMPLE Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find g given the period Tand the length Lof a pendulum. We can solve for g, assuming that the angle of deflection is less than 15 degrees. Recall that when the angle of deflection is less than 15 degrees, the pendulum is considered to be in simple harmonic motion, allowing us to use this equation. Solution 1. Square and solve for g. 2. Substitute known values into the new equation. 3. Calculate to find g. Discussion This method for determining g can be very accurate. This is why length and period are given to five digits in this example. WORKED EXAMPLE Hooke’s Law: How Stiff Are Car Springs? What is the force constant for the suspension system of a car, like that shown in Figure 5.41, that settles 1.20 cm when an 80.0-kg person gets in? Access for free at openstax.org. 5.5 • Simple Harmonic Motion 183 Figure 5.41 A car in a parking lot. (exfordy, Flickr) Strategy Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = −1.20×10−2 m. At that point, the springs supply a restoring force F equal to the person’s weight w = mg = (80.0 kg)(9.80 m/s2) = 784 N. We take this force to be F in Hooke’s law. Knowing F and x, we can then solve for the force constant k. Solution Solve Hooke’s law, F = −kx, for k. Substitute known values and solve for k. Discussion Note that F and x have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in, if it were not for the shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Practice Problems applied to a spring causes it to be displaced by . What is the force constant of the spring? 20. A force of a. b. c. d. 21. What is the force constant for the suspension system of a car that settles when a person gets in? a. b. 184 Chapter 5 • Motion in Two Dimensions c. d. Snap Lab Finding Gravity Using a Simple Pendulum Use a simple pendulum to find the acceleration due to gravity g in your home or classroom. •
• • 1 string 1 stopwatch 1 small dense object 1. Cut a piece of a string or dental floss so that it is about 1 m long. 2. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). 3. Starting at an angle of less than 10 degrees, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. 4. Calculate g. GRASP CHECK How accurate is this measurement for g? How might it be improved? a. Accuracy for value of gwill increase with an increase in the mass of a dense object. b. Accuracy for the value of gwill increase with increase in the length of the pendulum. c. The value of gwill be more accurate if the angle of deflection is more than 15°. d. The value of gwill be more accurate if it maintains simple harmonic motion. Check Your Understanding 22. What is deformation? a. Deformation is the magnitude of the restoring force. b. Deformation is the change in shape due to the application of force. c. Deformation is the maximum force that can be applied on a spring. d. Deformation is regaining the original shape upon the removal of an external force. 23. According to Hooke’s law, what is deformation proportional to? a. Force b. Velocity c. Displacement d. Force constant 24. What are oscillations? a. Motion resulting in small displacements b. Motion which repeats itself periodically c. Periodic, repetitive motion between two points d. motion that is the opposite to the direction of the restoring force 25. True or False—Oscillations can occur without force. a. True b. False Access for free at openstax.org. Chapter 5 • Key Terms 185 KEY TERMS air resistance a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero maximum height (of a projectile) the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile amplitude the maximum displacement from the oscillate moving back and forth regularly between two equilibrium position of an object oscillating around the equilibrium position analytical method the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities component (of a 2-dimensional vector) a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components points period time it takes to complete one oscillation periodic motion motion that repeats itself at regular time intervals projectile an object that travels through the air and experiences only acceleration due to gravity projectile motion the motion of an object that is subject only to the acceleration of gravity range the maximum horizontal distance that a projectile deformation displacement from equilibrium, or change in travels shape due to the application of force restoring force force acting in opposition to the force equilibrium position where an object would naturally rest caused by a deformation in the absence of force frequency number of events per unit of time graphical method drawing vectors on a graph to add them using the head-to-tail method head (of a vector) the end point of a vector; the location of the sum of the a collection of vectors resultant resultant vector simple harmonic motion the oscillatory motion in a the vector sum of two or more vectors system where the net force can be described by Hooke’s law the vector’s arrow; also referred to as the tip simple pendulum an object with a small mass suspended head-to-tail method a method of adding vectors in which from a light wire or string the tail of each vector is placed at the head of the previous vector Hooke’s law proportional relationship between the force F static friction a force that opposes the motion of two systems that are in contact and are not moving relative to one another on a material and the deformation it causes, tail the starting point of a vector; the point opposite to the kinetic friction a force that opposes the motion of two systems that are in contact and moving relative to one another SECTION SUMMARY 5.1 Vector Addition and Subtraction: Graphical Methods • The graphical method of adding vectors and involves drawing vectors on a graph and adding them by using the head-to-tail method. The resultant vector is defined such that A + B = R. The magnitude and direction of protractor. are then determined with a ruler and • The graphical method of subtracting vectors A and B involves adding the opposite of vector B, which is defined as −B. In this case, Next, use the head-totail method as for vector addition to obtain the resultant vector . • Addition of vectors is independent of the order in which they are added; A + B = B + A. • The head-to-tail method of adding vectors involves head or tip of the arrow trajectory the path of a projectile through the air vector addition adding together two or more vectors drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. • Variables in physics problems, such as force or velocity, can be represented with vectors by making the length of the vector proportional to the magnitude of the force or velocity. • Problems involving displacement, force, or velocity may be solved graphically by measuring the resultant vector’s magnitude with a ruler and measuring the direction with a protractor. 5.2 Vector Addition and Subtraction: Analytical Methods • The analytical method of vector addition and subtraction uses the Pythagorean theorem and trigonometric identities to determine the magnitude 186 Chapter 5 • Key Equations and direction of a resultant vector. • The steps to add vectors method are as follows: and using the analytical 1. Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations and 2. Add the horizontal and vertical components of each vector to determine the components resultant vector, and of the and 3. Use the Pythagorean theorem to determine the magnitude, , of the resultant vector • To solve projectile problems: choose a coordinate system; analyze the motion in the vertical and horizontal direction separately; then, recombine the horizontal and vertical components using vector addition equations. 5.4 Inclined Planes • Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. A normal force is always perpendicular to the contact surface between systems. Friction depends on both of the materials involved. • µs is the coefficient of static friction, which depends on both of the materials. • µk is the coefficient of kinetic friction, which also depends on both materials. • When objects rest on an inclined plane that makes an angle with the horizontal surface, the weight of the object can be broken into components that act perpendicular the plane. and parallel ( ) to the surface of 4. Use a trigonometric identity to determine the direction, • An oscillation is a back and forth motion of an object 5.5 Simple Harmonic Motion , of 5.3 Projectile Motion • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. • Projectile motion in the horizontal and vertical directions are independent of one another. • The maximum height of an projectile is the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile. • The range is the maximum horizontal distance traveled by a projectile. KEY EQUATIONS 5.2 Vector Addition and Subtraction: Analytical Methods resultant magnitude resultant direction x-component of a vector A (when an angle is given relative to the horizontal) Access for free at openstax.org. between two points of deformation. • An oscillation may create a wave, which is a disturbance that propagates from where it was created. • The simplest type of oscillations are related to systems that can be described by Hooke’s law. • Periodic motion is a repetitious oscillation. • The time for one oscillation is the period T. • The number of oscillations per unit time is the frequency • A mass msuspended by a wire of length Lis a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15 degrees. y-component of a vector A (when an angle is given relative to the horizontal) addition of vectors 5.3 Projectile Motion angle of displacement velocity angle of velocity maximum height range Chapter 5 • Chapter Review 187 perpendicular component of weight on an inclined plane parallel component of weight on an inclined plane 5.4 Inclined Planes Hooke’s law 5.5 Simple Harmonic Motion force of static friction force of kinetic friction period in simple harmonic motion frequency in simple harmonic motion period of a simple pendulum CHAPTER REVIEW Concept Items 5.1 Vector Addition and Subtraction: Graphical Methods d. 5.2 Vector Addition and Subtraction: Analytical Methods 1. There is a vector , with magnitude 5 units pointing 4. What is the angle between the x and y components of a , with magnitude 3 units, towards west and vector pointing towards south. Using vector addition, calculate the magnitude of the resultant vector. a. 4.0 b. 5.8 c. 6.3 d. 8.0 2. If you draw two vectors using the head-to-tail method, how can you then draw the resultant vector? a. By joining the head of the first vector to the head of the last vector? a. b. c. d. 5. Two vectors are equal in magnitude and opposite in direction. What is the magnitude of their resultant vector? a. The magnitude of the resultant vector will be zero. b. The magnitude of resultant vector will be twice the ma
gnitude of the original vector. b. By joining the head of the first vector with the tail of c. The magnitude of resultant vector will be same as the last magnitude of the original vector. c. By joining the tail of the first vector to the head of d. The magnitude of resultant vector will be half the the last magnitude of the original vector. d. By joining the tail of the first vector with the tail of the last 3. What is the global angle of south of west? a. b. c. 6. How can we express the x and y-components of a vector , and direction, global angle in terms of its magnitude, ? a. b. c. 188 Chapter 5 • Chapter Review d. 7. True or False—Every 2-D vector can be expressed as the c. d. 12. What equation gives the magnitude of kinetic friction? a. b. c. d. 5.5 Simple Harmonic Motion 13. Why is there a negative sign in the equation for Hooke’s law? a. The negative sign indicates that displacement decreases with increasing force. b. The negative sign indicates that the direction of the applied force is opposite to that of displacement. c. The negative sign indicates that the direction of the restoring force is opposite to that of displacement. d. The negative sign indicates that the force constant must be negative. 14. With reference to simple harmonic motion, what is the equilibrium position? a. The position where velocity is the minimum b. The position where the displacement is maximum c. The position where the restoring force is the maximum d. The position where the object rests in the absence of force 15. What is Hooke’s law? a. Restoring force is directly proportional to the displacement from the mean position and acts in the the opposite direction of the displacement. b. Restoring force is directly proportional to the displacement from the mean position and acts in the same direction as the displacement. c. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the opposite direction of the displacement. d. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the same direction as the displacement. displacement will be the same as it would have been had he followed directions correctly. a. True b. False product of its x and y-components. a. True b. False 5.3 Projectile Motion 8. Horizontal and vertical motions of a projectile are independent of each other. What is meant by this? a. Any object in projectile motion falls at the same rate as an object in freefall, regardless of its horizontal velocity. b. All objects in projectile motion fall at different rates, regardless of their initial horizontal velocities. c. Any object in projectile motion falls at the same rate as its initial vertical velocity, regardless of its initial horizontal velocity. d. All objects in projectile motion fall at different rates and the rate of fall of the object is independent of the initial velocity. 9. Using the conventional choice for positive and negative axes described in the text, what is the y-component of the acceleration of an object experiencing projectile motion? a. b. c. d. 5.4 Inclined Planes 10. True or False—Kinetic friction is less than the limiting static friction because once an object is moving, there are fewer points of contact, and the friction is reduced. For this reason, more force is needed to start moving an object than to keep it in motion. a. True b. False 11. When there is no motion between objects, what is the relationship between the magnitude of the static friction and the normal force ? a. b. Critical Thinking Items 5.1 Vector Addition and Subtraction: Graphical Methods 16. True or False—A person is following a set of directions. He has to walk 2 km east and then 1 km north. He takes a wrong turn and walks in the opposite direction for the second leg of the trip. The magnitude of his total Access for free at openstax.org. 5.2 Vector Addition and Subtraction: Analytical Methods 17. What is the magnitude of a vector whose x-component and whose angle is ? is a. b. c. d. 18. Vectors and are equal in magnitude and opposite in direction. Does have the same direction as vector or ? a. b. Chapter 5 • Chapter Review 189 force pushes against it and it starts moving when is applied to it. What can be said about the coefficient of kinetic friction between the box and the floor? a. b. c. d. 22. The component of the weight parallel to an inclined with the horizontal is plane of an object resting on an incline that makes an angle of the object’s mass? a. b. c. d. . What is 5.3 Projectile Motion 5.5 Simple Harmonic Motion 19. Two identical items, object 1 and object 2, are dropped building. Object 1 is dropped , while object 2 is thrown . from the top of a with an initial velocity of straight downward with an initial velocity of What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground? after object 2. a. Object 1 will hit the ground after object 2. b. Object 1 will hit the ground c. Object 1 will hit the ground at the same time as object 2. d. Object 1 will hit the ground after object 2. 20. An object is launched into the air. If the y-component of its acceleration is 9.8 m/s2, which direction is defined as positive? a. Vertically upward in the coordinate system b. Vertically downward in the coordinate system c. Horizontally to the right side of the coordinate system d. Horizontally to the left side of the coordinate system 5.4 Inclined Planes 21. A box weighing is at rest on the floor. A person Problems 5.1 Vector Addition and Subtraction: Graphical Methods 25. A person attempts to cross a river in a straight line by navigating a boat at . If the river flows at from his left to right, what would be the magnitude of the boat’s resultant velocity? In what direction would the boat go, relative to the straight line 23. Two springs are attached to two hooks. Spring A has a greater force constant than spring B. Equal weights are suspended from both. Which of the following statements is true? a. Spring A will have more extension than spring B. b. Spring B will have more extension than spring A. c. Both springs will have equal extension. d. Both springs are equally stiff. 24. Two simple harmonic oscillators are constructed by attaching similar objects to two different springs. The force constant of the spring on the left is that of the spring on the right is force is applied to both, which of the following statements is true? a. The spring on the left will oscillate faster than and . If the same spring on the right. b. The spring on the right will oscillate faster than the spring on the left. c. Both the springs will oscillate at the same rate. d. The rate of oscillation is independent of the force constant. across it? a. The resultant velocity of the boat will be The boat will go toward his right at an angle of to a line drawn across the river. b. The resultant velocity of the boat will be The boat will go toward his left at an angle of to a line drawn across the river. c. The resultant velocity of the boat will be The boat will go toward his right at an angle of . . . 190 Chapter 5 • Chapter Review to a line drawn across the river. d. The resultant velocity of the boat will be . The boat will go toward his left at an angle of to a line drawn across the river. 26. A river flows in a direction from south west to north east . A boat captain wants to cross at a velocity of this river to reach a point on the opposite shore due east of the boat’s current position. The boat moves at . Which direction should it head towards if the resultant velocity is a. b. c. d. It should head in a direction It should head in a direction It should head in a direction It should head in a direction east of south. south of east. east of south. south of east. ? 5.2 Vector Addition and Subtraction: Analytical Methods 30. A person wants to fire a water balloon cannon such that it hits a target 100 m away. If the cannon can only be launched at 45° above the horizontal, what should be the initial speed at which it is launched? 31.3 m/s a. 37.2 m/s b. c. 980.0 m/s 1,385.9 m/s d. 5.4 Inclined Planes 31. A coin is sliding down an inclined plane at constant to the horizontal, velocity. If the angle of the plane is what is the coefficient of kinetic friction? a. b. c. d. 27. A person walks 10.0 m north and then 2.00 m east. 32. A skier with a mass of 55 kg is skiing down a snowy slope Solving analytically, what is the resultant displacement of the person? a. b. c. d. = 10.2 m, θ = 78.7º east of north = 10.2 m, θ = 78.7º north of east = 12.0 m, θ = 78.7º east of north = 12.0 m, θ = 78.7º north of east that has an incline of 30°. Find the coefficient of kinetic friction for the skier if friction is known to be 25 N . a. b. c. d. 28. A person walks north of west for and 5.5 Simple Harmonic Motion south of west for . What is the magnitude 33. What is the time period of a long pendulum on of his displacement? Solve analytically. a. b. c. d. earth? a. b. c. d. 34. A simple harmonic oscillator has time period . If the , what is the force constant of mass of the system is the spring used? a. b. c. d. period 2 seconds). 5.3 Projectile Motion 29. A water balloon cannon is fired at at an angle of above the horizontal. How far away will it fall? a. b. c. d. Performance Task 5.5 Simple Harmonic Motion 35. Construct a seconds pendulum (pendulum with time Access for free at openstax.org. TEST PREP Multiple Choice 5.1 Vector Addition and Subtraction: Graphical Methods 36. True or False—We can use Pythagorean theorem to calculate the length of the resultant vector obtained from the addition of two vectors which are at right angles to each other. a. True b. False 37. True or False—The direction of the resultant vector depends on both the magnitude and direction of added vectors. a. True b. False with a headwind blowing . What is the resultant velocity 38. A plane flies north at from the north at of the plane
? a. b. c. d. north south north south 39. Two hikers take different routes to reach the same spot. southeast, then turns and goes The first one goes at south of east. The second hiker goes south. How far and in which direction must the second hiker travel now, in order to reach the first hiker's location destination? a. b. c. d. east south east south 5.2 Vector Addition and Subtraction: Analytical Methods 40. When will the x-component of a vector with angle be greater than its y-component? a. b. c. d. 41. The resultant vector of the addition of vectors and . The magnitudes of is are , respectively. Which of the following is true? a. , and , , , and b. c. Chapter 5 • Test Prep 191 d. 42. What is the dimensionality of vectors used in the study of atmospheric sciences? a. One-dimensional b. Two-dimensional c. Three-dimensional 5.3 Projectile Motion 43. After a projectile is launched in the air, in which direction does it experience constant, non-zero acceleration, ignoring air resistance? a. The x direction b. The y direction c. Both the x and y directions d. Neither direction 44. Which is true when the height of a projectile is at its maximum? a. b. c. 45. A ball is thrown in the air at an angle of 40°. If the maximum height it reaches is 10 m, what must be its initial speed? a. 17.46 m/s b. 21.78 m/s 304.92 m/s c. d. 474.37 m/s 46. A large rock is ejected from a volcano with a speed of above the horizontal. The and at an angle rock strikes the side of the volcano at an altitude of lower than its starting point. Calculate the horizontal displacement of the rock. a. b. c. d. 5.4 Inclined Planes 47. For objects of identical masses but made of different materials, which of the following experiences the most static friction? a. Shoes on ice b. Metal on wood c. Teflon on steel 48. If an object sits on an inclined plane and no other object makes contact with the object, what is typically equal in magnitude to the component of the weight perpendicular to the plane? 192 Chapter 5 • Test Prep a. The normal force b. The total weight c. The parallel force of weight 49. A 5 kg box is at rest on the floor. The coefficient of static friction between the box and the floor is 0.4. A horizontal force of 50 N is applied to the box. Will it move? a. No, because the applied force is less than the maximum limiting static friction. b. No, because the applied force is more than the maximum limiting static friction. c. Yes, because the applied force is less than the maximum limiting static friction. d. Yes, because the applied force is more than the maximum limiting static friction. 50. A skier with a mass of 67 kg is skiing down a snowy slope with an incline of 37°. Find the friction if the coefficient of kinetic friction is 0.07. a. 27.66 N 34.70 N b. 36.71 N c. d. 45.96 N 5.5 Simple Harmonic Motion 51. A change in which of the following is an example of deformation? a. Velocity b. Length c. Mass d. Weight 52. The units of amplitude are the same as those for which of the following measurements? a. Speed b. Displacement c. Acceleration d. Force 53. Up to approximately what angle is simple harmonic motion a good model for a pendulum? a. b. c. d. 54. How would simple harmonic motion be different in the absence of friction? a. Oscillation will not happen in the absence of friction. b. Oscillation will continue forever in the absence of friction. c. Oscillation will have changing amplitude in the absence of friction. d. Oscillation will cease after a certain amount of time in the absence of friction. 55. What mass needs to be attached to a spring with a force in order to make a simple harmonic constant of oscillator oscillate with a time period of a. b. c. d. ? Short Answer 5.1 Vector Addition and Subtraction: Graphical Methods 56. Find for the following vectors: the addition of these vectors? a. Zero b. Six c. Eight d. Twelve 108 cm, a. 108 cm, b. c. 206 cm, d. 206 cm, 57. Find for the following vectors: 108 cm, a. b. 108 cm, c. 232 cm, d. 232 cm, 59. Two people pull on ropes tied to a trolley, each applying 44 N of force. The angle the ropes form with each other is 39.5°. What is the magnitude of the net force exerted on the trolley? a. 0.0 N b. 79.6 N c. 82.8 N d. 88.0 N 5.2 Vector Addition and Subtraction: Analytical Methods 60. True or False—A vector can form the shape of a right 58. Consider six vectors of 2 cm each, joined from head to tail making a hexagon. What would be the magnitude of angle triangle with its x and y components. a. True Access for free at openstax.org. b. False 61. True or False—All vectors have positive x and y a. True b. False Chapter 5 • Test Prep 193 69. For what angle of a projectile is its range equal to zero? . What is in terms of and a. b. c. d. or or or or 5.4 Inclined Planes 70. What are the units of the coefficient of friction? . What is in terms of and a. b. c. d. unitless 71. Two surfaces in contact are moving slowly past each components. a. True b. False 62. Consider ? a. b. c. d. 63. Consider ? a. b. c. d. 64. When a three dimensional vector is used in the study of atmospheric sciences, what is z? a. Altitude b. Heat c. Temperature d. Wind speed 65. Which method is not an application of vector calculus? a. To find the rate of change in atmospheric temperature b. To study changes in wind speed and direction c. To predict changes in atmospheric pressure d. To measure changes in average rainfall 5.3 Projectile Motion 66. How can you express the velocity, terms of its initial velocity, time, ? a. b. c. d. 67. In the equation for the maximum height of a projectile, what does stand for? Initial velocity in the x direction a. b. Initial velocity in the y direction c. Final velocity in the x direction d. Final velocity in the y direction other. As the relative speed between the two surfaces in contact increases, what happens to the magnitude of their coefficient of kinetic friction? a. It increases with the increase in the relative motion. It decreases with the increase in the relative motion. It remains constant and is independent of the relative motion. b. c. 72. When will an object slide down an inclined plane at constant velocity? a. When the magnitude of the component of the weight along the slope is equal to the magnitude of the frictional force. b. When the magnitude of the component of the weight along the slope is greater than the magnitude of the frictional force. c. When the magnitude of the component of the weight perpendicular to the slope is less than the magnitude of the frictional force. 73. A box is sitting on an inclined plane. At what angle of incline is the perpendicular component of the box's weight at its maximum? a. b. c. d. 5.5 Simple Harmonic Motion 74. What is the term used for changes in shape due to the , of a projectile in , acceleration, , and d. When the magnitude of the component of the weight perpendicular to the slope is equal to the magnitude of the frictional force. 68. True or False—Range is defined as the maximum vertical distance travelled by a projectile. application of force? a. Amplitude 194 Chapter 5 • Test Prep b. Deformation c. Displacement d. Restoring force 75. What is the restoring force? a. The normal force on the surface of an object b. The weight of a mass attached to an object c. Force which is applied to deform an object from its original shape d. Force which brings an object back to its equilibrium position 76. For a given oscillator, what are the factors that affect its period and frequency? a. Mass only b. Force constant only c. Applied force and mass d. Mass and force constant 77. For an object in simple harmonic motion, when does the Extended Response 5.1 Vector Addition and Subtraction: Graphical Methods 80. True or False—For vectors the order of addition is important. a. True b. False 81. Consider five vectors a, b, c, d,and e.Is it true or false that their addition always results in a vector with a greater magnitude than if only two of the vectors were added? a. True b. False 5.2 Vector Addition and Subtraction: Analytical Methods 82. For what angle of a vector is it possible that its magnitude will be equal to its y-component? a. b. c. d. 83. True or False—If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False 84. True or false—We can find the magnitude and direction of the resultant vector if we know the angles of two vectors and the magnitude of one. Access for free at openstax.org. maximum speed occur? a. At the extreme positions b. At the equilibrium position c. At the moment when the applied force is removed d. Midway between the extreme and equilibrium positions 78. What is the equilibrium position of a pendulum? a. When the tension in the string is zero b. When the pendulum is hanging straight down c. When the tension in the string is maximum d. When the weight of the mass attached is minimum 79. If a pendulum is displaced by an angle θ, what is the net restoring force it experiences? a. mgsinθ b. mgcosθ c. –mgsinθ d. –mgcosθ a. True b. False 5.3 Projectile Motion 85. Ignoring drag, what is the x-component of the acceleration of a projectile? Why? a. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. b. The x component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. c. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. d. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. 86. What is the optimum angle at which a projectile should be launched in order to cover the maximum distance? a. b. c. d. 5.4 Inclined Planes 87. True or False—Friction varies from surface to surface because different substances have different degrees of rough
ness or smoothness. a. True b. False 88. As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. Both the perpendicular and the parallel component d. The force constant kis related to the friction in the system: The larger the force constant, the lower the friction in the system. Chapter 5 • Test Prep 195 will decrease. b. The perpendicular component will decrease and the parallel component will increase. c. The perpendicular component will increase and the parallel component will decrease. d. Both the perpendicular and the parallel component will increase. 5.5 Simple Harmonic Motion 89. What physical characteristic of a system is its force constant related to? a. The force constant kis related to the stiffness of a system: The larger the force constant, the stiffer the system. b. The force constant kis related to the stiffness of a system: The larger the force constant, the looser the system. c. The force constant kis related to the friction in the system: The larger the force constant, the greater the friction in the system. 90. How or why does a pendulum oscillate? a. A pendulum oscillates due to applied force. b. A pendulum oscillates due to the elastic nature of the string. c. A pendulum oscillates due to restoring force arising from gravity. d. A pendulum oscillates due to restoring force arising from tension in the string. 91. If a pendulum from earth is taken to the moon, will its frequency increase or decrease? Why? a. It will increase because on the Moon is less than on Earth. It will decrease because on the Moon is less than b. c. d. on Earth. It will increase because on the Moon is greater than on Earth. It will decrease because on the Moon is greater than on Earth. 196 Chapter 5 • Test Prep Access for free at openstax.org. CHAPTER 6 Circular and Rotational Motion Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly. The same physical principles are involved in both of these motions. (Richard Munckton). Chapter Outline 6.1 Angle of Rotation and Angular Velocity 6.2 Uniform Circular Motion 6.3 Rotational Motion You may recall learning about various aspects of motion along a straight line: kinematics (where we learned INTRODUCTION about displacement, velocity, and acceleration), projectile motion (a special case of two-dimensional kinematics), force, and Newton’s laws of motion. In some ways, this chapter is a continuation of Newton’s laws of motion. Recall that Newton’s first law tells us that objects move along a straight line at constant speed unless a net external force acts on them. Therefore, if an object moves along a circular path, such as the car in the photo, it must be experiencing an external force. In this chapter, we explore both circular motion and rotational motion. 198 Chapter 6 • Circular and Rotational Motion 6.1 Angle of Rotation and Angular Velocity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the angle of rotation and relate it to its linear counterpart • Describe angular velocity and relate it to its linear counterpart • Solve problems involving angle of rotation and angular velocity Section Key Terms angle of rotation angular velocity arc length circular motion radius of curvature rotational motion spin tangential velocity Angle of Rotation What exactly do we mean by circular motionor rotation? Rotational motion is the circular motion of an object about an axis of rotation. We will discuss specifically circular motion and spin. Circular motion is when an object moves in a circular path. Examples of circular motion include a race car speeding around a circular curve, a toy attached to a string swinging in a circle around your head, or the circular loop-the-loopon a roller coaster. Spin is rotation about an axis that goes through the center of mass of the object, such as Earth rotating on its axis, a wheel turning on its axle, the spin of a tornado on its path of destruction, or a figure skater spinning during a performance at the Olympics. Sometimes, objects will be spinning while in circular motion, like the Earth spinning on its axis while revolving around the Sun, but we will focus on these two motions separately. When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity, acceleration, and force) but take into account the curvature or rotation of the motion. Here, we define the angle of rotation, which is the angular equivalence of distance; and angular velocity, which is the angular equivalence of linear velocity. When objects rotate about some axis—for example, when the CD in Figure 6.2 rotates about its center—each point in the object follows a circular path. Figure 6.2 All points on a CD travel in circular paths. The pits (dots) along a line from the center to the edge all move through the same angle in time . The arc length, , is the distance traveled along a circular path. The radius of curvature, r, is the radius of the circular path. Both are shown in Figure 6.3. Access for free at openstax.org. 6.1 • Angle of Rotation and Angular Velocity 199 Figure 6.3 The radius (r) of a circle is rotated through an angle . The arc length, , is the distance covered along the circumference. Consider a line from the center of the CD to its edge. In a given time, each pit(used to record information) on this line moves through the same angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation is the arc length divided by the radius of curvature. The angle of rotation is often measured by using a unit called the radian. (Radians are actually dimensionless, because a radian is defined as the ratio of two distances, radius and arc length.) A revolution is one complete rotation, where every point on the radians (or 360 degrees), and therefore has an angle of rotation circle returns to its original position. One revolution covers of revolutions, and degrees using the relationship radians, and an arc length that is the same as the circumference of the circle. We can convert between radians, 1 revolution = rad = 360°. See Table 6.1 for the conversion of degrees to radians for some common angles. Degree Measures Radian Measures 6.1 Table 6.1 Commonly Used Angles in Terms of Degrees and Radians Angular Velocity How fast is an object rotating? We can answer this question by using the concept of angular velocity. Consider first the angular speed is the rate at which the angle of rotation changes. In equation form, the angular speed is 6.2 which means that an angular rotation given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s). occurs in a time, . If an object rotates through a greater angle of rotation in a Now let’s consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the 200 Chapter 6 • Circular and Rotational Motion angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counterclockwise, the angular velocity points toward you along the axis of rotation. Angular velocity (ω) is the angular version of linear velocity v. Tangential velocity is the instantaneous linear velocity of an object in rotational motion. To get the precise relationship between angular velocity and tangential velocity, consider again a pit on the rotating CD. This pit moves through an arc length so its tangential speedis in a shorttime From the definition of the angle of rotation, , we see that . Substituting this into the expression for vgives 6.3 says that the tangential speed vis proportional to the distance rfrom the center of rotation. Consequently, The equation tangential speed is greater for a point on the outer edge of the CD (with larger r) than for a point closer to the center of the CD (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation. See Figure 6.4. Figure 6.4 Points 1 and 2 rotate through the same angle ( ), but point 2 moves through a greater arc length ( ) because it is farther from the center of rotation. means large vbecause Now, consider another example: the tire of a moving car (see Figure 6.5). The faster the tire spins, the faster the car moves—large , will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time. . Similarly, a larger-radius tire rotating at the same angular velocity, Access for free at openstax.org. 6.1 • Angle of Rotation and Angular Velocity 201 Figure 6.5 A car moving at a velocity, v, to the right has a tire rotating with angular velocity . The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up and the wheels spinning without touching the road. Directly below the axle, where the tire touches the road, the tire tread moves backward with respect to the axle with tangential velocity , where ris the tire radius. Because the road is stationary with respect to this point of the tire, the car must move forward at the linear velocity v. A larger angular velocity for the tire means a greater linear velocity for the car. However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the
arc length through which the tire treads move is greater than the linear distance through which the car moves. It’s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast. TIPS FOR SUCCESS Angular velocity ω and tangential velocity v are vectors, so we must include magnitude and direction. The direction of the angular velocity is along the axis of rotation, and points away from you for an object rotating clockwise, and toward you for an object rotating counterclockwise. In mathematics this is described by the right-hand rule. Tangential velocity is usually described as up, down, left, right, north, south, east, or west, as shown in Figure 6.6. Figure 6.6 As the fly on the edge of an old-fashioned vinyl record moves in a circle, its instantaneous velocity is always at a tangent to the circle. The direction of the angular velocity is into the page this case. 202 Chapter 6 • Circular and Rotational Motion WATCH PHYSICS Relationship between Angular Velocity and Speed This video reviews the definition and units of angular velocity and relates it to linear speed. It also shows how to convert between revolutions and radians. Click to view content (https://www.youtube.com/embed/zAx61CO5mDw) GRASP CHECK For an object traveling in a circular path at a constant angular speed, would the linear speed of the object change if the radius of the path increases? a. Yes, because tangential speed is independent of the radius. b. Yes, because tangential speed depends on the radius. c. No, because tangential speed is independent of the radius. d. No, because tangential speed depends on the radius. Solving Problems Involving Angle of Rotation and Angular Velocity Snap Lab Measuring Angular Speed In this activity, you will create and measure uniform circular motion and then contrast it with circular motions with different radii. • One string (1 m long) • One object (two-hole rubber stopper) to tie to the end • One timer Procedure 1. Tie an object to the end of a string. 2. Swing the object around in a horizontal circle above your head (swing from your wrist). It is important that the circle be horizontal! 3. Maintain the object at uniform speed as it swings. 4. Measure the angular speed of the object in this manner. Measure the time it takes in seconds for the object to travel 10 revolutions. Divide that time by 10 to get the angular speed in revolutions per second, which you can convert to radians per second. 5. What is the approximate linear speed of the object? 6. Move your hand up the string so that the length of the string is 90 cm. Repeat steps 2–5. 7. Move your hand up the string so that its length is 80 cm. Repeat steps 2–5. 8. Move your hand up the string so that its length is 70 cm. Repeat steps 2–5. 9. Move your hand up the string so that its length is 60 cm. Repeat steps 2–5 10. Move your hand up the string so that its length is 50 cm. Repeat steps 2–5 11. Make graphs of angular speed vs. radius (i.e. string length) and linear speed vs. radius. Describe what each graph looks like. GRASP CHECK If you swing an object slowly, it may rotate at less than one revolution per second. What would be the revolutions per second for an object that makes one revolution in five seconds? What would be its angular speed in radians per second? . The angular speed of the object would be a. The object would spin at . b. The object would spin at . The angular speed of the object would be . Access for free at openstax.org. c. The object would spin at d. The object would spin at . The angular speed of the object would be . The angular speed of the object would be . . 6.1 • Angle of Rotation and Angular Velocity 203 Now that we have an understanding of the concepts of angle of rotation and angular velocity, we’ll apply them to the real-world situations of a clock tower and a spinning tire. WORKED EXAMPLE Angle of rotation at a Clock Tower The clock on a clock tower has a radius of 1.0 m. (a) What angle of rotation does the hour hand of the clock travel through when it moves from 12 p.m. to 3 p.m.? (b) What’s the arc length along the outermost edge of the clock between the hour hand at these two times? Strategy We can figure out the angle of rotation by multiplying a full revolution ( hour hand in going from 12 to 3. Once we have the angle of rotation, we can solve for the arc length by rearranging the equation radians) by the fraction of the 12 hours covered by the since the radius is given. Solution to (a) In going from 12 to 3, the hour hand covers 1/4 of the 12 hours needed to make a complete revolution. Therefore, the angle between the hour hand at 12 and at 3 is (i.e., 90 degrees). Solution to (b) Rearranging the equation we get Inserting the known values gives an arc length of 6.4 6.5 6.6 Discussion We were able to drop the radians from the final solution to part (b) because radians are actually dimensionless. This is because the radian is defined as the ratio of two distances (radius and arc length). Thus, the formula gives an answer in units of meters, as expected for an arc length. WORKED EXAMPLE How Fast Does a Car Tire Spin? Calculate the angular speed of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h). See Figure 6.5. Strategy In this case, the speed of the tire tread with respect to the tire axle is the same as the speed of the car with respect to the road, so we have v= 15.0 m/s. The radius of the tire is r= 0.300 m. Since we know vand r, we can rearrange the equation , to get and find the angular speed. Solution To find the angular speed, we use the relationship: . 204 Chapter 6 • Circular and Rotational Motion Inserting the known quantities gives Discussion When we cancel units in the above calculation, we get 50.0/s (i.e., 50.0 per second, which is usually written as 50.0 s−1). But the angular speed must have units of rad/s. Because radians are dimensionless, we can insert them into the answer for the angular speed because we know that the motion is circular. Also note that, if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular speed of 6.8 6.7 Practice Problems 1. What is the angle in degrees between the hour hand and the minute hand of a clock showing 9:00 a.m.? a. 0° b. 90° 180° c. 360° d. 2. What is the approximate value of the arc length between the hour hand and the minute hand of a clock showing 10:00 a.m if the radius of the clock is 0.2 m? a. 0.1 m b. 0.2 m c. 0.3 m d. 0.6 m Check Your Understanding 3. What is circular motion? a. Circular motion is the motion of an object when it follows a linear path. b. Circular motion is the motion of an object when it follows a zigzag path. c. Circular motion is the motion of an object when it follows a circular path. d. Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. 4. What is meant by radius of curvature when describing rotational motion? a. The radius of curvature is the radius of a circular path. b. The radius of curvature is the diameter of a circular path. c. The radius of curvature is the circumference of a circular path. d. The radius of curvature is the area of a circular path. 5. What is angular velocity? a. Angular velocity is the rate of change of the diameter of the circular path. b. Angular velocity is the rate of change of the angle subtended by the circular path. c. Angular velocity is the rate of change of the area of the circular path. d. Angular velocity is the rate of change of the radius of the circular path. 6. What equation defines angular velocity, ? Take that is the radius of curvature, is the angle, and is time. a. b. c. d. 7. Identify three examples of an object in circular motion. Access for free at openstax.org. 6.2 • Uniform Circular Motion 205 a. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a top spinning on its axis b. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head c. Earth spinning on its own axis, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head d. Earth spinning on its own axis, blades of a working ceiling fan, and a top spinning on its own axis 8. What is the relative orientation of the radius and tangential velocity vectors of an object in uniform circular motion? a. Tangential velocity vector is always parallel to the radius of the circular path along which the object moves. b. Tangential velocity vector is always perpendicular to the radius of the circular path along which the object moves. c. Tangential velocity vector is always at an acute angle to the radius of the circular path along which the object moves. d. Tangential velocity vector is always at an obtuse angle to the radius of the circular path along which the object moves. 6.2 Uniform Circular Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe centripetal acceleration and relate it to linear acceleration • Describe centripetal force and relate it to linear force • Solve problems involving centripetal acceleration and centripetal force Section Key Terms centrifugal force centripetal acceleration centripetal force uniform circular motion Centripetal Acceleration In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular mot
ion is always accelerating, even though the magnitude of its velocity is constant. You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes. Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine smaller, then the acceleration would point exactlytoward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking. becoming smaller and 206 Chapter 6 • Circular and Rotational Motion Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of , points exactly toward the center of the circle (but this is hard to draw). Because , the acceleration is also toward the center, so ac is called centripetal acceleration. Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed vin a circular path with radius r, the magnitude of centripetal acceleration is Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that ac is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h. We can also express ac in terms of the magnitude of angular velocity. Substituting into the equation above, we get . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is 6.9 TIPS FOR SUCCESS The equation expressed in the form ac = rω2 is useful for solving problems where you know the angular velocity rather than the tangential velocity. Virtual Physics Ladybug Motion in 2D In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion. Click to view content (https://archive.cnx.org/specials/317a2b1e-2fbd-11e5-99b5-e38ffb545fe6/ladybug-motion/) Access for free at openstax.org. 6.2 • Uniform Circular Motion 207 GRASP CHECK In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion? a. The angle between acceleration and velocity is 0°, and the body experiences linear acceleration. b. The angle between acceleration and velocity is 0°, and the body experiences centripetal acceleration. c. The angle between acceleration and velocity is 90°, and the body experiences linear acceleration. d. The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration. Centripetal Force Because an object in uniform circular motion undergoes constant acceleration (by changing direction), we know from Newton’s second law of motion that there must be a constant net external force acting on the object. Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac. Therefore, the magnitude of centripetal force, Fc, is . By using the two different forms of the equation for the magnitude of centripetal acceleration, get two expressions involving the magnitude of the centripetal force Fc. The first expression is in terms of tangential speed, the second is in terms of angular speed: , we and and . Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve. 208 Chapter 6 • Circular and Rotational Motion Figure 6.8 In this figure, the frictional force fserves as the centripetal force Fc. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force Fc, the smaller is the radius of curvature rand the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force Fc produces a smaller radius . WATCH PHYSICS Centripetal Force and Acceleration Intuition This video explains why a centripetal force creates centripetal acceleration and uniform circular motion. It also covers the difference between speed and velocity and shows examples of uniform circular motion. Click to view content (https://www.youtube.com/embed/vZOk8NnjILg) GRASP CHECK Imagine that you are swinging a yoyo in a vertical clockwise circle in front of you, perpendicular to the direction you are facing. Now, imagine that the string breaks just as the yoyo reaches its bottommost position, nearest the floor. Which of the following describes the path of the yoyo after the string breaks? a. The yoyo will fly upward in the direction of the centripetal force. b. The yoyo will fly downward in the direction of the centripetal force. Access for free at openstax.org. 6.2 • Uniform Circular Motion 209 c. The yoyo will fly to the left in the direction of the tangential velocity. d. The yoyo will fly to the right in the direction of the tangential velocity. Solving Centripetal Acceleration and Centripetal Force Problems To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve. Snap Lab Estimating Centripetal Acceleration In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is or . • One tennis racket or golf club • One timer • One ruler or tape measure Procedure 1. Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket. 2. Describe the motion of the swing—is this uniform circular motion? Why or why not? 3. Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make? 4. Measure the radius of curvature. What did you physically measure? 5. By using the timer, find either the linear or angular velocity, depending on which equation you decide to use. 6. What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate? GRASP CHECK Was it more useful to use the equation or in this activity? Why? a. It should be simpler to use because measuring angular velocity through observation would be easier. b. c. It should be simpler to use because measuring tangential velocity through observation would be easier. It should be simpler to use because measuring angular velocity through observation would be difficult. d. It should be simpler to use because measuring tangential velocity through observation would be difficult. WORKED EXAMPLE Comparing Centripetal Acceleration of a Car Rounding a Curve with Acceleration Due to Gravity A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle
curve taken at highway speed with acceleration due to gravity (g). 210 Chapter 6 • Circular and Rotational Motion Strategy Because linear rather than angular speed is given, it is most convenient to use the expression the centripetal acceleration. to find the magnitude of Solution Entering the given values of v= 25.0 m/s and r= 500 m into the expression for ac gives Discussion To compare this with the acceleration due to gravity (g= 9.80 m/s2), we take the ratio . Therefore, , which means that the centripetal acceleration is about one tenth the acceleration due to gravity. WORKED EXAMPLE Frictional Force on Car Tires Rounding a Curve a. Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s. b. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line. Access for free at openstax.org. 6.2 • Uniform Circular Motion 211 Strategy and Solution for (a) We know that . Therefore, Strategy and Solution for (b) The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve. Discussion Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since . Practice Problems 9. What is the centripetal acceleration of an object with speed going along a path of radius ? a. b. c. d. 10. Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s. a. b. c. d. 1 m/s 5 m/s 1 m/s2 5 m/s2 Check Your Understanding 11. What is uniform circular motion? 212 Chapter 6 • Circular and Rotational Motion a. Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity. b. Uniform circular motion is when an object travels on a circular path at a variable acceleration. c. Uniform circular motion is when an object travels on a circular path at a constant speed. d. Uniform circular motion is when an object travels on a circular path at a variable speed. 12. What is centripetal acceleration? a. The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit b. The acceleration of an object moving in a circular path and directed tangentially along the circular path c. The acceleration of an object moving in a linear path and directed in the direction of motion of the object d. The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object 13. Is there a net force acting on an object in uniform circular motion? a. Yes, the object is accelerating, so a net force must be acting on it. b. Yes, because there is no acceleration. c. No, because there is acceleration. d. No, because there is no acceleration. 14. Identify two examples of forces that can cause centripetal acceleration. a. The force of Earth’s gravity on the moon and the normal force b. The force of Earth’s gravity on the moon and the tension in the rope on an orbiting tetherball c. The normal force and the force of friction acting on a moving car d. The normal force and the tension in the rope on a tetherball 6.3 Rotational Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe rotational kinematic variables and equations and relate them to their linear counterparts • Describe torque and lever arm • Solve problems involving torque and rotational kinematics Section Key Terms angular acceleration kinematics of rotational motion lever arm tangential acceleration torque Rotational Kinematics In the section on uniform circular motion, we discussed motion in a circle at constant speed and, therefore, constant angular velocity. However, there are times when angular velocity is not constant—rotational motion can speed up, slow down, or reverse directions. Angular velocity is not constant when a spinning skater pulls in her arms, when a child pushes a merry-go-round to make it rotate, or when a CD slows to a halt when switched off. In all these cases, angular acceleration occurs because the angular velocity rate of change of angular velocity. In equation form, angular acceleration is changes. The faster the change occurs, the greater is the angular acceleration. Angular acceleration is the increases, then is the change in angular velocity and is the change in time. The units of angular acceleration are (rad/s)/s, or rad/ where s2. If is negative. Keep in mind that, by convention, counterclockwise is the positive direction and clockwise is the negative direction. For example, the skater in Figure 6.9 is rotating counterclockwise as seen from above, so her angular velocity is positive. Acceleration would be negative, for example, when an object that is rotating counterclockwise slows down. It would be positive when an object that is rotating counterclockwise speeds up. decreases, then is positive. If Access for free at openstax.org. 6.3 • Rotational Motion 213 Figure 6.9 A figure skater spins in the counterclockwise direction, so her angular velocity is normally considered to be positive. (Luu, Wikimedia Commons) The relationship between the magnitudes of tangential acceleration, a, and angular acceleration, 6.10 These equations mean that the magnitudes of tangential acceleration and angular acceleration are directly proportional to each other. The greater the angular acceleration, the larger the change in tangential acceleration, and vice versa. For example, consider riders in their pods on a Ferris wheel at rest. A Ferris wheel with greater angular acceleration will give the riders greater tangential acceleration because, as the Ferris wheel increases its rate of spinning, it also increases its tangential velocity. Note that the radius of the spinning object also matters. For example, for a given angular acceleration , a smaller Ferris wheel leads to a smaller tangential acceleration for the riders. TIPS FOR SUCCESS Tangential acceleration is sometimes denoted at. It is a linear acceleration in a direction tangent to the circle at the point of interest in circular or rotational motion. Remember that tangential acceleration is parallel to the tangential velocity (either in the same direction or in the opposite direction.) Centripetal acceleration is always perpendicular to the tangential velocity. So far, we have defined three rotational variables: Table 6.2 shows how they are related. , , and . These are the angular versions of the linear variables x, v, and a. Rotational Linear Relationship x Table 6.2 Rotational and Linear Variables 214 Chapter 6 • Circular and Rotational Motion Rotational Linear Relationship v a Table 6.2 Rotational and Linear Variables , and We can now begin to see how rotational quantities like , that starts at rest has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. Putting this in terms of the variables, if the wheel’s angular acceleration the final angular velocity undergoes a large linear acceleration, then it has a large final velocity and will have traveled a large distance. and angle of rotation are large. In the case of linear motion, if an object starts at rest and are related to each other. For example, if a motorcycle wheel is large for a long period of time t, then The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describesmotion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: (constant a). As in linear kinematics, we assume a is constant, which means that angular acceleration The equation for the kinematics relationship between , , and tis is also a constant, because . is the initial angular velocity. Notice that the equation is identical to the linear version, except with angular analogs of where the linear variables. In fact, all of the linear kinematics equations have rotational analogs, which are given in Table 6.3. These equations can be used to solve rotational or linear kinematics problem in which a and are constant. Rotational Linear constant , a constant , a constant , a Table 6.3 Equations for Rotational Kinematics In these equations, and are initial values, is zero, and the average angular velocity and average velocity are 6.11 Access for free at openstax.org. FUN IN PHYSICS Storm Chasing 6.3 • Rotational Motion 215 Figure 6.10 Tornadoes descend from clouds in funnel-like shapes that spin violently. (Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Storm chasers tend to fall into one of three groups: Amateurs chasing tornadoes as a hobby, atmospheric scientists gathering data for research, weather watchers for news media, or scientists having fun under the guise of work. Storm chasing is a dangerous pastime because tornadoes can change course rapidly with little warning. Since storm chasers follow in the wake of the destruction left by tornadoes, changing flat tires due to debris left on the highway is common. The most active part of the world for tornadoes, called tornado alley, is in the central United States, between the Rocky Mountains and Appalachian Mountains. Tornadoes are perfect examples of rotational motion in action in nature. They come out of severe thunderstorms called supercells, which have a c
olumn of air rotating around a horizontal axis, usually about four miles across. The difference in wind speeds between the strong cold winds higher up in the atmosphere in the jet stream and weaker winds traveling north from the Gulf of Mexico causes the column of rotating air to shift so that it spins around a vertical axis, creating a tornado. Tornadoes produce wind speeds as high as 500 km/h (approximately 300 miles/h), particularly at the bottom where the funnel is narrowest because the rate of rotation increases as the radius decreases. They blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. GRASP CHECK What is the physics term for the eye of the storm? Why would winds be weaker at the eye of the tornado than at its outermost edge? a. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to radius of curvature. b. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to radius of curvature. c. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to the square of the radius of curvature. d. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to the square of the radius of curvature. Torque If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity. The farther the force is applied from the pivot point (or fulcrum), the greater the angular acceleration. For example, a door opens slowly if you push too close to its hinge, but opens easily if you push far from the hinges. Furthermore, we know that the more 216 Chapter 6 • Circular and Rotational Motion massive the door is, the more slowly it opens; this is because angular acceleration is inversely proportional to mass. These relationships are very similar to the relationships between force, mass, and acceleration from Newton’s second law of motion. Since we have already covered the angular versions of distance, velocity and time, you may wonder what the angular version of force is, and how it relates to linear force. The angular version of force is torque , which is the turning effectiveness of a force. See Figure 6.11. The equation for the magnitude of torque is where ris the magnitude of the lever arm, F is the magnitude of the linear force, and is the angle between the lever arm and the force. The lever arm is the vector from the point of rotation (pivot point or fulcrum) to the location where force is applied. Since the magnitude of the lever arm is a distance, its units are in meters, and torque has units of N⋅m. Torque is a vector quantity and has the same direction as the angular acceleration that it produces. Figure 6.11 A man pushes a merry-go-round at its edge and perpendicular to the lever arm to achieve maximum torque. Applying a stronger torque will produce a greater angular acceleration. For example, the harder the man pushes the merry-goround in Figure 6.11, the faster it accelerates. Furthermore, the more massive the merry-go-round is, the slower it accelerates for the same torque. If the man wants to maximize the effect of his force on the merry-go-round, he should push as far from the center as possible to get the largest lever arm and, therefore, the greatest torque and angular acceleration. Torque is also maximized when the force is applied perpendicular to the lever arm. Solving Rotational Kinematics and Torque Problems Just as linear forces can balance to produce zero net force and no linear acceleration, the same is true of rotational motion. When two torques of equal magnitude act in opposing directions, there is no net torque and no angular acceleration, as you can see in the following video. If zero net torque acts on a system spinning at a constant angular velocity, the system will continue to spin at the same angular velocity. WATCH PHYSICS Introduction to Torque This video (https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/introduction-totorque) defines torque in terms of moment arm (which is the same as lever arm). It also covers a problem with forces acting in opposing directions about a pivot point. (At this stage, you can ignore Sal’s references to work and mechanical advantage.) GRASP CHECK Click to view content (https://www.openstax.org/l/28torque) If the net torque acting on the ruler from the example was positive instead of zero, what would this say about the angular Access for free at openstax.org. 6.3 • Rotational Motion 217 acceleration? What would happen to the ruler over time? a. The ruler is in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will be zero. b. The ruler is not in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will be zero. c. The ruler is not in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. d. The ruler is in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. Now let’s look at examples applying rotational kinematics to a fishing reel and the concept of torque to a merry-go-round. WORKED EXAMPLE Calculating the Time for a Fishing Reel to Stop Spinning A deep-sea fisherman uses a fishing rod with a reel of radius 4.50 cm. A big fish takes the bait and swims away from the boat, pulling the fishing line from his fishing reel. As the fishing line unwinds from the reel, the reel spins at an angular velocity of 220 rad/s. The fisherman applies a brake to the spinning reel, creating an angular acceleration of −300 rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time tfor the reel to come to a stop. The magnitude of the initial angular velocity is rad/s, rad/s2, and the magnitude of the final angular velocity . The signed magnitude of the angular acceleration is where the minus sign indicates that it acts in the direction opposite to the angular velocity. Looking at the rotational kinematic equations, we see all quantities but tare known in the equation problem. , making it the easiest equation to use for this Solution The equation to use is . We solve the equation algebraically for t, and then insert the known values. 6.12 Discussion The time to stop the reel is fairly small because the acceleration is fairly large. Fishing lines sometimes snap because of the forces involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration and therefore a smaller force. 218 Chapter 6 • Circular and Rotational Motion WORKED EXAMPLE Calculating the Torque on a Merry-Go-Round Consider the man pushing the playground merry-go-round in Figure 6.11. He exerts a force of 250 N at the edge of the merrygo-round and perpendicular to the radius, which is 1.50 m. How much torque does he produce? Assume that friction acting on the merry-go-round is negligible. Strategy To find the torque, note that the applied force is perpendicular to the radius and that friction is negligible. Solution 6.13 Discussion The man maximizes the torque by applying force perpendicular to the lever arm, so that maximizes his torque by pushing at the outer edge of the merry-go-round, so that he gets the largest-possible lever arm. and . The man also Practice Problems 15. How much torque does a person produce if he applies a force away from the pivot point, perpendicularly to the lever arm? a. b. c. d. 16. An object’s angular velocity changes from 3 rad/s clockwise to 8 rad/s clockwise in 5 s. What is its angular acceleration? a. 0.6 rad/s2 1.6 rad/s2 b. 1 rad/s2 c. 5 rad/s2 d. Check Your Understanding 17. What is angular acceleration? a. Angular acceleration is the rate of change of the angular displacement. b. Angular acceleration is the rate of change of the angular velocity. c. Angular acceleration is the rate of change of the linear displacement. d. Angular acceleration is the rate of change of the linear velocity. 18. What is the equation for angular acceleration, α? Assume θis the angle, ωis the angular velocity, and tis time. a. b. c. d. 19. Which of the following best describes torque? It is the rotational equivalent of a force. It is the force that affects linear motion. It is the rotational equivalent of acceleration. It is the acceleration that affects linear motion. a. b. c. d. 20. What is the equation for torque? Access for free at openstax.org. a. b. c. d. 6.3 • Rotational Motion 219 220 Chapter 6 • Key Terms KEY TERMS angle of rotation the ratio of the arc length to the radius of time curvature of a circular path lever arm the distance between the point of rotation (pivot angular acceleration the rate of change of angular velocity point) and the location where force is applied with time radius of curvature the distance between the center of a angular velocity ( ) the rate of change in the angular circular path and the path position of an object following a circular path rotational motion the circular motion of an object about an arc length ( ) the distance traveled by an object along a axis of rotation circular path spin rotation about an axis that goes through the center of centrifugal force a fictitious force that acts in the direction mass of the object opposite the centripetal acceleration tangential acceleration the acceleration in a direction centripetal acceleration the acceleration of an object moving in a circle, directed toward the center of the circle centripetal force any force causing uniform circular tangen
t to the circular path of motion and in the same direction or opposite direction as the tangential velocity tangential velocity the instantaneous linear velocity of an motion object in circular or rotational motion circular motion the motion of an object along a circular torque the effectiveness of a force to change the rotational path speed of an object kinematics of rotational motion the relationships between rotation angle, angular velocity, angular acceleration, and uniform circular motion the motion of an object in a circular path at constant speed that always points toward the center of rotation, perpendicular to the linear velocity, in the same direction as the net force, and in the direction opposite that of the radius vector. • The standard unit for centripetal acceleration is m/s2. • Centripetal force Fc is any net force causing uniform circular motion. 6.3 Rotational Motion • Kinematics is the description of motion. • The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. • Torque is the effectiveness of a force to change the rotational speed of an object. Torque is the rotational analog of force. • The lever arm is the distance between the point of rotation (pivot point) and the location where force is applied. • Torque is maximized by applying force perpendicular to the lever arm and at a point as far as possible from the pivot point or fulcrum. If torque is zero, angular acceleration is zero. SECTION SUMMARY 6.1 Angle of Rotation and Angular Velocity • Circular motion is motion in a circular path. • The angle of rotation is defined as the ratio of the arc length to the radius of curvature. • The arc length is the distance traveled along a circular path and ris the radius of curvature of the circular path. • The angle of rotation (rad), where • Angular velocity where a rotation is measured in units of radians revolution. is the rate of change of an angle, occurs in a time . • The units of angular velocity are radians per second (rad/s). • Tangential speed vand angular speed are related by , and tangential velocity has units of m/s. • The direction of angular velocity is along the axis of rotation, toward (away) from you for clockwise (counterclockwise) motion. 6.2 Uniform Circular Motion • Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. • Centripetal acceleration force is a center-seekingforce Access for free at openstax.org. KEY EQUATIONS 6.1 Angle of Rotation and Angular Velocity 6.3 Rotational Motion Chapter 6 • Key Equations 221 Angle of rotation Angular speed: Tangential speed: 6.2 Uniform Circular Motion Centripetal acceleration Centripetal force or , , Angular acceleration Rotational kinematic equations Tangential (linear) acceleration Torque , , , CHAPTER REVIEW Concept Items 6.2 Uniform Circular Motion 6.1 Angle of Rotation and Angular Velocity 4. What is the equation for centripetal acceleration in terms 1. One revolution is equal to how many radians? Degrees? a. b. c. d. 2. What is tangential velocity? a. Tangential velocity is the average linear velocity of an object in a circular motion. b. Tangential velocity is the instantaneous linear velocity of an object undergoing rotational motion. c. Tangential velocity is the average angular velocity of an object in a circular motion. d. Tangential velocity is the instantaneous angular velocity of an object in a circular motion. 3. What kind of motion is called spin? a. Spin is rotational motion of an object about an axis parallel to the axis of the object. of angular velocity and the radius? a. b. c. d. 5. How can you express centripetal force in terms of centripetal acceleration? a. b. c. d. 6. What is meant by the word centripetal? a. b. c. d. center-seeking center-avoiding central force central acceleration b. Spin is translational motion of an object about an 6.3 Rotational Motion axis parallel to the axis of the object. c. Spin is the rotational motion of an object about its center of mass. d. Spin is translational motion of an object about its own axis. 7. Conventionally, for which direction of rotation of an object is angular acceleration considered positive? a. b. c. d. the positive xdirection of the coordinate system the negative xdirection of the coordinate system the counterclockwise direction the clockwise direction 222 Chapter 6 • Chapter Review 8. When you push a door closer to the hinges, why does it 9. When is angular acceleration negative? open more slowly? a. It opens slowly, because the lever arm is shorter so the torque is large. It opens slowly because the lever arm is longer so the torque is large. It opens slowly, because the lever arm is shorter so the torque is less. It opens slowly, because the lever arm is longer so the torque is less. b. c. d. Critical Thinking Items 6.1 Angle of Rotation and Angular Velocity 10. When the radius of the circular path of rotational motion increases, what happens to the arc length for a given angle of rotation? a. The arc length is directly proportional to the radius of the circular path, and it increases with the radius. b. The arc length is inversely proportional to the radius of the circular path, and it decreases with the radius. c. The arc length is directly proportional to the radius of the circular path, and it decreases with the radius. d. The arc length is inversely proportional to the radius of the circular path, and it increases with the radius. 11. Consider a CD spinning clockwise. What is the sum of the instantaneous velocities of two points on both ends of its diameter? a. b. c. d. 6.2 Uniform Circular Motion 12. What are the directions of the velocity and acceleration of an object in uniform circular motion? a. Velocity is tangential, and acceleration is radially outward. b. Velocity is tangential, and acceleration is radially inward. c. Velocity is radially outward, and acceleration is tangential. d. Velocity is radially inward, and acceleration is tangential. 13. Suppose you have an object tied to a rope and are rotating it over your head in uniform circular motion. If Access for free at openstax.org. displacement and is negative when a. Angular acceleration is the rate of change of the increases. b. Angular acceleration is the rate of change of the decreases. displacement and is negative when c. Angular acceleration is the rate of change of angular velocity and is negative when increases. d. Angular acceleration is the rate of change of angular velocity and is negative when decreases. you increase the length of the rope, would you have to apply more or less force to maintain the same speed? a. More force is required, because the force is inversely proportional to the radius of the circular orbit. b. More force is required because the force is directly proportional to the radius of the circular orbit. c. Less force is required because the force is inversely proportional to the radius of the circular orbit. d. Less force is required because the force is directly proportional to the radius of the circular orbit. 6.3 Rotational Motion 14. Consider two spinning tops with different radii. Both have the same linear instantaneous velocities at their edges. Which top has a higher angular velocity? a. the top with the smaller radius because the radius of curvature is inversely proportional to the angular velocity the top with the smaller radius because the radius of curvature is directly proportional to the angular velocity the top with the larger radius because the radius of curvature is inversely proportional to the angular velocity b. c. d. The top with the larger radius because the radius of curvature is directly proportional to the angular velocity 15. A person tries to lift a stone by using a lever. If the lever arm is constant and the mass of the stone increases, what is true of the torque necessary to lift it? a. It increases, because the torque is directly proportional to the mass of the body. It increases because the torque is inversely proportional to the mass of the body. It decreases because the torque is directly proportional to the mass of the body. It decreases, because the torque is inversely proportional to the mass of the body. b. c. d. Chapter 6 • Test Prep 223 Problems d. 13, 333 N 6.1 Angle of Rotation and Angular Velocity 16. What is the angle of rotation (in degrees) between two hands of a clock, if the radius of the clock is the arc length separating the two hands is a. b. c. d. and ? 17. A clock has radius of . The outermost point on its minute hand travels along the edge. What is its tangential speed? a. b. c. d. 6.2 Uniform Circular Motion 18. What is the centripetal force exerted on a 1,600 kg car that rounds a 100 m radius curve at 12 m/s? 192 N a. b. 1, 111 N c. 2, 300 N Performance Task 6.3 Rotational Motion 22. Design a lever arm capable of lifting a 0.5 kg object such as a stone. The force for lifting should be provided by TEST PREP Multiple Choice 6.1 Angle of Rotation and Angular Velocity 23. What is 1 radian approximately in degrees? 57.3° 360° a. b. c. π° d. 2π° 24. If the following objects are spinning at the same angular velocities, the edge of which one would have the highest speed? a. Mini CD b. Regular CD c. Vinyl record 25. What are possible units for tangential velocity? a. b. 19. Find the frictional force between the tires and the road that allows a 1,000 kg car traveling at 30 m/s to round a 20 m radius curve. a. 22 N b. 667 N c. d. 45, 000 N 1, 500 N 6.3 Rotational Motion 20. An object’s angular acceleration is 36 rad/s2. If it were initially spinning with a velocity of 6.0 m/s, what would its angular velocity be after 5.0 s? 186 rad/s a. 190 rad/s2 b. c. −174 rad/s d. −174 rad/s2 21. When a fan is switched on, it undergoes an angular acceleration of 150 rad/s2. How long will it take to achieve its maximum angular velocity of 50 rad/s? a. −0.3 s b. 0.3 s 3.0 s c. placing coins on the other end of the
lever. How many coins would you need? What happens if you shorten or lengthen the lever arm? What does this say about torque? in radians? c. 26. What is a. b. c. d. 27. For a given object, what happens to the arc length as the angle of rotation increases? a. The arc length is directly proportional to the angle of rotation, so it increases with the angle of rotation. b. The arc length is inversely proportional to the angle of rotation, so it decreases with the angle of rotation. c. The arc length is directly proportional to the angle of rotation, so it decreases with the angle of rotation. 224 Chapter 6 • Test Prep d. The arc length is inversely proportional to the angle of rotation, so it increases with the angle of rotation. 6.2 Uniform Circular Motion 28. Which of these quantities is constant in uniform circular motion? a. Speed b. Velocity c. Acceleration d. Displacement 29. Which of these quantities impact centripetal force? a. Mass and speed only b. Mass and radius only c. Speed and radius only d. Mass, speed, and radius all impact centripetal force 30. An increase in the magnitude of which of these quantities causes a reduction in centripetal force? a. Mass b. Radius of curvature c. Speed 31. What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why? a. It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It increases, because the centripetal acceleration is directly proportional to the radius of curvature. It decreases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It decreases, because the centripetal acceleration is directly proportional to the radius of the curvature. b. c. d. 32. Why do we experience more sideways acceleration while driving around sharper curves? Short Answer 6.1 Angle of Rotation and Angular Velocity 37. What is the rotational analog of linear velocity? a. Angular displacement b. Angular velocity c. Angular acceleration d. Angular momentum 38. What is the rotational analog of distance? a. Rotational angle b. Torque c. Angular velocity d. Angular momentum Access for free at openstax.org. a. Centripetal acceleration is inversely proportional to the radius of curvature, so it increases as the radius of curvature decreases. b. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature decreases. c. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature increases. d. Centripetal acceleration is directly proportional to the radius of curvature, so it increases as the radius of curvature increases. 6.3 Rotational Motion 33. Which of these quantities is not described by the kinematics of rotational motion? a. Rotation angle b. Angular acceleration c. Centripetal force d. Angular velocity 34. In the equation , what is F? a. Linear force b. Centripetal force c. Angular force 35. What happens when two torques act equally in opposite directions? a. Angular velocity is zero. b. Angular acceleration is zero. 36. What is the mathematical relationship between angular and linear accelerations? a. b. c. d. 39. What is the equation that relates the linear speed of a point on a rotating object with the object's angular quantities? a. b. c. d. 40. As the angular velocity of an object increases, what happens to the linear velocity of a point on that object? It increases, because linear velocity is directly a. proportional to angular velocity. It increases, because linear velocity is inversely proportional to angular velocity. b. Chapter 6 • Test Prep 225 c. d. It decreases because linear velocity is directly proportional to angular velocity. It decreases because linear velocity is inversely proportional to angular velocity. a. b. c. d. 41. What is angular speed in terms of tangential speed and 47. What are the standard units for centripetal force? the radius? a. b. c. d. 42. Why are radians dimensionless? a. Radians are dimensionless, because they are defined as a ratio of distances. They are defined as the ratio of the arc length to the radius of the circle. b. Radians are dimensionless because they are defined as a ratio of distances. They are defined as the ratio of the area to the radius of the circle. c. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the arc length to the radius of the circle. d. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the area to the radius of the circle. 6.2 Uniform Circular Motion 43. What type of quantity is centripetal acceleration? a. Scalar quantity; centripetal acceleration has magnitude only but no direction b. Scalar quantity; centripetal acceleration has magnitude as well as direction c. Vector quantity; centripetal acceleration has magnitude only but no direction d. Vector quantity; centripetal acceleration has magnitude as well as direction 44. What are the standard units for centripetal acceleration? a. m/s b. c. d. 45. What is the angle formed between the vectors of tangential velocity and centripetal force? a. b. c. d. 46. What is the angle formed between the vectors of centripetal acceleration and centripetal force? a. m b. m/s c. m/s2 d. newtons 48. As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving at the same speed? a. It increases, because the centripetal force is directly proportional to the mass of the rotating body. It increases, because the centripetal force is inversely proportional to the mass of the rotating body. It decreases, because the centripetal force is directly proportional to the mass of the rotating body. It decreases, because the centripetal force is inversely proportional to the mass of the rotating body. b. c. d. 6.3 Rotational Motion 49. The relationships between which variables are described by the kinematics of rotational motion? a. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, and angular acceleration. b. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and angular momentum. c. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. d. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, torque, and time. 50. What is the kinematics relationship between , , and ? a. b. c. d. 51. What kind of quantity is torque? a. Scalar b. Vector 226 Chapter 6 • Test Prep c. Dimensionless d. Fundamental quantity 52. If a linear force is applied to a lever arm farther away from the pivot point, what happens to the resultant torque? a. b. c. d. It decreases. It increases. It remains the same. It changes the direction. 53. How can the same force applied to a lever produce different torques? a. By applying the force at different points of the lever Extended Response 6.1 Angle of Rotation and Angular Velocity 54. Consider two pits on a CD, one close to the center and one close to the outer edge. When the CD makes one full rotation, which pit would have gone through a greater angle of rotation? Which one would have covered a greater arc length? a. The one close to the center would go through the greater angle of rotation. The one near the outer edge would trace a greater arc length. b. The one close to the center would go through the greater angle of rotation. The one near the center would trace a greater arc length. c. Both would go through the same angle of rotation. The one near the outer edge would trace a greater arc length. d. Both would go through the same angle of rotation. The one near the center would trace a greater arc length. 55. Consider two pits on a CD, one close to the center and one close to the outer edge. For a given angular velocity of the CD, which pit has a higher angular velocity? Which has a higher tangential velocity? a. The point near the center would have the greater angular velocity and the point near the outer edge would have the higher linear velocity. arm along the length of the lever or by changing the angle between the lever arm and the applied force. b. By applying the force at the same point of the lever arm along the length of the lever or by changing the angle between the lever arm and the applied force. c. By applying the force at different points of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. d. By applying the force at the same point of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. the same? a. It increases because tangential velocity is directly proportional to the radius. It increases because tangential velocity is inversely proportional to the radius. It decreases because tangential velocity is directly proportional to the radius. It decreases because tangential velocity is inversely proportional to the radius. b. c. d. 6.2 Uniform Circular Motion 57. Is an object in uniform circular motion accelerating? Why or why not? a. Yes, because the velocity is not constant. b. No, because the velocity is not constant. c. Yes, because the velocity is constant. d. No, because the velocity is constant. 58. An object is in uniform circular motion. Suppose the centripetal force was removed. In which direction would the object now travel? a. b. In the direction of the centripetal force In the direction opposite to the direction of the centripetal force In the direction of the tangential velocity In the direction opposite to the direction of the tangential velocity c. d. b.
The point near the edge would have the greater 59. An object undergoes uniform circular motion. If the angular velocity and the point near the center would have the higher linear velocity. c. Both have the same angular velocity and the point near the outer edge would have the higher linear velocity. d. Both have the same angular velocity and the point near the center would have the higher linear velocity. 56. What happens to tangential velocity as the radius of an object increases provided the angular velocity remains Access for free at openstax.org. radius of curvature and mass of the object are constant, what is the centripetal force proportional to? a. b. c. d. 6.3 Rotational Motion 60. Why do tornadoes produce more wind speed at the Chapter 6 • Test Prep 227 bottom of the funnel? a. Wind speed is greater at the bottom because rate of rotation increases as the radius increases. b. The force should be applied perpendicularly to the lever arm as far as possible from the pivot point. c. The force should be applied parallel to the lever arm b. Wind speed is greater at the bottom because rate of as far as possible from the pivot point. rotation increases as the radius decreases. d. The force should be applied parallel to the lever arm c. Wind speed is greater at the bottom because rate of as close as possible from the pivot point. rotation decreases as the radius increases. d. Wind speed is greater at the bottom because rate of rotation decreases as the radius increases. 61. How can you maximize the torque applied to a given lever arm without applying more force? a. The force should be applied perpendicularly to the lever arm as close as possible from the pivot point. 62. When will an object continue spinning at the same angular velocity? a. When net torque acting on it is zero b. When net torque acting on it is non zero c. When angular acceleration is positive d. When angular acceleration is negative 228 Chapter 6 • Test Prep Access for free at openstax.org. CHAPTER 7 Newton's Law of Gravitation Figure 7.1 Johannes Kepler (left) showed how the planets move, and Isaac Newton (right) discovered that gravitational force caused them to move that way. ((left) unknown, Public Domain; (right) Sir Godfrey Kneller, Public Domain) Chapter Outline 7.1 Kepler's Laws of Planetary Motion 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity INTRODUCTION What do a falling apple and the orbit of the moon have in common? You will learn in this chapter that each is caused by gravitational force. The motion of all celestial objects, in fact, is determined by the gravitational force, which depends on their mass and separation. Johannes Kepler discovered three laws of planetary motion that all orbiting planets and moons follow. Years later, Isaac Newton found these laws useful in developing his law of universal gravitation. This law relates gravitational force to the masses of objects and the distance between them. Many years later still, Albert Einstein showed there was a little more to the gravitation story when he published his theory of general relativity. 7.1 Kepler's Laws of Planetary Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Explain Kepler’s three laws of planetary motion • Apply Kepler’s laws to calculate characteristics of orbits 230 Chapter 7 • Newton's Law of Gravitation Section Key Terms aphelion Copernican model eccentricity Kepler’s laws of planetary motion perihelion Ptolemaic model Concepts Related to Kepler’s Laws of Planetary Motion Examples of orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The moon’s orbit around Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets around the sun are no less interesting. If we look farther, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity. All these motions are governed by gravitational force. The orbital motions of objects in our own solar system are simple enough to describe with a few fairly simple laws. The orbits of planets and moons satisfy the following two conditions: • The mass of the orbiting object, m, is small compared to the mass of the object it orbits, M. • The system is isolated from other massive objects. Based on the motion of the planets about the sun, Kepler devised a set of three classical laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying these two conditions: 1. The orbit of each planet around the sun is an ellipse with the sun at one focus. 2. Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times. 3. The ratio of the squares of the periods of any two planets about the sun is equal to the ratio of the cubes of their average distances from the sun. These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630). He devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. Let’s look closer at each of these laws. Kepler’s First Law The orbit of each planet about the sun is an ellipse with the sun at one focus, as shown in Figure 7.2. The planet’s closest approach to the sun is called aphelion and its farthest distance from the sun is called perihelion. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 231 Figure 7.2 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci (f1 and f2) is constant. (b) For any closed orbit, mfollows an elliptical path with Mat one focus. (c) The aphelion (ra) is the closest distance between the planet and the sun, while the perihelion (rp) is the farthest distance from the sun. If you know the aphelion (ra) and perihelion (rp) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b). Figure 7.3 You can draw an ellipse as shown by putting a pin at each focus, and then placing a loop of string around a pen and the pins and tracing a line on the paper. 232 Chapter 7 • Newton's Law of Gravitation Kepler’s Second Law Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times, as shown in Figure 7.4. Figure 7.4 The shaded regions have equal areas. The time for mto go from A to B is the same as the time to go from C to D and from E to F. The mass mmoves fastest when it is closest to M. Kepler’s second law was originally devised for planets orbiting the sun, but it has broader validity. TIPS FOR SUCCESS Note that while, for historical reasons, Kepler’s laws are stated for planets orbiting the sun, they are actually valid for all bodies satisfying the two previously stated conditions. Kepler’s Third Law The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. In equation form, this is where Tis the period (time for one orbit) and ris the average distance (also called orbital radius). This equation is valid only for comparing two small masses orbiting a single large mass. Most importantly, this is only a descriptive equation; it gives no information about the cause of the equality. LINKS TO PHYSICS History: Ptolemy vs. Copernicus Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 7.5 (a). This is called the Ptolemaic model, named for the Greek philosopher Ptolemy who lived in the second century AD. The Ptolemaic model is characterized by a list of facts for the motions of planets, with no explanation of cause and effect. There tended to be a different rule for each heavenly body and a general lack of simplicity. Figure 7.5 (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all planetary motion in the solar system, but also all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 233 Figure 7.5 (a) The Ptolemaic model of the universe has Earth at the center with the moon, the planets, the sun, and the stars revolving about it in complex circular paths. This geocentric (Earth-centered) model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints about the causes of these motions. (b) The Copernican heliocentric (sun-centered) model is a simpler and more accurate model. Nicolaus Copernicus (1473–1543) first had the idea that the planets circle the sun, in about 1514. It took him almost 20 years to work out the mathematical details for his model. He waited another 10 years or so to publish his work. It is thought he hesitated because he was afraid people would make fun of his theory. Actually, the reaction of many people was more one of fear and anger. Many people felt the Copernican model threatened their basic belief system. About 100 years later, the astronomer Galileo was put under house arrest for providing evidence that planets, including Earth, orbited the sun. In all, it took almost 300 years for everyone to admit that Copernicus had been right all along. GRASP CHECK Explain why Earth does actually appear to be the center of the solar system. a. Earth appears to be the center of the solar system because Earth is at the center of the universe, and everything revolves around it in a circular orbit. b. Earth appears to be the center of the solar system
because, in the reference frame of Earth, the sun, moon, and planets all appear to move across the sky as if they were circling Earth. c. Earth appears to be at the center of the solar system because Earth is at the center of the solar system and all the heavenly bodies revolve around it. d. Earth appears to be at the center of the solar system because Earth is located at one of the foci of the elliptical orbit of the sun, moon, and other planets. Virtual Physics Acceleration This simulation allows you to create your own solar system so that you can see how changing distances and masses determines the orbits of planets. Click Helpfor instructions. Click to view content (https://archive.cnx.org/specials/ee816dff-0b5f-4e6f-8250-f9fb9e39d716/my-solar-system/) GRASP CHECK When the central object is off center, how does the speed of the orbiting object vary? a. The orbiting object moves fastest when it is closest to the central object and slowest when it is farthest away. b. The orbiting object moves slowest when it is closest to the central object and fastest when it is farthest away. 234 Chapter 7 • Newton's Law of Gravitation c. The orbiting object moves with the same speed at every point on the circumference of the elliptical orbit. d. There is no relationship between the speed of the object and the location of the planet on the circumference of the orbit. Calculations Related to Kepler’s Laws of Planetary Motion Kepler’s First Law Refer back to Figure 7.2 (a). Notice which distances are constant. The foci are fixed, so distance of an ellipse states that the sum of the distances perimeter of triangle distances of objects in a system that includes one object orbiting another. is also constant. These two facts taken together mean that the must also be constant. Knowledge of these constants will help you determine positions and is a constant. The definition Kepler’s Second Law Refer back to Figure 7.4. The second law says that the segments have equal area and that it takes equal time to sweep through each segment. That is, the time it takes to travel from A to B equals the time it takes to travel from C to D, and so forth. Velocity v equals distance ddivided by time t: , so distance divided by velocity is also a constant. For example, if we know the average velocity of Earth on June 21 and December 21, we can compare the distance Earth travels on those days. . Then, The degree of elongation of an elliptical orbit is called its eccentricity (e). Eccentricity is calculated by dividing the distance f from the center of an ellipse to one of the foci by half the long axis a. When , the ellipse is a circle. The area of an ellipse is given by sweeps out in a given period of time, you can calculate the fraction of the year that has elapsed. , where bis half the short axis. If you know the axes of Earth’s orbit and the area Earth 7.1 WORKED EXAMPLE Kepler’s First Law At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the other focus of its orbit, f2. The planet is focus f1 of the moon’s elliptical orbit. How far is the moon from the planet when it is 260,000 km from f2? Strategy Show and label the ellipse that is the orbit in your solution. Picture the triangle f1mf2 collapsed along the major axis and add up the lengths of the three sides. Find the length of the unknown side of the triangle when the moon is 260,000 km from f2. Solution Perimeter of Discussion The perimeter of triangle f1mf2 must be constant because the distance between the foci does not change and Kepler’s first law says the orbit is an ellipse. For any ellipse, the sum of the two sides of the triangle, which are f1mand mf2, is constant. WORKED EXAMPLE Kepler’s Second Law Figure 7.6 shows the major and minor axes of an ellipse. The semi-major and semi-minor axes are half of these, respectively. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 235 Figure 7.6 The major axis is the length of the ellipse, and the minor axis is the width of the ellipse. The semi-major axis is half the major axis, and the semi-minor axis is half the minor axis. Earth’s orbit is slightly elliptical, with a semi-major axis of 152 million km and a semi-minor axis of 147 million km. If Earth’s period is 365.26 days, what area does an Earth-to-sun line sweep past in one day? Strategy Each day, Earth sweeps past an equal-sized area, so we divide the total area by the number of days in a year to find the area swept past in one day. For total area use a year (i.e., its period). . Calculate A, the area inside Earth’s orbit and divide by the number of days in Solution 7.2 The area swept out in one day is thus . Discussion The answer is based on Kepler’s law, which states that a line from a planet to the sun sweeps out equal areas in equal times. Kepler’s Third Law Kepler’s third law states that the ratio of the squares of the periods of any two planets (T1, T2) is equal to the ratio of the cubes of their average orbital distance from the sun (r1, r2). Mathematically, this is represented by From this equation, it follows that the ratio r3/T2 is the same for all planets in the solar system. Later we will see how the work of Newton leads to a value for this constant. WORKED EXAMPLE Kepler’s Third Law Given that the moon orbits Earth each 27.3 days and that it is an average distance of calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface. Strategy The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form by from the center of Earth, 236 Chapter 7 • Newton's Law of Gravitation . Let us use the subscript 1 for the moon and the subscript 2 for the satellite. We are asked to find T2. The given information tells us that the orbital radius of the moon is , and that the period of the moon is . The height of the artificial satellite above Earth’s surface is given, so to get the distance r2 from the center of . Now all Earth we must add the height to the radius of Earth (6380 km). This gives quantities are known, so T2 can be found. Solution To solve for T2, we cross-multiply and take the square root, yielding 7.3 Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will complete one orbit in the same amount of time. Practice Problems 1. A planet with no axial tilt is located in another solar system. It circles its sun in a very elliptical orbit so that the temperature varies greatly throughout the year. If the year there has 612 days and the inhabitants celebrate the coldest day on day 1 of their calendar, when is the warmest day? a. Day 1 b. Day 153 c. Day 306 d. Day 459 2. A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). The ratio for the moon is . Calculate the radius of the orbit of such a satellite. a. b. c. d. Check Your Understanding 3. Are Kepler’s laws purely descriptive, or do they contain causal information? a. Kepler’s laws are purely descriptive. b. Kepler’s laws are purely causal. c. Kepler’s laws are descriptive as well as causal. d. Kepler’s laws are neither descriptive nor causal. 4. True or false—According to Kepler’s laws of planetary motion, a satellite increases its speed as it approaches its parent body and decreases its speed as it moves away from the parent body. a. True b. False 5. Identify the locations of the foci of an elliptical orbit. a. One focus is the parent body, and the other is located at the opposite end of the ellipse, at the same distance from the center as the parent body. b. One focus is the parent body, and the other is located at the opposite end of the ellipse, at half the distance from the center as the parent body. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 237 c. One focus is the parent body and the other is located outside of the elliptical orbit, on the line on which is the semi- major axis of the ellipse. d. One focus is on the line containing the semi-major axis of the ellipse, and the other is located anywhere on the elliptical orbit of the satellite. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Section Learning Objectives By the end of this section, you will be able to do the following: • Explain Newton’s law of universal gravitation and compare it to Einstein’s theory of general relativity • Perform calculations using Newton’s law of universal gravitation Section Key Terms Einstein’s theory of general relativity gravitational constant Newton’s universal law of gravitation Concepts Related to Newton’s Law of Universal Gravitation Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 7.7. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner, Galileo Galilei, had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph. It had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose an explanation of the mechanism that caused them to foll
ow these paths and not others. Figure 7.7 The popular legend that Newton suddenly discovered the law of universal gravitation when an apple fell from a tree and hit him on the head has an element of truth in it. A more probable account is that he was walking through an orchard and wondered why all the apples fell in the same direction with the same acceleration. Great importance is attached to it because Newton’s universal law of gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance 238 Chapter 7 • Newton's Law of Gravitation between them. Expressed in modern language, Newton’s universal law of gravitation states that every object in the universe attracts every other object with a force that is directed along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This attraction is illustrated by Figure 7.8. Figure 7.8 Gravitational attraction is along a line joining the centers of mass (CM) of the two bodies. The magnitude of the force on each body is the same, consistent with Newton’s third law (action-reaction). For two bodies having masses mand Mwith a distance rbetween their centers of mass, the equation for Newton’s universal law of gravitation is where F is the magnitude of the gravitational force and Gis a proportionality factor called the gravitational constant. Gis a universal constant, meaning that it is thought to be the same everywhere in the universe. It has been measured experimentally to be . If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth’s surface? Gis given above, Earth’s mass Mis 5.97 × 1024 kg, and the radius rof Earth is 6.38 × 106 m. Putting these values into Newton’s universal law of gravitation gives We can check this result with the relationship: You may remember that g, the acceleration due to gravity, is another important constant related to gravity. By substituting g for a in the equation for Newton’s second law of motion we get gravitation gives . Combining this with the equation for universal Cancelling the mass mon both sides of the equation and filling in the values for the gravitational constant and mass and radius of the Earth, gives the value of g,which may look familiar. This is a good point to recall the difference between mass and weight. Mass is the amount of matter in an object; weight is the Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 239 force of attraction between the mass within two objects. Weight can change because gis different on every moon and planet. An object’s mass mdoes not change but its weight mg can. Virtual Physics Gravity and Orbits Move the sun, Earth, moon and space station in this simulation to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies. Turn off gravity to see what would happen without it! Click to view content (https://archive.cnx.org/specials/a14085c8-96b8-4d04-bb5a-56d9ccbe6e69/gravity-and-orbits/) GRASP CHECK Why doesn’t the Moon travel in a smooth circle around the Sun? a. The Moon is not affected by the gravitational field of the Sun. b. The Moon is not affected by the gravitational field of the Earth. c. The Moon is affected by the gravitational fields of both the Earth and the Sun, which are always additive. d. The moon is affected by the gravitational fields of both the Earth and the Sun, which are sometimes additive and sometimes opposite. Snap Lab Take-Home Experiment: Falling Objects In this activity you will study the effects of mass and air resistance on the acceleration of falling objects. Make predictions (hypotheses) about the outcome of this experiment. Write them down to compare later with results. • Four sheets of -inch paper Procedure • Take four identical pieces of paper. ◦ Crumple one up into a small ball. ◦ Leave one uncrumpled. ◦ Take the other two and crumple them up together, so that they make a ball of exactly twice the mass of the other crumpled ball. ◦ Now compare which ball of paper lands first when dropped simultaneously from the same height. 1. Compare crumpled one-paper ball with crumpled two-paper ball. 2. Compare crumpled one-paper ball with uncrumpled paper. GRASP CHECK Why do some objects fall faster than others near the surface of the earth if all mass is attracted equally by the force of gravity? a. Some objects fall faster because of air resistance, which acts in the direction of the motion of the object and exerts more force on objects with less surface area. b. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with less surface area. c. Some objects fall faster because of air resistance, which acts in the direction of motion of the object and exerts more force on objects with more surface area. d. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with more surface area. It is possible to derive Kepler’s third law from Newton’s law of universal gravitation. Applying Newton’s second law of motion to 240 Chapter 7 • Newton's Law of Gravitation angular motion gives an expression for centripetal force, which can be equated to the expression for force in the universal gravitation equation. This expression can be manipulated to produce the equation for Kepler’s third law. We saw earlier that the expression r3/T2is a constant for satellites orbiting the same massive object. The derivation of Kepler’s third law from Newton’s law of universal gravitation and Newton’s second law of motion yields that constant: where Mis the mass of the central body about which the satellites orbit (for example, the sun in our solar system). The usefulness of this equation will be seen later. The universal gravitational constant Gis determined experimentally. This definition was first done accurately in 1798 by English scientist Henry Cavendish (1731–1810), more than 100 years after Newton published his universal law of gravitation. The measurement of Gis very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most) by using an apparatus like that in Figure 7.9. Remarkably, his value for Gdiffers by less than 1% from the modern value. Figure 7.9 Cavendish used an apparatus like this to measure the gravitational attraction between two suspended spheres (m) and two spheres on a stand (M) by observing the amount of torsion (twisting) created in the fiber. The distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity. Einstein’s Theory of General Relativity Einstein’s theory of general relativity explained some interesting properties of gravity not covered by Newton’s theory. Einstein based his theory on the postulate that acceleration and gravity have the same effect and cannot be distinguished from each other. He concluded that light must fall in both a gravitational field and in an accelerating reference frame. Figure 7.10 shows this effect (greatly exaggerated) in an accelerating elevator. In Figure 7.10(a), the elevator accelerates upward in zero gravity. In Figure 7.10(b), the room is not accelerating but is subject to gravity. The effect on light is the same: it “falls” downward in both situations. The person in the elevator cannot tell whether the elevator is accelerating in zero gravity or is stationary and subject to gravity. Thus, gravity affects the path of light, even though we think of gravity as acting between masses, while photons are massless. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 241 Figure 7.10 (a) A beam of light emerges from a flashlight in an upward-accelerating elevator. Since the elevator moves up during the time the light takes to reach the wall, the beam strikes lower than it would if the elevator were not accelerated. (b) Gravity must have the same effect on light, since it is not possible to tell whether the elevator is accelerating upward or is stationary and acted upon by gravity. Einstein’s theory of general relativity got its first verification in 1919 when starlight passing near the sun was observed during a solar eclipse. (See Figure 7.11.) During an eclipse, the sky is darkened and we can briefly see stars. Those on a line of sight nearest the sun should have a shift in their apparent positions. Not only was this shift observed, but it agreed with Einstein’s predictions well within experimental uncertainties. This discovery created a scientific and public sensation. Einstein was now a folk hero as well as a very great scientist. The bending of light by matter is equivalent to a bending of space itself, with light following the curve. This is another radical change in our concept of space and time. It is also another connection that any particle with mass orenergy (e.g., massless photons) is affected by gravity. Figure 7.11 This schematic shows how light passing near a massive body like the sun is curved toward it. The light that reaches the Earth then seems to be coming from different locations than the known positions of the originating stars. Not only was this effect observed, but the amount of be
nding was precisely what Einstein predicted in his general theory of relativity. To summarize the two views of gravity, Newton envisioned gravity as a tug of war along the line connecting any two objects in the universe. In contrast, Einstein envisioned gravity as a bending of space-time by mass. 242 Chapter 7 • Newton's Law of Gravitation BOUNDLESS PHYSICS NASA gravity probe B NASA’s Gravity Probe B (GP-B) mission has confirmed two key predictions derived from Albert Einstein’s general theory of relativity. The probe, shown in Figure 7.12 was launched in 2004. It carried four ultra-precise gyroscopes designed to measure two effects hypothesized by Einstein’s theory: • The geodetic effect, which is the warping of space and time by the gravitational field of a massive body (in this case, Earth) • The frame-dragging effect, which is the amount by which a spinning object pulls space and time with it as it rotates Figure 7.12 Artist concept of Gravity Probe B spacecraft in orbit around the Earth. (credit: NASA/MSFC) Both effects were measured with unprecedented precision. This was done by pointing the gyroscopes at a single star while orbiting Earth in a polar orbit. As predicted by relativity theory, the gyroscopes experienced very small, but measureable, changes in the direction of their spin caused by the pull of Earth’s gravity. The principle investigator suggested imagining Earth spinning in honey. As Earth rotates it drags space and time with it as it would a surrounding sea of honey. GRASP CHECK According to the general theory of relativity, a gravitational field bends light. What does this have to do with time and space? a. Gravity has no effect on the space-time continuum, and gravity only affects the motion of light. b. The space-time continuum is distorted by gravity, and gravity has no effect on the motion of light. c. Gravity has no effect on either the space-time continuum or on the motion of light. d. The space-time continuum is distorted by gravity, and gravity affects the motion of light. Calculations Based on Newton’s Law of Universal Gravitation TIPS FOR SUCCESS When performing calculations using the equations in this chapter, use units of kilograms for mass, meters for distances, newtons for force, and seconds for time. The mass of an object is constant, but its weight varies with the strength of the gravitational field. This means the value of g varies from place to place in the universe. The relationship between force, mass, and acceleration from the second law of motion can be written in terms of g. In this case, the force is the weight of the object, which is caused by the gravitational attraction of the planet or moon on which the object is located. We can use this expression to compare weights of an object on different moons and planets. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 243 WATCH PHYSICS Mass and Weight Clarification This video shows the mathematical basis of the relationship between mass and weight. The distinction between mass and weight are clearly explained. The mathematical relationship between mass and weight are shown mathematically in terms of the equation for Newton’s law of universal gravitation and in terms of his second law of motion. Click to view content (https://www.khanacademy.org/embed_video?v=IuBoeDihLUc) GRASP CHECK Would you have the same mass on the moon as you do on Earth? Would you have the same weight? a. You would weigh more on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. b. You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. c. You would weigh less on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. d. You would weigh more on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. Two equations involving the gravitational constant, G, are often useful. The first is Newton’s equation, . Several of the values in this equation are either constants or easily obtainable. F is often the weight of an object on the surface of a large object with mass M, which is usually known. The mass of the smaller object, m, is often known, and Gis a universal constant with the same value anywhere in the universe. This equation can be used to solve problems involving an object on or orbiting Earth or other massive celestial object. Sometimes it is helpful to equate the right-hand side of the equation to mg and cancel the mon both sides. The equation is also useful for problems involving objects in orbit. Note that there is no need to know the mass of the object. Often, we know the radius ror the period Tand want to find the other. If these are both known, we can use the equation to calculate the mass of a planet or star. WATCH PHYSICS Mass and Weight Clarification This video demonstrates calculations involving Newton’s universal law of gravitation. Click to view content (https://www.khanacademy.org/embed_video?v=391txUI76gM) GRASP CHECK and . are both the acceleration due to gravity Identify the constants a. b. c. d. and is acceleration due to gravity on Earth and is the gravitational constant and and are both the universal gravitational constant. is the universal gravitational constant. is the acceleration due to gravity on Earth. WORKED EXAMPLE Change in g The value of g on the planet Mars is 3.71 m/s2. If you have a mass of 60.0 kg on Earth, what would be your mass on Mars? What would be your weight on Mars? Strategy Weight equals acceleration due to gravity times mass: on Mars gMand weight on Mars WM. . An object’s mass is constant. Call acceleration due to gravity 244 Chapter 7 • Newton's Law of Gravitation Solution Mass on Mars would be the same, 60 kg. Discussion The value of g on any planet depends on the mass of the planet and the distance from its center. If the material below the surface varies from point to point, the value of g will also vary slightly. 7.4 WORKED EXAMPLE Earth’s g at the Moon Find the acceleration due to Earth’s gravity at the distance of the moon. Express the force of gravity in terms of g. Combine with the equation for universal gravitation. Solution Cancel mand substitute. 7.5 7.6 7.7 Discussion The value of g for the moon is 1.62 m/s2. Comparing this value to the answer, we see that Earth’s gravitational influence on an object on the moon’s surface would be insignificant. 7.8 Practice Problems 6. What is the mass of a person who weighs ? a. b. c. d. 7. Calculate Earth’s mass given that the acceleration due to gravity at the North Pole is and the radius of the Earth is from pole to center. a. b. c. d. Check Your Understanding 8. Some of Newton’s predecessors and contemporaries also studied gravity and proposed theories. What important advance did Newton make in the study of gravity that the other scientists had failed to do? a. He gave an exact mathematical form for the theory. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 245 b. He added a correction term to a previously existing formula. c. Newton found the value of the universal gravitational constant. d. Newton showed that gravitational force is always attractive. 9. State the law of universal gravitation in words only. a. Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the square of the distance between them. b. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. c. Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the distance between them. d. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the distance between them. 10. Newton’s law of universal gravitation explains the paths of what? a. A charged particle b. A ball rolling on a plane surface c. A planet moving around the sun d. A stone tied to a string and whirled at constant speed in a horizontal circle 246 Chapter 7 • Key Terms KEY TERMS aphelion closest distance between a planet and the sun (called apoapsis for other celestial bodies) Copernican model the model of the solar system where the sun is at the center of the solar system and all the planets orbit around it; this is also called the heliocentric model eccentricity a measure of the separation of the foci of an ellipse Johannes Kepler that describe the properties of all orbiting satellites Newton’s universal law of gravitation states that gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. perihelion farthest distance between a planet and the sun Einstein’s theory of general relativity the theory that (called periapsis for other celestial bodies) gravitational force results from the bending of spacetime by an object’s mass gravitational constant the proportionality constant in Newton’s law of universal gravitation Kepler’s laws of planetary motion three laws derived by SECTION SUMMARY 7.1 Kepler's Laws of Planetary Motion • All satellites follow elliptical orbits. • The line from the satellite to the parent body sweeps out equal areas in equal time. • The radius cubed divided by the period squared is a Ptolemaic model the model of the solar system where Earth is at the center of the solar system and the sun and all the planets orbit around it; this is also called the geocentric model 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity • Newton’s law of universal gravitation provides a mathematical basis for gravitational force and Kepler’s laws of planetary
motion. constant for all satellites orbiting the same parent body. • Einstein’s theory of general relativity shows that KEY EQUATIONS 7.1 Kepler's Laws of Planetary Motion Kepler’s third law eccentricity area of an ellipse semi-major axis of an ellipse semi-minor axis of an ellipse gravitational fields change the path of light and warp space and time. • An object’s mass is constant, but its weight changes when acceleration due to gravity, g, changes. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Newton’s second law of motion Newton’s universal law of gravitation acceleration due to gravity constant for satellites orbiting the same massive object CHAPTER REVIEW Concept Items 7.1 Kepler's Laws of Planetary Motion is different from other ellipses. a. The foci of a circle are at the same point and are located at the center of the circle. 1. A circle is a special case of an ellipse. Explain how a circle b. The foci of a circle are at the same point and are Access for free at openstax.org. located at the circumference of the circle. c. The foci of a circle are at the same point and are located outside of the circle. d. The foci of a circle are at the same point and are located anywhere on the diameter, except on its midpoint. 2. Comets have very elongated elliptical orbits with the sun at one focus. Using Kepler's Law, explain why a comet travels much faster near the sun than it does at the other end of the orbit. a. Because the satellite sweeps out equal areas in equal times b. Because the satellite sweeps out unequal areas in Chapter 7 • Chapter Review 247 illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. c. Gravity and acceleration have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. d. Gravity and acceleration have different effects and can be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. equal times 6. Titan, with a radius of , is the largest c. Because the satellite is at the other focus of the moon of the planet Saturn. If the mass of Titan is ellipse d. Because the square of the period of the satellite is proportional to the cube of its average distance from the sun 3. True or False—A planet-satellite system must be isolated from other massive objects to follow Kepler’s laws of planetary motion. a. True b. False 4. Explain why the string, pins, and pencil method works for drawing an ellipse. a. The string, pins, and pencil method works because the length of the two sides of the triangle remains constant as you are drawing the ellipse. b. The string, pins, and pencil method works because the area of the triangle remains constant as you are drawing the ellipse. c. The string, pins, and pencil method works because the perimeter of the triangle remains constant as you are drawing the ellipse. d. The string, pins, and pencil method works because the volume of the triangle remains constant as you are drawing the ellipse. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 5. Describe the postulate on which Einstein based the theory of general relativity and describe an everyday experience that illustrates this postulate. a. Gravity and velocity have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. b. Gravity and velocity have different effects and can be distinguished from each other. An acceptable , what is the acceleration due to gravity on the surface of this moon? a. b. c. d. 7. Saturn’s moon Titan has an orbital period of 15.9 days. If Saturn has a mass of 5.68×1023 kg, what is the average distance from Titan to the center of Saturn? 1.22×106 m a. b. 4.26×107 m 5.25×104 km c. d. 4.26×1010 km 8. Explain why doubling the mass of an object doubles its weight, but doubling its distance from the center of Earth reduces its weight fourfold. a. The weight is two times the gravitational force between the object and Earth. b. The weight is half the gravitational force between the object and Earth. c. The weight is equal to the gravitational force between the object and Earth, and the gravitational force is inversely proportional to the distance squared between the object and Earth. d. The weight is directly proportional to the square of the gravitational force between the object and Earth. 9. Explain why a star on the other side of the Sun might appear to be in a location that is not its true location. It can be explained by using the concept of a. atmospheric refraction. It can be explained by using the concept of the special theory of relativity. It can be explained by using the concept of the general theory of relativity. It can be explained by using the concept of light d. b. c. 248 Chapter 7 • Chapter Review scattering in the atmosphere. 10. The Cavendish experiment marked a milestone in the study of gravity. Part A. What important value did the experiment determine? Part B. Why was this so difficult in terms of the masses used in the apparatus and the strength of the gravitational force? a. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between very massive objects. b. Part A. The experiment measured the gravitational Critical Thinking Items 7.1 Kepler's Laws of Planetary Motion 11. In the figure, the time it takes for the planet to go from A to B, C to D, and E to F is the same. constant, G. Part B. Gravity is a very weak force but, despite this limitation, Cavendish was able to measure the attraction between very massive objects. c. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. d. Part A. The experiment measured the gravitational constant, G. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. a. Area X < Area Y; the speed is greater for area X. b. Area X > Area Y; the speed is greater for area Y. c. Area X = Area Y; the speed is greater for area X. d. Area X = Area Y; the speed is greater for area Y. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 14. Rhea, with a radius of 7.63×105 m, is the second-largest moon of the planet Saturn. If the mass of Rhea is 2.31×1021 kg, what is the acceleration due to gravity on the surface of this moon? a. 2.65×10−1 m/s b. 2.02×105 m/s c. 2.65×10−1 m/s2 d. 2.02×105 m/s2 15. Earth has a mass of 5.971×1024 kg and a radius of 6.371×106 m. Use the data to check the value of the gravitational constant. a. it matches the value of the gravitational constant G. it matches the value of the gravitational constant G. it matches the value of the b. c. gravitational constant G. Compare the areas A1, A2, and A3 in terms of size. a. A1 ≠ A2 ≠ A3 b. A1 = A2 = A3 c. A1 = A2 > A3 d. A1 > A2 = A3 12. A moon orbits a planet in an elliptical orbit. The foci of the ellipse are 50, 000 km apart. The closest approach of the moon to the planet is 400, 000 km. What is the length of the major axis of the orbit? a. 400, 000 km b. 450, 000, km c. 800, 000 km d. 850, 000 km 13. In this figure, if f1 represents the parent body, which set of statements holds true? Access for free at openstax.org. d. it matches the value of the gravitational constant G. 16. The orbit of the planet Mercury has a period of 88.0 days and an average radius of 5.791×1010 m. What is the mass of the sun? Problems 7.1 Kepler's Laws of Planetary Motion 17. The closest Earth comes to the sun is 1.47×108 km, and Earth’s farthest distance from the sun is 1.52×108 km. What is the area inside Earth’s orbit? a. 2.23×1016 km2 b. 6.79×1016 km2 7.02×1016 km2 c. 7.26×1016 km2 d. Performance Task 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 19. Design an experiment to test whether magnetic force is inversely proportional to the square of distance. Gravitational, magnetic, and electrical fields all act at a distance, but do they all follow the inverse square law? One difference in the forces related to these fields is that gravity is only attractive, but the other two can repel as well. In general, the inverse square law says that force F equals a constant Cdivided by the distance between objects, d, squared: Incorporate these materials into your design: . TEST PREP Multiple Choice 7.1 Kepler's Laws of Planetary Motion 20. A planet of mass m circles a sun of mass M. Which distance changes throughout the planet’s orbit? a. b. c. d. 21. The focal point of the elliptical orbit of a moon is from the center of the orbit. If the , what is the length of the eccentricity of the orbit is semi-major axis? a. b. c. d. Chapter 7 • Test Prep 249 3.43×1019 kg a. 1.99×1030 kg b. c. 2.56×1029 kg 1.48×1040 kg d. 18. Earth is 1.496×108 km from the sun, and Neptune is 4.490×109 km from the sun. What best represents the number of Earth years it takes for Neptune to complete one orbit around the sun? 10 years a. 30 years b. 160 years c. d. 900 years • Two strong, permanent bar magnets • A spring scale that can measure small forces • A short ruler calibrated in millimeters Use the magnets to study the relationship between attractive force and distance. a. What will be the independent variable? b. What will be the dependent variable? c. How will you measure each of these variables? If you p
lot the independent variable versus the d. dependent variable and the inverse square law is upheld, will the plot be a straight line? Explain. e. Which plot would be a straight line if the inverse square law were upheld? 22. An artificial satellite orbits the Earth at a distance of 1.45×104 km from Earth’s center. The moon orbits the Earth at a distance of 3.84×105 km once every 27.3 days. How long does it take the satellite to orbit the Earth? a. 0.200 days b. 3.07 days c. 243 days 3721 days d. 23. Earth is 1.496×108 km from the sun, and Venus is 1.08×108 km from the sun. One day on Venus is 243 Earth days long. What best represents the number of Venusian days in a Venusian year? a. 0.78 days b. 0.92 days 1.08 days c. 1.21 days d. 250 Chapter 7 • Test Prep 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 24. What did the Cavendish experiment measure? a. The mass of Earth b. The gravitational constant c. Acceleration due to gravity d. The eccentricity of Earth’s orbit 25. You have a mass of and you have just landed on one of the moons of Jupiter where you have a weight of , on . What is the acceleration due to gravity, the moon you are visiting? a. b. c. d. 26. A person is in an elevator that suddenly begins to descend. The person knows, intuitively, that the feeling of suddenly becoming lighter is because the elevator is accelerating downward. What other change would Short Answer 7.1 Kepler's Laws of Planetary Motion 27. Explain how the masses of a satellite and its parent body must compare in order to apply Kepler’s laws of planetary motion. a. The mass of the parent body must be much less than that of the satellite. b. The mass of the parent body must be much greater than that of the satellite. c. The mass of the parent body must be equal to the mass of the satellite. d. There is no specific relationship between the masses for applying Kepler’s laws of planetary motion. 28. Hyperion is a moon of the planet Saturn. Its orbit has an eccentricity of and a semi-major axis of . How far is the center of the orbit from the center of Saturn? a. b. c. d. 29. The orbits of satellites are elliptical. Define an ellipse. a. An ellipse is an open curve wherein the sum of the distance from the foci to any point on the curve is constant. b. An ellipse is a closed curve wherein the sum of the distance from the foci to any point on the curve is constant. Access for free at openstax.org. produce the same feeling? How does this demonstrate Einstein’s postulate on which he based the theory of general relativity? a. It would feel the same if the force of gravity suddenly became weaker. This illustrates Einstein’s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein’s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force of gravity suddenly became weaker. This illustrates Einstein’s postulates that gravity and acceleration are distinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein’s postulates that gravity and acceleration are distinguishable. b. c. d. c. An ellipse is an open curve wherein the distances from the two foci to any point on the curve are equal. d. An ellipse is a closed curve wherein the distances from the two foci to any point on the curve are equal. 30. Mars has two moons, Deimos and Phobos. The orbit of . The average radius of the and an average Deimos has a period of radius of . According to orbit of Phobos is Kepler’s third law of planetary motion, what is the period of Phobos? a. b. c. d. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 31. Newton’s third law of motion says that, for every action force, there is a reaction force equal in magnitude but that acts in the opposite direction. Apply this law to gravitational forces acting between the Washington Monument and Earth. a. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth’s weight. The situation can be represented with two force vectors of unequal magnitude and pointing in the same direction. b. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth’s weight. The situation can be represented with two force vectors of unequal magnitude but pointing in opposite directions. c. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude and pointing in the same direction. d. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude but pointing in opposite directions. 32. True or false—Gravitational force is the attraction of the mass of one object to the mass of another. Light, either Extended Response 7.1 Kepler's Laws of Planetary Motion 35. The orbit of Halley’'s Comet has an eccentricity of 0.967 and stretches to the edge of the solar system. Part A. Describe the shape of the comet’s orbit. Part B. Compare the distance traveled per day when it is near the sun to the distance traveled per day when it is at the edge of the solar system. Part C. Describe variations in the comet's speed as it completes an orbit. Explain the variations in terms of Kepler's second law of planetary motion. a. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it leaves the sun. b. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet accelerates as it approaches the sun and decelerates as it leaves the sun. c. Part A. The orbit is very elongated, with the sun near one end. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it moves away from the sun. 36. For convenience, astronomers often use astronomical units (AU) to measure distances within the solar system. One AU equals the average distance from Earth to the Chapter 7 • Test Prep 251 as a particle or a wave, has no rest mass. Despite this fact gravity bends a beam of light. a. True b. False 33. The average radius of Earth is . What is Earth’s mass? a. b. c. d. 34. What is the gravitational force between two apart? people sitting a. b. c. d. sun. Halley’s Comet returns once every 75.3 years. What is the average radius of the orbit of Halley’s Comet in AU? a. 0.002 AU b. 0.056 AU c. 17.8 AU d. 653 AU 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 37. It took scientists a long time to arrive at the understanding of gravity as explained by Galileo and Newton. They were hindered by two ideas that seemed like common sense but were serious misconceptions. First was the fact that heavier things fall faster than light things. Second, it was believed impossible that forces could act at a distance. Explain why these ideas persisted and why they prevented advances. a. Heavier things fall faster than light things if they have less surface area and greater mass density. In the Renaissance and before, forces that acted at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. b. Heavier things fall faster than light things because they have greater surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. c. Heavier things fall faster than light things because they have less surface area and greater mass density. In the Renaissance and before, forces that 252 Chapter 7 • Test Prep act at a distance were considered impossible, so people were quick to accept scientific theories that invoked such forces. d. Heavier things fall faster than light things because they have larger surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible because of people’s faith in scientific theories. 38. The masses of Earth and the moon are 5.97×1024 kg and 7.35×1022 kg, respectively. The distance from Earth to the moon is 3.80×105 km. At what point between the Earth and the moon are the opposing gravitational forces equal? (Use subscripts e and m to represent Earth and moon.) a. b. c. d. 3.42×105 km from the center of Earth 3.80×105 km from the center of Earth 3.42×106 km from the center of Earth 3.10×107 km from the center of Earth Access for free at openstax.org. CHAPTER 8 Momentum Figure 8.1 NFC defensive backs Ronde Barber and Roy Williams along with linebacker Jeremiah Trotter gang tackle AFC running back LaDainian Tomlinson during the 2006 Pro Bowl in Hawaii. (United States Marine Corps) Chapter Outline 8.1 Linear Momentum, Force, and Impulse 8.2 Conservation of Momentum 8.3 Elastic and Inelastic Collisions We know from everyday use of the word momentumthat it is a tendency to continue on course in the same INTRODUCTION direction. Newscasters speak of sports teams or politicians gaining, losing, or maintaining the momentum to win. As we learned when studying about inertia, which is Newton's first law of motion, every object or system has inertia—that is, a tendency for an object in motion to remain in motion or an object at rest to remain at rest. Mass is a u
seful variable that lets us quantify inertia. Momentum is mass in motion. Momentum is important because it is conserved in isolated systems; this fact is convenient for solving problems where objects collide. The magnitude of momentum grows with greater mass and/or speed. For example, look at the football players in the photograph (Figure 8.1). They collide and fall to the ground. During their collisions, momentum will play a large part. In this chapter, we will learn about momentum, the different types of collisions, and how to use momentum equations to solve collision problems. 254 Chapter 8 • Momentum 8.1 Linear Momentum, Force, and Impulse Section Learning Objectives By the end of this section, you will be able to do the following: • Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem • Describe Newton’s second law in terms of momentum • Solve problems using the impulse-momentum theorem Section Key Terms change in momentum impulse impulse–momentum theorem linear momentum Momentum, Impulse, and the Impulse-Momentum Theorem Linear momentum is the product of a system’s mass and its velocity. In equation form, linear momentum p is You can see from the equation that momentum is directly proportional to the object’s mass (m) and velocity (v). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object. Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s. Momentum is so important for understanding motion that it was called the quantity of motionby physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum. Recall our study of Newton’s second law of motion (Fnet = ma). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum. In equation form, this law is where Fnet is the net external force, is the change in momentum, and is the change in time. We can solve for by rearranging the equation to be is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force.Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. Newton’s Second Law in Terms of Momentum When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. is actually derived from the equation: Access for free at openstax.org. 8.1 • Linear Momentum, Force, and Impulse 255 For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of from by substituting the definitions of acceleration and momentum. The change in momentum is given by If the mass of the system is constant, then By substituting for , Newton’s second law of motion becomes for a constant mass. Because we can substitute to get the familiar equation when the mass of the system is constant. TIPS FOR SUCCESS We just showed how would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you would need to use Newton’s second law expressed in terms of momentum to account for the changing mass. applies only when the mass of the system is constant. An example of when this formula Snap Lab Hand Movement and Impulse In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse. • one ball • one tub filled with water Procedure: 1. Try catching a ball while givingwith the ball, pulling your hands toward your body. 2. Next, try catching a ball while keeping your hands still. 3. Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop. 4. After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive. 5. Explain what happens in each case and why. GRASP CHECK What are some other examples of motions that impulse affects? 256 Chapter 8 • Momentum a. a football player colliding with another, or a car moving at a constant velocity b. a car moving at a constant velocity, or an object moving in the projectile motion c. a car moving at a constant velocity, or a racket hitting a ball d. a football player colliding with another, or a racket hitting a ball LINKS TO PHYSICS Engineering: Saving Lives Using the Concept of Impulse Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as . Figure 8.2 Vehicles have safety features like airbags and seat belts installed. Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force see how increasing padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact. you can stays the same will decrease Fnet. This is another example of an inverse relationship. Similarly, a while Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident. GRASP CHECK You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem. a. Bending your knees increases the time of the impact, thus decreasing the force. b. Bending your knees decreases the time of the impact, thus decreasing the force. c. Bending your knees increases the time of the impact, thus increasing the force. d. Bending your knees decreases the time of the impact, thus increasing the force. Access for free at openstax.org. 8.1 • Linear Momentum, Force, and Impulse 257 Solving Problems Using the Impulse-Momentum Theorem WORKED EXAMPLE Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s. Strategy No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p. (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum: Solution for (a) To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation. Solution for (b) To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation. The ratio of the player’s momentum to the ball’s momentum is Discussion Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football. WORKED EXAMPLE Calculating Force: Venus Williams’ Racquet During the 2007 French Open, Venus Williams (Figure 8.3) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds). Figure 8.3 Venus Williams playing in the 2013 US Open (Edwin Martinez, Flickr) Strategy Recall that Newton’s second law stated in terms of momentum is 258 Chapter 8 • Momentum As noted above, when mass is constant, the change in momentum is given by whe
re vf is the final velocity and vi is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for to find the force. , we can use Solution To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above. Now we can find the magnitude of the net external force using 8.1 8.2 Discussion This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using Fnet = ma, but we would have had to do one more step. In this case, using momentum was a shortcut. Practice Problems 1. What is the momentum of a bowling ball with mass and velocity ? a. b. c. d. 2. What will be the change in momentum caused by a net force of acting on an object for seconds? a. b. c. d. Check Your Understanding 3. What is linear momentum? a. b. c. d. the sum of a system’s mass and its velocity the ratio of a system’s mass to its velocity the product of a system’s mass and its velocity the product of a system’s moment of inertia and its velocity 4. If an object’s mass is constant, what is its momentum proportional to? a. b. c. d. Its velocity Its weight Its displacement Its moment of inertia 5. What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is Access for free at openstax.org. 8.2 • Conservation of Momentum 259 constant? a. b. c. d. 6. Give an example of a system whose mass is not constant. a. A spinning top b. A baseball flying through the air c. A rocket launched from Earth d. A block sliding on a frictionless inclined plane 8.2 Conservation of Momentum Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the law of conservation of momentum verbally and mathematically Section Key Terms angular momentum isolated system law of conservation of momentum Conservation of Momentum It is important we realize that momentum is conserved during collisions, explosions, and other events involving objects in motion. To say that a quantity is conserved means that it is constant throughout the event. In the case of conservation of momentum, the total momentum in the system remains the same before and after the collision. You may have noticed that momentum was notconserved in some of the examples previously presented in this chapter. where forces acting on the objects produced large changes in momentum. Why is this? The systems of interest considered in those problems were not inclusive enough. If the systems were expanded to include more objects, then momentum would in fact be conserved in those sample problems. It is always possible to find a larger system where momentum is conserved, even though momentum changes for individual objects within the system. For example, if a football player runs into the goalpost in the end zone, a force will cause him to bounce backward. His momentum is obviously greatly changed, and considering only the football player, we would find that momentum is not conserved. However, the system can be expanded to contain the entire Earth. Surprisingly, Earth also recoils—conserving momentum—because of the force applied to it through the goalpost. The effect on Earth is not noticeable because it is so much more massive than the player, but the effect is real. Next, consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—in the example shown in Figure 8.4 of one car bumping into another. Both cars are coasting in the same direction when the lead car, labeled m2, is bumped by the trailing car, labeled m1. The only unbalanced force on each car is the force of the collision, assuming that the effects due to friction are negligible. Car m1 slows down as a result of the collision, losing some momentum, while car m2 speeds up and gains some momentum. If we choose the system to include both cars and assume that friction is negligible, then the momentum of the two-car system should remain constant. Now we will prove that the total momentum of the two-car system does in fact remain constant, and is therefore conserved. 260 Chapter 8 • Momentum Figure 8.4 Car of mass m1 moving with a velocity of v1 bumps into another car of mass m2 and velocity v2. As a result, the first car slows down to a velocity of v′1 and the second speeds up to a velocity of v′2. The momentum of each car is changed, but the total momentum ptot of the two cars is the same before and after the collision if you assume friction is negligible. Using the impulse-momentum theorem, the change in momentum of car 1 is given by where F1 is the force on car 1 due to car 2, and is the time the force acts, or the duration of the collision. Similarly, the change in momentum of car 2 is duration of the collision is the same for both cars. We know from Newton’s third law of motion that F2 = –F1, and so . where F2 is the force on car 2 due to car 1, and we assume the Therefore, the changes in momentum are equal and opposite, and . Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, where p′1 and p′2 are the momenta of cars 1 and 2 after the collision. This result that momentum is conserved is true not only for this example involving the two cars, but for any system where the net external force is zero, which is known as an isolated system. The law of conservation of momentum states that for an isolated system with any number of objects in it, the total momentum is conserved. In equation form, the law of conservation of momentum for an isolated system is written as or where ptot is the total momentum, or the sum of the momenta of the individual objects in the system at a given time, and p′tot is the total momentum some time later. The conservation of momentum principle can be applied to systems as diverse as a comet striking the Earth or a gas containing huge numbers of atoms and molecules. Conservation of momentum appears to be violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not. Access for free at openstax.org. 8.2 • Conservation of Momentum 261 TIPS FOR SUCCESS Momenta is the plural form of the word momentum. One object is said to have momentum, but two or more objects are said to have momenta. FUN IN PHYSICS Angular Momentum in Figure Skating So far we have covered linear momentum, which describes the inertia of objects traveling in a straight line. But we know that many objects in nature have a curved or circular path. Just as linear motion has linear momentum to describe its tendency to move forward, circular motion has the equivalent angular momentum to describe how rotational motion is carried forward. This is similar to how torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass or inertia. You may recall learning that the quantity mr2 is called the rotational inertia or moment of inertia of a point mass mat a distance rfrom the center of rotation. We already know the equation for linear momentum, p = mv. Since angular momentum is analogous to linear momentum, the moment of inertia (I) is analogous to mass, and angular velocity is analogous to linear velocity, it makes sense that angular momentum (L) is defined as Angular momentum is conserved when the net external torque ( external force is zero. ) is zero, just as linear momentum is conserved when the net Figure skaters take advantage of the conservation of angular momentum, likely without even realizing it. In Figure 8.5, a figure skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice, and because the friction is exerted very close to the pivot point. Both F and rare small, and so is negligibly small. Figure 8.5 (a) An ice skater is spinning on the tip of her skate with her arms extended. In the next image, (b), her rate of spin increases greatly when she pulls in her arms. Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that L = L′. Expressing this equation in terms of the moment of inertia, where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I′ is smaller, the angular velocity must increase to keep the angular momentum constant. This allows her to spin much faster without exerting any extra torque. A video (http://openstax.org/l/28figureskater) is also available that shows a real figure skater executing a spin. It discusses the physics of spins in figure skating. 262 Chapter 8 • Momentum GRASP CHECK Based on the equation L = Iω, how would you expect the moment of inertia of an object to affect angular momentum? How would angular velocity affect angular momentum? a. Large moment of inertia implies large angular momentum, and large angular velocity implies large angular momentum. b. Large moment of inertia implies small angular momentum, and large angular velocity implies small angular momentum. c. Large moment of inertia implies large angular momentum, and large angular velocity implies small angular momentum. d. Large moment of inertia implies small angular momentum, and large angular velocity implies large angular momentum. Check Your Understanding 7. When is momentum said to be conserved? a. When momentum is changing
during an event b. When momentum is increasing during an event c. When momentum is decreasing during an event d. When momentum is constant throughout an event 8. A ball is hit by a racket and its momentum changes. How is momentum conserved in this case? a. Momentum of the system can never be conserved in this case. b. Momentum of the system is conserved if the momentum of the racket is not considered. c. Momentum of the system is conserved if the momentum of the racket is also considered. d. Momentum of the system is conserved if the momenta of the racket and the player are also considered. 9. State the law of conservation of momentum. a. Momentum is conserved for an isolated system with any number of objects in it. b. Momentum is conserved for an isolated system with an even number of objects in it. c. Momentum is conserved for an interacting system with any number of objects in it. d. Momentum is conserved for an interacting system with an even number of objects in it. 8.3 Elastic and Inelastic Collisions Section Learning Objectives By the end of this section, you will be able to do the following: • Distinguish between elastic and inelastic collisions • Solve collision problems by applying the law of conservation of momentum Section Key Terms elastic collision inelastic collision point masses recoil Elastic and Inelastic Collisions When objects collide, they can either stick together or bounce off one another, remaining separate. In this section, we’ll cover these two different types of collisions, first in one dimension and then in two dimensions. In an elastic collision, the objects separate after impact and don’t lose any of their kinetic energy. Kinetic energy is the energy of motion and is covered in detail elsewhere. The law of conservation of momentum is very useful here, and it can be used whenever the net external force on a system is zero. Figure 8.6 shows an elastic collision where momentum is conserved. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 263 Figure 8.6 The diagram shows a one-dimensional elastic collision between two objects. An animation of an elastic collision between balls can be seen by watching this video (http://openstax.org/l/28elasticball) . It replicates the elastic collisions between balls of varying masses. Perfectly elastic collisions can happen only with subatomic particles. Everyday observable examples of perfectly elastic collisions don’t exist—some kinetic energy is always lost, as it is converted into heat transfer due to friction. However, collisions between everyday objects are almost perfectly elastic when they occur with objects and surfaces that are nearly frictionless, such as with two steel blocks on ice. Now, to solve problems involving one-dimensional elastic collisions between two objects, we can use the equation for conservation of momentum. First, the equation for conservation of momentum for two objects in a one-dimensional collision is Substituting the definition of momentum p = mv for each initial and final momentum, we get where the primes (') indicate values after the collision; In some texts, you may see ifor initial (before collision) and ffor final (after collision). The equation assumes that the mass of each object does not change during the collision. WATCH PHYSICS Momentum: Ice Skater Throws a Ball This video covers an elastic collision problem in which we find the recoilvelocityof an ice skater who throws a ball straight forward. To clarify, Sal is using the equation Click to view content (https://www.khanacademy.org/embed_video?v=vPkkCOlGND4) . GRASP CHECK The resultant vector of the addition of vectors respectively. Which of the following is true? a. b. and is . The magnitudes of , , and are , , and , 264 Chapter 8 • Momentum c. d. Now, let us turn to the second type of collision. An inelastic collision is one in which objects stick together after impact, and kinetic energy is notconserved. This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere. For inelastic collisions, kinetic energy may be lost in the form of heat. Figure 8.7 shows an example of an inelastic collision. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide. Some of the energy of motion gets converted to thermal energy, or heat. Figure 8.7 A one-dimensional inelastic collision between two objects. Momentum is conserved, but kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward each other at the same speed. (b) The objects stick together, creating a perfectly inelastic collision. In the case shown in this figure, the combined objects stop; This is not true for all inelastic collisions. Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from to for inelastic collisions, where v′ is the final velocity for both objects as they are stuck together, either in motion or at rest. WATCH PHYSICS Introduction to Momentum This video reviews the definitions of momentum and impulse. It also covers an example of using conservation of momentum to solve a problem involving an inelastic collision between a car with constant velocity and a stationary truck. Note that Sal accidentally gives the unit for impulse as Joules; it is actually N s or k gm/s. Click to view content (https://www.khanacademy.org/embed_video?v=XFhntPxow0U) GRASP CHECK How would the final velocity of the car-plus-truck system change if the truck had some initial velocity moving in the same direction as the car? What if the truck were moving in the opposite direction of the car initially? Why? a. If the truck was initially moving in the same direction as the car, the final velocity would be greater. If the truck was initially moving in the opposite direction of the car, the final velocity would be smaller. If the truck was initially moving in the same direction as the car, the final velocity would be smaller. If the truck was initially moving in the opposite direction of the car, the final velocity would be greater. b. c. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either Access for free at openstax.org. direction, the final velocity would be smaller. d. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either direction, the final velocity would be greater. 8.3 • Elastic and Inelastic Collisions 265 Snap Lab Ice Cubes and Elastic Collisions In this activity, you will observe an elastic collision by sliding an ice cube into another ice cube on a smooth surface, so that a negligible amount of energy is converted to heat. • Several ice cubes (The ice must be in the form of cubes.) • A smooth surface Procedure 1. Find a few ice cubes that are about the same size and a smooth kitchen tabletop or a table with a glass top. 2. Place the ice cubes on the surface several centimeters away from each other. 3. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. 4. Explain the speeds and directions of the ice cubes using momentum. GRASP CHECK Was the collision elastic or inelastic? a. perfectly elastic b. perfectly inelastic c. Nearly perfect elastic d. Nearly perfect inelastic TIPS FOR SUCCESS Here’s a trick for remembering which collisions are elastic and which are inelastic: Elastic is a bouncy material, so when objects bounceoff one another in the collision and separate, it is an elastic collision. When they don’t, the collision is inelastic. Solving Collision Problems The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In onedimensional collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with twodimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its xand y components. One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.8. Because momentum is conserved, the components of momentum along the x- and y-axes, displayed as px and py, will also be conserved. With the chosen coordinate system, pyis initially zero and pxis the momentum of the incoming particle. 266 Chapter 8 • Momentum Figure 8.8 A two-dimensional collision with the coordinate system chosen so that m2 is initially at rest and v1 is parallel to the x-axis. Now, we will take the conservation of momentum equation, p1 + p2 = p′1 + p′2 and break it into its xand ycomponents. Along the x-axis, the equation for conservation of momentum is In terms of masses and velocities, this equation is But because particle 2 is initially at rest, this equ
ation becomes 8.3 8.4 The components of the velocities along the x-axis have the form v cos θ. Because particle 1 initially moves along the x-axis, we find v1x= v1. Conservation of momentum along the x-axis gives the equation where and are as shown in Figure 8.8. Along the y-axis, the equation for conservation of momentum is or But v1yis zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v2yis also zero. The equation for conservation of momentum along the y-axis becomes 8.5 8.6 8.7 The components of the velocities along the y-axis have the form v sin . Therefore, conservation of momentum along the y-axis gives the following equation: Virtual Physics Collision Lab In this simulation, you will investigate collisions on an air hockey table. Place checkmarks next to the momentum vectors Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 267 and momenta diagram options. Experiment with changing the masses of the balls and the initial speed of ball 1. How does this affect the momentum of each ball? What about the total momentum? Next, experiment with changing the elasticity of the collision. You will notice that collisions have varying degrees of elasticity, ranging from perfectly elastic to perfectly inelastic. Click to view content (https://archive.cnx.org/specials/2c7acb3c-2fbd-11e5-b2d9-e7f92291703c/collision-lab/) GRASP CHECK If you wanted to maximize the velocity of ball 2 after impact, how would you change the settings for the masses of the balls, the initial speed of ball 1, and the elasticity setting? Why? Hint—Placing a checkmark next to the velocity vectors and removing the momentum vectors will help you visualize the velocity of ball 2, and pressing the More Data button will let you take readings. a. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 50 percent. b. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 100 percent. c. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 100 percent. d. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 50 percent. WORKED EXAMPLE Calculating Velocity: Inelastic Collision of a Puck and a Goalie Find the recoil velocity of a 70 kg ice hockey goalie who catches a 0.150-kg hockey puck slapped at him at a velocity of 35 m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible (see Figure 8.9). Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward in an inelastic collision. Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. Therefore, we can use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Solution For an inelastic collision, conservation of momentum is where v′ is the velocity of both the goalie and the puck after impact. Because the goalie is initially at rest, we know v2 = 0. This simplifies the equation to 8.8 8.9 Solving for v′ yields 268 Chapter 8 • Momentum Entering known values in this equation, we get 8.10 8.11 Discussion This recoil velocity is small and in the same direction as the puck’s original velocity. WORKED EXAMPLE Calculating Final Velocity: Elastic Collision of Two Carts Two hard, steel carts collide head-on and then ricochet off each other in opposite directions on a frictionless surface (see Figure 8.10). Cart 1 has a mass of 0.350 kg and an initial velocity of 2 m/s. Cart 2 has a mass of 0.500 kg and an initial velocity of −0.500 m/s. After the collision, cart 1 recoils with a velocity of −4 m/s. What is the final velocity of cart 2? Figure 8.10 Two carts collide with each other in an elastic collision. Strategy Since the track is frictionless, Fnet = 0 and we can use conservation of momentum to find the final velocity of cart 2. Solution As before, the equation for conservation of momentum for a one-dimensional elastic collision in a two-object system is The only unknown in this equation is v′2. Solving for v′2 and substituting known values into the previous equation yields 8.12 8.13 Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 269 WORKED EXAMPLE Calculating Final Velocity in a Two-Dimensional Collision Suppose the following experiment is performed (Figure 8.11). An object of mass 0.250 kg (m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object of mass 0.400 kg (m2). The 0.250 kg object emerges from the room at an angle of 45º with its incoming direction. The speed of the 0.250 kg object is originally 2 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′2 and ) of the 0.400 kg object after the collision. Figure 8.11 The incoming object of mass m1 is scattered by an initially stationary object. Only the stationary object’s mass m2 is known. By measuring the angle and speed at which the object of mass m1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision. Strategy Momentum is conserved because the surface is frictionless. We chose the coordinate system so that the initial velocity is parallel to the x-axis, and conservation of momentum along the x- and y-axes applies. Everything is known in these equations except v′2 and θ2, which we need to find. We can find two unknowns because we have two independent equations—the equations describing the conservation of momentum in the xand ydirections. Solution First, we’ll solve both conservation of momentum equations ( ) for v′2 sin . For conservation of momentum along x-axis, let’s substitute sin comes from rearranging the definition of the trigonometric identity tan = sin /cos for cos /tan Solving for v′2 sin yields For conservation of momentum along y-axis, solving for v′2 sin yields and so that terms may cancel out later on. This . This gives us 8.14 8.15 8.16 270 Chapter 8 • Momentum Since both equations equal v′2 sin , we can set them equal to one another, yielding Solving this equation for tan , we get Entering known values into the previous equation gives Therefore, Since angles are defined as positive in the counterclockwise direction, m2 is scattered to the right. We’ll use the conservation of momentum along the y-axis equation to solve for v′2. Entering known values into this equation gives Therefore, 8.17 8.18 8.19 8.20 8.21 8.22 8.23 Discussion Either equation for the x- or y-axis could have been used to solve for v′2, but the equation for the y-axis is easier because it has fewer terms. Practice Problems 10. In an elastic collision, an object with momentum collides with another object moving to the right that has a . After the collision, both objects are still moving to the right, but the first object’s momentum . What is the final momentum of the second object? momentum changes to a. b. c. d. 11. In an elastic collision, an object with momentum 25 kg ⋅ m/s collides with another that has a momentum 35 kg ⋅ m/s. The first object’s momentum changes to 10 kg ⋅ m/s. What is the final momentum of the second object? 10 kg ⋅ m/s a. b. 20 kg ⋅ m/s 35 kg ⋅ m/s c. 50 kg ⋅ m/s d. Check Your Understanding 12. What is an elastic collision? a. An elastic collision is one in which the objects after impact are deformed permanently. b. An elastic collision is one in which the objects after impact lose some of their internal kinetic energy. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 271 c. An elastic collision is one in which the objects after impact do not lose any of their internal kinetic energy. d. An elastic collision is one in which the objects after impact become stuck together and move with a common velocity. 13. Are perfectly elastic collisions possible? a. Perfectly elastic collisions are not possible. b. Perfectly elastic collisions are possible only with subatomic particles. c. Perfectly elastic collisions are possible only when the objects stick together after impact. d. Perfectly elastic collisions are possible if the objects and surfaces are nearly frictionless. 14. What is the equation for conservation of momentum for two objects in a one-dimensional collision? a. p1 + p1′ = p2 + p2′ b. p1 + p2 = p1′ + p2′ c. p1 − p2 = p1′ − p2′ d. p1 + p2 + p1′ + p2′ = 0 272 Chapter 8 • Key Terms KEY TERMS angular momentum the product of the moment of inertia and angular velocity change in momentum the difference between the final and initial values of momentum; the mass times the change in velocity elastic collision collision in which objects separate after after impact and kinetic energy is not conserved isolated system system in which the net external force is zero law of conservation of momentum when the net external force is zero, the total momentum of the system is conserved or constant impact and kinetic energy is conserved linear momentum the product of a system's mass and impulse average net external force multiplied by the time the force acts; equal to the change in momentum impulse–momentum theorem the impulse, or change in momentum, is the product of the net external force and the time over which the force acts inelastic collision collision in which objects stick together SECTION SUMMARY 8.1 Linear Momentum, Force, and Impulse • Linear momentum, often referenced as momentumfor short, is defined as the product of a system’s mass multiplied by its velocity, p = mv. • The SI unit for momentum is kg m/s. • Newton’s second law of motion in terms of mom
entum states that the net external force equals the change in momentum of a system divided by the time over which it changes, . • Impulse is the average net external force multiplied by the time this force acts, and impulse equals the change in momentum, . • Forces are usually not constant over a period of time, so we use the average of the force over the time it acts. 8.2 Conservation of Momentum velocity point masses spin structureless particles that cannot rotate or recoil backward movement of an object caused by the transfer of momentum from another object in a collision In an isolated system, the net external force is zero. • • Conservation of momentum applies only when the net external force is zero, within the defined system. 8.3 Elastic and Inelastic Collisions • If objects separate after impact, the collision is elastic; If they stick together, the collision is inelastic. • Kinetic energy is conserved in an elastic collision, but not in an inelastic collision. • The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle. • Two-dimensional collisions of point masses, where mass 2 is initially at rest, conserve momentum along the initial direction of mass 1, or the x-axis, and along the direction perpendicular to the initial direction, or the y-axis. • The law of conservation of momentum is written ptot = • Point masses are structureless particles that cannot constant or ptot = p′tot (isolated system), where ptot is the initial total momentum and p′tot is the total momentum some time later. spin. KEY EQUATIONS 8.1 Linear Momentum, Force, and Impulse Newton’s second law in terms of momentum impulse impulse–momentum theorem linear momentum Access for free at openstax.org. 8.2 Conservation of Momentum law of conservation of momentum ptot = constant, or ptot = p′tot Chapter 8 • Chapter Review 273 conservation of momentum along x-axis for 2D collisions conservation of momentum along y-axis for 2D collisions conservation of momentum for two objects p1 + p2 = constant, or p1 + p2 = p′1 + p′2 angular momentum L = I 8.3 Elastic and Inelastic Collisions conservation of momentum in an elastic collision conservation of momentum in an inelastic collision CHAPTER REVIEW Concept Items 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 5. What is angular momentum? 1. What is impulse? a. Change in velocity b. Change in momentum c. Rate of change of velocity d. Rate of change of momentum 2. In which equation of Newton’s second law is mass assumed to be constant? a. b. c. d. 3. What is the SI unit of momentum? a. b. c. d. 4. What is the equation for linear momentum? a. b. c. d. a. The sum of moment of inertia and angular velocity b. The ratio of moment of inertia to angular velocity c. The product of moment of inertia and angular velocity d. Half the product of moment of inertia and square of angular velocity 6. What is an isolated system? a. A system in which the net internal force is zero b. A system in which the net external force is zero c. A system in which the net internal force is a nonzero constant d. A system in which the net external force is a nonzero constant 8.3 Elastic and Inelastic Collisions 7. In the equation p1 + p2 = p'1 + p'2 for the collision of two objects, what is the assumption made regarding the friction acting on the objects? a. Friction is zero. b. Friction is nearly zero. c. Friction acts constantly. d. Friction before and after the impact remains the same. 8. What is an inelastic collision? 274 Chapter 8 • Chapter Review a. when objects stick together after impact, and their c. when objects stick together after impact, and always internal energy is not conserved come to rest instantaneously after collision b. when objects stick together after impact, and their d. when objects stick together after impact, and their internal energy is conserved internal energy increases Critical Thinking Items 8.1 Linear Momentum, Force, and Impulse 9. Consider two objects of the same mass. If a force of acts on the first for a duration of and on the , which of the following other for a duration of statements is true? a. The first object will acquire more momentum. b. The second object will acquire more momentum. c. Both objects will acquire the same momentum. d. Neither object will experience a change in momentum. 10. Cars these days have parts that can crumple or collapse in the event of an accident. How does this help protect the passengers? a. It reduces injury to the passengers by increasing the time of impact. It reduces injury to the passengers by decreasing the time of impact. It reduces injury to the passengers by increasing the change in momentum. It reduces injury to the passengers by decreasing the change in momentum. b. c. d. 11. How much force would be needed to cause a 17 kg ⋅ m/s change in the momentum of an object, if the force acted for 5 seconds? 3.4 N a. b. 12 N c. 22 N d. 85 N 8.2 Conservation of Momentum 12. A billiards ball rolling on the table has momentum p1. It hits another stationary ball, which then starts rolling. Considering friction to be negligible, what will happen to the momentum of the first ball? Problems 8.1 Linear Momentum, Force, and Impulse is applied to an object for , and it , what could be the mass 16. If a force of changes its velocity by of the object? a. b. Access for free at openstax.org. a. b. c. d. It will decrease. It will increase. It will become zero. It will remain the same. 13. A ball rolling on the floor with momentum p1 collides with a stationary ball and sets it in motion. The momentum of the first ball becomes p'1, and that of the second becomes p'2. Compare the magnitudes of p1 and p'2. a. Momenta p1 and p'2 are the same in magnitude. b. The sum of the magnitudes of p1 and p'2 is zero. c. The magnitude of p1 is greater than that of p'2. d. The magnitude of p'2 is greater than that of p1. 14. Two cars are moving in the same direction. One car with momentum p1 collides with another, which has momentum p2. Their momenta become p'1 and p'2 respectively. Considering frictional losses, compare (p'1 + p'2 ) with (p1 + p2). a. The value of (p'1 + p'2 ) is zero. b. The values of (p1 + p2) and (p'1 + p'2 ) are equal. c. The value of (p1 + p2) will be greater than (p'1 + p'2 ). d. The value of (p'1 + p'2 ) will be greater than (p1 + p2). 8.3 Elastic and Inelastic Collisions 15. Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil. a. The first person will gain more velocity as a result of recoil. b. The second person will gain more velocity as a result of recoil. c. Both people will gain the same velocity as a result of recoil. d. The velocity of both people will be zero as a result of recoil. c. d. 17. For how long should a force of 130 N be applied to an object of mass 50 kg to change its speed from 20 m/s to 60 m/s? a. 0.031 s b. 0.065 s 15.4 s c. Chapter 8 • Test Prep 275 d. 40 s d. 50.0 m/s 8.3 Elastic and Inelastic Collisions 18. If a man with mass 70 kg, standing still, throws an object with mass 5 kg at 50 m/s, what will be the recoil velocity of the man, assuming he is standing on a frictionless surface? a. −3.6 m/s b. 0 m/s c. 3.6 m/s 19. Find the recoil velocity of a ice hockey goalie who hockey puck slapped at him at a . Assume that the goalie is at rest catches a velocity of before catching the puck, and friction between the ice and the puck-goalie system is negligible. a. b. c. d. Performance Task 8.3 Elastic and Inelastic Collisions 20. You will need the following: • balls of different weights • a ruler or wooden strip • some books • a paper cup Make an inclined plane by resting one end of a ruler on a stack of books. Place a paper cup on the other end. Roll TEST PREP Multiple Choice 8.1 Linear Momentum, Force, and Impulse 21. What kind of quantity is momentum? a. Scalar b. Vector 22. When does the net force on an object increase? a. When Δp decreases b. When Δtincreases c. When Δtdecreases 23. In the equation Δp = m(vf − vi), which quantity is considered to be constant? a. Initial velocity b. Final velocity c. Mass d. Momentum 24. For how long should a force of be applied to change the momentum of an object by a. b. c. d. 8.2 Conservation of Momentum 25. In the equation L = Iω, what is I? a ball from the top of the ruler so that it hits the paper cup. Measure the displacement of the paper cup due to the collision. Now use increasingly heavier balls for this activity and see how that affects the displacement of the cup. Plot a graph of mass vs. displacement. Now repeat the same activity, but this time, instead of using different balls, change the incline of the ruler by varying the height of the stack of books. This will give you different velocities of the ball. See how this affects the displacement of the paper cup. a. Linear momentum b. Angular momentum c. Torque d. Moment of inertia 26. Give an example of an isolated system. a. A cyclist moving along a rough road b. A figure skater gliding in a straight line on an ice rink c. A baseball player hitting a home run d. A man drawing water from a well 8.3 Elastic and Inelastic Collisions 27. In which type of collision is kinetic energy conserved? a. Elastic b. Inelastic 28. In physics, what are structureless particles that cannot ? rotate or spin called? a. Elastic particles b. Point masses c. Rigid masses 29. Two objects having equal masses and velocities collide with each other and come to a rest. What type of a collision is this and why? a. Elastic collision, because internal kinetic energy is conserved 276 Chapter 8 • Test Prep b. Inelastic collision, because internal kinetic energy is not conserved c. Elastic collision, because internal kinetic ener
gy is d. not conserved Inelastic collision, because internal kinetic energy is conserved 30. Two objects having equal masses and velocities collide with each other and come to a rest. Is momentum conserved in this case? a. Yes b. No Short Answer 8.1 Linear Momentum, Force, and Impulse 31. If an object’s velocity is constant, what is its momentum proportional to? a. b. c. d. Its shape Its mass Its length Its breadth 32. If both mass and velocity of an object are constant, what can you tell about its impulse? a. b. c. d. Its impulse would be constant. Its impulse would be zero. Its impulse would be increasing. Its impulse would be decreasing. 33. When the momentum of an object increases with respect to time, what is true of the net force acting on it? a. It is zero, because the net force is equal to the rate of change of the momentum. It is zero, because the net force is equal to the product of the momentum and the time interval. It is nonzero, because the net force is equal to the rate of change of the momentum. It is nonzero, because the net force is equal to the product of the momentum and the time interval. b. c. d. 34. How can you express impulse in terms of mass and velocity when neither of those are constant? a. b. c. d. 35. How can you express impulse in terms of mass and initial and final velocities? a. b. c. d. 36. Why do we use average force while solving momentum problems? How is net force related to the momentum of the object? a. Forces are usually constant over a period of time, Access for free at openstax.org. and net force acting on the object is equal to the rate of change of the momentum. b. Forces are usually not constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. c. Forces are usually constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. d. Forces are usually not constant over a period of time, and net force acting on the object is equal to the rate of change of the momentum. 8.2 Conservation of Momentum 37. Under what condition(s) is the angular momentum of a system conserved? a. When net torque is zero b. When net torque is not zero c. When moment of inertia is constant d. When both moment of inertia and angular momentum are constant 38. If the moment of inertia of an isolated system increases, what happens to its angular velocity? a. b. c. d. It increases. It decreases. It stays constant. It becomes zero. 39. If both the moment of inertia and the angular velocity of a system increase, what must be true of the force acting on the system? a. Force is zero. b. Force is not zero. c. Force is constant. d. Force is decreasing. 8.3 Elastic and Inelastic Collisions 40. Two objects collide with each other and come to a rest. How can you use the equation of conservation of momentum to describe this situation? a. m1v1 + m2v2 = 0 b. m1v1 − m2v2 = 0 c. m1v1 + m2v2 = m1v1′ d. m1v1 + m2v2 = m1v2 41. What is the difference between momentum and impulse? a. Momentum is the sum of mass and velocity. Impulse is the change in momentum. b. Momentum is the sum of mass and velocity. Impulse is the rate of change in momentum. c. Momentum is the product of mass and velocity. Impulse is the change in momentum. d. Momentum is the product of mass and velocity. Impulse is the rate of change in momentum. 42. What is the equation for conservation of momentum along the x-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? Chapter 8 • Test Prep 277 a. m1v1 = m1v1′cos θ1 b. m1v1 = m1v1′cos θ1 + m2v2′cos θ2 c. m1v1 = m1v1′cos θ1 − m2v2′cos θ2 d. m1v1 = m1v1′sin θ1 + m2v2′sin θ2 43. What is the equation for conservation of momentum along the y-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? a. 0 = m1v1′sin θ1 b. 0 = m1v1′sin θ1 + m2v2′sin θ2 c. 0 = m1v1′sin θ1 − m2v2′sin θ2 d. 0 = m1v1′cos θ1 + m2v2′cos θ2 Extended Response 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 47. Why does a figure skater spin faster if he pulls his arms 44. Can a lighter object have more momentum than a heavier one? How? a. No, because momentum is independent of the velocity of the object. b. No, because momentum is independent of the mass of the object. c. Yes, if the lighter object’s velocity is considerably high. d. Yes, if the lighter object’s velocity is considerably low. 45. Why does it hurt less when you fall on a softer surface? a. The softer surface increases the duration of the impact, thereby reducing the effect of the force. b. The softer surface decreases the duration of the impact, thereby reducing the effect of the force. c. The softer surface increases the duration of the and legs in? a. Due to an increase in moment of inertia b. Due to an increase in angular momentum c. Due to conservation of linear momentum d. Due to conservation of angular momentum 8.3 Elastic and Inelastic Collisions 48. A driver sees another car approaching him from behind. He fears it is going to collide with his car. Should he speed up or slow down in order to reduce damage? a. He should speed up. b. He should slow down. c. He should speed up and then slow down just before the collision. d. He should slow down and then speed up just before the collision. impact, thereby increasing the effect of the force. 49. What approach would you use to solve problems d. The softer surface decreases the duration of the impact, thereby increasing the effect of the force. involving 2D collisions? a. Break the momenta into components and then 46. Can we use the equation when the mass is constant? a. No, because the given equation is applicable for the variable mass only. b. No, because the given equation is not applicable for the constant mass. c. Yes, and the resultant equation is F = mv d. Yes, and the resultant equation is F = ma choose a coordinate system. b. Choose a coordinate system and then break the momenta into components. c. Find the total momenta in the x and y directions, and then equate them to solve for the unknown. d. Find the sum of the momenta in the x and y directions, and then equate it to zero to solve for the unknown. 278 Chapter 8 • Test Prep Access for free at openstax.org. CHAPTER 9 Work, Energy, and Simple Machines Figure 9.1 People on a roller coaster experience thrills caused by changes in types of energy. (Jonrev, Wikimedia Commons) Chapter Outline 9.1 Work, Power, and the Work–Energy Theorem 9.2 Mechanical Energy and Conservation of Energy 9.3 Simple Machines Roller coasters have provided thrills for daring riders around the world since the nineteenth century. INTRODUCTION Inventors of roller coasters used simple physics to build the earliest examples using railroad tracks on mountainsides and old mines. Modern roller coaster designers use the same basic laws of physics to create the latest amusement park favorites. Physics principles are used to engineer the machines that do the work to lift a roller coaster car up its first big incline before it is set loose to roll. Engineers also have to understand the changes in the car’s energy that keep it speeding over hills, through twists, turns, and even loops. What exactly is energy? How can changes in force, energy, and simple machines move objects like roller coaster cars? How can machines help us do work? In this chapter, you will discover the answer to this question and many more, as you learn about 280 Chapter 9 • Work, Energy, and Simple Machines work, energy, and simple machines. 9.1 Work, Power, and the Work–Energy Theorem Section Learning Objectives By the end of this section, you will be able to do the following: • Describe and apply the work–energy theorem • Describe and calculate work and power Section Key Terms energy gravitational potential energy joule kinetic energy mechanical energy potential energy power watt work work–energy theorem The Work–Energy Theorem In physics, the term work has a very specific definition. Work is application of force, the direction that the force is applied. Work, W, is described by the equation , to move an object over a distance, d, in Some things that we typically consider to be work are not work in the scientific sense of the term. Let’s consider a few examples. Think about why each of the following statements is true. • Homework is notwork. • Lifting a rock upwards off the ground iswork. • Carrying a rock in a straight path across the lawn at a constant speed is notwork. The first two examples are fairly simple. Homework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. Recall from the laws of motion that force is notrequired to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity. Work and energy are closely related. When you do work to move an object, you change the object’s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy, which comes in two forms: kinetic energy and potential energy. • Kinetic energy is also called energy of motion. A moving object has kinetic energy. • Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy. Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a roc
k off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground. The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g. The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PEe. Kinetic energy depends on the mass of an object and its velocity, v. Access for free at openstax.org. 9.1 • Work, Power, and the Work–Energy Theorem 281 When we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity . This is a statement of the work–energy theorem, which is expressed mathematically as The subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown in Figure 9.2. Does the name Joule sound familiar? The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N∙m, the units of force multiplied by distance. 1.0 N = 1.0 k∙m/s2, so 1.0 J = 1.0 k∙m2/s2. Analyzing the units of the term (1/2)mv2 will produce the same units for joules. Figure 9.2 The joule is named after physicist James Joule (1818–1889). (C. H. Jeens, Wikimedia Commons) WATCH PHYSICS Work and Energy This video explains the work energy theorem and discusses how work done on an object increases the object’s KE. Click to view content (https://www.khanacademy.org/embed_video?v=2WS1sG9fhOk) GRASP CHECK True or false—The energy increase of an object acted on only by a gravitational force is equal to the product of the object's weight and the distance the object falls. a. True b. False Calculations Involving Work and Power In applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let’s take a look at how to calculate the time it takes to do work. Recall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time. Power is calculated by dividing the work done by the time it took to do the work. Let’s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in Figure 9.3 lifting the TV with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five 282 Chapter 9 • Work, Energy, and Simple Machines minutes to arrive at the same place. They have done the same amount of work mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t, is smaller. (For simplicity’s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.) on the TV, because they have moved the same Figure 9.3 No matter how you move a TV to the fourth floor, the amount of work performed and the potential energy gain are the same. Power can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed. The watt unit was named after James Watt (1736–1819) (see Figure 9.4). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines. Figure 9.4 Is James Watt thinking about watts? (Carl Frederik von Breda, Wikimedia Commons) Access for free at openstax.org. 9.1 • Work, Power, and the Work–Energy Theorem 283 LINKS TO PHYSICS Watt’s Steam Engine James Watt did not invent the steam engine, but by the time he was finished tinkering with it, it was more useful. The first steam engines were not only inefficient, they only produced a back and forth, or reciprocal, motion. This was natural because pistons move in and out as the pressure in the chamber changes. This limitation was okay for simple tasks like pumping water or mashing potatoes, but did not work so well for moving a train. Watt was able build a steam engine that converted reciprocal motion to circular motion. With that one innovation, the industrial revolution was off and running. The world would never be the same. One of Watt's steam engines is shown in Figure 9.5. The video that follows the figure explains the importance of the steam engine in the industrial revolution. Figure 9.5 A late version of the Watt steam engine. (Nehemiah Hawkins, Wikimedia Commons) WATCH PHYSICS Watt's Role in the Industrial Revolution This video demonstrates how the watts that resulted from Watt's inventions helped make the industrial revolution possible and allowed England to enter a new historical era. Click to view content (https://www.youtube.com/embed/zhL5DCizj5c) GRASP CHECK Which form of mechanical energy does the steam engine generate? a. Potential energy b. Kinetic energy c. Nuclear energy d. Solar energy Before proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done. WORKED EXAMPLE Applying the Work–Energy Theorem An ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater? 284 Chapter 9 • Work, Energy, and Simple Machines Strategy The work–energy theorem can be applied to the problem. Write the equation for the theorem and simplify it if possible. Solution Identify the variables. m= 50 kg, Substitute. 9.1 9.2 Discussion Work done on an object or system increases its energy. In this case, the increase is to the skater’s kinetic energy. It follows that the increase in energy must be the difference in KE before and after the push. TIPS FOR SUCCESS This problem illustrates a general technique for approaching problems that require you to apply formulas: Identify the unknown and the known variables, express the unknown variables in terms of the known variables, and then enter all the known values. Practice Problems 1. How much work is done when a weightlifter lifts a barbell from the floor to a height of ? a. b. c. d. 2. Identify which of the following actions generates more power. Show your work. • • carrying a carrying a TV to the second floor in or watermelon to the second floor in ? a. Carrying a TV generates more power than carrying a watermelon to the same height because power is defined as work done times the time interval. b. Carrying a TV generates more power than carrying a watermelon to the same height because power is defined as the ratio of work done to the time interval. c. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as work done times the time interval. d. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as the ratio of work done and the time interval. Check Your Understanding 3. Identify two properties that are expressed in units of joules. a. work and force b. energy and weight c. work and energy d. weight and force Access for free at openstax.org. 9.2 • Mechanical Energy and Conservation of Energy 285 4. When a coconut falls from a tree, work Wis done on it as it falls to the beach. This work is described by the equation 9.3 Identify the quantities F, d, m, v1, and v2 in this event. a. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. b. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. c. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. d. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. 9.2 Mechanical Energy and Conservation of Energy Section Learning Objectives By the end of this section, you will be able to do the following: • Explain the law of conservation of energy in terms of kinetic and potential energy • Perform calculations related to kinetic and potential energy. Apply the law of conservation of energy Section Key Terms law of conservation of energy Mechanical Energy and Conservation of Energy We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy re
mains constant. Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potentialpart of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules. Refer back to . The amount of work required to raise the TV from point A to point B is equal to the amount of gravitational potential energy the TV gains from its height above the ground. This is generally true for any object raised above the ground. If all the work done on an object is used to raise the object above the ground, the amount work equals the object’s gain in gravitational potential energy. However, note that because of the work done by friction, these energy–work transformations are never perfect. Friction causes the loss of some useful energy. In the discussions to follow, we will use the approximation that transformations are frictionless. Now, let’s look at the roller coaster in Figure 9.6. Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PEis converted to KE. At the low point much of the original PEhas been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KEis transformed back into PEand the car slows down. 286 Chapter 9 • Work, Energy, and Simple Machines Figure 9.6 During this roller coaster ride, there are conversions between potential and kinetic energy. Virtual Physics Energy Skate Park Basics This simulation shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KEand PEby clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. Click to view content (http://phet.colorado.edu/sims/html/energy-skate-park-basics/latest/energy-skate-parkbasics_en.html) GRASP CHECK This simulation (http://phet.colorado.edu/en/simulation/energy-skate-park-basics (http://phet.colorado.edu/en/ simulation/energy-skate-park-basics) ) shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KE and PE by clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. The bar graphs show how KE and PE are transformed back and forth. Which statement best explains what happens to the mechanical energy of the system as speed is increasing? a. The mechanical energy of the system increases, provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. b. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. c. The mechanical energy of the system increases provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. d. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved—that is, it is constant. Using subscripts 1 and 2 to represent initial and final energy, this law is expressed as Either side equals the total mechanical energy. The phrase in aclosed systemmeans we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption. For the roller coaster, this assumption introduces some inaccuracy to the calculation. Access for free at openstax.org. 9.2 • Mechanical Energy and Conservation of Energy 287 Calculations involving Mechanical Energy and Conservation of Energy TIPS FOR SUCCESS When calculating work or energy, use units of meters for distance, newtons for force, kilograms for mass, and seconds for time. This will assure that the result is expressed in joules. WATCH PHYSICS Conservation of Energy This video discusses conversion of PEto KEand conservation of energy. The scenario is very similar to the roller coaster and the skate park. It is also a good explanation of the energy changes studied in the snap lab. Click to view content (https://www.khanacademy.org/embed_video?v=kw_4Loo1HR4) GRASP CHECK Did you expect the speed at the bottom of the slope to be the same as when the object fell straight down? Which statement best explains why this is not exactly the case in real-life situations? a. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. b. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. c. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. d. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. WORKED EXAMPLE Applying the Law of Conservation of Energy A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m? Strategy Choose the equation. 9.4 9.5 9.6 9.7 List the knowns. m= 10 kg, v1 = 0, g = 9.80 h1 = 20 m, h2 = 10 m Identify the unknowns. KE2 and PE2 Substitute the known values into the equation and solve for the unknown variables. 288 Chapter 9 • Work, Energy, and Simple Machines Solution 9.8 9.9 Discussion Alternatively, conservation of energy equation could be solved for v2 and KE2 could be calculated. Note that mcould also be eliminated. TIPS FOR SUCCESS Note that we can solve many problems involving conversion between KEand PEwithout knowing the mass of the object in question. This is because kinetic and potential energy are both proportional to the mass of the object. In a situation where KE= PE, we know that mgh= (1/2)mv2. Dividing both sides by mand rearranging, we have the relationship 2gh= v2. Practice Problems 5. A child slides down a playground slide. If the slide is 3 m high and the child weighs 300 N, how much potential energy does ) the child have at the top of the slide? (Round gto a. 0 J 100 J b. c. 300 J d. 900 J 6. A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground? a. 0 m/s b. 2 m/s 10 m/s c. 50 m/s d. Snap Lab Converting Potential Energy to Kinetic Energy In this activity, you will calculate the potential energy of an object and predict the object’s speed when all that potential energy has been converted to kinetic energy. You will then check your prediction. You will be dropping objects from a height. Be sure to stay a safe distance from the edge. Don’t lean over the railing too far. Make sure that you do not drop objects into an area where people or vehicles pass by. Make sure that dropping objects will not cause damage. You will need the following: Materials for each pair of students: • Four marbles (or similar small, dense objects) • Stopwatch Materials for class: • Metric measuring tape long enough to measure the chosen height • A scale Instructions Access for free at openstax.org. 9.2 • Mechanical Energy and Conservation of Energy 289 Procedure 1. Work with a partner. Find and record the mass of four small, dense objects per group. 2. Choose a location where the objects can be safely dropped from a height of at least 15 meters. A bridge over water with a safe pedestrian walkway will work well. 3. Measure the distance the object will fall. 4. Calculate the potential energy of the object before you drop it using PE= mgh= (9.80)mh. 5. Predict the kinetic energy and velocity of the object when it lands using PE= KEand so, 6. One partner drops the object while the other measures the time it takes to fall. 7. Take turns being the dropper and the timer until you have made four measurements. 8. Average your drop multiplied by and calculate the velocity of the object when it landed using v = at= gt= (9.80)t. 9. Compare your results to your prediction. GRASP CHECK Galileo’s experiments proved that, contrary to popular belief, heavy objects do not fall faster than light objects. How do the equations you used support this fact? a. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resi
stance. b. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. c. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. d. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. Check Your Understanding 7. Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens. a. Kinetic energy is being transformed into potential energy. b. Potential energy is being transformed into kinetic energy. c. Work is being transformed into kinetic energy. d. Kinetic energy is being transformed into work. 8. True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. a. True b. False 9. Identify equivalent terms for stored energyand energy of motion. a. Stored energy is potential energy, and energy of motion is kinetic energy. b. Energy of motion is potential energy, and stored energy is kinetic energy. c. Stored energy is the potential as well as the kinetic energy of the system. d. Energy of motion is the potential as well as the kinetic energy of the system. 290 Chapter 9 • Work, Energy, and Simple Machines 9.3 Simple Machines Section Learning Objectives By the end of this section, you will be able to do the following: • Describe simple and complex machines • Calculate mechanical advantage and efficiency of simple and complex machines Section Key Terms complex machine efficiency output ideal mechanical advantage inclined plane input work lever mechanical advantage output work pulley screw simple machine wedge wheel and axle Simple Machines Simple machines make work easier, but they do not decrease the amount of work you have to do. Why can’t simple machines change the amount of work that you do? Recall that in closed systems the total amount of energy is conserved. A machine cannot increase the amount of energy you put into it. So, why is a simple machine useful? Although it cannot change the amount of work you do, a simple machine can change the amount of force you must apply to an object, and the distance over which you apply the force. In most cases, a simple machine is used to reduce the amount of force you must exert to do work. The down side is that you must exert the force over a greater distance, because the product of force and distance, fd, (which equals work) does not change. Let’s examine how this works in practice. In Figure 9.7(a), the worker uses a type of lever to exert a small force over a large distance, while the pry bar pulls up on the nail with a large force over a small distance. Figure 9.7(b) shows the how a lever works mathematically. The effort force, applied at Fe, lifts the load (the resistance force) which is pushing down at Fr. The triangular pivot is called the fulcrum; the part of the lever between the fulcrum and Feis the effort arm, Le; and the part to the left is the resistance arm, Lr. The mechanical advantage is a number that tells us how many times a simple machine multiplies the effort force. The ideal mechanical advantage, IMA, is the mechanical advantage of a perfect machine with no loss of useful work caused by friction between moving parts. The equation for IMAis shown in Figure 9.7(b). Figure 9.7 (a) A pry bar is a type of lever. (b) The ideal mechanical advantage equals the length of the effort arm divided by the length of the resistance arm of a lever. In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Getting back to conservation of energy, for any simple machine, the work put into the machine, Wi, equals the work the machine puts out, Wo. Combining this with the information in the paragraphs above, we can write Access for free at openstax.org. 9.3 • Simple Machines 291 The equations show how a simple machine can output the same amount of work while reducing the amount of effort force by increasing the distance over which the effort force is applied. WATCH PHYSICS Introduction to Mechanical Advantage This video shows how to calculate the IMAof a lever by three different methods: (1) from effort force and resistance force; (2) from the lengths of the lever arms, and; (3) from the distance over which the force is applied and the distance the load moves. Click to view content (https://www.youtube.com/embed/pfzJ-z5Ij48) GRASP CHECK Two children of different weights are riding a seesaw. How do they position themselves with respect to the pivot point (the fulcrum) so that they are balanced? a. The heavier child sits closer to the fulcrum. b. The heavier child sits farther from the fulcrum. c. Both children sit at equal distance from the fulcrum. d. Since both have different weights, they will never be in balance. Some levers exert a large force to a short effort arm. This results in a smaller force acting over a greater distance at the end of the resistance arm. Examples of this type of lever are baseball bats, hammers, and golf clubs. In another type of lever, the fulcrum is at the end of the lever and the load is in the middle, as in the design of a wheelbarrow. The simple machine shown in Figure 9.8 is called a wheel and axle. It is actually a form of lever. The difference is that the effort arm can rotate in a complete circle around the fulcrum, which is the center of the axle. Force applied to the outside of the wheel causes a greater force to be applied to the rope that is wrapped around the axle. As shown in the figure, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle. Any crank-operated device is an example of a wheel and axle. Figure 9.8 Force applied to a wheel exerts a force on its axle. An inclined plane and a wedge are two forms of the same simple machine. A wedge is simply two inclined planes back to back. Figure 9.9 shows the simple formulas for calculating the IMAs of these machines. All sloping, paved surfaces for walking or driving are inclined planes. Knives and axe heads are examples of wedges. 292 Chapter 9 • Work, Energy, and Simple Machines Figure 9.9 An inclined plane is shown on the left, and a wedge is shown on the right. The screw shown in Figure 9.10 is actually a lever attached to a circular inclined plane. Wood screws (of course) are also examples of screws. The lever part of these screws is a screw driver. In the formula for IMA, the distance between screw threads is called pitchand has the symbol P. Figure 9.10 The screw shown here is used to lift very heavy objects, like the corner of a car or a house a short distance. Figure 9.11 shows three different pulley systems. Of all simple machines, mechanical advantage is easiest to calculate for pulleys. Simply count the number of ropes supporting the load. That is the IMA. Once again we have to exert force over a longer distance to multiply force. To raise a load 1 meter with a pulley system you have to pull Nmeters of rope. Pulley systems are often used to raise flags and window blinds and are part of the mechanism of construction cranes. Figure 9.11 Three pulley systems are shown here. Access for free at openstax.org. 9.3 • Simple Machines 293 WATCH PHYSICS Mechanical Advantage of Inclined Planes and Pulleys The first part of this video shows how to calculate the IMAof pulley systems. The last part shows how to calculate the IMAof an inclined plane. Click to view content (https://www.khanacademy.org/embed_video?v=vSsK7Rfa3yA) GRASP CHECK How could you use a pulley system to lift a light load to great height? a. Reduce the radius of the pulley. b. Increase the number of pulleys. c. Decrease the number of ropes supporting the load. Increase the number of ropes supporting the load. d. A complex machine is a combination of two or more simple machines. The wire cutters in Figure 9.12 combine two levers and two wedges. Bicycles include wheel and axles, levers, screws, and pulleys. Cars and other vehicles are combinations of many machines. Figure 9.12 Wire cutters are a common complex machine. Calculating Mechanical Advantage and Efficiency of Simple Machines In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Refer back to the discussions of each simple machine for the specific equations for the IMAfor each type of machine. No simple or complex machines have the actual mechanical advantages calculated by the IMAequations. In real life, some of the applied work always ends up as wasted heat due to friction between moving parts. Both the input work (Wi) and output work (Wo) are the result of a force, F, acting over a distance, d. The efficiency output of a machine is simply the output work divided by the input work, and is usually multiplied by 100 so that it is expressed as a percent. Look back at the pictures of the simple machines and think about which would have the highest efficiency. Efficiency is related to friction, and friction depends on the smoothness of surfaces and on the area of the surfac
es in contact. How would lubrication affect the efficiency of a simple machine? WORKED EXAMPLE Efficiency of a Lever The input force of 11 N acting on the effort arm of a lever moves 0.4 m, which lifts a 40 N weight resting on the resistance arm a 294 Chapter 9 • Work, Energy, and Simple Machines distance of 0.1 m. What is the efficiency of the machine? Strategy State the equation for efficiency of a simple machine, are the product Fd. Solution = (11)(0.4) = 4.4 J and = (40)(0.1) = 4.0 J, then and calculate Woand Wi. Both work values Discussion Efficiency in real machines will always be less than 100 percent because of work that is converted to unavailable heat by friction and air resistance. Woand Wican always be calculated as a force multiplied by a distance, although these quantities are not always as obvious as they are in the case of a lever. Practice Problems 10. What is the IMA of an inclined plane that is long and high? a. b. c. d. 11. If a pulley system can lift a 200N load with an effort force of 52 N and has an efficiency of almost 100 percent, how many ropes are supporting the load? 1 rope is required because the actual mechanical advantage is 0.26. a. b. 1 rope is required because the actual mechanical advantage is 3.80. c. 4 ropes are required because the actual mechanical advantage is 0.26. d. 4 ropes are required because the actual mechanical advantage is 3.80. Check Your Understanding 12. True or false—The efficiency of a simple machine is always less than 100 percent because some small fraction of the input work is always converted to heat energy due to friction. a. True b. False 13. The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1 cm and the IMA of the machine is 6, what is the radius of the handle? A. 0.08 cm B. 0.17 cm C. 3.0 cm D. 6.0 cm Access for free at openstax.org. Chapter 9 • Key Terms 295 KEY TERMS complex machine a machine that combines two or more output work output force multiplied by the distance over simple machines which it acts efficiency output work divided by input work energy the ability to do work gravitational potential energy energy acquired by doing potential energy stored energy power pulley a simple machine consisting of a rope that passes the rate at which work is done work against gravity ideal mechanical advantage the mechanical advantage of an idealized machine that loses no energy to friction inclined plane a simple machine consisting of a slope input work effort force multiplied by the distance over which it is applied over one or more grooved wheels screw a simple machine consisting of a spiral inclined plane simple machine a machine that makes work easier by changing the amount or direction of force required to move an object joule the metric unit for work and energy; equal to 1 watt the metric unit of power; equivalent to joules per newton meter (N∙m) second kinetic energy energy of motion law of conservation of energy states that energy is neither wedge a simple machine consisting of two back-to-back inclined planes created nor destroyed wheel and axle a simple machine consisting of a rod fixed lever a simple machine consisting of a rigid arm that pivots to the center of a wheel on a fulcrum mechanical advantage the number of times the input force is multiplied mechanical energy kinetic or potential energy work force multiplied by distance work–energy theorem states that the net work done on a system equals the change in kinetic energy SECTION SUMMARY 9.1 Work, Power, and the Work–Energy Theorem • Doing work on a system or object changes its energy. • The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object.The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object. • Power is the rate at which work is done. 9.2 Mechanical Energy and Conservation of Energy • Mechanical energy may be either kinetic (energy of KEY EQUATIONS 9.1 Work, Power, and the Work–Energy Theorem equation for work force work equivalencies motion) or potential (stored energy). • Doing work on an object or system changes its energy. • Total energy in a closed, isolated system is constant. 9.3 Simple Machines • The six types of simple machines make work easier by changing the fdterm so that force is reduced at the expense of increased distance. • The ratio of output force to input force is a machine’s mechanical advantage. • Combinations of two or more simple machines are called complex machines. • The ratio of output work to input work is a machine’s efficiency. kinetic energy work–energy theorem power 296 Chapter 9 • Chapter Review 9.2 Mechanical Energy and Conservation of Energy conservation of energy 9.3 Simple Machines ideal mechanical advantage (general) ideal mechanical advantage (lever) ideal mechanical advantage (wheel and axle) CHAPTER REVIEW Concept Items 9.1 Work, Power, and the Work–Energy Theorem 1. Is it possible for the sum of kinetic energy and potential energy of an object to change without work having been done on the object? Explain. a. No, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. b. No, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant. c. Yes, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. d. Yes, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant. 2. Define work for one-dimensional motion. a. Work is defined as the ratio of the force over the distance. b. Work is defined as the sum of the force and the distance. c. Work is defined as the square of the force over the Access for free at openstax.org. ideal mechanical advantage (inclined plane) ideal mechanical advantage (wedge) ideal mechanical advantage (pulley) ideal mechanical advantage (screw) input work output work efficiency output distance. d. Work is defined as the product of the force and the distance. 3. A book with a mass of 0.30 kg falls 2 m from a shelf to the floor. This event is described by the work–energy theorem: Explain why this is enough information to calculate the speed with which the book hits the floor. a. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg . v1 is the initial velocity and v2 is the final velocity. The final velocity is the only unknown quantity. b. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg . v1 is the final velocity and v2 is the initial velocity. The final velocity is the only unknown quantity. c. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg . v1 is the initial velocity and v2 is the final velocity. The final velocity and the initial velocities are the only unknown quantities. d. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg . v1 is the final velocity and v2 is the initial velocity. The final velocity and the initial velocities are the only unknown quantities. Chapter 9 • Chapter Review 297 9.2 Mechanical Energy and Conservation of Energy 4. Describe the changes in KE and PE of a person jumping up and down on a trampoline. a. While going up, the person’s KE would change to PE. While coming down, the person’s PE would change to KE. b. While going up, the person’s PE would change to KE. While coming down, the person’s KE would change to PE. c. While going up, the person’s KE would not change, but while coming down, the person’s PE would change to KE. d. While going up, the person’s PE would change to KE, but while coming down, the person’s KE would not change. 6. The starting line of a cross country foot race is at the bottom of a hill. Which form(s) of mechanical energy of the runners will change when the starting gun is fired? a. Kinetic energy only b. Potential energy only c. Both kinetic and potential energy d. Neither kinetic nor potential energy 9.3 Simple Machines 7. How does a simple machine make work easier? a. b. c. d. It reduces the input force and the output force. It reduces the input force and increases the output force. It increases the input force and reduces the output force. It increases the input force and the output force. 5. You know the height from which an object is dropped. 8. Which type of simple machine is a knife? Which equation could you use to calculate the velocity as the object hits the ground? a. b. c. d. a. A ramp b. A wedge c. A pulley d. A screw Critical Thinking Items 9.1 Work, Power, and the Work–Energy Theorem 9. Which activity requires a person to exert force on an object that causes the object to move but does not change the kinetic or potential energy of the object? a. Moving an object to a greater height with acceleration b. Moving an object to a greater height without acceleration c. Carrying an object with acceleration at the same height 9.2 Mechanical Energy and Conservation of Energy 11. True or false—A cyclist coasts down one hill and up another hill until she comes to a stop. The point at which the bicycle stops is lower than the point at which it started coasting because part of the original potential energy has been converted to a quantity of heat and this makes the tires of the bicycle warm. a. True b. False 9.3 Simple Machines d. Carrying an object without acceleration at the same 12. We think of levers being used to decrease
effort force. height 10. Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the books or doing any work? a. By moving the book without acceleration and keeping the height of the book constant b. By moving the book with acceleration and keeping the height of the book constant c. By moving the book without acceleration and changing the height of the book d. By moving the book with acceleration and changing the height of the book Which of the following describes a lever that requires a large effort force which causes a smaller force to act over a large distance and explains how it works? a. Anything that is swung by a handle, such as a hammer or racket. Force is applied near the fulcrum over a short distance, which makes the other end move rapidly over a long distance. b. Anything that is swung by a handle, such as a hammer or racket. Force is applied far from the fulcrum over a large distance, which makes the other end move rapidly over a long distance. c. A lever used to lift a heavy stone. Force is applied near the fulcrum over a short distance, which 298 Chapter 9 • Chapter Review makes the other end lift a heavy object easily. d. A lever used to lift a heavy stone. Force is applied far from the fulcrum over a large distance, which makes the other end lift a heavy object easily 13. A baseball bat is a lever. Which of the following explains how a baseball bat differs from a lever like a pry bar? a. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a long distance. Problems 9.1 Work, Power, and the Work–Energy Theorem 14. A baseball player exerts a force of on a ball for a as he throws it. If the ball has a mass , what is its velocity as it leaves his hand? distance of of a. b. c. d. 15. A boy pushes his little sister on a sled. The sled accelerates from 0 to 3.2 m/s . If the combined mass of his sister and the sled is 40.0 kg and 18 W of power were generated, how long did the boy push the sled? a. 205 s b. 128 s c. 23 s 11 s d. 9.2 Mechanical Energy and Conservation of Energy 16. What is the kinetic energy of a bullet traveling ? at a velocity of a. b. c. d. 17. A marble rolling across a flat, hard surface at rolls and no energy is up a ramp. Assuming that lost to friction, what will be the vertical height of the marble when it comes to a stop before rolling back down? Ignore effects due to the rotational kinetic energy. a. b. c. d. 18. The potential energy stored in a compressed spring is Access for free at openstax.org. b. c. d. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a short distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a long distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a short distance. , where kis the force constant and xis the distance the spring is compressed from the equilibrium position. Four experimental setups described below can be used to determine the force constant of a spring. Which one(s) require measurement of the fewest number of variables to determine k? Assume the acceleration due to gravity is known. I. An object is propelled vertically by a compressed spring. II. An object is propelled horizontally on a frictionless surface by a compressed spring. III. An object is statically suspended from a spring. IV. An object suspended from a spring is set into oscillatory motion. a. b. c. d. I only III only I and II only III and IV only 9.3 Simple Machines 19. A man is using a wedge to split a block of wood by hitting the wedge with a hammer. This drives the wedge into the wood creating a crack in the wood. When he hits the wedge with a force of 400 N it travels 4 cm into the wood. This caused the wedge to exert a force of 1,400 N sideways increasing the width of the crack by 1 cm . What is the efficiency of the wedge? a. 0.875 percent b. 0.14 c. 0.751 d. 87.5 percent 20. An electrician grips the handles of a wire cutter, like the one shown, 10 cm from the pivot and places a wire between the jaws 2 cm from the pivot. If the cutter blades are 2 cm wide and 0.3 cm thick, what is the overall IMA of this complex machine? Performance Task 9.3 Simple Machines 21. Conservation of Energy and Energy Transfer; Cause and Effect; and S&EP, Planning and Carrying Out Investigations Plan an investigation to measure the mechanical advantage of simple machines and compare to the IMA of the machine. Also measure the efficiency of each machine studied. Design an investigation to make these measurements for these simple machines: lever, inclined plane, wheel and axle and a pulley system. In addition to these machines, include a spring scale, a tape measure, and a weight with a loop on top that can be attached to the hook on the spring scale. A spring scale is shown in the image. Chapter 9 • Test Prep 299 a. b. c. d. 1.34 1.53 33.3 33.5 LEVER: Beginning with the lever, explain how you would measure input force, output force, effort arm, and resistance arm. Also explain how you would find the distance the load travels and the distance over which the effort force is applied. Explain how you would use this data to determine IMAand efficiency. INCLINED PLANE: Make measurements to determine IMAand efficiency of an inclined plane. Explain how you would use the data to calculate these values. Which property do you already know? Note that there are no effort and resistance arm measurements, but there are height and length measurements. WHEEL AND AXLE: Again, you will need two force measurements and four distance measurements. Explain how you would use these to calculate IMAand efficiency. SCREW: You will need two force measurements, two distance traveled measurements, and two length measurements. You may describe a screw like the one shown in Figure 9.10 or you could use a screw and screw driver. (Measurements would be easier for the former). Explain how you would use these to calculate IMAand efficiency. PULLEY SYSTEM: Explain how you would determine the IMAand efficiency of the four-pulley system shown in Figure 9.11. Why do you only need two distance measurements for this machine? Design a table that compares the efficiency of the five simple machines. Make predictions as to the most and least efficient machines. A spring scale measures weight, not mass. TEST PREP Multiple Choice 9.1 Work, Power, and the Work–Energy Theorem 22. Which expression represents power? a. b. c. d. 23. The work–energy theorem states that the change in the kinetic energy of an object is equal to what? a. The work done on the object b. The force applied to the object c. The loss of the object’s potential energy d. The object’s total mechanical energy minus its kinetic energy 24. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s . If the runner has a mass of 60 kg, how long did it take him to reach that speed? a. 0.33 s b. 0.83 s c. d. 1.2 s 3.0 s 300 Chapter 9 • Test Prep 25. A car’s engine generates 100,000 W of power as it exerts a force of 10,000 N. How long does it take the car to travel 100 m? a. 0.001 s b. 0.01 s 10 s c. 1,000 s d. 9.2 Mechanical Energy and Conservation of Energy 26. Why is this expression for kinetic energy incorrect? is missing. . a. The constant b. The term should not be squared. c. The expression should be divided by . d. The energy lost to friction has not been subtracted. 27. What is the kinetic energy of a object moving at ? a. b. c. d. 28. Which statement best describes the PE-KE transformations for a javelin, starting from the instant the javelin leaves the thrower's hand until it hits the ground. a. Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, KE is transformed into PE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, PE is transformed into KE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. b. c. d. 29. At the beginning of a roller coaster ride, the roller coaster car has an initial energy mostly in the form of PE. Which statement explains why the fastest speeds of the car will be at the lowest points in the ride? a. At the bottom of the slope kinetic energy is at its Access for free at openstax.org. maximum value and potential energy is at its minimum value. b. At the bottom of the slope potential energy is at its maximum value and kinetic energy is at its minimum value. c. At the bottom of the slope both kinetic and potential energy reach their maximum values d. At the bottom of the slope both kinetic and potential energy reach their minimum values. 9.3 Simple Machines 30. A large radius divided by a small radius is the expression used to calculate the IMA of what? a. A screw b. A pulley c. A wheel and axle d. An inclined plane. 31. What is the IMA of a wedge that is long and thick? a. b. c. d. 32. Which statement correctly describes the simple machines, like the crank in the image, that make up an Archimedes screw and the forces it applies? a. The crank is a wedge in which the IMA is the length of the tube
divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. b. The crank is an inclined plane in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. c. The crank is a wheel and axle. The effort force of the crank becomes the resistance force of the screw. d. The crank is a wheel and axle. The resistance force of the crank becomes the effort force of the screw. 33. Refer to the pulley system on right in the image. Assume this pulley system is an ideal machine. How hard would you have to pull on the rope to lift a 120 N Chapter 9 • Test Prep 301 load? How many meters of rope would you have to pull out of the system to lift the load 1 m? a. 480 N 4 m b. 480 N c. d. m 30 N 4 m 30 N m Short Answer 9.1 Work, Power, and the Work–Energy Theorem 34. Describe two ways in which doing work on an object can increase its mechanical energy. a. Raising an object to a higher elevation does work as it increases its PE; increasing the speed of an object does work as it increases its KE. b. Raising an object to a higher elevation does work as it increases its KE; increasing the speed of an object does work as it increases its PE. c. Raising an object to a higher elevation does work as it increases its PE; decreasing the speed of an object does work as it increases its KE. d. Raising an object to a higher elevation does work as it increases its KE; decreasing the speed of an object does work as it increases its PE. 35. True or false—While riding a bicycle up a gentle hill, it is fairly easy to increase your potential energy, but to increase your kinetic energy would make you feel exhausted. a. True b. False 36. Which statement best explains why running on a track with constant speed at 3 m/s is not work, but climbing a mountain at 1 m/s is work? a. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy. b. At constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy. c. At constant speed, change in the kinetic energy is finite, but climbing a mountain produces no change in the potential energy. d. At constant speed, change in the potential energy is finite, but climbing a mountain produces no change in the kinetic energy. 37. You start at the top of a hill on a bicycle and coast to the bottom without applying the brakes. By the time you reach the bottom of the hill, work has been done on you and your bicycle, according to the equation: If is the mass of you and your bike, what are a. and ? is your speed at the top of the hill, and is your speed at the bottom. b. c. d. is your speed at the bottom of the hill, and is your speed at the top. is your displacement at the top of the hill, and is your displacement at the bottom. is your displacement at the bottom of the hill, and is your displacement at the top. 9.2 Mechanical Energy and Conservation of Energy 38. True or false—The formula for gravitational potential energy can be used to explain why joules, J, are equivalent to kg × mg2 / s2 . Show your work. a. True b. False 39. Which statement best explains why accelerating a car from a. Because kinetic energy is directly proportional to quadruples its kinetic energy? to the square of the velocity. b. Because kinetic energy is inversely proportional to the square of the velocity. c. Because kinetic energy is directly proportional to the fourth power of the velocity. d. Because kinetic energy is inversely proportional to 302 Chapter 9 • Test Prep the fourth power of the velocity. plane. 40. A coin falling through a vacuum loses no energy to friction, and yet, after it hits the ground, it has lost all its potential and kinetic energy. Which statement best explains why the law of conservation of energy is still valid in this case? a. When the coin hits the ground, the ground gains potential energy that quickly changes to thermal energy. b. When the coin hits the ground, the ground gains kinetic energy that quickly changes to thermal energy. c. When the coin hits the ground, the ground gains thermal energy that quickly changes to kinetic energy. d. When the coin hits the ground, the ground gains thermal energy that quickly changes to potential energy. 41. True or false—A marble rolls down a slope from height h1 and up another slope to height h2, where (h2 < h1). The difference mg(h1 – h2) is equal to the heat lost due to the friction. a. True b. False 9.3 Simple Machines 42. Why would you expect the lever shown in the top image to have a greater efficiency than the inclined plane shown in the bottom image? b. The effort arm is shorter in case of the inclined plane. c. The area of contact is greater in case of the inclined plane. 43. Why is the wheel on a wheelbarrow not a simple machine in the same sense as the simple machine in the image? a. The wheel on the wheelbarrow has no fulcrum. b. The center of the axle is not the fulcrum for the wheels of a wheelbarrow. c. The wheelbarrow differs in the way in which load is attached to the axle. d. The wheelbarrow has less resistance force than a wheel and axle design. 44. A worker pulls down on one end of the rope of a pulley system with a force of 75 N to raise a hay bale tied to the other end of the rope. If she pulls the rope down 2.0 m and the bale raises 1.0 m, what else would you have to know to calculate the efficiency of the pulley system? a. b. c. d. the weight of the worker the weight of the hay bale the radius of the pulley the height of the pulley from ground 45. True or false—A boy pushed a box with a weight of 300 N up a ramp. He said that, because the ramp was 1.0 m high and 3.0 m long, he must have been pushing with force of exactly 100 N. a. True b. False a. The resistance arm is shorter in case of the inclined Extended Response 9.1 Work, Power, and the Work–Energy Theorem 46. Work can be negative as well as positive because an object or system can do work on its surroundings as well as have work done on it. Which of the following statements describes: a situation in which an object does work on its surroundings by decreasing its velocity and a situation in which an object can do work on its surroundings by decreasing its altitude? a. A gasoline engine burns less fuel at a slower speed. Solar cells capture sunlight to generate electricity. Access for free at openstax.org. Chapter 9 • Test Prep 303 b. A hybrid car charges its batteries as it decelerates. Falling water turns a turbine to generate electricity. c. Airplane flaps use air resistance to slow down for landing. Rising steam turns a turbine to generate electricity. d. An electric train requires less electrical energy as it decelerates. A parachute captures air to slow a skydiver’s fall. 47. A boy is pulling a girl in a child’s wagon at a constant speed. He begins to pull harder, which increases the speed of the wagon. Which of the following describes two ways you could calculate the change in energy of the wagon and girl if you had all the information you needed? a. Calculate work done from the force and the velocity. Calculate work done from the change in the potential energy of the system. b. Calculate work done from the force and the displacement. Calculate work done from the change in the potential energy of the system. c. Calculate work done from the force and the velocity. Calculate work done from the change in the kinetic energy of the system. be 12 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. d. The velocity of the rock as it hits the ground would be 12 m/s. Due to the lack of air friction, there would be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. 49. A boulder rolls from the top of a mountain, travels across a valley below, and rolls part way up the ridge on the opposite side. Describe all the energy transformations taking place during these events and identify when they happen. a. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. b. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. d. Calculate work done from the force and the c. As the boulder rolls down the mountainside, PE is displacement. Calculate work done from the change in the kinetic energy of the system. 9.2 Mechanical Energy and Conservation of Energy 48. Acceleration due to gravity on the moon is 1.6 m/s2 or about 16% of the value of gon Earth. If an astronaut on the moon threw a moon rock to a height of 7.8 m, what would be its velocity as it struck the moon’s surface? How would the fact that the moon has no atmosphere affect the velocity of the falling moon rock? Explain your answer. a. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. b. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. c. The velocity of the rock as it hits the ground would converted to KE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energ
y due to friction. d. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. 9.3 Simple Machines 50. To dig a hole, one holds the handles together and thrusts the blades of a posthole digger, like the one in the image, into the ground. Next, the handles are pulled apart, which squeezes the dirt between them, making it possible to remove the dirt from the hole. This complex machine is composed of two pairs of two different simple machines. Identify and describe the parts that are simple machines and explain how you would find the IMA of each type of simple machine. a. Each handle and its attached blade is a lever with the 304 Chapter 9 • Test Prep fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. b. Each handle and its attached to blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. c. Each handle and its attached blade is a lever with the fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its width. d. Each handle and its attached blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its width. 51. A wooden crate is pulled up a ramp that is 1.0 m high and 6.0 m long. The crate is attached to a rope that is wound around an axle with a radius of 0.020 m . The axle is turned by a 0.20 m long handle. What is the overall IMA of the complex machine? A. 6 B. 10 16 C. D. 60 Access for free at openstax.org. CHAPTER 10 Special Relativity Figure 10.1 Special relativity explains why travel to other star systems, such as these in the Orion Nebula, is unlikely using our current level of technology. (s58y, Flickr) Chapter Outline 10.1 Postulates of Special Relativity 10.2 Consequences of Special Relativity Have you ever dreamed of traveling to other planets in faraway star systems? The trip might seem possible by INTRODUCTION traveling fast enough, but you will read in this chapter why it is not. In 1905, Albert Einstein developed the theory of special relativity. Einstein developed the theory to help explain inconsistencies between the equations describing electromagnetism and Newtonian mechanics, and to explain why the ether did not exist. This theory explains the limit on an object’s speed among other implications. Relativity is the study of how different observers moving with respect to one another measure the same events. Galileo and Newton developed the first correct version of classical relativity. Einstein developed the modern theory of relativity. Modern relativity is divided into two parts. Special relativity deals with observers moving at constant velocity. General relativity deals with observers moving at constant acceleration. Einstein’s theories of relativity made revolutionary predictions. Most importantly, his predictions have been verified by experiments. In this chapter, you learn how experiments and puzzling contradictions in existing theories led to the development of the theory of special relativity. You will also learn the simple postulates on which the theory was based; a postulate is a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument. 10.1 Postulates of Special Relativity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the experiments and scientific problems that led Albert Einstein to develop the special theory of relativity • Understand the postulates on which the special theory of relativity was based 306 Chapter 10 • Special Relativity Section Key Terms ether frame of reference inertial reference frame general relativity postulate relativity simultaneity special relativity Scientific Experiments and Problems Relativity is not new. Way back around the year 1600, Galileo explained that motion is relative. Wherever you happen to be, it seems like you are at a fixed point and that everything moves with respect to you. Everyone else feels the same way. Motion is always measured with respect to a fixed point. This is called establishing a frame of reference. But the choice of the point is arbitrary, and all frames of reference are equally valid. A passenger in a moving car is not moving with respect to the driver, but they are both moving from the point of view of a person on the sidewalk waiting for a bus. They are moving even faster as seen by a person in a car coming toward them. It is all relative. TIPS FOR SUCCESS A frame of reference is not a complicated concept. It is just something you decide is a fixed point or group of connected points. It is completely up to you. For example, when you look up at celestial objects in the sky, you choose the earth as your frame of reference, and the sun, moon, etc., seem to move across the sky. Light is involved in the discussion of relativity because theories related to electromagnetism are inconsistent with Galileo’s and Newton’s explanation of relativity. The true nature of light was a hot topic of discussion and controversy in the late 19th century. At the time, it was not generally believed that light could travel across empty space. It was known to travel as waves, and all other types of energy that propagated as waves needed to travel though a material medium. It was believed that space was filled with an invisible medium that light waves traveled through. This imaginary (as it turned out) material was called the ether (also spelled aether). It was thought that everything moved through this mysterious fluid. In other words, ether was the one fixed frame of reference. The Michelson–Morley experiment proved it was not. In 1887, Albert Michelson and Edward Morley designed the interferometer shown in Figure 10.2 to measure the speed of Earth through the ether. A light beam is split into two perpendicular paths and then recombined. Recombining the waves produces an inference pattern, with a bright fringe at the locations where the two waves arrive in phase; that is, with the crests of both waves arriving together and the troughs arriving together. A dark fringe appears where the crest of one wave coincides with a trough of the other, so that the two cancel. If Earth is traveling through the ether as it orbits the sun, the peaks in one arm would take longer than in the other to reach the same location. The places where the two waves arrive in phase would change, and the interference pattern would shift. But, using the interferometer, there was no shift seen! This result led to two conclusions: that there is no ether and that the speed of light is the same regardless of the relative motion of source and observer. The Michelson–Morley investigation has been called the most famous failed experiment in history. Access for free at openstax.org. 10.1 • Postulates of Special Relativity 307 Figure 10.2 This is a diagram of the instrument used in the Michelson–Morley experiment. To see what Michelson and Morley expected to find when they measured the speed of light in two directions, watch this animation (http://openstax.org/l/28MMexperiment) . In the video, two people swimming in a lake are represented as an analogy to light beams leaving Earth as it moves through the ether (if there were any ether). The swimmers swim away from and back to a platform that is moving through the water. The swimmers swim in different directions with respect to the motion of the platform. Even though they swim equal distances at the same speed, the motion of the platform causes them to arrive at different times. Einstein’s Postulates The results described above left physicists with some puzzling and unsettling questions such as, why doesn’t light emitted by a fast-moving object travel faster than light from a street lamp? A radical new theory was needed, and Albert Einstein, shown in Figure 10.3, was about to become everyone’s favorite genius. Einstein began with two simple postulates based on the two things we have discussed so far in this chapter. 1. The laws of physics are the same in all inertial reference frames. 2. The speed of light is the same in all inertial reference frames and is not affected by the speed of its source. Figure 10.3 Albert Einstein (1879–1955) developed modern relativity and also made fundamental contributions to the foundations of quantum mechanics. (The Library of Congress) The speed of light is given the symbol cand is equal to exactly 299,792,458 m/s. This is the speed of light in vacuum; that is, in the absence of air. For most purposes, we round this number off to The term inertial reference frame simply 308 Chapter 10 • Special Relativity refers to a frame of reference where all objects follow Newton’s first law of motion: Objects at rest remain at rest, and objects in motion remain in motion at a constant velocity in a straight line, unless acted upon by an external force. The inside of a car moving along a road at constant velocity and the inside of a stationary house are inertial reference frames. WATCH PHYSICS The Speed of Light This lecture on light summarizes the most important facts about the speed of light. If you are interested, you can watch the whole video, but the parts relevant to this chapter are found between 3:25 and 5:10, which you find by running your cursor along the
bottom of the video. Click to view content (https://www.youtube.com/embed/rLNM8zI4Q_M) GRASP CHECK An airliner traveling at 200 m/s emits light from the front of the plane. Which statement describes the speed of the light? a. b. c. d. It travels at a speed of c+ 200 m/s. It travels at a speed of c– 200 m/s. It travels at a speed c, like all light. It travels at a speed slightly less than c. Snap Lab Measure the Speed of Light In this experiment, you will measure the speed of light using a microwave oven and a slice of bread. The waves generated by a microwave oven are not part of the visible spectrum, but they are still electromagnetic radiation, so they travel at the speed of light. If we know the wavelength, λ, and frequency, f, of a wave, we can calculate its speed, v, using the equation v = λf. You can measure the wavelength. You will find the frequency on a label on the back of a microwave oven. The wave in a microwave is a standing wave with areas of high and low intensity. The high intensity sections are one-half wavelength apart. • High temperature: Very hot temperatures are encountered in this lab. These can cause burns. • a microwave oven • one slice of plain white bread • a centimeter ruler • a calculator 1. Work with a partner. 2. Turn off the revolving feature of the microwave oven or remove the wheels under the microwave dish that make it turn. It is important that the dish does not turn. 3. Place the slice of bread on the dish, set the microwave on high, close the door, run the microwave for about 15 seconds. 4. A row of brown or black marks should appear on the bread. Stop the microwave as soon as they appear. Measure the distance between two adjacent burn marks and multiply the result by 2. This is the wavelength. 5. The frequency of the waves is written on the back of the microwave. Look for something like “2,450 MHz.” Hz is the unit hertz, which means per second. The M represents mega, which stands for million, so multiply the number by 106. 6. Express the wavelength in meters and multiply it times the frequency. If you did everything correctly, you will get a number very close to the speed of light. Do not eat the bread. It is a general laboratory safety rule never to eat anything in the lab. GRASP CHECK How does your measured value of the speed of light compare to the accepted value (% error)? Access for free at openstax.org. 10.1 • Postulates of Special Relativity 309 a. The measured value of speed will be equal to c. b. The measured value of speed will be slightly less than c. c. The measured value of speed will be slightly greater than c. d. The measured value of speed will depend on the frequency of the microwave. Einstein’s postulates were carefully chosen, and they both seemed very likely to be true. Einstein proceeded despite realizing that these two ideas taken together and applied to extreme conditions led to results that contradict Newtonian mechanics. He just took the ball and ran with it. In the traditional view, velocities are additive. If you are running at 3 m/s and you throw a ball forward at a speed of 10 m/s, the ball should have a net speed of 13 m/s. However, according to relativity theory, the speed of a moving light source is not added to the speed of the emitted light. In addition, Einstein’s theory shows that if you were moving forward relative to Earth at nearly c(the speed of light) and could throw a ball forward at c, an observer at rest on the earth would not see the ball moving at nearly twice the speed of light. The observer would see it moving at a speed that is still less than c. This result conforms to both of Einstein’s postulates: The speed of light has a fixed maximum and neither reference frame is privileged. Consider how we measure elapsed time. If we use a stopwatch, for example, how do we know when to start and stop the watch? One method is to use the arrival of light from the event, such as observing a light turn green to start a drag race. The timing will be more accurate if some sort of electronic detection is used, avoiding human reaction times and other complications. Now suppose we use this method to measure the time interval between two flashes of light produced by flash lamps on a moving train. (See Figure 10.4) 310 Chapter 10 • Special Relativity Figure 10.4 Light arriving to observer A as seen by two different observers. A woman (observer A) is seated in the center of a rail car, with two flash lamps at opposite sides equidistant from her. Multiple light rays that are emitted from the flash lamps move towards observer A, as shown with arrows. A velocity vector arrow for the rail car is shown towards the right. A man (observer B) standing on the platform is facing the woman and also observes the flashes of light. Observer A moves with the lamps on the rail car as the rail car moves towards the right of observer B. Observer B receives the light flashes simultaneously, and sees the bulbs as both having flashed at the same time. However, he sees observer A receive the flash from the right first. Because the pulse from the right reaches her first, in her frame of reference she sees the bulbs as not having flashed simultaneously. Here, a relative velocity between observers affects whether two events at well-separated locations are observed to be simultaneous. Simultaneity, or whether different events occur at the same instant, depends on the frame of reference of the observer. Remember that velocity equals distance divided by time, so t = d/v. If velocity appears to be different, then duration of time appears to be different. This illustrates the power of clear thinking. We might have guessed incorrectly that, if light is emitted simultaneously, then two observers halfway between the sources would see the flashes simultaneously. But careful analysis shows this not to be the case. Einstein was brilliant at this type of thought experiment (in German, Gedankenexperiment). He very carefully considered how an observation is made and disregarded what might seem obvious. The validity of thought experiments, of course, is determined by actual observation. The genius of Einstein is evidenced by the fact that experiments have repeatedly confirmed his theory of relativity. No experiments after that of Michelson and Morley were able to detect any ether medium. We will describe later how experiments also confirmed other predictions of special relativity, such as the distance between two objects and the time interval of two events being different for two observers moving with respect to each other. In summary: Two events are defined to be simultaneous if an observer measures them as occurring at the same time (such as by receiving light from the events). Two events are not necessarily simultaneous to all observers. Access for free at openstax.org. 10.1 • Postulates of Special Relativity 311 The discrepancies between Newtonian mechanics and relativity theory illustrate an important point about how science advances. Einstein’s theory did not replace Newton’s but rather extended it. It is not unusual that a new theory must be developed to account for new information. In most cases, the new theory is built on the foundation of older theory. It is rare that old theories are completely replaced. In this chapter, you will learn about the theory of special relativity, but, as mentioned in the introduction, Einstein developed two relativity theories: special and general. Table 10.1 summarizes the differences between the two theories. Special Relativity General Relativity Published in 1905 Final form published in 1916 A theory of space-time A theory of gravity Applies to observers moving at constant speed Applies to observers that are accelerating Most useful in the field of nuclear physics Most useful in the field of astrophysics Accepted quickly and put to practical use by nuclear physicists and quantum chemists Largely ignored until 1960 when new mathematical techniques made the theory more accessible and astronomers found some important applications Also note that the theory of general relativity includes the theory of special relativity. Table 10.1 Comparing Special Relativity and General Relativity WORKED EXAMPLE Calculating the Time it Takes Light to Travel a Given Distance The sun is 1.50 × 108 km from Earth. How long does it take light to travel from the sun to Earth in minutes and seconds? Strategy Identify knowns. Identify unknowns. Time Find the equation that relates knowns and unknowns. Be sure to use consistent units. Solution 10.1 10.2 Discussion The answer is written as 5.00 × 102 rather than 500 in order to show that there are three significant figures. When astronomers witness an event on the sun, such as a sunspot, it actually happened minutes earlier. Compare 8 light minutesto the distance to stars, which are light yearsaway. Any events on other stars happened years ago. 312 Chapter 10 • Special Relativity Practice Problems 1. Light travels through 1.00 m of water in 4.42×10-9 s. What is the speed of light in water? a. 4.42×10-9 m/s b. 4.42×109 m/s c. 2.26×108 m/s d. 226×108 m/s 2. An astronaut on the moon receives a message from mission control on Earth. The signal is sent by a form of electromagnetic radiation and takes 1.28 s to travel the distance between Earth and the moon. What is the distance from Earth to the moon? a. 2.34×105 km b. 2.34×108 km 3.84×105 km c. 3.84×108 km d. Check Your Understanding 3. Explain what is meant by a frame of reference. a. A frame of reference is a graph plotted between distance and time. b. A frame of reference is a graph plotted between speed and time. c. A frame of reference is the velocity of an object through empty space without regard to its surroundings. d. A frame of reference is an arbitrarily fixed point with respect to which motion of other points is measured. 4. Two people swim away from a raft that is floating downstream. One swims upstream and returns, and the other swims across the current and back
. If this scenario represents the Michelson–Morley experiment, what do (i) the water, (ii) the swimmers, and (iii) the raft represent? the ether rays of light Earth a. rays of light the ether Earth b. c. the ether Earth rays of light d. Earth rays of light the ether 5. If Michelson and Morley had observed the interference pattern shift in their interferometer, what would that have indicated? a. The speed of light is the same in all frames of reference. b. The speed of light depends on the motion relative to the ether. c. The speed of light changes upon reflection from a surface. d. The speed of light in vacuum is less than 3.00×108 m/s. 6. If you designate a point as being fixed and use that point to measure the motion of surrounding objects, what is the point called? a. An origin b. A frame of reference c. A moving frame d. A coordinate system 10.2 Consequences of Special Relativity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the relativistic effects seen in time dilation, length contraction, and conservation of relativistic momentum • Explain and perform calculations involving mass-energy equivalence Section Key Terms binding energy length contraction mass defect time dilation Access for free at openstax.org. 10.2 • Consequences of Special Relativity 313 proper length relativistic relativistic momentum relativistic energy relativistic factor rest mass Relativistic Effects on Time, Distance, and Momentum Consideration of the measurement of elapsed time and simultaneity leads to an important relativistic effect. Time dilation is the phenomenon of time passing more slowly for an observer who is moving relative to another observer. For example, suppose an astronaut measures the time it takes for light to travel from the light source, cross her ship, bounce off a mirror, and return. (See Figure 10.5.) How does the elapsed time the astronaut measures compare with the elapsed time measured for the same event by a person on the earth? Asking this question (another thought experiment) produces a profound result. We find that the elapsed time for a process depends on who is measuring it. In this case, the time measured by the astronaut is smaller than the time measured by the earth bound observer. The passage of time is different for the two observers because the distance the light travels in the astronaut’s frame is smaller than in the earth bound frame. Light travels at the same speed in each frame, and so it will take longer to travel the greater distance in the earth bound frame. Figure 10.5 (a) An astronaut measures the time for light to cross her ship using an electronic timer. Light travels a distance in the astronaut’s frame. (b) A person on the earth sees the light follow the longer path and take a longer time The relationship between Δtand Δto is given by where is the relativistic factor given by and vand care the speeds of the moving observer and light, respectively. 314 Chapter 10 • Special Relativity TIPS FOR SUCCESS Try putting some values for vinto the expression for the relativistic factor ( a difference and when is so close to 1 that it can be ignored. Try 225 m/s, the speed of an airliner; 2.98 × 104 m/s, the speed of Earth in its orbit; and 2.990 × 108 m/s, the speed of a particle in an accelerator. ). Observe at which speeds this factor will make Notice that when the velocity vis small compared to the speed of light c, then v/cbecomes small, and becomes close to 1. When this happens, time measurements are the same in both frames of reference. Relativistic effects, meaning those that have to do with special relativity, usually become significant when speeds become comparable to the speed of light. This is seen to be the case for time dilation. You may have seen science fiction movies in which space travelers return to Earth after a long trip to find that the planet and everyone on it has aged much more than they have. This type of scenario is a based on a thought experiment, known as the twin paradox, which imagines a pair of twins, one of whom goes on a trip into space while the other stays home. When the space traveler returns, she finds her twin has aged much more than she. This happens because the traveling twin has been in two frames of reference, one leaving Earth and one returning. Time dilation has been confirmed by comparing the time recorded by an atomic clock sent into orbit to the time recorded by a clock that remained on Earth. GPS satellites must also be adjusted to compensate for time dilation in order to give accurate positioning. Have you ever driven on a road, like that shown in Figure 10.6, that seems like it goes on forever? If you look ahead, you might say you have about 10 km left to go. Another traveler might say the road ahead looks like it is about 15 km long. If you both measured the road, however, you would agree. Traveling at everyday speeds, the distance you both measure would be the same. You will read in this section, however, that this is not true at relativistic speeds. Close to the speed of light, distances measured are not the same when measured by different observers moving with respect to one other. Figure 10.6 People might describe distances differently, but at relativistic speeds, the distances really are different. (Corey Leopold, Flickr) One thing all observers agree upon is their relative speed. When one observer is traveling away from another, they both see the other receding at the same speed, regardless of whose frame of reference is chosen. Remember that speed equals distance divided by time: v = d/t. If the observers experience a difference in elapsed time, they must also observe a difference in distance traversed. This is because the ratio d/tmust be the same for both observers. The shortening of distance experienced by an observer moving with respect to the points whose distance apart is measured is called length contraction. Proper length, L0, is the distance between two points measured in the reference frame where the observer and the points are at rest. The observer in motion with respect to the points measures L. These two lengths are related by the equation Because is the same expression used in the time dilation equation above, the equation becomes Access for free at openstax.org. 10.2 • Consequences of Special Relativity 315 To see how length contraction is seen by a moving observer, go to this simulation (http://openstax.org/l/28simultaneity) . Here you can also see that simultaneity, time dilation, and length contraction are interrelated phenomena. This link is to a simulation that illustrates the relativity of simultaneous events. In classical physics, momentum is a simple product of mass and velocity. When special relativity is taken into account, objects that have mass have a speed limit. What effect do you think mass and velocity have on the momentum of objects moving at relativistic speeds; i.e., speeds close to the speed of light? Momentum is one of the most important concepts in physics. The broadest form of Newton’s second law is stated in terms of momentum. Momentum is conserved in classical mechanics whenever the net external force on a system is zero. This makes momentum conservation a fundamental tool for analyzing collisions. We will see that momentum has the same importance in modern physics. Relativistic momentum is conserved, and much of what we know about subatomic structure comes from the analysis of collisions of accelerator-produced relativistic particles. One of the postulates of special relativity states that the laws of physics are the same in all inertial frames. Does the law of conservation of momentum survive this requirement at high velocities? The answer is yes, provided that the momentum is defined as follows. Relativistic momentum, p, is classical momentum multiplied by the relativistic factor is the rest mass of the object (that is, the mass measured at rest, without any as before, is the relativistic factor. We use the mass of the object as measured at rest because we cannot where to an observer, and determine its mass while it is moving. factor involved), is its velocity relative 10.3 for velocity here to distinguish it from relative velocity between observers. Only one observer is being Note that we use considered here. With defined in this way, Again we see that the relativistic quantity becomes virtually the same as the classical at low velocities. That is, relativistic momentum is conserved whenever the net external force is zero, just as in classical physics. is very nearly equal to 1 at low velocities. at low velocities, because becomes the classical Relativistic momentum has the same intuitive feel as classical momentum. It is greatest for large masses moving at high velocities. Because of the factor approaching infinity as speed of light. If it did, its momentum would become infinite, which is an unreasonable value. however, relativistic momentum behaves differently from classical momentum by (See Figure 10.7.) This is another indication that an object with mass cannot reach the approaches Figure 10.7 Relativistic momentum approaches infinity as the velocity of an object approaches the speed of light. Relativistic momentum is defined in such a way that the conservation of momentum will hold in all inertial frames. Whenever the net external force on a system is zero, relativistic momentum is conserved, just as is the case for classical momentum. This 316 Chapter 10 • Special Relativity has been verified in numerous experiments. Mass-Energy Equivalence Let us summarize the calculation of relativistic effects on objects moving at speeds near the speed of light. In each case we will need to calculate the relativistic factor, given by where v and care as defined earlier. We use u as the velocity of a particle or an object in one frame of reference, and v for the velocity of one frame of reference with respect to another. Time Dilation Elapsed tim
e on a moving object, on the moving object when it is taken to be the frame or reference. as seen by a stationary observer is given by Length Contraction Length measured by a person at rest with respect to a moving object, L, is given by where is the time observed where L0 is the length measured on the moving object. Relativistic Momentum Momentum, p, of an object of mass, m, traveling at relativistic speeds is given by object as seen by a stationary observer. where u is velocity of a moving Relativistic Energy The original source of all the energy we use is the conversion of mass into energy. Most of this energy is generated by nuclear reactions in the sun and radiated to Earth in the form of electromagnetic radiation, where it is then transformed into all the forms with which we are familiar. The remaining energy from nuclear reactions is produced in nuclear power plants and in Earth’s interior. In each of these cases, the source of the energy is the conversion of a small amount of mass into a large amount of energy. These sources are shown in Figure 10.8. Figure 10.8 The sun (a) and the Susquehanna Steam Electric Station (b) both convert mass into energy. ((a) NASA/Goddard Space Flight Center, Scientific Visualization Studio; (b) U.S. government) The first postulate of relativity states that the laws of physics are the same in all inertial frames. Einstein showed that the law of conservation of energy is valid relativistically, if we define energy to include a relativistic factor. The result of his analysis is that a particle or object of mass mmoving at velocity u has relativistic energy given by This is the expression for the total energy of an object of mass mat any speed u and includes both kinetic and potential energy. Look back at the equation for and you will see that it is equal to 1 when u is 0; that is, when an object is at rest. Then the rest Access for free at openstax.org. 10.2 • Consequences of Special Relativity 317 energy, E0, is simply This is the correct form of Einstein’s famous equation. This equation is very useful to nuclear physicists because it can be used to calculate the energy released by a nuclear reaction. This is done simply by subtracting the mass of the products of such a reaction from the mass of the reactants. The difference is the min Here is a simple example: A positron is a type of antimatter that is just like an electron, except that it has a positive charge. When a positron and an electron collide, their masses are completely annihilated and converted to energy in the form of gamma rays. Because both particles have a rest mass of 9.11 × 10–31 kg, we multiply the mc2 term by 2. So the energy of the gamma rays is 10.4 where we have the expression for the joule (J) in terms of its SI base units of kg, m, and s. In general, the nuclei of stable isotopes have less mass then their constituent subatomic particles. The energy equivalent of this difference is called the binding energy of the nucleus. This energy is released during the formation of the isotope from its constituent particles because the product is more stable than the reactants. Expressed as mass, it is called the mass defect. For example, a helium nucleus is made of two neutrons and two protons and has a mass of 4.0003 atomic mass units (u). The sum of the masses of two protons and two neutrons is 4.0330 u. The mass defect then is 0.0327 u. Converted to kg, the mass defect is 5.0442 × 10–30 kg. Multiplying this mass times c2 gives a binding energy of 4.540 × 10–12 J. This does not sound like much because it is only one atom. If you were to make one gram of helium out of neutrons and protons, it would release 683,000,000,000 J. By comparison, burning one gram of coal releases about 24 J. BOUNDLESS PHYSICS The RHIC Collider Figure 10.9 shows the Brookhaven National Laboratory in Upton, NY. The circular structure houses a particle accelerator called the RHIC, which stands for Relativistic Heavy Ion Collider. The heavy ions in the name are gold nuclei that have been stripped of their electrons. Streams of ions are accelerated in several stages before entering the big ring seen in the figure. Here, they are accelerated to their final speed, which is about 99.7 percent the speed of light. Such high speeds are called relativistic. All the relativistic phenomena we have been discussing in this chapter are very pronounced in this case. At this speed = 12.9, so that relativistic time dilates by a factor of about 13, and relativistic length contracts by the same factor. Figure 10.9 Brookhaven National Laboratory. The circular structure houses the RHIC. (energy.gov, Wikimedia Commons) Two ion beams circle the 2.4-mile long track around the big ring in opposite directions. The paths can then be made to cross, thereby causing ions to collide. The collision event is very short-lived but amazingly intense. The temperatures and pressures produced are greater than those in the hottest suns. At 4 trillion degrees Celsius, this is the hottest material ever created in a 318 Chapter 10 • Special Relativity laboratory But what is the point of creating such an extreme event? Under these conditions, the neutrons and protons that make up the gold nuclei are smashed apart into their components, which are called quarks and gluons. The goal is to recreate the conditions that theorists believe existed at the very beginning of the universe. It is thought that, at that time, matter was a sort of soup of quarks and gluons. When things cooled down after the initial bang, these particles condensed to form protons and neutrons. Some of the results have been surprising and unexpected. It was thought the quark-gluon soup would resemble a gas or plasma. Instead, it behaves more like a liquid. It has been called a perfectliquid because it has virtually no viscosity, meaning that it has no resistance to flow. GRASP CHECK Calculate the relativistic factor γ, for a particle traveling at 99.7 percent of the speed of light. a. 0.08 b. 0.71 1.41 c. 12.9 d. WORKED EXAMPLE The Speed of Light One night you are out looking up at the stars and an extraterrestrial spaceship flashes across the sky. The ship is 50 meters long and is travelling at 95 percent of the speed of light. What would the ship’s length be when measured from your earthbound frame of reference? Strategy List the knowns and unknowns. Knowns: proper length of the ship, L0= 50 m; velocity, v, = 0.95c Unknowns: observed length of the ship accounting for relativistic length contraction, L. Choose the relevant equation. Solution Discussion Calculations of sure to also square the decimal representing the percentage before subtracting from 1. Note that the aliens will still see the length as L0 because they are moving with the frame of reference that is the ship. can usually be simplified in this way when vis expressed as a percentage of cbecause the c2 terms cancel. Be Practice Problems 7. Calculate the relativistic factor, γ, for an object traveling at 2.00×108 m/s. a. 0.74 b. 0.83 c. d. 1.2 1.34 8. The distance between two points, called the proper length, L0, is 1.00 km. An observer in motion with respect to the frame of Access for free at openstax.org. 10.2 • Consequences of Special Relativity 319 reference of the two points measures 0.800 km, which is L. What is the relative speed of the frame of reference with respect to the observer? 1.80×108 m/s a. b. 2.34×108 m/s 3.84×108 m/s c. 5.00×108 m/s d. 9. Consider the nuclear fission reaction . If a neutron has a rest mass of 1.009u, has rest mass of 136.907u, and has a rest mass of 96.937u, what is the value of Ein has a rest mass of 235.044u, joules? a. b. c. d. J J J J Solution The correct answer is (b). The mass deficit in the reaction is Converting that mass to kg and applying to find the energy equivalent of the mass deficit gives or 0.191u. 10. Consider the nuclear fusion reaction . If has a rest mass of 2.014u, has a rest mass of 3.016u, and has a rest mass of 1.008u, what is the value of Ein joules? a. b. c. d. J J J J Solution The correct answer is (a). The mass deficit in the reaction is mass to kg and applying to find the energy equivalent of the mass deficit gives , or 0.004u. Converting that Check Your Understanding 11. Describe time dilation and state under what conditions it becomes significant. a. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame. b. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame. c. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame. d. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame. 12. The equation used to calculate relativistic momentum is p= γ · m · u. Define the terms to the right of the equal sign and state how mand uare measured. a. γis the relativistic factor, mis the rest mass measured when the object is at rest in the frame of reference, and uis the velocity of the frame. b. γis the relativistic factor, mis the rest mass measured when the object is at rest in the frame of reference, and uis the velocity relative to an observer. c. γis the relativistic factor, mis the relativistic mass measured when the object is moving in the frame of reference, and uis the velocity of the frame. 320 Chapter 10 • Special Relativity d. γis the relativistic factor, mis the relativistic mass measured when the object is moving in the frame of reference, and uis the velocity relative to an observ
er. 13. Describe length contraction and state when it occurs. a. When the speed of an object becomes the speed of light, its length appears to shorten when viewed by a stationary observer. b. When the speed of an object approaches the speed of light, its length appears to shorten when viewed by a stationary observer. c. When the speed of an object becomes the speed of light, its length appears to increase when viewed by a stationary observer. d. When the speed of an object approaches the speed of light, its length appears to increase when viewed by a stationary observer. Access for free at openstax.org. Chapter 10 • Key Terms 321 KEY TERMS binding energy the energy equivalent of the difference between the mass of a nucleus and the masses of its nucleons effects that become significant only when an object is to be moving close enough to the speed of light for significantly greater than 1 ether scientists once believed there was a medium that carried light waves; eventually, experiments proved that ether does not exist relativistic energy the total energy of a moving object or which includes both its rest energy particle mc2 and its kinetic energy frame of reference the point or collection of points relativistic factor , where u is the velocity of a arbitrarily chosen, which motion is measured in relation to general relativity the theory proposed to explain gravity and acceleration inertial reference frame a frame of reference where all objects follow Newton’s first law of motion length contraction the shortening of an object as seen by an observer who is moving relative to the frame of reference of the object mass defect the difference between the mass of a nucleus and the masses of its nucleons postulate a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument proper length the length of an object within its own frame of reference, as opposed to the length observed by an observer moving relative to that frame of reference relativistic having to do with modern relativity, such as the SECTION SUMMARY 10.1 Postulates of Special Relativity • One postulate of special relativity theory is that the laws of physics are the same in all inertial frames of reference. • The other postulate is that the speed of light in a vacuum is the same in all inertial frames. • Einstein showed that simultaneity, or lack of it, depends on the frame of reference of the observer. KEY EQUATIONS 10.1 Postulates of Special Relativity speed of light constant value for the speed of light moving object and cis the speed of light relativistic momentum p = γmu, where is the relativistic factor, mis rest mass of an object, and u is the velocity relative to an observer relativity the explanation of how objects move relative to one another rest mass the mass of an object that is motionless with respect to its frame of reference simultaneity the property of events that occur at the same time special relativity the theory proposed to explain the consequences of requiring the speed of light and the laws of physics to be the same in all inertial frames time dilation the contraction of time as seen by an observer in a frame of reference that is moving relative to the observer 10.2 Consequences of Special Relativity • Time dilates, length contracts, and momentum increases as an object approaches the speed of light. • Energy and mass are interchangeable, according to the relationship E = mc2. The laws of conservation of mass and energy are combined into the law of conservation of mass-energy. 10.2 Consequences of Special Relativity elapsed time relativistic factor length contraction relativistic momentum 322 Chapter 10 • Chapter Review relativistic energy rest energy CHAPTER REVIEW Concept Items 10.1 Postulates of Special Relativity 1. Why was it once believed that light must travel through a medium and could not propagate across empty space? a. The longitudinal nature of light waves implies this. b. Light shows the phenomenon of diffraction. c. The speed of light is the maximum possible speed. d. All other wave energy needs a medium to travel. 2. Describe the relative motion of Earth and the sun: 1. 2. if Earth is taken as the inertial frame of reference and if the sun is taken as the inertial frame of reference. 1. Earth is at rest and the sun orbits Earth. a. 2. The sun is at rest and Earth orbits the sun. b. c. d. 1. The sun is at rest and Earth orbits the sun. 2. Earth is at rest and the sun orbits Earth. 1. The sun is at rest and Earth orbits the sun. 2. The sun is at rest and Earth orbits the sun. 1. Earth is at rest and the sun orbits Earth. 2. Earth is at rest and the sun orbits Earth. 10.2 Consequences of Special Relativity 3. A particle (a free electron) is speeding around the track Critical Thinking Items 10.1 Postulates of Special Relativity 6. Explain how the two postulates of Einstein’s theory of special relativity, when taken together, could lead to a situation that seems to contradict the mechanics and laws of motion as described by Newton. a. In Newtonian mechanics, velocities are multiplicative but the speed of a moving light source cannot be multiplied to the speed of light because, according to special relativity, the speed of light is the maximum speed possible. In Newtonian mechanics, velocities are additive but the speed of a moving light source cannot be added to the speed of light because the speed of light is the maximum speed possible. b. c. An object that is at rest in one frame of reference may appear to be in motion in another frame of reference, while in Newtonian mechanics such a situation is not possible. Access for free at openstax.org. in a cyclotron, rapidly gaining speed. How will the particle’s momentum change as its speed approaches the speed of light? Explain. a. The particle’s momentum will rapidly decrease. b. The particle’s momentum will rapidly increase. c. The particle’s momentum will remain constant. d. The particle’s momentum will approach zero. 4. An astronaut goes on a long space voyage at near the speed of light. When she returns home, how will her age compare to the age of her twin who stayed on Earth? a. Both of them will be the same age. b. This is a paradox and hence the ages cannot be compared. c. The age of the twin who traveled will be less than the age of her twin. d. The age of the twin who traveled will be greater than the age of her twin. 5. A comet reaches its greatest speed as it travels near the sun. True or false— Relativistic effects make the comet’s tail look longer to an observer on Earth. a. True b. False d. The postulates of Einstein’s theory of special relativity do not contradict any situation that Newtonian mechanics explains. 7. It takes light to travel from the sun to the planet Venus. How far is Venus from the sun? a. b. c. d. 8. In 2003, Earth and Mars were the closest they had been in 50,000 years. The two planets were aligned so that Earth was between Mars and the sun. At that time it took light from the sun 500 s to reach Earth and 687 s to get to Mars. What was the distance from Mars to Earth? 5.6×107 km a. 5.6×1010 km b. c. 6.2×106 km d. 6.2×1012 km 9. Describe two ways in which light differs from all other forms of wave energy. a. 1. Light travels as a longitudinal wave. 2. Light travels through a medium that fills up the empty space in the universe. b. 1. Light travels as a transverse wave. 2. Light travels through a medium that fills up the empty space in the universe. c. 1. Light travels at the maximum possible speed in the universe. 2. Light travels through a medium that fills up the empty space in the universe. d. 1. Light travels at the maximum possible speed in the universe. 2. Light does not require any material medium to travel. 10. Use the postulates of the special relativity theory to explain why the speed of light emitted from a fastmoving light source cannot exceed 3.00×108 m/s. a. The speed of light is maximum in the frame of reference of the moving object. b. The speed of light is minimum in the frame of reference of the moving object. c. The speed of light is the same in all frames of reference, including in the rest frame of its source. d. Light always travels in a vacuum with a speed less than 3.00×108 m/s, regardless of the speed of the Problems 10.2 Consequences of Special Relativity 13. Deuterium (2 H) is an isotope of hydrogen that has one proton and one neutron in its nucleus. The binding energy of deuterium is 3.56×10-13 J. What is the mass defect of deuterium? 3.20×10-4 kg a. 1.68×10-6 kg b. 1.19×10-21 kg c. 3.96×10-30 kg d. 14. The sun orbits the center of the galaxy at a speed of 2.3×105 m/s. The diameter of the sun is 1.391684×109 m. An observer is in a frame of reference that is stationary with respect to the center of the galaxy. True or false—The sun is moving fast enough for the observer to notice length contraction of the sun’s diameter. a. True b. False 15. Consider the nuclear fission reaction Chapter 10 • Chapter Review 323 source. 10.2 Consequences of Special Relativity 11. Halley’s Comet comes near Earth every 75 years as it travels around its 22 billion km orbit at a speed of up to 700, 000 m/s. If it were possible to put a clock on the comet and read it each time the comet passed, which part of special relativity theory could be tested? What would be the expected result? Explain. a. It would test time dilation. The clock would appear to be slightly slower. It would test time dilation. The clock would appear to be slightly faster. It would test length contraction. The length of the orbit would appear to be shortened from Earth’s frame of reference. It would test length contraction. The length of the orbit would appear to be shortened from the comet’s frame of reference. b. c. d. 12. The nucleus of the isotope fluorine-18 (18 F) has mass defect of 2.44×10-28 kg. What is the binding energy of 18F? a. 2.2×10-11 J 7.3×10-20 J b. c. 2.2×10-20 J d. 2.4×10-28 J has a rest mass of 1.009u, ha
s a rest mass of . If a neutron has rest mass of 143.923u, and 235.044u, has a rest mass of 88.918u, what is the value of Ein joules? a. b. c. d. J J J J 16. Consider the nuclear fusion reaction . If has a rest mass of has a rest has a rest mass of 3.016u, 2.014u, mass of 4.003u, and a neutron has a rest mass of 1.009u, what is the value of Ein joules? a. b. c. d. J J J J 324 Chapter 10 • Test Prep Performance Task 10.2 Consequences of Special Relativity 17. People are fascinated by the possibility of traveling across the universe to discover intelligent life on other planets. To do this, we would have to travel enormous distances. Suppose we could somehow travel at up to 90 percent of the speed of light. The closest star is Alpha Centauri, which is 4.37 light years away. (A light year is the distance light travels in one year.) TEST PREP Multiple Choice 10.1 Postulates of Special Relativity 18. What was the purpose of the Michelson–Morley experiment? a. To determine the exact speed of light b. To analyze the electromagnetic spectrum c. To establish that Earth is the true frame of reference a. How long, from the point of view of people on Earth, would it take a space ship to travel to Alpha Centauri and back at 0.9c? b. How much would the astronauts on the spaceship have aged by the time they got back to Earth? c. Discuss the problems related to travel to stars that are 20 or 30 light years away. Assume travel speeds near the speed of light. been in 50,000 years. People looking up saw Mars as a very bright red light on the horizon. If Mars was 2.06×108 km from the sun, how long did the reflected light people saw take to travel from the sun to Earth? a. b. c. d. 14 min and 33 s 12 min and 15 s 11 min and 27 s 3 min and 7 s d. To learn how the ether affected the propagation of 10.2 Consequences of Special Relativity light 19. What is the speed of light in a vacuum to three 23. What does this expression represent: significant figures? a. b. c. d. 20. How far does light travel in ? a. b. c. d. a. b. c. d. time dilation relativistic factor relativistic energy length contraction 24. What is the rest energy, E0, of an object with a mass of 1.00 g ? 3.00×105 J a. 3.00×1011 J b. c. 9.00×1013 J d. 9.00×1016 J 21. Describe what is meant by the sentence, “Simultaneity is 25. The fuel rods in a nuclear reactor must be replaced from not absolute.” a. Events may appear simultaneous in all frames of reference. b. Events may not appear simultaneous in all frames of reference. c. The speed of light is not the same in all frames of reference. d. The laws of physics may be different in different inertial frames of reference. 22. In 2003, Earth and Mars were aligned so that Earth was between Mars and the sun. Earth and Mars were 5.6×107 km from each other, which was the closest they had time to time because so much of the radioactive material has reacted that they can no longer produce energy. How would the mass of the spent fuel rods compare to their mass when they were new? Explain your answer. a. The mass of the spent fuel rods would decrease. b. The mass of the spent fuel rods would increase. c. The mass of the spent fuel rods would remain the same. d. The mass of the spent fuel rods would become close to zero. Access for free at openstax.org. Chapter 10 • Test Prep 325 Short Answer 10.2 Consequences of Special Relativity 10.1 Postulates of Special Relativity 30. What is the relationship between the binding energy 26. What is the postulate having to do with the speed of light on which the theory of special relativity is based? a. The speed of light remains the same in all inertial frames of reference. b. The speed of light depends on the speed of the source emitting the light. c. The speed of light changes with change in medium through which it travels. d. The speed of light does not change with change in medium through which it travels. 27. What is the postulate having to do with reference frames on which the theory of special relativity is based? a. The frame of reference chosen is arbitrary as long as it is inertial. and the mass defect of an atomic nucleus? a. The binding energy is the energy equivalent of the mass defect, as given by E0 = mc. b. The binding energy is the energy equivalent of the mass defect, as given by E0 = mc2. c. The binding energy is the energy equivalent of the mass defect, as given by d. The binding energy is the energy equivalent of the mass defect, as given by 31. True or false—It is possible to just use the relationships F= maand E= Fdto show that both sides of the equation E0 = mc2 have the same units. a. True b. False b. The frame of reference is chosen to have constant 32. Explain why the special theory of relativity caused the nonzero acceleration. c. The frame of reference is chosen in such a way that the object under observation is at rest. d. The frame of reference is chosen in such a way that the object under observation is moving with a constant speed. 28. If you look out the window of a moving car at houses going past, you sense that you are moving. What have you chosen as your frame of reference? the car a. the sun b. c. a house 29. Why did Michelson and Morley orient light beams at right angles to each other? a. To observe the particle nature of light b. To observe the effect of the passing ether on the speed of light c. To obtain a diffraction pattern by combination of light d. To obtain a constant path difference for interference of light Extended Response 10.1 Postulates of Special Relativity 34. Explain how Einstein’s conclusion that nothing can travel faster than the speed of light contradicts an older concept about the speed of an object propelled from another, already moving, object. a. The older concept is that speeds are subtractive. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the law of conservation of energy to be modified. a. The law of conservation of energy is not valid in relativistic mechanics. b. The law of conservation of energy has to be modified because of time dilation. c. The law of conservation of energy has to be modified because of length contraction. d. The law of conservation of energy has to be modified because of mass-energy equivalence. 33. The sun loses about 4 × 109 kg of mass every second. Explain in terms of special relativity why this is happening. a. The sun loses mass because of its high temperature. b. The sun loses mass because it is continuously releasing energy. c. The Sun loses mass because the diameter of the sun is contracted. d. The sun loses mass because the speed of the sun is very high and close to the speed of light. speed at which the person was running minus the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves faster than the speed of light emitted from a stationary source. b. The older concept is that speeds are additive. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the speed at which the person was running plus the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves no 326 Chapter 10 • Test Prep faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. c. The older concept is that speeds are multiplicative. For example, if a person throws a ball while running, the speed of the ball relative to the ground is the speed at which the person was running multiplied by the speed of the throw. A relativistic example is when light is emitted from car headlights, it moves no faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. d. The older concept is that speeds are frame independent. For example, if a person throws a ball while running, the speed of the ball relative to the ground has nothing to do with the speed at which the person was running. A relativistic example is when light is emitted from car headlights, it moves no faster than the speed of light emitted from a stationary source. The car's speed does not affect the speed of light. 35. A rowboat is drifting downstream. One person swims 20 m toward the shore and back, and another, leaving at the same time, swims upstream 20 m and back to the boat. The swimmer who swam toward the shore gets back first. Explain how this outcome is similar to the outcome expected in the Michelson–Morley experiment. a. The rowboat represents Earth, the swimmers are beams of light, and the water is acting as the ether. Light going against the current of the ether would get back later because, by then, Earth would have moved on. b. The rowboat represents the beam of light, the swimmers are the ether, and water is acting as Earth. Light going against the current of the ether would get back later because, by then, Earth would have moved on. c. The rowboat represents the ether, the swimmers are ray of light, and the water is acting as the earth. Light going against the current of the ether would get back later because, by then, Earth would have moved on. d. The rowboat represents the Earth, the swimmers are the ether, and the water is acting as the rays of light. Light going against the current of the ether would get back later because, by then, Earth would have moved on. 10.2 Consequences of Special Relativity 36. A helium-4 nucleus is made up of two neutrons and two protons. The binding energy of helium-4 is 4.53×10-12 J. What is the difference in the mass of this helium nucleus and the sum of the masses of two neutrons and two protons? Which weighs more, the nucleus or its constituents? a. b. c. d. 1.51×10-20 kg; the constituents weigh more 5.03×10-29 kg; the constituents weigh more 1.51×10-29 kg; the nucleus weighs more 5.03×10-29 kg; the nucleus weighs more 37. Use the equation for length contraction to explain the relationship between the length of an object perceived by a sta
tionary observer who sees the object as moving, and the proper length of the object as measured in the frame of reference where it is at rest. a. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches zero. For other speeds, the length perceived is always less than the proper length. b. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches zero. For other speeds, the length perceived is always greater than the proper length. c. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches infinity. For other speeds, the length perceived is always less than the proper length. d. As the speed vof an object moving with respect to a stationary observer approaches c, the length perceived by the observer approaches infinity. For other speeds, the length perceived is always greater than the proper length. Access for free at openstax.org. CHAPTER 11 Thermal Energy, Heat, and Work Figure 11.1 The welder’s gloves and helmet protect the welder from the electric arc, which transfers enough thermal energy to melt the rod, spray sparks, and emit high-energy electromagnetic radiation that can burn the retina of an unprotected eye. The thermal energy can be felt on exposed skin a few meters away, and its light can be seen for kilometers (Kevin S. O’Brien, U.S. Navy) Chapter Outline 11.1 Temperature and Thermal Energy 11.2 Heat, Specific Heat, and Heat Transfer 11.3 Phase Change and Latent Heat Heat is something familiar to all of us. We feel the warmth of the summer sun, the hot vapor rising up out of INTRODUCTION a cup of hot cocoa, and the cooling effect of our sweat. When we feel warmth, it means that heat is transferring energy toour bodies; when we feel cold, that means heat is transferring energy away fromour bodies. Heat transfer is the movement of thermal energy from one place or material to another, and is caused by temperature differences. For example, much of our weather is caused by Earth evening out the temperature across the planet through wind and violent storms, which are driven by heat transferring energy away from the equator towards the cold poles. In this chapter, we’ll explore the precise meaning of heat, how it relates to temperature as well as to other forms of energy, and its connection to work. 11.1 Temperature and Thermal Energy Section Learning Objectives By the end of this section, you will be able to do the following: • Explain that temperature is a measure of internal kinetic energy • Interconvert temperatures between Celsius, Kelvin, and Fahrenheit scales 328 Chapter 11 • Thermal Energy, Heat, and Work Section Key Terms absolute zero Celsius scale degree Celsius thermal energy degree Fahrenheit Fahrenheit scale heat kelvin (K) Kelvin scale temperature Temperature What is temperature? It’s one of those concepts so ingrained in our everyday lives that, although we know what it means intuitively, it can be hard to define. It is tempting to say that temperature measures heat, but this is not strictly true. Heat is the transfer of energy due to a temperature difference. Temperature is defined in terms of the instrument we use to tell us how hot or cold an object is, based on a mechanism and scale invented by people. Temperature is literally defined as what we measure on a thermometer. Heat is often confused with temperature. For example, we may say that the heat was unbearable, when we actually mean that the temperature was high. This is because we are sensitive to the flow of energy by heat, rather than the temperature. Since heat, like work, transfers energy, it has the SI unit of joule (J). Atoms and molecules are constantly in motion, bouncing off one another in random directions. Recall that kinetic energy is the energy of motion, and that it increases in proportion to velocity squared. Without going into mathematical detail, we can say that thermal energy—the energy associated with heat—is the average kinetic energy of the particles (molecules or atoms) in a substance. Faster moving molecules have greater kinetic energies, and so the substance has greater thermal energy, and thus a higher temperature. The total internal energy of a system is the sum of the kinetic and potential energies of its atoms and molecules. Thermal energy is one of the subcategories of internal energy, as is chemical energy. To measure temperature, some scale must be used as a standard of measurement. The three most commonly used temperature scales are the Fahrenheit, Celsius, and Kelvin scales. Both the Fahrenheit scale and Celsius scale are relative temperature scales, meaning that they are made around a reference point. For example, the Celsius scale uses the freezing point of water as its reference point; all measurements are either lower than the freezing point of water by a given number of degrees (and have a negative sign), or higher than the freezing point of water by a given number of degrees (and have a positive sign). The boiling point of water is 100 for the Celsius scale, and its unit is the degree Celsius ). On the Fahrenheit scale, the freezing point of water is at 32 . The unit of temperature on ). Note that the difference in degrees between the freezing and boiling points is greater this scale is the degree Fahrenheit for the Fahrenheit scale than for the Celsius scale. Therefore, a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Since 100 Celsius degrees span the same range as 180 Fahrenheit degrees, one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale (because , and the boiling point is at 212 ). This relationship can be used to convert between temperatures in Fahrenheit and Celsius (see Figure 11.2). Access for free at openstax.org. 11.1 • Temperature and Thermal Energy 329 Figure 11.2 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales, rounded to the nearest degree. The relative sizes of the scales are also shown. The Kelvin scale is the temperature scale that is commonly used in science because it is an absolute temperature scale. This means that the theoretically lowest-possible temperature is assigned the value of zero. Zero degrees on the Kelvin scale is known as absolute zero; it is theoretically the point at which there is no molecular motion to produce thermal energy. On the original Kelvin scale first created by Lord Kelvin, all temperatures have positive values, making it useful for scientific work. The official temperature unit on this scale is the kelvin, which is abbreviated as K. The freezing point of water is 273.15 K, and the boiling point of water is 373.15 K. Although absolute zero is possible in theory, it cannot be reached in practice. The lowest temperature ever created and measured K, at Helsinki University of Technology in Finland. In comparison, the coldest during a laboratory experiment was recorded temperature for a place on Earth’s surface was 183 K (–89 °C ), at Vostok, Antarctica, and the coldest known place (outside the lab) in the universe is the Boomerang Nebula, with a temperature of 1 K. Luckily, most of us humans will never have to experience such extremes. The average normal body temperature is 98.6 ). to 111 ranging from 75 to 44 (24 (37.0 ), but people have been known to survive with body temperatures WATCH PHYSICS Comparing Celsius and Fahrenheit Temperature Scales This video shows how the Fahrenheit and Celsius temperature scales compare to one another. Click to view content (https://www.openstax.org/l/02celfahtemp) GRASP CHECK Even without the number labels on the thermometer, you could tell which side is marked Fahrenheit and which is Celsius by how the degree marks are spaced. Why? a. The separation between two consecutive divisions on the Fahrenheit scale is greater than a similar separation on the Celsius scale, because each degree Fahrenheit is equal to degrees Celsius. b. The separation between two consecutive divisions on the Fahrenheit scale is smaller than the similar separation on the Celsius scale, because each degree Celsius is equal to degrees Fahrenheit. c. The separation between two consecutive divisions on the Fahrenheit scale is greater than a similar separation on the Celsius scale, because each degree Fahrenheit is equal to degrees Celsius. d. The separation between two consecutive divisions on the Fahrenheit scale is smaller than a similar separation on the Celsius scale, because each degree Celsius is equal to degrees Fahrenheit. 330 Chapter 11 • Thermal Energy, Heat, and Work Converting Between Celsius, Kelvin, and Fahrenheit Scales While the Fahrenheit scale is still the most commonly used scale in the United States, the majority of the world uses Celsius, and scientists prefer Kelvin. It’s often necessary to convert between these scales. For instance, if the TV meteorologist gave the local weather report in kelvins, there would likely be some confused viewers! Table 11.1 gives the equations for conversion between the three temperature scales. To Convert From… Use This Equation Celsius to Fahrenheit Fahrenheit to Celsius Celsius to Kelvin Kelvin to Celsius Fahrenheit to Kelvin Kelvin to Fahrenheit Table 11.1 Temperature Conversions WORKED EXAMPLE Room temperatureis generally defined to be 25 (a) What is room temperature in (b) What is it in K? STRATEGY To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values. Solution for (a) 1. Choose the right equation. To convert from to , use the equation 2. Plug the known value into the equation and solve. Solution for (b) 1. Choose the right equation. To convert from to K, use the equation 2. Plug the known value into the equation and solve. 11.1 11.2 11.3 11.4 Discussion Living in the
United States, you are likely to have more of a sense of what the temperature feels like if it’s described as 77 as 25 (or 298 K, for that matter). than Access for free at openstax.org. WORKED EXAMPLE 11.1 • Temperature and Thermal Energy 331 Converting Between Temperature Scales: The Reaumur Scale The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur If “room temperature” is 25 temperature scale, the freezing point of water is 0 the Celsius scale, what is it on the Reaumur scale? STRATEGY To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is 80 scales start at 0 for freezing, so we can create a simple formula to convert between temperatures on the two scales. and the boiling temperature is 80 . On the Celsius scale, it is 100 . Therefore, 100 . Both on Solution 1. Derive a formula to convert from one scale to the other. 2. Plug the known value into the equation and solve. 11.5 11.6 Discussion As this example shows, relative temperature scales are somewhat arbitrary. If you wanted, you could create your own temperature scale! Practice Problems 1. What is 12.0 °C in kelvins? a. 112.0 K b. 273.2 K c. 12.0 K d. 285.2 K 2. What is 32.0 °C in degrees Fahrenheit? a. 57.6 °F b. 25.6 °F c. 305.2 °F d. 89.6 °F TIPS FOR SUCCESS Sometimes it is not so easy to guess the temperature of the air accurately. Why is this? Factors such as humidity and wind speed affect how hot or cold we feel. Wind removes thermal energy from our bodies at a faster rate than usual, making us feel colder than we otherwise would; on a cold day, you may have heard the TV weather person refer to the wind chill. On humid summer days, people tend to feel hotter because sweat doesn’t evaporate from the skin as efficiently as it does on dry days, when the evaporation of sweat cools us off. Check Your Understanding 3. What is thermal energy? a. The thermal energy is the average potential energy of the particles in a system. b. The thermal energy is the total sum of the potential energies of the particles in a system. c. The thermal energy is the average kinetic energy of the particles due to the interaction among the particles in a system. d. The thermal energy is the average kinetic energy of the particles in a system. 4. What is used to measure temperature? 332 Chapter 11 • Thermal Energy, Heat, and Work a. a galvanometer b. a manometer c. a thermometer d. a voltmeter 11.2 Heat, Specific Heat, and Heat Transfer Section Learning Objectives By the end of this section, you will be able to do the following: • Explain heat, heat capacity, and specific heat • Distinguish between conduction, convection, and radiation • Solve problems involving specific heat and heat transfer Section Key Terms conduction convection heat capacity radiation specific heat Heat Transfer, Specific Heat, and Heat Capacity We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance’s temperature is higher. If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors—the change in the substance’s temperature, the mass of the substance, and certain physical properties related to the phase of the substance. The equation for heat transfer Qis 11.7 where mis the mass of the substance and ΔTis the change in its temperature, in units of Celsius or Kelvin. The symbol cstands for specific heat, and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 ºC. The specific heat cis a property of the substance; its SI unit is J/(kg K) or J/(kg The temperature change ( closely related to the concept of heat capacity. Heat capacity is the amount of heat necessary to change the temperature of a , where mis mass and cis specific heat. Note that heat substance by 1.00 capacity is the same as specific heat, but without any dependence on mass. Consequently, two objects made up of the same material but with different masses will have different heat capacities. This is because the heat capacity is a property of an object, but specific heat is a property of anyobject made of the same material. ) is the same in units of kelvins and degrees Celsius (but not degrees Fahrenheit). Specific heat is . In equation form, heat capacity Cis ). Values of specific heat must be looked up in tables, because there is no simple way to calculate them. Table 11.2 gives the values of specific heat for a few substances as a handy reference. We see from this table that the specific heat of water is five times that of glass, which means that it takes five times as much heat to raise the temperature of 1 kg of water than to raise the temperature of 1 kg of glass by the same number of degrees. Substances Specific Heat (c) Solids Aluminum J/(kg ) 900 Table 11.2 Specific Heats of Various Substances. Access for free at openstax.org. 11.2 • Heat, Specific Heat, and Heat Transfer 333 Substances Specific Heat (c) Asbestos Concrete, granite (average) Copper Glass Gold Human body (average) Ice (average) Iron, steel Lead Silver Wood Liquids Benzene Ethanol Glycerin Mercury Water Gases (at 1 atm constant pressure) Air (dry) Ammonia Carbon dioxide Nitrogen Oxygen Steam 800 840 387 840 129 3500 2090 452 128 235 1700 1740 2450 2410 139 4186 1015 2190 833 1040 913 2020 Table 11.2 Specific Heats of Various Substances. 334 Chapter 11 • Thermal Energy, Heat, and Work Snap Lab Temperature Change of Land and Water What heats faster, land or water? You will answer this question by taking measurements to study differences in specific heat capacity. • Open flame—Tie back all loose hair and clothing before igniting an open flame. Follow all of your teacher's instructions on how to ignite the flame. Never leave an open flame unattended. Know the location of fire safety equipment in the laboratory. • Sand or soil • Water • Oven or heat lamp • Two small jars • Two thermometers Instructions Procedure 1. Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can get equal masses by using 50 percent more water by volume.) 2. Heat both substances (using an oven or a heat lamp) for the same amount of time. 3. Record the final temperatures of the two masses. 4. Now bring both jars to the same temperature by heating for a longer period of time. 5. Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes. GRASP CHECK Did it take longer to heat the water or the sand/soil to the same temperature? Which sample took longer to cool? What does this experiment tell us about how the specific heat of water compared to the specific heat of land? a. The sand/soil will take longer to heat as well as to cool. This tells us that the specific heat of land is greater than that of water. b. The sand/soil will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land. c. The water will take longer to heat as well as to cool. This tells us that the specific heat of land is greater than that of water. d. The water will take longer to heat as well as to cool. This tells us that the specific heat of water is greater than that of land. Conduction, Convection, and Radiation Whenever there is a temperature difference, heat transfer occurs. Heat transfer may happen rapidly, such as through a cooking pan, or slowly, such as through the walls of an insulated cooler. There are three different heat transfer methods: conduction, convection, and radiation. At times, all three may happen simultaneously. See Figure 11.3. Access for free at openstax.org. 11.2 • Heat, Specific Heat, and Heat Transfer 335 Figure 11.3 In a fireplace, heat transfer occurs by all three methods: conduction, convection, and radiation. Radiation is responsible for most of the heat transferred into the room. Heat transfer also occurs through conduction into the room, but at a much slower rate. Heat transfer by convection also occurs through cold air entering the room around windows and hot air leaving the room by rising up the chimney. Conduction is heat transfer through direct physical contact. Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction. Sometimes, we try to control the conduction of heat to make ourselves more comfortable. Since the rate of heat transfer is different for different materials, we choose fabrics, such as a thick wool sweater, that slow down the transfer of heat away from our bodies in winter. As you walk barefoot across the living room carpet, your feet feel relatively comfortable…until you step onto the kitchen’s tile floor. Since the carpet and tile floor are both at the same temperature, why does one feel colder than the other? This is explained by different rates of heat transfer: The tile material removes heat from your skin at a greater rate than the carpeting, which makes it feelcolder. Some materials simply conduct thermal energy faster than others. In general, metals (like copper, aluminum, gold, and si
lver) are good heat conductors, whereas materials like wood, plastic, and rubber are poor heat conductors. Figure 11.4 shows particles (either atoms or molecules) in two bodies at different temperatures. The (average) kinetic energy of a particle in the hot body is higher than in the colder body. If two particles collide, energy transfers from the particle with greater kinetic energy to the particle with less kinetic energy. When two bodies are in contact, many particle collisions occur, resulting in a net flux of heat from the higher-temperature body to the lower-temperature body. The heat flux depends on the temperature difference water. . Therefore, you will get a more severe burn from boiling water than from hot tap Figure 11.4 The particles in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a particle in the lower- temperature region (right side) has low kinetic energy before collision, but its kinetic energy increases after colliding with the contact 336 Chapter 11 • Thermal Energy, Heat, and Work surface. In contrast, a particle in the higher-temperature region (left side) has more kinetic energy before collision, but its energy decreases after colliding with the contact surface. Convection is heat transfer by the movement of a fluid. This type of heat transfer happens, for example, in a pot boiling on the stove, or in thunderstorms, where hot air rises up to the base of the clouds. TIPS FOR SUCCESS In everyday language, the term fluidis usually taken to mean liquid. For example, when you are sick and the doctor tells you to “push fluids,” that only means to drink more beverages—not to breath more air. However, in physics, fluid means a liquid or a gas. Fluids move differently than solid material, and even have their own branch of physics, known as fluid dynamics, that studies how they move. As the temperature of fluids increase, they expand and become less dense. For example, Figure 11.4 could represent the wall of a balloon with different temperature gases inside the balloon than outside in the environment. The hotter and thus faster moving gas particles inside the balloon strike the surface with more force than the cooler air outside, causing the balloon to expand. This decrease in density relative to its environment creates buoyancy (the tendency to rise). Convection is driven by buoyancy—hot air rises because it is less dense than the surrounding air. Sometimes, we control the temperature of our homes or ourselves by controlling air movement. Sealing leaks around doors with weather stripping keeps out the cold wind in winter. The house in Figure 11.5 and the pot of water on the stove in Figure 11.6 are both examples of convection and buoyancy by human design. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another, and are examples of natural convection. Figure 11.5 Air heated by the so-called gravity furnace expands and rises, forming a convective loop that transfers energy to other parts of the room. As the air is cooled at the ceiling and outside walls, it contracts, eventually becoming denser than room air and sinking to the floor. A properly designed heating system like this one, which uses natural convection, can be quite efficient in uniformly heating a home. Figure 11.6 Convection plays an important role in heat transfer inside this pot of water. Once conducted to the inside fluid, heat transfer to other parts of the pot is mostly by convection. The hotter water expands, decreases in density, and rises to transfer heat to other regions of the water, while colder water sinks to the bottom. This process repeats as long as there is water in the pot. Radiation is a form of heat transfer that occurs when electromagnetic radiation is emitted or absorbed. Electromagnetic Access for free at openstax.org. 11.2 • Heat, Specific Heat, and Heat Transfer 337 radiation includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, all of which have different wavelengths and amounts of energy (shorter wavelengths have higher frequency and more energy). You can feel the heat transfer from a fire and from the sun. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. Another example is thermal radiation from the human body; people are constantly emitting infrared radiation, which is not visible to the human eye, but is felt as heat. Radiation is the only method of heat transfer where no medium is required, meaning that the heat doesn’t need to come into direct contact with or be transported by any matter. The space between Earth and the sun is largely empty, without any possibility of heat transfer by convection or conduction. Instead, heat is transferred by radiation, and Earth is warmed as it absorbs electromagnetic radiation emitted by the sun. Figure 11.7 Most of the heat transfer from this fire to the observers is through infrared radiation. The visible light transfers relatively little thermal energy. Since skin is very sensitive to infrared radiation, you can sense the presence of a fire without looking at it directly. (Daniel X. O’Neil) All objects absorb and emit electromagnetic radiation (see Figure 11.7). The rate of heat transfer by radiation depends mainly on the color of the object. Black is the most effective absorber and radiator, and white is the least effective. People living in hot climates generally avoid wearing black clothing, for instance. Similarly, black asphalt in a parking lot will be hotter than adjacent patches of grass on a summer day, because black absorbs better than green. The reverse is also true—black radiates better than green. On a clear summer night, the black asphalt will be colder than the green patch of grass, because black radiates energy faster than green. In contrast, white is a poor absorber and also a poor radiator. A white object reflects nearly all radiation, like a mirror. Virtual Physics Energy Forms and Changes Click to view content (http://www.openstax.org/l/28energyForms) In this animation, you will explore heat transfer with different materials. Experiment with heating and cooling the iron, brick, and water. This is done by dragging and dropping the object onto the pedestal and then holding the lever either to Heat or Cool. Drag a thermometer beside each object to measure its temperature—you can watch how quickly it heats or cools in real time. Now let’s try transferring heat between objects. Heat the brick and then place it in the cool water. Now heat the brick again, but then place it on top of the iron. What do you notice? Selecting the fast forward option lets you speed up the heat transfers, to save time. GRASP CHECK Compare how quickly the different materials are heated or cooled. Based on these results, what material do you think has the greatest specific heat? Why? Which has the smallest specific heat? Can you think of a real-world situation where you would want to use an object with large specific heat? a. Water will take the longest, and iron will take the shortest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. 338 Chapter 11 • Thermal Energy, Heat, and Work b. Water will take the shortest, and iron will take the longest time to heat, as well as to cool. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. c. Brick will take shortest and iron will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. d. Water will take shortest and brick will take longest time to heat up as well as to cool down. Objects with greater specific heat would be desirable for insulation. For instance, woolen clothes with large specific heat would prevent heat loss from the body. Solving Heat Transfer Problems WORKED EXAMPLE Calculating the Required Heat: Heating Water in an Aluminum Pan A 0.500 kg aluminum pan on a stove is used to heat 0.250 L of water from 20.0 What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water? STRATEGY The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for heat transfer for the given temperature change and masses of water and aluminum. The specific heat values for water and aluminum are given in the previous table. . (a) How much heat is required? to 80.0 Solution to (a) Because the water is in thermal contact with the aluminum, the pan and the water are at the same temperature. 1. Calculate the temperature difference. 11.8 2. Calculate the mass of water using the relationship between density, mass, and volume. Density is mass per unit volume, or . Rearranging this equation, solve for the mass of water. 3. Calculate the heat transferred to the water. Use the specific heat of water in the previous table. 4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in the previous table. 5. Find the total transferred heat. Solution to (b) The percentage of heat going into heating the pan is Solution to (c) The percentage of heat going into heating the water is 11.9 11.10 11.11 11.12 11.13 11.14 Discussion In this example, most of the total heat transferred is used to heat the water, even though the pan has twice as much mass. This is Access for free at openstax.org. 11.2 • H
eat, Specific Heat, and Heat Transfer 339 because the specific heat of water is over four times greater than the specific heat of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water than for the aluminum pan. Water can absorb a tremendous amount of energy with very little resulting temperature change. This property of water allows for life on Earth because it stabilizes temperatures. Other planets are less habitable because wild temperature swings make for a harsh environment. You may have noticed that climates closer to large bodies of water, such as oceans, are milder than climates landlocked in the middle of a large continent. This is due to the climate-moderating effect of water’s large heat capacity—water stores large amounts of heat during hot weather and releases heat gradually when it’s cold outside. WORKED EXAMPLE Calculating Temperature Increase: Truck Brakes Overheat on Downhill Runs When a truck headed downhill brakes, the brakes must do work to convert the gravitational potential energy of the truck to internal energy of the brakes. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck, and keeps the truck from speeding up and losing control. The increased internal energy of the brakes raises their temperature. When the hill is especially steep, the temperature increase may happen too quickly and cause the brakes to overheat. Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg truck descending 75.0 m (in vertical displacement) at a constant speed. from a 10,000 kg STRATEGY We first calculate the gravitational potential energy (Mgh) of the truck, and then find the temperature increase produced in the brakes. Solution 1. Calculate the change in gravitational potential energy as the truck goes downhill. 2. Calculate the temperature change from the heat transferred by rearranging the equation to solve for 11.15 11.16 where mis the mass of the brake material (not the entire truck). Insert the values Q= 7.35×106 J (since the heat transfer is equal to the change in gravitational potential energy), m 100 kg, and c 800 J/kg to find 11.17 Discussion This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, which would be hard on the brakes. This is why truck drivers sometimes use a different technique for called “engine braking” to avoid burning their brakes during steep descents. Engine braking is using the slowing forces of an engine in low gear rather than brakes to slow down. 340 Chapter 11 • Thermal Energy, Heat, and Work Practice Problems 5. How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 °C ? a. 84 J b. 42 J c. 84 kJ d. 42 kJ 6. Calculate the change in temperature of 1.0 kg of water that is initially at room temperature if 3.0 kJ of heat is added. 358 °C a. 716 °C b. c. 0.36 °C d. 0.72 °C Check Your Understanding 7. What causes heat transfer? a. The mass difference between two objects causes heat transfer. b. The density difference between two objects causes heat transfer. c. The temperature difference between two systems causes heat transfer. d. The pressure difference between two objects causes heat transfer. 8. When two bodies of different temperatures are in contact, what is the overall direction of heat transfer? a. The overall direction of heat transfer is from the higher-temperature object to the lower-temperature object. b. The overall direction of heat transfer is from the lower-temperature object to the higher-temperature object. c. The direction of heat transfer is first from the lower-temperature object to the higher-temperature object, then back again to the lower-temperature object, and so-forth, until the objects are in thermal equilibrium. d. The direction of heat transfer is first from the higher-temperature object to the lower-temperature object, then back again to the higher-temperature object, and so-forth, until the objects are in thermal equilibrium. 9. What are the different methods of heat transfer? conduction, radiation, and reflection conduction, reflection, and convection convection, radiation, and reflection conduction, radiation, and convection a. b. c. d. 10. True or false—Conduction and convection cannot happen simultaneously a. True b. False 11.3 Phase Change and Latent Heat Section Learning Objectives By the end of this section, you will be able to do the following: • Explain changes in heat during changes of state, and describe latent heats of fusion and vaporization • Solve problems involving thermal energy changes when heating and cooling substances with phase changes Section Key Terms condensation freezing latent heat sublimation latent heat of fusion latent heat of vaporization melting vaporization phase change phase diagram plasma Access for free at openstax.org. 11.3 • Phase Change and Latent Heat 341 Phase Changes So far, we have learned that adding thermal energy by heat increases the temperature of a substance. But surprisingly, there are situations where adding energy does not change the temperature of a substance at all! Instead, the additional thermal energy acts to loosen bonds between molecules or atoms and causes a phase change. Because this energy enters or leaves a system during a phase change without causing a temperature change in the system, it is known as latent heat (latent means hidden). The three phases of matter that you frequently encounter are solid, liquid and gas (see Figure 11.8). Solid has the least energetic state; atoms in solids are in close contact, with forces between them that allow the particles to vibrate but not change position with neighboring particles. (These forces can be thought of as springs that can be stretched or compressed, but not easily broken.) Liquid has a more energetic state, in which particles can slide smoothly past one another and change neighbors, although they are still held together by their mutual attraction. Gas has a more energetic state than liquid, in which particles are broken free of their bonds. Particles in gases are separated by distances that are large compared with the size of the particles. The most energetic state of all is plasma. Although you may not have heard much about plasma, it is actually the most common state of matter in the universe—stars are made up of plasma, as is lightning. The plasma state is reached by heating a gas to the point where particles are pulled apart, separating the electrons from the rest of the particle. This produces an ionized gas that is a combination of the negatively charged free electrons and positively charged ions, known as plasma. Figure 11.8 (a) Particles in a solid always have the same neighbors, held close by forces represented here by springs. These particles are essentially in contact with one another. A rock is an example of a solid. This rock retains its shape because of the forces holding its atoms or molecules together. (b) Particles in a liquid are also in close contact but can slide over one another. Forces between them strongly resist attempts to push them closer together and also hold them in close contact. Water is an example of a liquid. Water can flow, but it also remains in an open container because of the forces between its molecules. (c) Particles in a gas are separated by distances that are considerably larger than the size of the particles themselves, and they move about freely. A gas must be held in a closed container to prevent it from moving out into its surroundings. (d) The atmosphere is ionized in the extreme heat of a lightning strike. During a phase change, matter changes from one phase to another, either through the addition of energy by heat and the transition to a more energetic state, or from the removal of energy by heat and the transition to a less energetic state. Phase changes to a more energetic state include the following: • Melting—Solid to liquid • Vaporization—Liquid to gas (included boiling and evaporation) • Sublimation—Solid to gas Phase changes to a less energetic state are as follows: • Condensation—Gas to liquid • Freezing—Liquid to solid Energy is required to melt a solid because the bonds between the particles in the solid must be broken. Since the energy involved in a phase changes is used to break bonds, there is no increase in the kinetic energies of the particles, and therefore no rise in temperature. Similarly, energy is needed to vaporize a liquid to overcome the attractive forces between particles in the liquid. There is no temperature change until a phase change is completed. The temperature of a cup of soda and ice that is initially at 0 stays at 0 until all of the ice has melted. In the reverse of these processes—freezing and condensation—energy is released 342 Chapter 11 • Thermal Energy, Heat, and Work from the latent heat (see Figure 11.9). Figure 11.9 (a) Energy is required to partially overcome the attractive forces between particles in a solid to form a liquid. That same energy must be removed for freezing to take place. (b) Particles are separated by large distances when changing from liquid to vapor, requiring significant energy to overcome molecular attraction. The same energy must be removed for condensation to take place. There is no temperature change until a phase change is completed. (c) Enough energy is added that the liquid state is skipped over completely as a substance undergoes sublimation. The heat, Q, required to change the phase of a sample of mass mis (for melting/freezing), (for vaporization/condensation), is the latent heat of fusion, and where needed to cause a phase change between sol
id and liquid. The latent heat of vaporization is the amount of heat needed to cause a is the latent heat of vaporization. The latent heat of fusion is the amount of heat Access for free at openstax.org. 11.3 • Phase Change and Latent Heat 343 phase change between liquid and gas. strength of intermolecular forces, and both have standard units of J/kg. See Table 11.3 for values of substances. are coefficients that vary from substance to substance, depending on the of different and and Substance Melting Point ( ) Lf (kJ/kg) Boiling Point ( ) Lv (kJ/kg) Helium ‒269.7 Hydrogen ‒259.3 Nitrogen ‒210.0 Oxygen ‒218.8 Ethanol ‒114 Ammonia ‒78 Mercury ‒38.9 Water 0.00 Sulfur Lead 119 327 Antimony 631 Aluminum 660 Silver Gold Copper 961 1063 1083 Uranium 1133 Tungsten 3410 5.23 58.6 25.5 13.8 104 332 11.8 334 38.1 24.5 165 380 88.3 64.5 134 84 184 ‒268.9 ‒252.9 ‒195.8 ‒183.0 78.3 ‒33.4 357 100.0 444.6 1750 1440 2520 2193 2660 2595 3900 5900 20.9 452 201 213 854 1370 272 2256 326 871 561 11400 2336 1578 5069 1900 4810 Table 11.3 Latent Heats of Fusion and Vaporization, along with Melting and Boiling Points Let’s consider the example of adding heat to ice to examine its transitions through all three phases—solid to liquid to gas. A phase diagram indicating the temperature changes of water as energy is added is shown in Figure 11.10. The ice starts out at −20 , and its temperature rises linearly, absorbing heat at a constant rate until it reaches 0 Once at this temperature, the ice gradually melts, absorbing 334 kJ/kg. The temperature remains constant at 0 melted, the temperature of the liquid water rises, absorbing heat at a new constant rate. At 100 the temperature again remains constant while the water absorbs 2256 kJ/kg during this phase change. When all the liquid has become steam, the temperature rises again at a constant rate. during this phase change. Once all the ice has , the water begins to boil and 344 Chapter 11 • Thermal Energy, Heat, and Work Figure 11.10 A graph of temperature versus added energy. The system is constructed so that no vapor forms while ice warms to become liquid water, and so when vaporization occurs, the vapor remains in the system. The long stretches of constant temperature values at 0 and 100 reflect the large latent heats of melting and vaporization, respectively. We have seen that vaporization requires heat transfer to a substance from its surroundings. Condensation is the reverse process, where heat in transferred away froma substance toits surroundings. This release of latent heat increases the temperature of the surroundings. Energy must be removed from the condensing particles to make a vapor condense. This is why condensation occurs on cold surfaces: the heat transfers energy away from the warm vapor to the cold surface. The energy is exactly the same as that required to cause the phase change in the other direction, from liquid to vapor, and so it can be calculated from . Latent heat is also released into the environment when a liquid freezes, and can be calculated from . FUN IN PHYSICS Making Ice Cream Figure 11.11 With the proper ingredients, some ice and a couple of plastic bags, you could make your own ice cream in five minutes. (ElinorD, Wikimedia Commons) Ice cream is certainly easy enough to buy at the supermarket, but for the hardcore ice cream enthusiast, that may not be satisfying enough. Going through the process of making your own ice cream lets you invent your own flavors and marvel at the physics firsthand (Figure 11.11). The first step to making homemade ice cream is to mix heavy cream, whole milk, sugar, and your flavor of choice; it could be as Access for free at openstax.org. 11.3 • Phase Change and Latent Heat 345 simple as cocoa powder or vanilla extract, or as fancy as pomegranates or pistachios. The next step is to pour the mixture into a container that is deep enough that you will be able to churn the mixture without it spilling over, and that is also freezer-safe. After placing it in the freezer, the ice cream has to be stirred vigorously every 45 minutes for four to five hours. This slows the freezing process and prevents the ice cream from turning into a solid block of ice. Most people prefer a soft creamy texture instead of one giant popsicle. As it freezes, the cream undergoes a phase change from liquid to solid. By now, we’re experienced enough to know that this means that the cream must experience a loss of heat. Where does that heat go? Due to the temperature difference between the freezer and the ice cream mixture, heat transfers thermal energy from the ice cream to the air in the freezer. Once the temperature in the freezer rises enough, the freezer is cooled by pumping excess heat outside into the kitchen. A faster way to make ice cream is to chill it by placing the mixture in a plastic bag, surrounded by another plastic bag half full of ice. (You can also add a teaspoon of salt to the outer bag to lower the temperature of the ice/salt mixture.) Shaking the bag for five minutes churns the ice cream while cooling it evenly. In this case, the heat transfers energy out of the ice cream mixture and into the ice during the phase change. This video (http://www.openstax.org/l/28icecream) gives a demonstration of how to make home-made ice cream using ice and plastic bags. GRASP CHECK Why does the ice bag method work so much faster than the freezer method for making ice cream? a. Ice has a smaller specific heat than the surrounding air in a freezer. Hence, it absorbs more energy from the ice-cream mixture. Ice has a smaller specific heat than the surrounding air in a freezer. Hence, it absorbs less energy from the ice-cream mixture. Ice has a greater specific heat than the surrounding air in a freezer. Hence, it absorbs more energy from the ice-cream mixture. Ice has a greater specific heat than the surrounding air in a freezer. Hence, it absorbs less energy from the ice-cream mixture. b. c. d. Solving Thermal Energy Problems with Phase Changes WORKED EXAMPLE Calculating Heat Required for a Phase Change Calculate a) how much energy is needed to melt 1.000 kg of ice at 0 vaporize 1.000 kg of water at 100 STRATEGY FOR (A) Using the equation for the heat required for melting, and the value of the latent heat of fusion of water from the previous table, we can solve for part (a). (freezing point), and b) how much energy is required to (boiling point). Solution to (a) The energy to melt 1.000 kg of ice is STRATEGY FOR (B) To solve part (b), we use the equation for heat required for vaporization, along with the latent heat of vaporization of water from the previous table. 11.18 Solution to (b) The energy to vaporize 1.000 kg of liquid water is 11.19 346 Chapter 11 • Thermal Energy, Heat, and Work Discussion The amount of energy need to melt a kilogram of ice (334 kJ) is the same amount of energy needed to raise the temperature of 1.000 kg of liquid water from 0 energy associated with temperature changes. It also demonstrates that the amount of energy needed for vaporization is even greater. . This example shows that the energy for a phase change is enormous compared to to 79.8 WORKED EXAMPLE and with a mass of Calculating Final Temperature from Phase Change: Cooling Soda with Ice Cubes Ice cubes are used to chill a soda at 20 cubes is 0.018 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all of the ice has melted. STRATEGY The ice cubes are at the melting temperature of 0 occurs in two steps: first, the phase change occurs and solid (ice) transforms into liquid water at the melting temperature; then, , so more heat is transferred from the soda to this water until the temperature of this water rises. Melting yields water at 0 they are the same temperature. Since the amount of heat leaving the soda is the same as the amount of heat transferred to the ice. . Heat is transferred from the soda to the ice for melting. Melting of ice and the total mass of the ice . The ice is at 0 11.20 The heat transferred to the ice goes partly toward the phase change (melting), and partly toward raising the temperature after melting. Recall from the last section that the relationship between heat and temperature change is temperature change is . The total heat transferred to the ice is therefore . For the ice, the Since the soda doesn’t change phase, but only temperature, the heat given off by the soda is Since , 11.21 11.22 11.23 Bringing all terms involving to the left-hand-side of the equation, and all other terms to the right-hand-side, we can solve for . Substituting the known quantities 11.24 11.25 Discussion This example shows the enormous energies involved during a phase change. The mass of the ice is about 7 percent the mass of the soda, yet it causes a noticeable change in the soda’s temperature. TIPS FOR SUCCESS If the ice were not already at the freezing point, we would also have to factor in how much energy would go into raising its temperature up to 0 often below 0 , before the phase change occurs. This would be a realistic scenario, because the temperature of ice is . Access for free at openstax.org. Practice Problems 11. How much energy is needed to melt 2.00 kg of ice at 0 °C ? 11.3 • Phase Change and Latent Heat 347 334 kJ a. 336 kJ b. c. 167 kJ d. 668 kJ 12. If a. b. c. d. of energy is just enough to melt of a substance, what is the substance’s latent heat of fusion? Check Your Understanding 13. What is latent heat? a. b. c. d. It is the heat that must transfer energy to or from a system in order to cause a mass change with a slight change in the temperature of the system. It is the heat that must transfer energy to or from a system in order to cause a mass change without a temperature change in the system. It is the heat that must transfer energy to or from a system in order to cause a phase change with a slight change in the temperature of
the system. It is the heat that must transfer energy to or from a system in order to cause a phase change without a temperature change in the system. 14. In which phases of matter are molecules capable of changing their positions? a. gas, liquid, solid liquid, plasma, solid b. c. liquid, gas, plasma d. plasma, gas, solid 348 Chapter 11 • Key Terms KEY TERMS absolute zero lowest possible temperature; the temperature at which all molecular motion ceases Kelvin scale temperature scale in which 0 K is the lowest possible temperature, representing absolute zero Celsius scale temperature scale in which the freezing point latent heat heat related to the phase change of a substance of water is 0 at 1 atm of pressure and the boiling point of water is 100 condensation phase change from gas to liquid conduction heat transfer through stationary matter by physical contact convection heat transfer by the movement of fluid degree Celsius unit on the Celsius temperature scale degree Fahrenheit unit on the Fahrenheit temperature scale Fahrenheit scale temperature scale in which the freezing and the boiling point of water is point of water is 32 212 freezing phase change from liquid to solid heat transfer of thermal (or internal) energy due to a temperature difference heat capacity amount of heat necessary to change the rather than a change of temperature latent heat of fusion amount of heat needed to cause a phase change between solid and liquid latent heat of vaporization amount of heat needed to cause a phase change between liquid and gas melting phase change from solid to liquid phase change transition between solid, liquid, or gas states of a substance plasma ionized gas that is a combination of the negatively charged free electrons and positively charged ions radiation energy transferred by electromagnetic waves specific heat amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 sublimation phase change from solid to gas temperature quantity measured by a thermometer thermal energy average random kinetic energy of a temperature of a substance by 1.00 molecule or an atom Kelvin unit on the Kelvin temperature scale; note that it is vaporization phase change from liquid to gas never referred to in terms of “degrees” Kelvin SECTION SUMMARY 11.1 Temperature and Thermal Energy • Temperature is the quantity measured by a thermometer. • Temperature is related to the average kinetic energy of atoms and molecules in a system. • Absolute zero is the temperature at which there is no molecular motion. • There are three main temperature scales: Celsius, Fahrenheit, and Kelvin. • Temperatures on one scale can be converted into temperatures on another scale. 11.2 Heat, Specific Heat, and Heat Transfer • Heat is thermal (internal) energy transferred due to a temperature difference. • The transfer of heat Qthat leads to a change temperature of a body with mass m is where cis the specific heat of the material. • Heat is transferred by three different methods: in the , conduction, convection, and radiation. • Heat conduction is the transfer of heat between two objects in direct contact with each other. • Convection is heat transfer by the movement of mass. • Radiation is heat transfer by electromagnetic waves. 11.3 Phase Change and Latent Heat • Most substances have four distinct phases: solid, liquid, gas, and plasma. • Gas is the most energetic state and solid is the least. • During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. • Heat is added to a substance during melting and vaporization. • Latent heat is released by a substance during condensation and freezing. • Phase changes occur at fixed temperatures called boiling and freezing (or melting) points for a given substance. Access for free at openstax.org. KEY EQUATIONS 11.1 Temperature and Thermal Energy 11.2 Heat, Specific Heat, and Heat Transfer Chapter 11 • Key Equations 349 Celsius to Fahrenheit conversion Fahrenheit to Celsius conversion Celsius to Kelvin conversion Kelvin to Celsius conversion Fahrenheit to Kelvin conversion Kelvin to Fahrenheit conversion heat transfer density 11.3 Phase Change and Latent Heat heat transfer for melting/freezing phase change heat transfer for vaporization/ condensation phase change CHAPTER REVIEW Concept Items 11.1 Temperature and Thermal Energy 1. A glass of water has a temperature of 31 degrees Celsius. solid liquid What state of matter is it in? a. b. c. gas d. plasma 2. What is the difference between thermal energy and internal energy? a. The thermal energy of the system is the average kinetic energy of the system’s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies and the potential energies of its constituent particles. b. The thermal energy of the system is the average potential energy of the system’s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies and the potential energies of its constituent particles. c. The thermal energy of the system is the average kinetic energy of the system’s constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies of its constituent particles. d. The thermal energy of the system is the average potential energy of the systems’ constituent particles due to their motion. The total internal energy of the system is the sum of the kinetic energies of its constituent particles. 3. What does the Celsius scale use as a reference point? a. The boiling point of mercury b. The boiling point of wax c. The freezing point of water d. The freezing point of wax 11.2 Heat, Specific Heat, and Heat Transfer 4. What are the SI units of specific heat? a. b. c. d. 5. What is radiation? a. The transfer of energy through emission and absorption of the electromagnetic waves is known as radiation. b. The transfer of energy without any direct physical 350 Chapter 11 • Chapter Review contact between any two substances. c. The transfer of energy through direct physical contact between any two substances. d. The transfer of energy by means of the motion of fluids at different temperatures and with different densities. 11.3 Phase Change and Latent Heat 6. Why is there no change in temperature during a phase change, even if energy is absorbed by the system? a. The energy is used to break bonds between particles, and so does not increase the potential energy of the system’s particles. b. The energy is used to break bonds between particles, Critical Thinking Items 11.1 Temperature and Thermal Energy 8. The temperature of two equal quantities of water needs to be raised - the first container by degrees Celsius and the second by degrees Fahrenheit. Which one would require more heat? a. The heat required by the first container is more than the second because each degree Celsius is equal to degrees Fahrenheit. b. The heat required by the first container is less than the second because each degree Fahrenheit is equal to degrees Celsius. c. The heat required by the first container is more than the second because each degree Celsius is equal to degrees Fahrenheit. d. The heat required by the first container is less than the second because each degree Fahrenheit is equal to degrees Celsius. 9. What is 100.00 °C in kelvins? a. 212.00 K b. 100.00 K c. 473.15 K 373.15 K d. 11.2 Heat, Specific Heat, and Heat Transfer 10. The value of specific heat is the same whether the units are J/kg⋅K or J/kg⋅ºC. How? a. Temperature difference is dependent on the chosen temperature scale. b. Temperature change is different in units of kelvins and degrees Celsius. c. Reading of temperatures in kelvins and degree Celsius are the same. Access for free at openstax.org. and so increases the potential energy of the system’s particles. c. The energy is used to break bonds between particles, and so does not increase the kinetic energy of the system’s particles. d. The energy is used to break bonds between particles, and so increases the kinetic energy of the system’s particles. 7. In which two phases of matter do atoms and molecules have the most distance between them? a. gas and solid b. gas and liquid c. gas and plasma d. liquid and plasma d. The temperature change is the same in units of kelvins and degrees Celsius. 11. If the thermal energy of a perfectly black object is increased by conduction, will the object remain black in appearance? Why or why not? a. No, the energy of the radiation increases as the temperature increases, and the radiation becomes visible at certain temperatures. b. Yes, the energy of the radiation decreases as the temperature increases, and the radiation remains invisible at those energies. c. No, the energy of the radiation decreases as the temperature increases, until the frequencies of the radiation are the same as those of visible light. d. Yes, as the temperature increases, and the energy is transferred from the object by other mechanisms besides radiation, so that the energy of the radiation does not increase. 12. What is the specific heat of a substance that requires 5.00 kJ of heat to raise the temperature of 3.00 kg by 5.00 °F? 3.33×103 J/kg ⋅° C a. b. 6.00×103 J/kg ⋅° C 3.33×102 J/kg ⋅ ° C c. d. 6.00×102 J/kg ⋅ ° C 11.3 Phase Change and Latent Heat 13. Assume 1.0 kg of ice at 0 °C starts to melt. It absorbs 300 kJ of energy by heat. What is the temperature of the water afterwards? a. 10 °C b. 20 °C c. 5 °C d. 0 °C Problems 11.1 Temperature and Thermal Energy 14. What is 35.0 °F in kelvins? 1.67 K a. 35.0 K b. c. -271.5 K d. 274.8 K 15. Design a temperature scale where the freezing point of water is 0 degrees and its boiling point is 70 degrees. What would be the room temperature on this scale? a. If room temperature is 25.0 °C, the temperature on the new scale will be 17.5 °. If room temperature is 25.0 °C, the temperatur
e on the new scale will be 25.0°. If the room temperature is 25.0 °C, the temperature on the new scale will be 35.7°. If the room temperature is 25.0 °C, the temperature on the new scale will be 50.0°. b. c. d. 11.2 Heat, Specific Heat, and Heat Transfer 16. A certain quantity of water is given 4.0 kJ of heat. This raises its temperature by 30.0 °F. What is the mass of the water in grams? 5.7 g a. 570 g b. Performance Task 11.3 Phase Change and Latent Heat 20. You have been tasked with designing a baking pan that will bake batter the fastest. There are four materials available for you to test. • Four pans of similar design, consisting of aluminum, iron (steel), copper, and glass • Oven or similar heating source • Device for measuring high temperatures • Balance for measuring mass Instructions Procedure 1. Design a safe experiment to test the specific heat of each material (i.e., no extreme temperatures TEST PREP Multiple Choice 11.1 Temperature and Thermal Energy 21. The temperature difference of is the same as Chapter 11 • Test Prep 351 c. d. 5700 g 57 g 17. 5290 J of heat is given to 0.500 kg water at 15.00 °C. What will its final temperature be? a. 15.25° C 12.47 ° C b. c. 40.3° C 17.53° C d. 11.3 Phase Change and Latent Heat 18. How much energy would it take to heat 1.00 kg of ice at 0 °C to water at 15.0 °C? a. 271 kJ b. 334 kJ c. 62.8 kJ 397 kJ d. 19. Ice cubes are used to chill a soda with a mass msoda = 0.300 kg at 15.0 °C. The ice is at 0 °C, and the total mass of the ice cubes is 0.020 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all ice has melted. a. 19.02 °C b. 90.3 °C c. 0.11 °C d. 9.03 °C should be used) 2. Write down the materials needed for your experiment and the procedure you will follow. Make sure that you include every detail, so that the experiment can be repeated by others. 3. Carry out the experiment and record any data collected. 4. Review your results and make a recommendation as to which metal should be used for the pan. a. What physical quantities do you need to measure to determine the specific heats for the different materials? b. How does the glass differ from the metals in terms of thermal properties? c. What are your sources of error? a. b. c. d. degree Celsius degree Fahrenheit degrees Celsius degrees Fahrenheit 352 Chapter 11 • Test Prep 22. What is the preferred temperature scale used in celsius fahrenheit scientific laboratories? a. b. c. kelvin d. rankine 11.2 Heat, Specific Heat, and Heat Transfer 23. Which phase of water has the largest specific heat? solid liquid a. b. c. gas 24. What kind of heat transfer requires no medium? a. b. c. d. conduction convection reflection radiation 25. Which of these substances has the greatest specific heat? a. copper b. mercury c. aluminum d. wood 26. Give an example of heat transfer through convection. a. The energy emitted from the filament of an electric bulb b. The energy coming from the sun c. A pan on a hot burner d. Water boiling in a pot 11.3 Phase Change and Latent Heat 27. What are the SI units of latent heat? Short Answer 11.1 Temperature and Thermal Energy 31. What is absolute zeroon the Fahrenheit scale? a. 0 °F 32 °F b. -273.15 °F c. -459.67 °F d. 32. What is absolute zeroon the Celsius scale? a. 0 °C b. 273.15 °C c. d. -459.67 °C -273.15 °C 33. A planet’s atmospheric pressure is such that water there boils at a lower temperature than it does at sea level on Access for free at openstax.org. a. b. c. d. 28. Which substance has the largest latent heat of fusion? a. gold b. water c. mercury tungsten d. 29. In which phase changes does matter undergo a transition to a more energetic state? a. freezing and vaporization b. melting and sublimation c. melting and vaporization d. melting and freezing 30. A room has a window made from thin glass. The room is colder than the air outside. There is some condensation on the glass window. On which side of the glass would the condensation most likely be found? a. Condensation is on the outside of the glass when the cool, dry air outside the room comes in contact with the cold pane of glass. b. Condensation is on the outside of the glass when the warm, moist air outside the room comes in contact with the cold pane of glass. c. Condensation is on the inside of the glass when the cool, dry air inside the room comes in contact with the cold pane of glass. d. Condensation is on the inside of the glass when the warm, moist air inside the room comes in contact with the cold pane of glass. Earth. If a Celsius scale is derived on this planet, will it be the same as that on Earth? a. The Celsius scale derived on the planet will be the same as that on Earth, because the Celsius scale is independent of the freezing and boiling points of water. b. The Celsius scale derived on that planet will not be the same as that on Earth, because the Celsius scale is dependent and derived by using the freezing and boiling points of water. c. The Celsius scale derived on the planet will be the same as that on Earth, because the Celsius scale is an absolute temperature scale based on molecular motion, which is independent of pressure. d. The Celsius scale derived on the planet will not be the same as that on Earth, but the Fahrenheit scale Chapter 11 • Test Prep 353 will be the same, because its reference temperatures are not based on the freezing and boiling points of water. b. 63 °C c. d. 6.3 °C 1.8×10-2 °C 34. What is the difference between the freezing point and 40. Aluminum has a specific heat of 900 J/kg·ºC. How much boiling point of water on the Reaumur scale? a. The boiling point of water is 80° on the Reaumur scale. b. Reaumur scale is less than 120°. c. 100° d. 80° energy would it take to change the temperature of 2 kg aluminum by 3 ºC? a. 1.3 kJ b. 0.60 kJ 54 kJ c. 5.4 kJ d. 11.2 Heat, Specific Heat, and Heat Transfer 35. In the specific heat equation what does cstand for? a. Total heat b. Specific heat c. Specific temperature d. Specific mass 36. Specific heat may be measured in J/kg · K, J/kg · °C. What other units can it be measured in? a. kg/kcal · °C b. kcal · °C/kg c. kg · °C/kcal d. kcal/kg · °C 37. What is buoyancy? a. Buoyancy is a downward force exerted by a solid that opposes the weight of an object. b. Buoyancy is a downward force exerted by a fluid that opposes the weight of an immersed object. c. Buoyancy is an upward force exerted by a solid that opposes the weight an object. d. Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. 38. Give an example of convection found in nature. a. heat transfer through metallic rod b. heat transfer from the sun to Earth c. heat transfer through ocean currents d. heat emitted by a light bulb into its environment 39. Calculate the temperature change in a substance with specific heat 735 J/kg · °C when 14 kJ of heat is given to a 3.0-kg sample of that substance. a. 57 °C 11.3 Phase Change and Latent Heat 41. Upon what does the required amount of heat removed to freeze a sample of a substance depend? a. The mass of the substance and its latent heat of vaporization b. The mass of the substance and its latent heat of fusion c. The mass of the substance and its latent heat of sublimation d. The mass of the substance only 42. What do latent heats, Lf and Lv, depend on? a. Lf and Lv depend on the forces between the particles in the substance. b. Lf and Lv depend on the mass of the substance. c. Lf and Lv depend on the volume of the substance. d. Lf and Lv depend on the temperature of the substance. 43. How much energy is required to melt 7.00 kg a block of aluminum that is at its melting point? (Latent heat of fusion of aluminum is 380 kJ/kg.) 54.3 kJ a. b. 2.66 kJ c. 0.0184 kJ d. 2.66×103 kJ 44. A 3.00 kg sample of a substance is at its boiling point. If 5,360 kJ of energy are enough to boil away the entire substance, what is its latent heat of vaporization? a. 2,685 kJ/kg b. 3,580 kJ/kg c. 895 kJ/kg d. 1,790 kJ/kg Extended Response 11.1 Temperature and Thermal Energy 45. What is the meaning of absolute zero? a. It is the temperature at which the internal energy of the system is maximum, because the speed of its b. constituent particles increases to maximum at this point. It is the temperature at which the internal energy of the system is maximum, because the speed of its constituent particles decreases to zero at this point. 354 Chapter 11 • Test Prep c. d. It is the temperature at which the internal energy of the system approaches zero, because the speed of its constituent particles increases to a maximum at this point. It is that temperature at which the internal energy of the system approaches zero, because the speed of its constituent particles decreases to zero at this point. 46. Why does it feel hotter on more humid days, even though there is no difference in temperature? a. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is lower, which reduces the evaporation rate from the skin’s surface. b. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is higher, which reduces the evaporation rate from the skin’s surface. c. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is lower, which increases the evaporation rate from the skin’s surface. d. On hot, dry days, the evaporation of the sweat from the skin cools the body, whereas on humid days the concentration of water in the atmosphere is higher, which increases the evaporation rate from the skin’s surface. 11.2 Heat, Specific Heat, and Heat Transfer 47. A hot piece of metal needs to be cooled. If you were to put the metal in ice or in cold water, such that the ice d
id not melt and the temperature of either changed by the same amount, which would reduce the metal’s temperature more? Why? a. Water would reduce the metal’s temperature more, because water has a greater specific heat than ice. b. Water would reduce the metal’s temperature more, because water has a smaller specific heat than ice. Ice would reduce the metal’s temperature more, because ice has a smaller specific heat than water. c. d. Ice would reduce the metal’s temperature more, because ice has a greater specific heat than water. 48. On a summer night, why does a black object seem colder than a white one? a. The black object radiates energy faster than the white one, and hence reaches a lower temperature in less time. b. The black object radiates energy slower than the white one, and hence reaches a lower temperature in less time. c. The black object absorbs energy faster than the white one, and hence reaches a lower temperature in less time. d. The black object absorbs energy slower than the white one, and hence reaches a lower temperature in less time. 49. Calculate the difference in heat required to raise the temperatures of 1.00 kg of gold and 1.00 kg of aluminum by 1.00 °C. (The specific heat of aluminum equals 900 J/kg · °C; the specific heat of gold equals 129 J/ kg · °C.) 771 J a. b. 129 J c. 90 J d. 900 J 11.3 Phase Change and Latent Heat 50. True or false—You have an ice cube floating in a glass of water with a thin thread resting across the cube. If you cover the ice cube and thread with a layer of salt, they will stick together, so that you are able to lift the icecube when you pick up the thread. a. True b. False 51. Suppose the energy required to freeze 0.250 kg of water were added to the same mass of water at an initial temperature of 1.0 °C. What would be the final temperature of the water? -69.8 °C a. 79.8 °C b. c. -78.8 °C d. 80.8 °C Access for free at openstax.org. CHAPTER 12 Thermodynamics Figure 12.1 A steam engine uses energy transfer by heat to do work. (Modification of work by Gerald Friedrich, Pixabay) Chapter Outline 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 12.2 First law of Thermodynamics: Thermal Energy and Work 12.3 Second Law of Thermodynamics: Entropy 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators Energy can be transferred to or from a system, either through a temperature difference between it and INTRODUCTION another system (i.e., by heat) or by exerting a force through a distance (work). In these ways, energy can be converted into other forms of energy in other systems. For example, a car engine burns fuel for heat transfer into a gas. Work is done by the gas as it exerts a force through a distance by pushing a piston outward. This work converts the energy into a variety of other forms—into an increase in the car’s kinetic or gravitational potential energy; into electrical energy to run the spark plugs, radio, and lights; and back into stored energy in the car’s battery. But most of the thermal energy transferred by heat from the fuel burning in the engine does not do work on the gas. Instead, much of this energy is released into the surroundings at lower temperature (i.e., lost through heat), which is quite inefficient. Car engines are only about 25 to 30 percent efficient. This inefficiency leads to increased fuel costs, so there is great interest in improving fuel efficiency. However, it is common knowledge that modern gasoline engines cannot be made much more efficient. The same is true about the conversion to electrical energy in large power stations, whether they are coal, oil, natural gas, or nuclear powered. Why is this the case? The answer lies in the nature of heat. Basic physical laws govern how heat transfer for doing work takes place and limit the 356 Chapter 12 • Thermodynamics maximum possible efficiency of the process. This chapter will explore these laws as well their applications to everyday machines. These topics are part of thermodynamics—the study of heat and its relationship to doing work. 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium Section Learning Objectives By the end of this section, you will be able to do the following: • Explain the zeroth law of thermodynamics Section Key Terms thermal equilibrium zeroth law of thermodynamics We learned in the previous chapter that when two objects (or systems) are in contact with one another, heat will transfer thermal energy from the object at higher temperature to the one at lower temperature until they both reach the same temperature. The objects are then in thermal equilibrium, and no further temperature changes will occur if they are isolated from other systems. The systems interact and change because their temperatures are different, and the changes stop once their temperatures are the same. Thermal equilibrium is established when two bodies are in thermal contactwith each other—meaning heat transfer (i.e., the transfer of energy by heat) can occur between them. If two systems cannot freely exchange energy, they will not reach thermal equilibrium. (It is fortunate that empty space stands between Earth and the sun, because a state of thermal equilibrium with the sun would be too toasty for life on this planet!) If two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. This statement may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics. TIPS FOR SUCCESS The zeroth law of thermodynamics is very similar to the transitive property of equality in mathematics: If a = b and b = c, then a = c. You may be wondering at this point, why the wacky name? Shouldn’t this be called the firstlaw of thermodynamics rather than the zeroth? The explanation is that this law was discovered after the first and second laws of thermodynamics but is so fundamental that scientists decided it should logically come first. As an example of the zeroth law in action, consider newborn babies in neonatal intensive-care units in hospitals. Prematurely born or sick newborns are placed in special incubators. These babies have very little covering while in the incubators, so to an observer, they look as though they may not be warm enough. However, inside the incubator, the temperature of the air, the cot, and the baby are all the same—that is, they are in thermal equilibrium. The ambient temperature is just high enough to keep the baby safe and comfortable. WORK IN PHYSICS Thermodynamics Engineer Thermodynamics engineers apply the principles of thermodynamics to mechanical systems so as to create or test products that rely on the interactions between heat, work, pressure, temperature, and volume. This type of work typically takes place in the aerospace industry, chemical manufacturing companies, industrial manufacturing plants, power plants (Figure 12.2), engine manufacturers, or electronics companies. Access for free at openstax.org. 12.1 • Zeroth Law of Thermodynamics: Thermal Equilibrium 357 Figure 12.2 An engineer makes a site visit to the Baghdad South power plant. The need for energy creates quite a bit of demand for thermodynamics engineers, because both traditional energy companies and alternative (green) energy startups rely on interactions between heat and work and so require the expertise of thermodynamics engineers. Traditional energy companies use mainly nuclear energy and energy from burning fossil fuels, such as coal. Alternative energy is finding new ways to harness renewable and, often, more readily available energy sources, such as solar, water, wind, and bio-energy. A thermodynamics engineer in the energy industry can find the most efficient way to turn the burning of a biofuel or fossil fuel into energy, store that energy for times when it’s needed most, or figure out how to best deliver that energy from where it’s produced to where it’s used: in homes, factories, and businesses. Additionally, he or she might also design pollution-control equipment to remove harmful pollutants from the smoke produced as a by-product of burning fuel. For example, a thermodynamics engineer may develop a way to remove mercury from burning coal in a coal-fired power plant. Thermodynamics engineering is an expanding field, where employment opportunities are expected to grow by as much as 27 percent between 2012 and 2022, according to the U.S. Bureau of Labor Statistics. To become a thermodynamics engineer, you must have a college degree in chemical engineering, mechanical engineering, environmental engineering, aerospace engineering, civil engineering, or biological engineering (depending on which type of career you wish to pursue), with coursework in physics and physical chemistry that focuses on thermodynamics. GRASP CHECK What would be an example of something a thermodynamics engineer would do in the aeronautics industry? a. Test the fuel efficiency of a jet engine b. Test the functioning of landing gear c. Test the functioning of a lift control device d. Test the autopilot functions Check Your Understanding 1. What is thermal equilibrium? a. When two objects in contact with each other are at the same pressure, they are said to be in thermal equilibrium. b. When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium. c. When two objects in contact with each other are at the same temperature, they are said to be in thermal equilibrium. d. When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium. 2. What is the zeroth law of thermodynamics? 358 Chapter 12 • Thermodynamics a. Energy can neither be created nor destroyed in a chemical reaction. b. If two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system, C, then
A is also in thermal equilibrium with C. c. Entropy of any isolated system not in thermal equilibrium always increases. d. Entropy of a system approaches a constant value as temperature approaches absolute zero. 12.2 First law of Thermodynamics: Thermal Energy and Work Section Learning Objectives By the end of this section, you will be able to do the following: • Describe how pressure, volume, and temperature relate to one another and to work, based on the ideal gas law • Describe pressure–volume work • Describe the first law of thermodynamics verbally and mathematically • Solve problems involving the first law of thermodynamics Section Key Terms Boltzmann constant first law of thermodynamics ideal gas law internal energy pressure Pressure, Volume, Temperature, and the Ideal Gas Law Before covering the first law of thermodynamics, it is first important to understand the relationship between pressure, volume, and temperature. Pressure, P, is defined as where Fis a force applied to an area, A, that is perpendicular to the force. Depending on the area over which it is exerted, a given force can have a significantly different effect, as shown in Figure 12.3. 12.1 Figure 12.3 (a) Although the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the same force applied to an area the size of the sharp end of a needle is great enough to break the skin. The SI unit for pressure is the pascal, where Pressure is defined for all states of matter but is particularly important when discussing fluids (such as air). You have probably heard the word pressurebeing used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. The relationship between the pressure, volume, and temperature for an ideal gas is given by the ideal gas law. A gas is considered ideal at low pressure and fairly high temperature, and forces between its component particles can be ignored. The ideal gas law states that 12.2 where Pis the pressure of a gas, Vis the volume it occupies, Nis the number of particles (atoms or molecules) in the gas, and Tis Access for free at openstax.org. 12.2 • First law of Thermodynamics: Thermal Energy and Work 359 its absolute temperature. The constant kis called the Boltzmann constant and has the value purposes of this chapter, we will not go into calculations using the ideal gas law. Instead, it is important for us to notice from the equation that the following are true for a given mass of gas: For the • When volume is constant, pressure is directly proportional to temperature. • When temperature is constant, pressure is inversely proportional to volume. • When pressure is constant, volume is directly proportional to temperature. This last point describes thermal expansion—the change in size or volume of a given mass with temperature. What is the underlying cause of thermal expansion? An increase in temperature means that there’s an increase in the kinetic energy of the individual atoms. Gases are especially affected by thermal expansion, although liquids expand to a lesser extent with similar increases in temperature, and even solids have minor expansions at higher temperatures. This is why railroad tracks and bridges have expansion joints that allow them to freely expand and contract with temperature changes. To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into a deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If you continue to pump air into tire (which now has a nearly constant volume), the pressure increases with increasing temperature (see Figure 12.4). Figure 12.4 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases as more air is added. (c) Once the tire is inflated fully, its pressure increases with temperature. Pressure–Volume Work Pressure–volume workis the work that is done by the compression or expansion of a fluid. Whenever there is a change in volume and external pressure remains constant, pressure–volume work is taking place. During a compression, a decrease in volume increases the internal pressure of a system as work is done onthe system. During an expansion (Figure 12.5), an increase in volume decreases the internal pressure of a system as the system doeswork. 360 Chapter 12 • Thermodynamics Figure 12.5 An expansion of a gas requires energy transfer to keep the pressure constant. Because pressure is constant, the work done is . Recall that the formula for work is in terms of pressure. We can rearrange the definition of pressure, to get an expression for force Substituting this expression for force into the definition of work, we get Because area multiplied by displacement is the change in volume, pressure–volume work is , the mathematical expression for 12.3 12.4 12.5 Just as we say that work is force acting over a distance, for fluids, we can say that work is the pressure acting through the change in volume. For pressure–volume work, pressure is analogous to force, and volume is analogous to distance in the traditional definition of work. WATCH PHYSICS Work from Expansion This video describes work from expansion (or pressure–volume work). Sal combines the equations and to get . Click to view content (https://www.openstax.org/l/28expansionWork) GRASP CHECK If the volume of a system increases while pressure remains constant, is the value of work done by the system Wpositive or negative? Will this increase or decrease the internal energy of the system? a. Positive; internal energy will decrease b. Positive; internal energy will increase c. Negative; internal energy will decrease d. Negative; internal energy will increase The First Law of Thermodynamics Heat (Q) and work (W) are the two ways to add or remove energy from a system. The processes are very different. Heat is driven Access for free at openstax.org. 12.2 • First law of Thermodynamics: Thermal Energy and Work 361 by temperature differences, while work involves a force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air’s temperature. Similarly, work can be done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat or work. Heat and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy, U, of a system. Internal energy is the sum of the kinetic and potential energies of a system’s atoms and molecules. It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P, V, and T), not on how the energy enters or leaves the system. In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics. The first law of thermodynamics applies the conservation of energyprinciple to systems where heat and work are the methods of transferring energy into and out of the systems. It can also be used to describe how energy transferred by heat is converted and transferred again by work. TIPS FOR SUCCESS Recall that the principle of conservation of energy states that energy cannot be created or destroyed, but it can be altered from one form to another. The first law of thermodynamics states that the change in internal energy of a closed system equals the net heat transfer intothe system minus the net work done bythe system. In equation form, the first law of thermodynamics is 12.6 is the change in internal energy, U, of the system. As shown in Figure 12.6, Qis the net heat transferred into the Here, system—that is, Qis the sum of all heat transfers into and out of the system. Wis the net work done by the system—that is, Wis the sum of all work done on or by the system. By convention, if Qis positive, then there is a net heat transfer into the system; if Wis positive, then there is net work done by the system. So positive Qadds energy to the system by heat, and positive Wtakes energy from the system by work. Note that if heat transfers more energy into the system than that which is done by work, the difference is stored as internal energy. Figure 12.6 The first law of thermodynamics is the conservation of energyprinciple stated for a system, where heat and work are the methods of transferring energy to and from a system. Qrepresents the net heat transfer—it is the sum of all transfers of energy by heat into and out of the system. Qis positive for net heat transfer intothe system. is the work done bythe system, and is the work done on the system. Wis the total work done on or bythe system. Wis positive when more work is done bythe system than onit. The change in the internal energy of the system, , is related to heat and work by the first law of thermodynamics: It follows also that negative Qindicates that energy is transferred awayfrom the system by heat and so decreases the system’s internal energy, whereas negative Wis work done onthe system, which increases the internal energy. WATCH PHYSICS First Law of Thermodynamics/Internal Energy This video explains the first law of thermodynamics, conservation of energy, and internal energy. It goes over an example of energy transforming between kinetic energy, potential energy, and heat transfer due to air resistance. Click to view
content (https://www.openstax.org/l/28FirstThermo) 362 Chapter 12 • Thermodynamics GRASP CHECK Consider the example of tossing a ball when there’s air resistance. As air resistance increases, what would you expect to happen to the final velocity and final kinetic energy of the ball? Why? a. Both will decrease. Energy is transferred to the air by heat due to air resistance. b. Both will increase. Energy is transferred from the air to the ball due to air resistance. c. Final velocity will increase, but final kinetic energy will decrease. Energy is transferred by heat to the air from the ball through air resistance. d. Final velocity will decrease, but final kinetic energy will increase. Energy is transferred by heat from the air to the ball through air resistance. WATCH PHYSICS More on Internal Energy This video goes into further detail, explaining internal energy and how to use the equation the equation system. , where Wis the work done onthe system, whereas we use Wto represent work done bythe Note that Sal uses Click to view content (https://www.openstax.org/l/28IntrnEnergy) GRASP CHECK are taken away by heat from the system, and the system does If system? a. b. c. d. of work, what is the change in internal energy of the LINKS TO PHYSICS Biology: Biological Thermodynamics We often think about thermodynamics as being useful for inventing or testing machinery, such as engines or steam turbines. However, thermodynamics also applies to living systems, such as our own bodies. This forms the basis of the biological thermodynamics (Figure 12.7). Figure 12.7 (a) The first law of thermodynamics applies to metabolism. Heat transferred out of the body (Q) and work done by the body (W) remove internal energy, whereas food intake replaces it. (Food intake may be considered work done on the body.) (b) Plants convert part of Access for free at openstax.org. 12.2 • First law of Thermodynamics: Thermal Energy and Work 363 the radiant energy in sunlight into stored chemical energy, a process called photosynthesis. Life itself depends on the biological transfer of energy. Through photosynthesis, plants absorb solar energy from the sun and use this energy to convert carbon dioxide and water into glucose and oxygen. Photosynthesis takes in one form of energy—light—and converts it into another form—chemical potential energy (glucose and other carbohydrates). Human metabolismis the conversion of food into energy given off by heat, work done by the body’s cells, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. Eating increases the internal energy of the body by adding chemical potential energy; this is an unromantic view of a good burrito. The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Exercise helps you lose weight, because it provides energy transfer from your body by both heat and work and raises your metabolic rate even when you are at rest. Biological thermodynamics also involves the study of transductions between cells and living organisms. Transductionis a process where genetic material—DNA—is transferred from one cell to another. This often occurs during a viral infection (e.g., influenza) and is how the virus spreads, namely, by transferring its genetic material to an increasing number of previously healthy cells. Once enough cells become infected, you begin to feel the effects of the virus (flu symptoms—muscle weakness, coughing, and congestion). Energy is transferred along with the genetic material and so obeys the first law of thermodynamics. Energy is transferred—not created or destroyed—in the process. When work is done on a cell or heat transfers energy to a cell, the cell’s internal energy increases. When a cell does work or loses heat, its internal energy decreases. If the amount of work done by a cell is the same as the amount of energy transferred in by heat, or the amount of work performed on a cell matches the amount of energy transferred out by heat, there will be no net change in internal energy. GRASP CHECK Based on what you know about heat transfer and the first law of thermodynamics, do you need to eat more or less to maintain a constant weight in colder weather? Explain why. a. more; as more energy is lost by the body in colder weather, the need to eat increases so as to maintain a constant weight b. more; eating more food means accumulating more fat, which will insulate the body from colder weather and will reduce c. d. the energy loss less; as less energy is lost by the body in colder weather, the need to eat decreases so as to maintain a constant weight less; eating less food means accumulating less fat, so less energy will be required to burn the fat, and, as a result, weight will remain constant Solving Problems Involving the First Law of Thermodynamics WORKED EXAMPLE Calculating Change in Internal Energy Suppose 40.00 J of energy is transferred by heat to a system, while the system does 10.00 J of work. Later, heat transfers 25.00 J out of the system, while 4.00 J is done by work on the system. What is the net change in the system’s internal energy? STRATEGY You must first calculate the net heat and net work. Then, using the first law of thermodynamics, change in internal energy. find the Solution The net heat is the transfer into the system by heat minus the transfer out of the system by heat, or The total work is the work done by the system minus the work done on the system, or 12.7 12.8 364 Chapter 12 • Thermodynamics The change in internal energy is given by the first law of thermodynamics. Discussion A different way to solve this problem is to find the change in internal energy for each of the two steps separately and then add the two changes to get the total change in internal energy. This approach would look as follows: For 40.00 J of heat in and 10.00 J of work out, the change in internal energy is 12.9 For 25.00 J of heat out and 4.00 J of work in, the change in internal energy is The total change is 12.10 12.11 12.12 No matter whether you look at the overall process or break it into steps, the change in internal energy is the same. WORKED EXAMPLE Calculating Change in Internal Energy: The Same Change in Uis Produced by Two Different Processes What is the change in the internal energy of a system when a total of 150.00 J is transferred by heat from the system and 159.00 J is done by work on the system? STRATEGY The net heat and work are already given, so simply use these values in the equation Solution Here, the net heat and total work are given directly as so that 12.13 Access for free at openstax.org. Discussion 12.2 • First law of Thermodynamics: Thermal Energy and Work 365 Figure 12.8 Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work takes out a total of 6.00 J. The change in internal energy is ΔU = Q – W = 9.00 J. (b) Heat transfer removes 150.00 J from the system while work puts 159.00 J into it, producing an increase of 9.00 J in internal energy. If the system starts out in the same state in (a) and (b), it will end up in the same final state in either case—its final state is related to internal energy, not how that energy was acquired. A very different process in this second worked example produces the same 9.00 J change in internal energy as in the first worked example. Note that the change in the system in both parts is related to system ends up in the samestate in both problems. Note that, as usual, in Figure 12.8 above, and and not to the individual Q’s or W’s involved. The is work done onthe system. is work done bythe system, Practice Problems 3. What is the pressure-volume work done by a system if a pressure of causes a change in volume of ? a. b. c. d. 4. What is the net heat out of the system when is transferred by heat into the system and is transferred out of it? a. b. c. d. 366 Chapter 12 • Thermodynamics Check Your Understanding 5. What is pressure? a. Pressure is force divided by length. b. Pressure is force divided by area. c. Pressure is force divided by volume. d. Pressure is force divided by mass. 6. What is the SI unit for pressure? a. pascal, or N/m3 coulomb b. c. newton d. pascal, or N/m2 7. What is pressure-volume work? a. b. c. d. It is the work that is done by the compression or expansion of a fluid. It is the work that is done by a force on an object to produce a certain displacement. It is the work that is done by the surface molecules of a fluid. It is the work that is done by the high-energy molecules of a fluid. 8. When is pressure-volume work said to be done ON a system? a. When there is an increase in both volume and internal pressure. b. When there is a decrease in both volume and internal pressure. c. When there is a decrease in volume and an increase in internal pressure. d. When there is an increase in volume and a decrease in internal pressure. 9. What are the ways to add energy to or remove energy from a system? a. Transferring energy by heat is the only way to add energy to or remove energy from a system. b. Doing compression work is the only way to add energy to or remove energy from a system. c. Doing expansion work is the only way to add energy to or remove energy from a system. d. Transferring energy by heat or by doing work are the ways to add energy to or remove energy from a system. 10. What is internal energy? a. b. c. d. It is the sum of the kinetic energies of a system’s atoms and molecules. It is the sum of the potential energies of a system’s atoms and molecules. It is the sum of the kinetic and potential energies of a system’s atoms and molecules. It is the difference between the magnitudes of the kinetic and potential energies of a system’s atoms and molecules. 12.3 Second Law of Thermodynamics: Entropy Section Learning Objectives By the end o
f this section, you will be able to do the following: • Describe entropy • Describe the second law of thermodynamics • Solve problems involving the second law of thermodynamics Section Key Terms entropy second law of thermodynamics Entropy Recall from the chapter introduction that it is not even theoretically possible for engines to be 100 percent efficient. This phenomenon is explained by the second law of thermodynamics, which relies on a concept known as entropy. Entropy is a measure of the disorder of a system. Entropy also describes how much energy is notavailable to do work. The more disordered a system and higher the entropy, the less of a system's energy is available to do work. Access for free at openstax.org. 12.3 • Second Law of Thermodynamics: Entropy 367 Although all forms of energy can be used to do work, it is not possible to use the entire available energy for work. Consequently, not all energy transferred by heat can be converted into work, and some of it is lost in the form of waste heat—that is, heat that does not go toward doing work. The unavailability of energy is important in thermodynamics; in fact, the field originated from efforts to convert heat to work, as is done by engines. The equation for the change in entropy, , is where Qis the heat that transfers energy during a process, and Tis the absolute temperature at which the process takes place. Qis positive for energy transferred intothe system by heat and negative for energy transferred out ofthe system by heat. In SI, entropy is expressed in units of joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take Tto be the average temperature in order to avoid trickier math (calculus). TIPS FOR SUCCESS Absolute temperature is the temperature measured in Kelvins. The Kelvin scale is an absolute temperature scale that is measured in terms of the number of degrees above absolute zero. All temperatures are therefore positive. Using temperatures from another, nonabsolute scale, such as Fahrenheit or Celsius, will give the wrong answer. Second Law of Thermodynamics Have you ever played the card game 52 pickup? If so, you have been on the receiving end of a practical joke and, in the process, learned a valuable lesson about the nature of the universe as described by the second law of thermodynamics. In the game of 52 pickup, the prankster tosses an entire deck of playing cards onto the floor, and you get to pick them up. In the process of picking up the cards, you may have noticed that the amount of work required to restore the cards to an orderly state in the deck is much greater than the amount of work required to toss the cards and create the disorder. The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases.An important implication of this law is that heat transfers energy spontaneously from higher- to lower-temperature objects, but never spontaneously in the reverse direction. This is because entropy increases for heat transfer of energy from hot to cold (Figure 12.9). Because the change in entropy is Q/T, there is a larger change in at lower temperatures (smaller T). The decrease in entropy of the hot (larger T) object is therefore less than the increase in entropy of the cold (smaller T) object, producing an overall increase in entropy for the system. Figure 12.9 The ice in this drink is slowly melting. Eventually, the components of the liquid will reach thermal equilibrium, as predicted by the second law of thermodynamics—that is, after heat transfers energy from the warmer liquid to the colder ice. (Jon Sullivan, PDPhoto.org) Another way of thinking about this is that it is impossible for any process to have, as its sole result, heat transferring energy from a cooler to a hotter object. Heat cannot transfer energy spontaneously from colder to hotter, because the entropy of the 368 Chapter 12 • Thermodynamics overall system would decrease. Suppose we mix equal masses of water that are originally at two different temperatures, say will be water at an intermediate temperature of has become unavailable to do work, and the system has become less orderly. Let us think about each of these results. . The result . Three outcomes have resulted: entropy has increased, some energy and First, why has entropy increased? Mixing the two bodies of water has the same effect as the heat transfer of energy from the higher-temperature substance to the lower-temperature substance. The mixing decreases the entropy of the hotter water but increases the entropy of the colder water by a greater amount, producing an overall increase in entropy. Second, once the two masses of water are mixed, there is no more temperature difference left to drive energy transfer by heat and therefore to do work. The energy is still in the water, but it is now unavailableto do work. Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have a single mass with a broad distribution of molecular speeds, the average of which yields an intermediate temperature. These three results—entropy, unavailability of energy, and disorder—not only are related but are, in fact, essentially equivalent. Heat transfer of energy from hot to cold is related to the tendency in nature for systems to become disordered and for less energy to be available for use as work. Based on this law, what cannot happen? A cold object in contact with a hot one never spontaneously transfers energy by heat to the hot object, getting colder while the hot object gets hotter. Nor does a hot, stationary automobile ever spontaneously cool off and start moving. Another example is the expansion of a puff of gas introduced into one corner of a vacuum chamber. The gas expands to fill the chamber, but it never regroups on its own in the corner. The random motion of the gas molecules could take them all back to the corner, but this is never observed to happen (Figure 12.10). Access for free at openstax.org. 12.3 • Second Law of Thermodynamics: Entropy 369 Figure 12.10 Examples of one-way processes in nature. (a) Heat transfer occurs spontaneously from hot to cold, but not from cold to hot. (b) The brakes of this car convert its kinetic energy to increase their internal energy (temperature), and heat transfers this energy to the environment. The reverse process is impossible. (c) The burst of gas released into this vacuum chamber quickly expands to uniformly fill every part of the chamber. The random motions of the gas molecules will prevent them from returning altogether to the corner. We've explained that heat never transfers energy spontaneously from a colder to a hotter object. The key word here is spontaneously. If we do workon a system, it ispossible to transfer energy by heat from a colder to hotter object. We'll learn more about this in the next section, covering refrigerators as one of the applications of the laws of thermodynamics. Sometimes people misunderstand the second law of thermodynamics, thinking that based on this law, it is impossible for entropy to decrease at any particular location. But, it actually ispossible for the entropy of one partof the universe to decrease, as long as the total change in entropy of the universe increases. In equation form, we can write this as Based on this equation, we see that can be negative as long as is positive and greater in magnitude. How is it possible for the entropy of a system to decrease? Energy transfer is necessary. If you pick up marbles that are scattered about the room and put them into a cup, your work has decreased the entropy of that system. If you gather iron ore from the ground and convert it into steel and build a bridge, your work has decreased the entropy of that system. Energy coming from the sun can decrease the entropy of local systems on Earth—that is, universe increases by a greater amount—that is, although you made the system of the bridge and steel more structured, you did so at the expense of the universe. Altogether, the entropy of the universe is increased by the disorder created by digging up the ore and converting it to steel. Therefore, is positive and greater in magnitude. In the case of the iron ore, is negative. But the overall entropy of the rest of the 12.14 370 Chapter 12 • Thermodynamics and the second law of thermodynamics is notviolated. Every time a plant stores some solar energy in the form of chemical potential energy, or an updraft of warm air lifts a soaring bird, Earth experiences local decreases in entropy as it uses part of the energy transfer from the sun into deep space to do work. There is a large total increase in entropy resulting from this massive energy transfer. A small part of this energy transfer by heat is stored in structured systems on Earth, resulting in much smaller, local decreases in entropy. Solving Problems Involving the Second Law of Thermodynamics Entropy is related not only to the unavailability of energy to do work; it is also a measure of disorder. For example, in the case of a melting block of ice, a highly structured and orderly system of water molecules changes into a disorderly liquid, in which molecules have no fixed positions (Figure 12.11). There is a large increase in entropy for this process, as we'll see in the following worked example. Figure 12.11 These ice floes melt during the Arctic summer. Some of them refreeze in the winter, but the second law of thermodynamics predicts that it would be extremely unlikely for the water molecules contained in these particular floes to reform in the distinctive alligator- like shape they possessed when this picture was taken in the summer of 2009. (Patrick Kelley, U.S. Coast Guard, U.S. Geological Survey) WORKED EXAMPLE Entropy
Associated with Disorder Find the increase in entropy of 1.00 kg of ice that is originally at STRATEGY The change in entropy can be calculated from the definition of and melts to form water at . once we find the energy, Q, needed to melt the ice. Solution The change in entropy is defined as Here, Qis the heat necessary to melt 1.00 kg of ice and is given by where mis the mass and is the latent heat of fusion. for water, so Because Qis the amount of energy heat adds to the ice, its value is positive, and Tis the melting temperature of ice, So the change in entropy is 12.15 12.16 12.17 12.18 Access for free at openstax.org. Discussion 12.3 • Second Law of Thermodynamics: Entropy 371 Figure 12.12 When ice melts, it becomes more disordered and less structured. The systematic arrangement of molecules in a crystal structure is replaced by a more random and less orderly movement of molecules without fixed locations or orientations. Its entropy increases because heat transfer occurs into it. Entropy is a measure of disorder. The change in entropy is positive, because heat transfers energy intothe ice to cause the phase change. This is a significant increase in entropy, because it takes place at a relatively low temperature. It is accompanied by an increase in the disorder of the water molecules. Practice Problems are added by heat to water at , what is the change in entropy? 11. If a. b. c. d. 12. What is the increase in entropy when of ice at melt to form water at ? a. b. c. d. Check Your Understanding 13. What is entropy? a. Entropy is a measure of the potential energy of a system. b. Entropy is a measure of the net work done by a system. c. Entropy is a measure of the disorder of a system. d. Entropy is a measure of the heat transfer of energy into a system. 14. Which forms of energy can be used to do work? a. Only work is able to do work. b. Only heat is able to do work. c. Only internal energy is able to do work. d. Heat, work, and internal energy are all able to do work. 15. What is the statement for the second law of thermodynamics? a. All the spontaneous processes result in decreased total entropy of a system. b. All the spontaneous processes result in increased total entropy of a system. c. All the spontaneous processes result in decreased or constant total entropy of a system. d. All the spontaneous processes result in increased or constant total entropy of a system. 16. For heat transferring energy from a high to a low temperature, what usually happens to the entropy of the whole system? It decreases. It must remain constant. a. b. c. The entropy of the system cannot be predicted without specific values for the temperatures. 372 Chapter 12 • Thermodynamics d. It increases. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators Section Learning Objectives By the end of this section, you will be able to do the following: • Explain how heat engines, heat pumps, and refrigerators work in terms of the laws of thermodynamics • Describe thermal efficiency • Solve problems involving thermal efficiency Section Key Terms cyclical process heat engine heat pump thermal efficiency Heat Engines, Heat Pumps, and Refrigerators In this section, we’ll explore how heat engines, heat pumps, and refrigerators operate in terms of the laws of thermodynamics. One of the most important things we can do with heat is to use it to do work for us. A heat engine does exactly this—it makes use of the properties of thermodynamics to transform heat into work. Gasoline and diesel engines, jet engines, and steam turbines that generate electricity are all examples of heat engines. Figure 12.13 illustrates one of the ways in which heat transfers energy to do work. Fuel combustion releases chemical energy that heat transfers throughout the gas in a cylinder. This increases the gas temperature, which in turn increases the pressure of the gas and, therefore, the force it exerts on a movable piston. The gas does work on the outside world, as this force moves the piston through some distance. Thus, heat transfer of energy to the gas in the cylinder results in work being done. Access for free at openstax.org. 12.4 • Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 373 Figure 12.13 (a) Heat transfer to the gas in a cylinder increases the internal energy of the gas, creating higher pressure and temperature. (b) The force exerted on the movable cylinder does work as the gas expands. Gas pressure and temperature decrease during expansion, indicating that the gas’s internal energy has decreased as it does work. (c) Heat transfer of energy to the environment further reduces pressure in the gas, so that the piston can more easily return to its starting position. To repeat this process, the piston needs to be returned to its starting point. Heat now transfers energy from the gas to the surroundings, so that the gas’s pressure decreases, and a force is exerted by the surroundings to push the piston back through some distance. A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. All heat engines use cyclical processes. , from the high-temperature object (or hot reservoir), whereas heat transfers unused energy, Heat engines do work by using part of the energy transferred by heat from some source. As shown in Figure 12.14, heat transfers energy, temperature object (or cold reservoir), and the work done by the engine is W. In physics, a reservoiris defined as an infinitely large mass that can take in or put out an unlimited amount of heat, depending upon the needs of the system. The temperature of the hot reservoir is and the temperature of the cold reservoir is , into the low- . 374 Chapter 12 • Thermodynamics Figure 12.14 (a) Heat transfers energy spontaneously from a hot object to a cold one, as is consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the energy transferred by heat to do work. The hot and cold objects are called the hot and cold reservoirs. Qh is the heat out of the hot reservoir, Wis the work output, and Qc is the unused heat into the cold reservoir. As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system’s internal energy, U, is the same at the beginning and end of every cycle—that is, . The first law of thermodynamics states that where Qis the netheat transfer during the cycle, and Wis the network done by the system. The net heat transfer is the energy transferred in by heat from the hot reservoir minus the amount that is transferred out to the cold reservoir ( ). Because there is no change in internal energy for a complete cycle ( ), we have so that Therefore, the net work done by the system equals the net heat into the system, or for a cyclical process. 12.19 12.20 12.21 Because the hot reservoir is heated externally, which is an energy-intensive process, it is important that the work be done as efficiently as possible. In fact, we want Wto equal Unfortunately, this is impossible. According to the second law of thermodynamics, heat engines cannot have perfect conversion of heat into work. Recall that entropy is a measure of the disorder of a system, which is also how much energy is unavailable to do work. The second law of thermodynamics requires that the total entropy of a system either increases or remains constant in that cannot be used for work. The amount of heat rejected to the cold any process. Therefore, there is a minimum amount of reservoir, , the smaller the value of depends upon the efficiency of the heat engine. The smaller the increase in entropy, , and for there to be no heat to the environment (that is, ). , and the more heat energy is available to do work. Heat pumps, air conditioners, and refrigerators utilize heat transfer of energy from low to high temperatures, which is the opposite of what heat engines do. Heat transfers energy into a hot one. This requires work input, W, which produces a transfer of energy by heat. Therefore, the total heat transfer to the hot reservoir is from a cold reservoir and delivers energy 12.22 The purpose of a heat pump is to transfer energy by heat to a warm environment, such as a home in the winter. The great advantage of using a heat pump to keep your home warm rather than just burning fuel in a fireplace or furnace is that a heat pump supplies You only pay for W, and you get an additional heat transfer of much energy is transferred to the heated space as is used to run the heat pump. When you burn fuel to keep warm, you pay for all of it. The disadvantage to a heat pump is that the work input (required by the second law of thermodynamics) is sometimes comes from the outside air, even at a temperature below freezing, to the indoor space. from the outside at no cost. In many cases, at least twice as . Heat Access for free at openstax.org. 12.4 • Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 375 more expensive than simply burning fuel, especially if the work is provided by electrical energy. The basic components of a heat pump are shown in Figure 12.15. A working fluid, such as a refrigerant, is used. In the outdoor coils (the evaporator), heat enters the working fluid from the cold outdoor air, turning it into a gas. Figure 12.15 A simple heat pump has four basic components: (1) an evaporator, (2) a compressor, (3) a condenser, and (4) an expansion valve. In the heating mode, heat transfers to the working fluid in the evaporator (1) from the colder, outdoor air, turning it into a gas. The electrically driven compressor (2) increases the temperature and pressure of the gas and forces it into the condenser coils (3) inside the heated space. Because the temperature of the gas is higher than the temperature in the room, heat transfers energy from the gas to the room as the gas condenses into a liquid. The working fluid is
then cooled as it flows back through an expansion valve (4) to the outdoor evaporator coils. The electrically driven compressor (work input W) raises the temperature and pressure of the gas and forces it into the condenser coils that are inside the heated space. Because the temperature of the gas is higher than the temperature inside the room, heat transfers energy to the room, and the gas condenses into a liquid. The liquid then flows back through an expansion (pressure-reducing) valve. The liquid, having been cooled through expansion, returns to the outdoor evaporator coils to resume the cycle. The quality of a heat pump is judged by how much energy is transferred by heat into the warm space ( much input work (W) is required. ) compared with how Figure 12.16 Heat pumps, air conditioners, and refrigerators are heat engines operated backward. Almost every home contains a refrigerator. Most people don’t realize that they are also sharing their homes with a heat pump. Air conditioners and refrigerators are designed to cool substances by transferring energy by heat to a warmer one, where heat is given up. In the case of a refrigerator, heat is moved out of the inside of the fridge into the out of a cool environment 376 Chapter 12 • Thermodynamics surrounding room. For an air conditioner, heat is transferred outdoors from inside a home. Heat pumps are also often used in a reverse setting to cool rooms in the summer. As with heat pumps, work input is required for heat transfer of energy from cold to hot. The quality of air conditioners and refrigerators is judged by how much energy is removed by heat W, is required. So, what is considered the energy benefit in a heat pump, is considered waste heat in a refrigerator. from a cold environment, compared with how much work, Thermal Efficiency In the conversion of energy into work, we are always faced with the problem of getting less out than we put in. The problem is that, in all processes, there is some heat that. A way to quantify how efficiently a machine runs is through a quantity called thermal efficiency. that transfers energy to the environment—and usually a very significant amount at We define thermal efficiency, Eff, to be the ratio of useful energy output to the energy input (or, in other words, the ratio of what we get to what we spend). The efficiency of a heat engine is the output of net work, W, divided by heat-transferred energy, into the engine; that is , An efficiency of 1, or 100 percent, would be possible only if there were no heat to the environment ( ). TIPS FOR SUCCESS All values of heat ( plus or minus sign. For example, and ) are positive; there is no such thing as negative heat. The directionof heat is indicated by a is out of the system, so it is preceded by a minus sign in the equation for net heat. 12.23 Solving Thermal Efficiency Problems WORKED EXAMPLE Daily Work Done by a Coal-Fired Power Station and Its Efficiency A coal-fired power station is a huge heat engine. It uses heat to transfer energy from burning coal to do work to turn turbines, which are used then to generate electricity. In a single day, a large coal power station transfers burning coal and transfers What is the efficiency of the power station? STRATEGY We can use water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators and then condensed back to water to start the cycle again. by heat from by heat into the environment. (a) What is the work done by the power station? (b) to find the work output, W, assuming a cyclical process is used in the power station. In this process, 12.24 12.25 , because is given, and work, W, was calculated in the first part of this Solution Work output is given by Substituting the given values, STRATEGY The efficiency can be calculated with example. Solution Efficiency is given by Access for free at openstax.org. 12.4 • Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 377 The work, W, is found to be , and is given ( ), so the efficiency is 12.26 12.27 Discussion The efficiency found is close to the usual value of 42 percent for coal-burning power stations. It means that fully 59.2 percent of the energy is transferred by heat to the environment, which usually results in warming lakes, rivers, or the ocean near the power station and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants—including plants fired by nuclear fuel, oil, and natural gas—the energy transferred by heat to the environment could be, and sometimes is, used for heating homes or for industrial processes. Practice Problems 17. A heat engine is given by heat and releases by heat to the environment. What is the amount of work done by the system? a. b. c. d. 18. A heat engine takes in 6.0 kJ from heat and produces waste heat of 4.8 kJ. What is its efficiency? a. 25 percent b. 2.50 percent c. 2.00 percent d. 20 percent Check Your Understanding 19. What is a heat engine? a. A heat engine converts mechanical energy into thermal energy. b. A heat engine converts thermal energy into mechanical energy. c. A heat engine converts thermal energy into electrical energy. d. A heat engine converts electrical energy into thermal energy. 20. Give an example of a heat engine. a. A generator b. A battery c. A water pump d. A car engine 21. What is thermal efficiency? a. Thermal efficiency is the ratio of work input to the energy input. b. Thermal efficiency is the ratio of work output to the energy input. c. Thermal efficiency is the ratio of work input to the energy output. d. Thermal efficiency is the ratio of work output to the energy output. 22. What is the mathematical expression for thermal efficiency? a. b. c. d. 378 Chapter 12 • Key Terms KEY TERMS Boltzmann constant constant with the value k= 1.38×10−23 of the gas J/K, which is used in the ideal gas law cyclical process process in which a system is brought back internal energy sum of the kinetic and potential energies of a system’s constituent particles (atoms or molecules) to its original state at the end of every cycle pressure force per unit area perpendicular to the force, entropy measurement of a system's disorder and how much energy is not available to do work in a system states that the change in first law of thermodynamics internal energy of a system equals the net energy transfer by heat intothe system minus the net work done bythe system over which the force acts second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases thermal efficiency ratio of useful energy output to the energy input heat engine machine that uses energy transfer by heat to thermal equilibrium condition in which heat no longer do work heat pump machine that generates the heat transfer of energy from cold to hot transfers energy between two objects that are in contact; the two objects have the same temperature zeroth law of thermodynamics states that if two objects ideal gas law physical law that relates the pressure and volume of a gas to the number of gas molecules or atoms, or number of moles of gas, and the absolute temperature are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object SECTION SUMMARY 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium • Systems are in thermal equilibrium when they have the same temperature. • Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy. • The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. 12.2 First law of Thermodynamics: Thermal Energy and Work • Pressure is the force per unit area over which the force is applied perpendicular to the area. • Thermal expansion is the increase, or decrease, of the size (length, area, or volume) of a body due to a change in temperature. • The ideal gas law relates the pressure and volume of a gas to the number of gas particles (atoms or molecules) and the absolute temperature of the gas. • Heat and work are the two distinct methods of energy transfer. • Heat is energy transferred solely due to a temperature difference. • The first law of thermodynamics is given as , where is the change in internal energy of a system, Qis the net energy transfer into the system by heat (the sum of all transfers by heat into and out of the system), and Wis the net work done by the Access for free at openstax.org. system (the sum of all energy transfers by work out of or into the system). • Both Qand Wrepresent energy in transit; only represents an independent quantity of energy capable of being stored. • The internal energy Uof a system depends only on the state of the system, and not how it reached that state. 12.3 Second Law of Thermodynamics: Entropy • Entropy is a measure of a system's disorder: the greater the disorder, the larger the entropy. • Entropy is also the reduced availability of energy to do work. • The second law of thermodynamics states that, for any spontaneous process, the total entropy of a system either increases or remains constant; it never decreases. • Heat transfers energy spontaneously from higher- to lower-temperature bodies, but never spontaneously in the reverse direction. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators • Heat engines use the heat transfer of energy to do work. • Cyclical processes are processes that return to their original state at the end of every cycle. • The thermal efficiency of a heat engine is the ratio of work output divided by the amount of energy input. • The amount of work a heat engine can do is determined by the net heat transfer of energy during a cycle; more waste hea
t leads to less work output. • Heat pumps draw energy by heat from cold outside air and use it to heat an interior room. • A refrigerator is a type of heat pump; it takes energy KEY EQUATIONS 12.2 First law of Thermodynamics: Thermal Energy and Work Chapter 12 • Key Equations 379 from the warm air from the inside compartment and transfers it to warmer exterior air. 12.3 Second Law of Thermodynamics: Entropy ideal gas law change in entropy first law of thermodynamics pressure pressure–volume work CHAPTER REVIEW Concept Items 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 1. When are two bodies in thermal equilibrium? a. When they are in thermal contact and are at different pressures b. When they are not in thermal contact but are at the same pressure c. When they are not in thermal contact but are at different temperatures d. When they are in thermal contact and are at the same temperature 2. What is thermal contact? a. Two objects are said to be in thermal contact when they are in contact with each other in such a way that the transfer of energy by heat can occur between them. b. Two objects are said to be in thermal contact when they are in contact with each other in such a way that the transfer of energy by mass can occur between them. c. Two objects are said to be in thermal contact when they neither lose nor gain energy by heat. There is no transfer of energy between them. d. Two objects are said to be in thermal contact when they only gain energy by heat. There is transfer of energy between them. 3. To which mathematical property is the zeroth law of 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators thermal efficiency of a heat engine work output for a cyclical process thermodynamics similar? a. Associative property b. Commutative property c. Distributive property d. Transitive property 12.2 First law of Thermodynamics: Thermal Energy and Work 4. Why does thermal expansion occur? a. An increase in temperature causes intermolecular distances to increase. b. An increase in temperature causes intermolecular distances to decrease. c. An increase in temperature causes an increase in the work done on the system. d. An increase in temperature causes an increase in the work done by the system. 5. How does pressure-volume work relate to heat and internal energy of a system? a. The energy added to a system by heat minus the change in the internal energy of that system is equal to the pressure-volume work done by the system. b. The sum of the energy released by a system by heat and the change in the internal energy of that system is equal to the pressure-volume work done by the system. c. The product of the energy added to a system by heat and the change in the internal energy of that system 380 Chapter 12 • Chapter Review d. is equal to the pressure-volume work done by the system. If the energy added to a system by heat is divided by the change in the internal energy of that system, the quotient is equal to the pressure-volume work done by the system. 6. On what does internal energy depend? a. The path of energy changes in the system b. The state of the system c. The size of the system d. The shape of the system 7. The first law of thermodynamics helps us understand the relationships among which three quantities? a. Heat, work, and internal energy b. Heat, work, and external energy c. Heat, work, and enthalpy d. Heat, work, and entropy 12.3 Second Law of Thermodynamics: Entropy 8. Air freshener is sprayed from a bottle. The molecules spread throughout the room and cannot make their way back into the bottle. Why is this the case? a. The entropy of the molecules increases. b. The entropy of the molecules decreases. c. The heat content (enthalpy, or total energy available 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 10. What is the quality by which air conditioners are judged? a. The amount of energy generated by heat from a hot environment, compared with the required work input b. The amount of energy transferred by heat from a cold environment, compared with the required work input c. The amount of energy transferred by heat from a hot environment, compared with the required work output d. The amount of energy transferred by heat from a cold environment, compared with the required work output 11. Why is the efficiency of a heat engine never 100 percent? a. Some energy is always gained by heat from the environment. b. Some energy is always lost by heat to the environment. c. Work output is always greater than energy input. d. Work output is infinite for any energy input. 12. What is a cyclic process? a. A process in which the system returns to its original for heat) of the molecules increases. state at the end of the cycle d. The heat content (enthalpy, or total energy available b. A process in which the system does not return to its for heat) of the molecules decreases. 9. Give an example of entropy as experienced in everyday rotation of Earth formation of a solar eclipse life. a. b. c. filling a car tire with air d. motion of a pendulum bob original state at the end of the cycle c. A process in which the system follows the same path for every cycle d. A process in which the system follows a different path for every cycle Critical Thinking Items 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 13. What are the necessary conditions for energy transfer by heat to occur between two bodies through the process of conduction? a. They should be at the same temperature, and they should be in thermal contact. b. They should be at the same temperature, and they should not be in thermal contact. c. They should be at different temperatures, and they should be in thermal contact. should not be in thermal contact. 14. Oil is heated in a pan on a hot plate. The pan is in thermal equilibrium with the hot plate and also with the oil. The temperature of the hot plate is 150 °C . What is the temperature of the oil? a. b. c. d. 160 °C 150 °C 140 °C 130 °C 12.2 First law of Thermodynamics: Thermal Energy and Work d. They should be at different temperatures, and they 15. When an inflated balloon experiences a decrease in size, Access for free at openstax.org. the air pressure inside the balloon remains nearly constant. If there is no transfer of energy by heat to or from the balloon, what physical change takes place in the balloon? a. The average kinetic energy of the gas particles decreases, so the balloon becomes colder. b. The average kinetic energy of the gas particles increases, so the balloon becomes hotter. c. The average potential energy of the gas particles decreases, so the balloon becomes colder. d. The average potential energy of the gas particles increases, so the balloon becomes hotter. 16. When heat adds energy to a system, what is likely to happen to the pressure and volume of the system? a. Pressure and volume may both decrease with added energy. b. Pressure and volume may both increase with added energy. c. Pressure must increase with added energy, while volume must remain constant. d. Volume must decrease with added energy, while pressure must remain constant. 17. If more energy is transferred into the system by net heat as compared to the net work done by the system, what happens to the difference in energy? a. b. c. d. It is transferred back to its surroundings. It is stored in the system as internal energy. It is stored in the system as potential energy. It is stored in the system as entropy. 18. Air is pumped into a car tire, causing its temperature to increase. In another tire, the temperature increase is due to exposure to the sun. Is it possible to tell what caused the temperature increase in each tire? Explain your answer. a. No, because it is a chemical change, and the cause of that change does not matter; the final state of both systems are the same. b. Although the final state of each system is identical, the source is different in each case. c. No, because the changes in energy for both systems are the same, and the cause of that change does not matter; the state of each system is identical. d. Yes, the changes in the energy for both systems are the same, but the causes of that change are different, so the states of each system are not identical. 19. How does the transfer of energy from the sun help plants? a. Plants absorb solar energy from the sun and utilize it during the fertilization process. b. Plants absorb solar energy from the sun and utilize Chapter 12 • Chapter Review 381 it during the process of photosynthesis to turn it into plant matter. c. Plants absorb solar energy from the sun and utilize it to increase the temperature inside them. d. Plants absorb solar energy from the sun and utilize it during the shedding of their leaves and fruits. 12.3 Second Law of Thermodynamics: Entropy 20. If an engine were constructed to perform such that there would be no losses due to friction, what would be its efficiency? a. b. c. d. It would be 0 percent. It would be less than 100 percent. It would be 100 percent. It would be greater than 100 percent. 21. Entropy never decreases in a spontaneous process. Give an example to support this statement. a. The transfer of energy by heat from colder bodies to hotter bodies is a spontaneous process in which the entropy of the system of bodies increases. b. The melting of an ice cube placed in a room causes an increase in the entropy of the room. c. The dissolution of salt in water is a spontaneous process in which the entropy of the system increases. d. A plant uses energy from the sun and converts it into sugar molecules by the process of photosynthesis. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 22. What is the advantage of a heat pump as opposed to burning fuel (as in a fireplace) for keeping warm? a. A heat pump supplies energy by heat from the cold, outside air. b. A heat pump supplies energy generated by the work done. c. A heat pump supplies e
nergy by heat from the cold, outside air and also from the energy generated by the work done. d. A heat pump supplies energy not by heat from the cold, outside air, nor from the energy generated by the work done, but from more accessible sources. 23. What is thermal efficiency of an engine? Can it ever be 100 percent? Why or why not? a. Thermal efficiency is the ratio of the output (work) to the input (heat). It is always 100 percent. b. Thermal efficiency is the ratio of the output (heat) 382 Chapter 12 • Chapter Review to the input (work). It is always 100 percent. c. Thermal efficiency is the ratio of the output (heat) to the input (work). It is never 100 percent. environment b. When mass transferred to the environment is zero c. When mass transferred to the environment is at a d. Thermal efficiency is the ratio of the output (work) maximum to the input (heat). It is never 100 percent. d. When no energy is transferred by heat to the 24. When would 100 percent thermal efficiency be possible? a. When all energy is transferred by heat to the environment Problems 12.2 First law of Thermodynamics: Thermal Energy and Work 25. Some amount of energy is transferred by heat into a , while . What is the system. The net work done by the system is the increase in its internal energy is amount of net heat? a. b. c. d. 26. Eighty joules are added by heat to a system, while it are added by heat to the of work. What is the change in of work. Later, does system, and it does the system’s internal energy? a. b. c. d. Performance Task 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 29. You have been tasked to design and construct a thermometer that works on the principle of thermal expansion. There are four materials available for you to test, each of which will find use under different sets of conditions and temperature ranges: Materials • Four sample materials with similar mass or volume: copper, steel, water, and alcohol (ethanol or isopropanol) • Oven or similar heating source • Instrument (e.g., meter ruler, Vernier calipers, or micrometer) for measuring changes in dimension • Balance for measuring mass Procedure 1. Design a safe experiment to analyze the thermal Access for free at openstax.org. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 27. A coal power station functions at 40.0 percent efficiency. What is the amount of work it does if it takes in 1.20×1012 J by heat? 3×1010 J a. b. 4.8×1011 J 3×1012 J c. d. 4.8×1013 J 28. A heat engine functions with 70.7 percent thermal efficiency and consumes 12.0 kJ from heat daily. If its efficiency were raised to 75.0 percent, how much energy from heat would be saved daily, while providing the same output? a. −10.8 kJ b. −1.08 kJ c. 0.7 kJ 7 kJ d. expansion properties of each material. 2. Write down the materials needed for your experiment and the procedure you will follow. Make sure that you include every detail so that the experiment can be repeated by others. 3. Select an appropriate material to measure temperature over a predecided temperature range, and give reasons for your choice. 4. Calibrate your instrument to measure temperature changes accurately. a. Which physical quantities are affected by temperature change and thermal expansion? b. How do such properties as specific heat and thermal conductivity affect the use of each material as a thermometer? c. Does a change of phase take place for any of the tested materials over the temperature range to be examined? d. What are your independent and dependent variables for this series of tests? Which variables need to be controlled in the experiment? e. What are your sources of error? f. Can all the tested materials be used effectively in the same ranges of temperature? Which applications might be suitable for one or more of the tested substances but not the others? Chapter 12 • Test Prep 383 TEST PREP Multiple Choice 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 30. Which law of thermodynamics describes thermal equilibrium? a. zeroth b. first c. d. second third 31. Name any two industries in which the principles of thermodynamics are used. a. aerospace and information technology (IT) industries industrial manufacturing and aerospace b. c. mining and textile industries d. mining and agriculture industries 12.2 First law of Thermodynamics: Thermal Energy and Work 32. What is the value of the Boltzmann constant? a. b. c. d. 33. Which of the following involves work done BY a system? increasing internal energy compression a. b. c. expansion cooling d. 34. Which principle does the first law of thermodynamics state? a. b. c. d. the ideal gas law the transitive property of equality the law of conservation of energy the principle of thermal equilibrium a. A real gas behaves like an ideal gas at high temperature and low pressure. b. A real gas behaves like an ideal gas at high temperature and high pressure. c. A real gas behaves like an ideal gas at low temperature and low pressure. d. A real gas behaves like an ideal gas at low temperature and high pressure. 12.3 Second Law of Thermodynamics: Entropy 37. In an engine, what is the unused energy converted into? internal energy a. b. pressure c. work d. heat 38. It is natural for systems in the universe to _____ spontaneously. a. become disordered b. become ordered c. produce heat d. do work 39. If is and is , what is the change in entropy? a. b. c. d. 40. Why does entropy increase during a spontaneous process? a. Entropy increases because energy always transfers spontaneously from a dispersed state to a concentrated state. b. Entropy increases because energy always transfers spontaneously from a concentrated state to a dispersed state. c. Entropy increases because pressure always 35. What is the change in internal energy of a system when increases spontaneously. and ? a. b. c. d. 36. When does a real gas behave like an ideal gas? d. Entropy increases because temperature of any system always increases spontaneously. 41. A system consists of ice melting in a glass of water. What happens to the entropy of this system? a. The entropy of the ice decreases, while the entropy of the water cannot be predicted without more 384 Chapter 12 • Test Prep specific information. b. The entropy of the system remains constant. c. The entropy of the system decreases. d. The entropy of the system increases. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 42. Which equation represents the net work done by a system in a cyclic process? a. b. c. d. 43. Which of these quantities needs to be zero for efficiency to be 100 percent? a. ΔU b. W c. Qh d. Qc 44. Which of the following always has the greatest value in a system having 80 percent thermal efficiency? a. ΔU Short Answer 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 47. What does greenenergy development entail? a. Green energy involves finding new ways to harness clean and renewable alternative energy sources. b. Green energy involves finding new ways to conserve alternative energy sources. b. W c. Qh d. Qc 45. In the equation Q= Qh − Qc, what does the negative sign indicate? a. Heat transfer of energy is always negative. b. Heat transfer can only occur in one direction. c. Heat is directed into the system from the surroundings outside the system. d. Heat is directed out of the system. 46. What is the purpose of a heat pump? a. A heat pump uses work to transfer energy by heat from a colder environment to a warmer environment. b. A heat pump uses work to transfer energy by heat from a warmer environment to a colder environment. c. A heat pump does work by using heat to convey energy from a colder environment to a warmer environment. d. A heat pump does work by using heat to convey energy from a warmer environment to a colder environment. volume, which variable relates to pressure, and what is that relation? a. Temperature; inverse proportionality b. Temperature, direct proportionality to square root c. Temperature; direct proportionality d. Temperature; direct proportionality to square c. Green energy involves decreasing the efficiency of 50. When is volume directly proportional to temperature? nonrenewable energy resources. d. Green energy involves finding new ways to harness nonrenewable energy resources. 48. Why are the sun and Earth not in thermal equilibrium? a. The mass of the sun is much greater than the mass of Earth. b. There is a vast amount of empty space between the sun and Earth. c. The diameter of the sun is much greater than the diameter of Earth. d. The sun is in thermal contact with Earth. 12.2 First law of Thermodynamics: Thermal Energy and Work 49. If a fixed quantity of an ideal gas is held at a constant a. when the pressure of the gas is variable b. when the pressure of the gas is constant c. when the mass of the gas is variable d. when the mass of the gas is constant 51. For fluids, what can work be defined as? a. pressure acting over the change in depth b. pressure acting over the change in temperature c. d. pressure acting over the change in volume temperature acting over the change in volume , what does 52. In the equation indicate? a. b. c. d. the work done on the system the work done by the system the heat into the system the heat out of the system Access for free at openstax.org. Chapter 12 • Test Prep 385 53. By convention, if Qis positive, what is the direction in which heat transfers energy with regard to the system? a. The direction of the heat transfer of energy depends on the changes in W, regardless of the sign of Q. a. Entropy depends on the change of phase of a system, but not on any other state conditions. b. Entropy does not depend on how the final state is reached from the initial state. c. Entropy is least when the path between the initial b. The direction of Qcannot be determined from just state and the final state is the shortest. the sign of Q. d. Entropy is least when the path between the initial
c. The direction of net heat transfer of energy will be state and the final state is the longest. the work was done 61. What is the change in entropy caused by melting 5.00 kg out of the system. d. The direction of net heat transfer of energy will be into the system. 54. What is net transfer of energy by heat? a. b. c. d. It is the sum of all energy transfers by heat into the system. It is the product of all energy transfers by heat into the system. It is the sum of all energy transfers by heat into and out of the system. It is the product of all energy transfers by heat into and out of the system. 55. Three hundred ten joules of heat enter a system, after which the system does of work. What is the change in its internal energy? Would this amount change if the energy transferred by heat were added after the work was done instead of before? a. ; this would change if heat added energy after the work was done ; this would change if heat added energy after ; this would not change even if heat added energy after the work was done ; this would not change even if heat added energy after the work was done 56. Ten joules are transferred by heat into a system, . What is the change in the followed by another system’s internal energy? What would be the difference in this change if the system at once? a. ; the change in internal energy would be same of energy were added by heat to even if the heat added the energy at once ; the change in internal energy would be same even if the heat added the energy at once ; the change in internal energy would be more if the heat added the energy at once ; the change in internal energy would be more b. c. d. if the heat added the energy at once 12.3 Second Law of Thermodynamics: Entropy 57. How does the entropy of a system depend on how the system reaches a given state? b. c. d. 58. Which sort of thermal energy do molecules in a solid possess? a. electric potential energy b. gravitational potential energy c. translational kinetic energy d. vibrational kinetic energy 59. A cold object in contact with a hot one never spontaneously transfers energy by heat to the hot object. Which law describes this phenomenon? the first law of thermodynamics a. the second law of thermodynamics b. the third law of thermodynamics c. the zeroth law of thermodynamics d. 60. How is it possible for us to transfer energy by heat from cold objects to hot ones? a. by doing work on the system b. by having work done by the system c. by increasing the specific heat of the cold body d. by increasing the specific heat of the hot body of ice at 0 °C ? a. 0 J/K b. 6.11×103 J/K c. 6.11×104 J/K d. ∞J/K 62. What is the amount of heat required to cause a change in the entropy of a system at ? of a. b. c. d. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 63. In a refrigerator, what is the function of an evaporator? a. The evaporator converts gaseous refrigerant into liquid. b. The evaporator converts solid refrigerant into liquid. c. The evaporator converts solid refrigerant into gas. 386 Chapter 12 • Test Prep d. The evaporator converts liquid refrigerant into gas. 64. Which component of an air conditioner converts gas into liquid? a. b. c. d. the condenser the compressor the evaporator the thermostat 65. What is one example for which calculating thermal efficiency is of interest? a. A wind turbine b. An electric pump c. A bicycle d. A car engine 66. How is the efficiency of a refrigerator or heat pump expressed? a. Extended Response 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium 69. What is the meaning of efficiency in terms of a car engine? a. An engine’s efficiency equals the sum of useful energy (work) and the input energy. b. An engine’s efficiency equals the proportion of useful energy (work) to the input energy. c. An engine’s efficiency equals the product of useful energy (work) and the input energy. d. An engine’s efficiency equals the difference between the useful energy (work) and the input energy. 12.2 First law of Thermodynamics: Thermal Energy and Work 70. Why does a bridge have expansion joints? a. because the bridge expands and contracts with the change in temperature b. because the bridge expands and contracts with the change in motion of objects moving on the bridge c. because the bridge expands and contracts with the change in total load on the bridge d. because the bridge expands and contracts with the change in magnitude of wind blowing 71. Under which conditions will the work done by the gas in a system increase? a. It will increase when a large amount of energy is added to the system, and that energy causes an increase in the gas’s volume, its pressure, or both. Access for free at openstax.org. b. c. d. 67. How can you mathematically express thermal efficiency in terms of and ? a. b. c. d. 68. How can you calculate percentage efficiency? a. percentage efficiency b. percentage efficiency c. percentage efficiency d. percentage efficiency b. c. d. It will increase when a large amount of energy is extracted from the system, and that energy causes an increase in the gas’s volume, its pressure, or both. It will increase when a large amount of energy is added to the system, and that energy causes a decrease in the gas’s volume, its pressure, or both. It will increase when a large amount of energy is extracted from the system, and that energy causes a decrease in the gas’s volume, its pressure, or both. 72. How does energy transfer by heat aid in body metabolism? a. The energy is given to the body through the work done by the body (W) and through the intake of food, which may also be considered as the work done on the body. The transfer of energy out of the body is by heat (−Q) . b. The energy given to the body is by the intake of food, which may also be considered as the work done on the body. The transfer of energy out of the body is by heat (−Q) and the work done by the body (W) . c. The energy given to the body is by the transfer of energy by heat (Q) into the body, which may also be considered as the work done on the body. The transfer of energy out of the body is the work done by the body (W) . d. The energy given to the body is by the transfer of energy by heat (Q) inside the body. The transfer of energy out of the body is by the intake of food and the work done by the body (W) . 73. Two distinct systems have the same amount of stored internal energy. Five hundred joules are added by heat to the first system, and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy? a. b. c. d. 700 J; 0 J 300 J; 300 J 700 J; 300 J 300 J; 0 J 12.3 Second Law of Thermodynamics: Entropy 74. Why is it not possible to convert all available energy into work? a. Due to the entropy of a system, some energy is always unavailable for work. b. Due to the entropy of a system, some energy is always available for work. c. Due to the decrease in internal energy of a system, some energy is always made unavailable for work. d. Due to the increase in internal energy of a system, some energy is always made unavailable for work. 75. Why does entropy increase when ice melts into water? a. Melting converts the highly ordered solid structure into a disorderly liquid, thereby increasing entropy. b. Melting converts the highly ordered liquid into a disorderly solid structure, thereby increasing entropy. c. Melting converts the highly ordered solid structure into a disorderly solid structure, thereby increasing entropy. d. Melting converts the highly ordered liquid into a disorderly liquid, thereby increasing entropy. 76. Why is change in entropy lower for higher temperatures? a. Increase in the disorder in the substance is low for high temperature. Chapter 12 • Test Prep 387 b. Increase in the disorder in the substance is high for high temperature. c. Decrease in the disorder in the substance is low for high temperature. d. Decrease in the disorder in the substance is high for high temperature. 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators 77. In the equation W= Qh − Qc, if the value of Qc were equal to zero, what would it signify? a. The efficiency of the engine is 75 percent. b. The efficiency of the engine is 25 percent. c. The efficiency of the engine is 0 percent. d. The efficiency of the engine is 100 percent. 78. Can the value of thermal efficiency be greater than 1? Why or why not? a. No, according to the first law of thermodynamics, energy output can never be more than the energy input. b. No, according to the second law of thermodynamics, energy output can never be more than the energy input. c. Yes, according to the first law of thermodynamics, energy output can be more than the energy input. d. Yes, according to the second law of thermodynamics, energy output can be more than the energy input. 79. A coal power station transfers 3.0×1012 J by heat from burning coal and transfers 1.5×1012 J by heat into the environment. What is the efficiency of the power station? a. 0.33 b. 0.5 c. 0.66 1 d. 388 Chapter 12 • Test Prep Access for free at openstax.org. CHAPTER 13 Waves and Their Properties Figure 13.1 Waves in the ocean behave similarly to all other types of waves. (Steve Jurveston, Flickr) Chapter Outline 13.1 Types of Waves 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 13.3 Wave Interaction: Superposition and Interference Recall from the chapter on Motion in Two Dimensions that oscillations—the back-and-forth movement INTRODUCTION between two points—involve force and energy. Some oscillations create waves, such as the sound waves created by plucking a guitar string. Other examples of waves include earthquakes and visible light. Even subatomic particles, such as electrons, can behave like waves. You can make water waves in a swimming pool by slapping the water with your hand.
Some of these waves, such as water waves, are visible; others, such as sound waves, are not. But every wave is a disturbance that moves from its source and carries energy. In this chapter, we will learn about the different types of waves, their properties, and how they interact with one another. 390 Chapter 13 • Waves and Their Properties 13.1 Types of Waves Section Learning Objectives By the end of this section, you will be able to do the following: • Define mechanical waves and medium, and relate the two • Distinguish a pulse wave from a periodic wave • Distinguish a longitudinal wave from a transverse wave and give examples of such waves Section Key Terms Mechanical Waves longitudinal wave mechanical wave medium wave periodic wave pulse wave transverse wave What do we mean when we say something is a wave? A wave is a disturbance that travels or propagatesfrom the place where it was created. Waves transfer energy from one place to another, but they do not necessarily transfer any mass. Light, sound, and waves in the ocean are common examples of waves. Sound and water waves are mechanical waves; meaning, they require a medium to travel through. The medium may be a solid, a liquid, or a gas, and the speed of the wave depends on the material properties of the medium through which it is traveling. However, light is not a mechanical wave; it can travel through a vacuum such as the empty parts of outer space. A familiar wave that you can easily imagine is the water wave. For water waves, the disturbance is in the surface of the water, an example of which is the disturbance created by a rock thrown into a pond or by a swimmer splashing the water surface repeatedly. For sound waves, the disturbance is caused by a change in air pressure, an example of which is when the oscillating cone inside a speaker creates a disturbance. For earthquakes, there are several types of disturbances, which include the disturbance of Earth’s surface itself and the pressure disturbances under the surface. Even radio waves are most easily understood using an analogy with water waves. Because water waves are common and visible, visualizing water waves may help you in studying other types of waves, especially those that are not visible. Water waves have characteristics common to all waves, such as amplitude, period, frequency, and energy, which we will discuss in the next section. Pulse Waves and Periodic Waves If you drop a pebble into the water, only a few waves may be generated before the disturbance dies down, whereas in a wave pool, the waves are continuous. A pulse wave is a sudden disturbance in which only one wave or a few waves are generated, such as in the example of the pebble. Thunder and explosions also create pulse waves. A periodic wave repeats the same oscillation for several cycles, such as in the case of the wave pool, and is associated with simple harmonic motion. Each particle in the medium experiences simple harmonic motion in periodic waves by moving back and forth periodically through the same positions. Consider the simplified water wave in Figure 13.2. This wave is an up-and-down disturbance of the water surface, characterized by a sine wave pattern. The uppermost position is called the crestand the lowest is the trough. It causes a seagull to move up and down in simple harmonic motion as the wave crests and troughs pass under the bird. Figure 13.2 An idealized ocean wave passes under a seagull that bobs up and down in simple harmonic motion. Access for free at openstax.org. 13.1 • Types of Waves 391 Longitudinal Waves and Transverse Waves Mechanical waves are categorized by their type of motion and fall into any of two categories: transverse or longitudinal. Note that both transverse and longitudinal waves can be periodic. A transverse wave propagates so that the disturbance is perpendicular to the direction of propagation. An example of a transverse wave is shown in Figure 13.3, where a woman moves a toy spring up and down, generating waves that propagate away from herself in the horizontal direction while disturbing the toy spring in the vertical direction. Figure 13.3 In this example of a transverse wave, the wave propagates horizontally and the disturbance in the toy spring is in the vertical direction. In contrast, in a longitudinal wave, the disturbance is parallel to the direction of propagation. Figure 13.4 shows an example of a longitudinal wave, where the woman now creates a disturbance in the horizontal direction—which is the same direction as the wave propagation—by stretching and then compressing the toy spring. Figure 13.4 In this example of a longitudinal wave, the wave propagates horizontally and the disturbance in the toy spring is also in the horizontal direction. TIPS FOR SUCCESS Longitudinal waves are sometimes called compression wavesor compressional waves, and transverse waves are sometimes called shear waves. Waves may be transverse, longitudinal, or a combination of the two. The waves on the strings of musical instruments are transverse (as shown in Figure 13.5), and so are electromagnetic waves, such as visible light. Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. 392 Chapter 13 • Waves and Their Properties Figure 13.5 The wave on a guitar string is transverse. However, the sound wave coming out of a speaker rattles a sheet of paper in a direction that shows that such sound wave is longitudinal. Sound in solids can be both longitudinal and transverse. Essentially, water waves are also a combination of transverse and longitudinal components, although the simplified water wave illustrated in Figure 13.2 does not show the longitudinal motion of the bird. Earthquake waves under Earth’s surface have both longitudinal and transverse components as well. The longitudinal waves in an earthquake are called pressure or P-waves, and the transverse waves are called shear or S-waves. These components have important individual characteristics; for example, they propagate at different speeds. Earthquakes also have surface waves that are similar to surface waves on water. WATCH PHYSICS Introduction to Waves This video explains wave propagation in terms of momentum using an example of a wave moving along a rope. It also covers the differences between transverse and longitudinal waves, and between pulse and periodic waves. Click to view content (https://openstax.org/l/02introtowaves) GRASP CHECK In a longitudinal sound wave, after a compression wave moves through a region, the density of molecules briefly decreases. Why is this? a. After a compression wave, some molecules move forward temporarily. b. After a compression wave, some molecules move backward temporarily. c. After a compression wave, some molecules move upward temporarily. d. After a compression wave, some molecules move downward temporarily. FUN IN PHYSICS The Physics of Surfing Many people enjoy surfing in the ocean. For some surfers, the bigger the wave, the better. In one area off the coast of central California, waves can reach heights of up to 50 feet in certain times of the year (Figure 13.6). Access for free at openstax.org. 13.1 • Types of Waves 393 Figure 13.6 A surfer negotiates a steep take-off on a winter day in California while his friend watches. (Ljsurf, Wikimedia Commons) How do waves reach such extreme heights? Other than unusual causes, such as when earthquakes produce tsunami waves, most huge waves are caused simply by interactions between the wind and the surface of the water. The wind pushes up against the surface of the water and transfers energy to the water in the process. The stronger the wind, the more energy transferred. As waves start to form, a larger surface area becomes in contact with the wind, and even more energy is transferred from the wind to the water, thus creating higher waves. Intense storms create the fastest winds, kicking up massive waves that travel out from the origin of the storm. Longer-lasting storms and those storms that affect a larger area of the ocean create the biggest waves since they transfer more energy. The cycle of the tides from the Moon’s gravitational pull also plays a small role in creating waves. Actual ocean waves are more complicated than the idealized model of the simple transverse wave with a perfect sinusoidal shape. Ocean waves are examples of orbital progressive waves, where water particles at the surface follow a circular path from the crest to the trough of the passing wave, then cycle back again to their original position. This cycle repeats with each passing wave. As waves reach shore, the water depth decreases and the energy of the wave is compressed into a smaller volume. This creates higher waves—an effect known as shoaling. Since the water particles along the surface move from the crest to the trough, surfers hitch a ride on the cascading water, gliding along the surface. If ocean waves work exactly like the idealized transverse waves, surfing would be much less exciting as it would simply involve standing on a board that bobs up and down in place, just like the seagull in the previous figure. Additional information and illustrations about the scientific principles behind surfing can be found in the “Using Science to Surf Better!” (http://www.openstax.org/l/28Surf) video. GRASP CHECK If we lived in a parallel universe where ocean waves were longitudinal, what would a surfer’s motion look like? a. The surfer would move side-to-side/back-and-forth vertically with no horizontal motion. b. The surfer would forward and backward horizontally with no vertical motion. Check Your Understanding 1. What is a wave? a. A wave is a force that propagates from the place where it was created. b. A wave is a disturbance that propagates from the place where it was created. c. A wave is matter that provides volume to an object. d. A wave is matter that provides mass to an object. 2. Do all waves requir
e a medium to travel? Explain. a. No, electromagnetic waves do not require any medium to propagate. b. No, mechanical waves do not require any medium to propagate. c. Yes, both mechanical and electromagnetic waves require a medium to propagate. d. Yes, all transverse waves require a medium to travel. 3. What is a pulse wave? a. A pulse wave is a sudden disturbance with only one wave generated. b. A pulse wave is a sudden disturbance with only one or a few waves generated. 394 Chapter 13 • Waves and Their Properties c. A pulse wave is a gradual disturbance with only one or a few waves generated. d. A pulse wave is a gradual disturbance with only one wave generated. 4. Is the following statement true or false? A pebble dropped in water is an example of a pulse wave. a. False b. True 5. What are the categories of mechanical waves based on the type of motion? a. Both transverse and longitudinal waves b. Only longitudinal waves c. Only transverse waves d. Only surface waves 6. In which direction do the particles of the medium oscillate in a transverse wave? a. Perpendicular to the direction of propagation of the transverse wave b. Parallel to the direction of propagation of the transverse wave 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period Section Learning Objectives By the end of this section, you will be able to do the following: • Define amplitude, frequency, period, wavelength, and velocity of a wave • Relate wave frequency, period, wavelength, and velocity • Solve problems involving wave properties Section Key Terms wavelength wave velocity Wave Variables In the chapter on motion in two dimensions, we defined the following variables to describe harmonic motion: • Amplitude—maximum displacement from the equilibrium position of an object oscillating around such equilibrium position • Frequency—number of events per unit of time • Period—time it takes to complete one oscillation For waves, these variables have the same basic meaning. However, it is helpful to word the definitions in a more specific way that applies directly to waves: • Amplitude—distance between the resting position and the maximum displacement of the wave • Frequency—number of waves passing by a specific point per second • Period—time it takes for one wave cycle to complete In addition to amplitude, frequency, and period, their wavelength and wave velocity also characterize waves. The wavelength is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. The wave velocity is the speed at which the disturbance moves. TIPS FOR SUCCESS Wave velocity is sometimes also called the propagation velocityor propagation speedbecause the disturbance propagates from one location to another. Consider the periodic water wave in Figure 13.7. Its wavelength is the distance from crest to crest or from trough to trough. The wavelength can also be thought of as the distance a wave has traveled after one complete cycle—or one period. The time for one complete up-and-down motion is the simple water wave’s period T. In the figure, the wave itself moves to the right with a wave Access for free at openstax.org. 13.2 • Wave Properties: Speed, Amplitude, Frequency, and Period 395 velocity vw. Its amplitude Xis the distance between the resting position and the maximum displacement—either the crest or the trough—of the wave. It is important to note that this movement of the wave is actually the disturbancemoving to the right, not the water itself; otherwise, the bird would move to the right. Instead, the seagull bobs up and down in place as waves pass underneath, traveling a total distance of 2Xin one cycle. However, as mentioned in the text feature on surfing, actual ocean waves are more complex than this simplified example. Figure 13.7 The wave has a wavelength λ, which is the distance between adjacent identical parts of the wave. The up-and-down disturbance of the surface propagates parallel to the surface at a speed vw. WATCH PHYSICS Amplitude, Period, Frequency, and Wavelength of Periodic Waves This video is a continuation of the video “Introduction to Waves” from the "Types of Waves" section. It discusses the properties of a periodic wave: amplitude, period, frequency, wavelength, and wave velocity. Click to view content (https://www.openstax.org/l/28wavepro) TIPS FOR SUCCESS The crest of a wave is sometimes also called the peak. GRASP CHECK If you are on a boat in the trough of a wave on the ocean, and the wave amplitude is position? a. b. c. d. , what is the wave height from your The Relationship between Wave Frequency, Period, Wavelength, and Velocity Since wave frequency is the number of waves per second, and the period is essentially the number of seconds per wave, the relationship between frequency and period is or 13.1 13.2 just as in the case of harmonic motion of an object. We can see from this relationship that a higher frequency means a shorter period. Recall that the unit for frequency is hertz (Hz), and that 1 Hz is one cycle—or one wave—per second. The speed of propagation vw is the distance the wave travels in a given time, which is one wavelength in a time of one period. In 396 Chapter 13 • Waves and Their Properties equation form, it is written as or 13.3 13.4 From this relationship, we see that in a medium where vw is constant, the higher the frequency, the smaller the wavelength. See Figure 13.8. Figure 13.8 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high- frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, while the higher-frequency sounds are emitted by the small speaker, called a tweeter. These fundamental relationships hold true for all types of waves. As an example, for water waves, vw is the speed of a surface wave; for sound, vw is the speed of sound; and for visible light, vw is the speed of light. The amplitude Xis completely independent of the speed of propagation vw and depends only on the amount of energy in the wave. Snap Lab Waves in a Bowl In this lab, you will take measurements to determine how the amplitude and the period of waves are affected by the transfer of energy from a cork dropped into the water. The cork initially has some potential energy when it is held above the water—the greater the height, the higher the potential energy. When it is dropped, such potential energy is converted to kinetic energy as the cork falls. When the cork hits the water, that energy travels through the water in waves. • Large bowl or basin • Water • Cork (or ping pong ball) • Stopwatch • Measuring tape Instructions Procedure 1. Fill a large bowl or basin with water and wait for the water to settle so there are no ripples. 2. Gently drop a cork into the middle of the bowl. 3. Estimate the wavelength and the period of oscillation of the water wave that propagates away from the cork. You can estimate the period by counting the number of ripples from the center to the edge of the bowl while your partner times it. This information, combined with the bowl measurement, will give you the wavelength when the correct formula is used. 4. Remove the cork from the bowl and wait for the water to settle again. 5. Gently drop the cork at a height that is different from the first drop. 6. Repeat Steps 3 to 5 to collect a second and third set of data, dropping the cork from different heights and recording the resulting wavelengths and periods. Access for free at openstax.org. 13.2 • Wave Properties: Speed, Amplitude, Frequency, and Period 397 7. Interpret your results. GRASP CHECK A cork is dropped into a pool of water creating waves. Does the wavelength depend upon the height above the water from which the cork is dropped? a. No, only the amplitude is affected. b. Yes, the wavelength is affected. LINKS TO PHYSICS Geology: Physics of Seismic Waves Figure 13.9 The destructive effect of an earthquake is a palpable evidence of the energy carried in the earthquake waves. The Richter scale rating of earthquakes is related to both their amplitude and the energy they carry. (Petty Officer 2nd Class Candice Villarreal, U.S. Navy) Geologists rely heavily on physics to study earthquakes since earthquakes involve several types of wave disturbances, including disturbance of Earth’s surface and pressure disturbances under the surface. Surface earthquake waves are similar to surface waves on water. The waves under Earth’s surface have both longitudinal and transverse components. The longitudinal waves in an earthquake are called pressure waves (P-waves) and the transverse waves are called shear waves (S-waves). These two types of waves propagate at different speeds, and the speed at which they travel depends on the rigidity of the medium through which they are traveling. During earthquakes, the speed of P-waves in granite is significantly higher than the speed of S-waves. Both components of earthquakes travel more slowly in less rigid materials, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves have speeds of 2 to 5 km/s, but both are faster in more rigid materials. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. For that reason, the time difference between the P- and S-waves is used to determine the distance to their source, the epicenter of the earthquake. We know from seismic waves produced by earthquakes that parts of the interior of Earth are liquid. Shear or transverse waves cannot travel through a liquid and are not transmitted through Earth’s core. In contrast, compression or longitudinal waves can pass through a liquid and they do go through the core. All waves carry energy, and the energy of earthquake waves is easy to observe based on the amount of damage left behind after the ground has stopped moving. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. The amount of energy in a w
ave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements and greater damage. As earthquake waves spread out, their amplitude decreases, so there is less damage the farther they get from the source. GRASP CHECK What is the relationship between the propagation speed, frequency, and wavelength of the S-waves in an earthquake? a. The relationship between the propagation speed, frequency, and wavelength is b. The relationship between the propagation speed, frequency, and wavelength is c. The relationship between the propagation speed, frequency, and wavelength is 398 Chapter 13 • Waves and Their Properties d. The relationship between the propagation speed, frequency, and wavelength is Virtual Physics Wave on a String Click to view content (http://www.openstax.org/l/28wavestring) In this animation, watch how a string vibrates in slow motion by choosing the Slow Motion setting. Select the No End and Manual options, and wiggle the end of the string to make waves yourself. Then switch to the Oscillate setting to generate waves automatically. Adjust the frequency and the amplitude of the oscillations to see what happens. Then experiment with adjusting the damping and the tension. GRASP CHECK Which of the settings—amplitude, frequency, damping, or tension—changes the amplitude of the wave as it propagates? What does it do to the amplitude? a. Frequency; it decreases the amplitude of the wave as it propagates. b. Frequency; it increases the amplitude of the wave as it propagates. c. Damping; it decreases the amplitude of the wave as it propagates. d. Damping; it increases the amplitude of the wave as it propagates. Solving Wave Problems WORKED EXAMPLE Calculate the Velocity of Wave Propagation: Gull in the Ocean Calculate the wave velocity of the ocean wave in the previous figure if the distance between wave crests is 10.0 m and the time for a seagull to bob up and down is 5.00 s. STRATEGY The values for the wavelength can use are given and we are asked to find Therefore, we to find the wave velocity. and the period Solution Enter the known values into Discussion This slow speed seems reasonable for an ocean wave. Note that in the figure, the wave moves to the right at this speed, which is different from the varying speed at which the seagull bobs up and down. 13.5 WORKED EXAMPLE Calculate the Period and the Wave Velocity of a Toy Spring The woman in creates two waves every second by shaking the toy spring up and down. (a)What is the period of each wave? (b) If each wave travels 0.9 meters after one complete wave cycle, what is the velocity of wave propagation? STRATEGY FOR (A) To find the period, we solve for , given the value of the frequency Access for free at openstax.org. 13.2 • Wave Properties: Speed, Amplitude, Frequency, and Period 399 Solution for (a) Enter the known value into 13.6 STRATEGY FOR (B) Since one definition of wavelength is the distance a wave has traveled after one complete cycle—or one period—the values for the wavelength as well as the frequency are given. Therefore, we can use to find the wave velocity. Solution for (b) Enter the known values into Discussion We could have also used the equation to solve for the wave velocity since we already know the value of the period from our calculation in part (a), and we would come up with the same answer. Practice Problems 7. The frequency of a wave is 10 Hz. What is its period? a. The period of the wave is 100 s. b. The period of the wave is 10 s. c. The period of the wave is 0.01 s. d. The period of the wave is 0.1 s. 8. What is the velocity of a wave whose wavelength is 2 m and whose frequency is 5 Hz? a. 20 m/s b. 2.5 m/s c. 0.4 m/s 10 m/s d. Check Your Understanding 9. What is the amplitude of a wave? a. A quarter of the total height of the wave b. Half of the total height of the wave c. Two times the total height of the wave d. Four times the total height of the wave 10. What is meant by the wavelength of a wave? a. The wavelength is the distance between adjacent identical parts of a wave, parallel to the direction of propagation. b. The wavelength is the distance between adjacent identical parts of a wave, perpendicular to the direction of propagation. c. The wavelength is the distance between a crest and the adjacent trough of a wave, parallel to the direction of propagation. d. The wavelength is the distance between a crest and the adjacent trough of a wave, perpendicular to the direction of propagation. 11. How can you mathematically express wave frequency in terms of wave period? a. b. c. d. 12. When is the wavelength directly proportional to the period of a wave? 400 Chapter 13 • Waves and Their Properties a. When the velocity of the wave is halved b. When the velocity of the wave is constant c. When the velocity of the wave is doubled d. When the velocity of the wave is tripled 13.3 Wave Interaction: Superposition and Interference Section Learning Objectives By the end of this section, you will be able to do the following: • Describe superposition of waves • Describe interference of waves and distinguish between constructive and destructive interference of waves • Describe the characteristics of standing waves • Distinguish reflection from refraction of waves Section Key Terms antinode constructive interference destructive interference inversion nodes reflection refraction standing wave superposition Superposition of Waves Most waves do not look very simple. They look more like the waves in Figure 13.10, rather than the simple water wave considered in the previous sections, which has a perfect sinusoidal shape. Figure 13.10 These waves result from the superposition of several waves from different sources, producing a complex pattern. (Waterborough, Wikimedia Commons) Most waves appear complex because they result from two or more simple waves that combine as they come together at the same place at the same time—a phenomenon called superposition. Waves superimpose by adding their disturbances; each disturbance corresponds to a force, and all the forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves, that is, their amplitudes add. Wave Interference The two special cases of superposition that produce the simplest results are pure constructive interference and pure destructive interference. Pure constructive interference occurs when two identical waves arrive at the same point exactly in phase. When waves are exactly in phase, the crests of the two waves are precisely aligned, as are the troughs. Refer to Figure 13.11. Because the disturbances add, the pure constructive interference of two waves with the same amplitude produces a wave that has twice the amplitude of the two individual waves, but has the same wavelength. Access for free at openstax.org. 13.3 • Wave Interaction: Superposition and Interference 401 Figure 13.11 The pure constructive interference of two identical waves produces a wave with twice the amplitude but the same wavelength. Figure 13.12 shows two identical waves that arrive exactly outof phase—that is, precisely aligned crest to trough—producing pure destructive interference. Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference; that is, the waves completely cancel out each other. Figure 13.12 The pure destructive interference of two identical waves produces zero amplitude, or complete cancellation. While pure constructive interference and pure destructive interference can occur, they are not very common because they require precisely aligned identical waves. The superposition of most waves that we see in nature produces a combination of constructive and destructive interferences. Waves that are not results of pure constructive or destructive interference can vary from place to place and time to time. The sound from a stereo, for example, can be loud in one spot and soft in another. The varying loudness means that the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers that create sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of jet engines heard by a stationary passenger. The volume of the combined sound can fluctuate up and down as the sound from the two engines varies in time from constructive to destructive. The two previous examples considered waves that are similar—both stereo speakers generate sound waves with the same amplitude and wavelength, as do the jet engines. But what happens when two waves that are not similar, that is, having different amplitudes and wavelengths, are superimposed? An example of the superposition of two dissimilar waves is shown in Figure 13.13. Here again, the disturbances add and subtract, but they produce an even more complicated-looking wave. The resultant wave from the combined disturbances of two dissimilar waves looks much different than the idealized sinusoidal shape of a periodic wave. 402 Chapter 13 • Waves and Their Properties Figure 13.13 The superposition of nonidentical waves exhibits both constructive and destructive interferences. Virtual Physics Wave Interference Click to view content (http://www.openstax.org/l/28interference) In this simulation, make waves with a dripping faucet, an audio speaker, or a laser by switching between the water, sound, and light tabs. Contrast and compare how the different types of waves behave. Try rotating the view from top to side to make observations. Then experiment with adding a second source or a pair of slits to create an interference pattern. GRASP CHECK In the water tab, compare the waves generated by one drip versus two drips. What happens to the amplitude of the waves when there are two drips? Is this constructive
or destructive interference? Why would this be the case? a. The amplitude of the water waves remains same because of the destructive interference as the drips of water hit the surface at the same time. b. The amplitude of the water waves is canceled because of the destructive interference as the drips of water hit the surface at the same time. c. The amplitude of water waves remains same because of the constructive interference as the drips of water hit the surface at the same time. d. The amplitude of water waves doubles because of the constructive interference as the drips of water hit the surface at the same time. Standing Waves Sometimes waves do not seem to move and they appear to just stand in place, vibrating. Such waves are called standing waves and are formed by the superposition of two or more waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. Standing waves created by the superposition of two identical waves moving in opposite directions are illustrated in Figure 13.14. Figure 13.14 A standing wave is created by the superposition of two identical waves moving in opposite directions. The oscillations are at fixed locations in space and result from alternating constructive and destructive interferences. As an example, standing waves can be seen on the surface of a glass of milk in a refrigerator. The vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. The two waves that Access for free at openstax.org. 13.3 • Wave Interaction: Superposition and Interference 403 produce standing waves may be due to the reflections from the side of the glass. Earthquakes can create standing waves and cause constructive and destructive interferences. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. As a result, areas closer to the epicenter are not damaged while areas farther from the epicenter are damaged. Standing waves are also found on the strings of musical instruments and are due to reflections of waves from the ends of the string. Figure 13.15 and Figure 13.16 show three standing waves that can be created on a string that is fixed at both ends. When the wave reaches the fixed end, it has nowhere else to go but back where it came from, causing the reflection. The nodes are the points where the string does not move; more generally, the nodes are the points where the wave disturbance is zero in a standing wave. The fixed ends of strings must be nodes, too, because the string cannot move there. The antinode is the location of maximum amplitude in standing waves. The standing waves on a string have a frequency that is of the disturbance on the string. The wavelength is determined by the distance between related to the propagation speed the points where the string is fixed in place. Figure 13.15 The figure shows a string oscillating with its maximum disturbance as the antinode. Figure 13.16 The figure shows a string oscillating with multiple nodes. Reflection and Refraction of Waves As we saw in the case of standing waves on the strings of a musical instrument, reflection is the change in direction of a wave when it bounces off a barrier, such as a fixed end. When the wave hits the fixed end, it changes direction, returning to its source. As it is reflected, the wave experiences an inversion, which means that it flips vertically. If a wave hits the fixed end with a crest, it will return as a trough, and vice versa (Henderson 2015). Refer to Figure 13.17. Figure 13.17 A wave is inverted after reflection from a fixed end. TIPS FOR SUCCESS If the end is not fixed, it is said to be a free end, and no inversion occurs. When the end is loosely attached, it reflects without 404 Chapter 13 • Waves and Their Properties inversion, and when the end is not attached to anything, it does not reflect at all. You may have noticed this while changing the settings from Fixed End to Loose End to No End in the Waves on a String PhET simulation. Rather than encountering a fixed end or barrier, waves sometimes pass from one medium into another, for instance, from air into water. Different types of media have different properties, such as density or depth, that affect how a wave travels through them. At the boundary between media, waves experience refraction—they change their path of propagation. As the wave bends, it also changes its speed and wavelength upon entering the new medium. Refer to Figure 13.18. Figure 13.18 A wave refracts as it enters a different medium. For example, water waves traveling from the deep end to the shallow end of a swimming pool experience refraction. They bend in a path closer to perpendicular to the surface of the water, propagate slower, and decrease in wavelength as they enter shallower water. Check Your Understanding 13. What is the superposition of waves? a. When a single wave splits into two different waves at a point b. When two waves combine at the same place at the same time 14. How do waves superimpose on one another? a. By adding their frequencies b. By adding their wavelengths c. By adding their disturbances d. By adding their speeds 15. What is interference of waves? a. b. c. d. Interference is a superposition of two waves to form a resultant wave with higher or lower frequency. Interference is a superposition of two waves to form a wave of larger or smaller amplitude. Interference is a superposition of two waves to form a resultant wave with higher or lower velocity. Interference is a superposition of two waves to form a resultant wave with longer or shorter wavelength. 16. Is the following statement true or false? The two types of interference are constructive and destructive interferences. a. True b. False 17. What are standing waves? a. Waves that appear to remain in one place and do not seem to move b. Waves that seem to move along a trajectory 18. How are standing waves formed? a. Standing waves are formed by the superposition of two or more waves moving in opposite directions. b. Standing waves are formed by the superposition of two or more waves moving in the same direction. c. Standing waves are formed by the superposition of two or more waves moving in perpendicular directions. d. Standing waves are formed by the superposition of two or more waves moving in any arbitrary directions. 19. What is the reflection of a wave? a. The reflection of a wave is the change in amplitude of a wave when it bounces off a barrier. b. The reflection of a wave is the change in frequency of a wave when it bounces off a barrier. c. The reflection of a wave is the change in velocity of a wave when it bounces off a barrier. d. The reflection of a wave is the change in direction of a wave when it bounces off a barrier. Access for free at openstax.org. 13.3 • Wave Interaction: Superposition and Interference 405 20. What is inversion of a wave? a. b. c. d. Inversion occurs when a wave reflects off a fixed end, and the wave amplitude changes sign. Inversion occurs when a wave reflects off a loose end, and the wave amplitude changes sign. Inversion occurs when a wave reflects off a fixed end without the wave amplitude changing sign. Inversion occurs when a wave reflects off a loose end without the wave amplitude changing sign. 406 Chapter 13 • Key Terms KEY TERMS antinode location of maximum amplitude in standing motion waves pulse wave sudden disturbance with only one wave or a few constructive interference when two waves arrive at the waves generated same point exactly in phase; that is, the crests of the two waves are precisely aligned, as are the troughs reflection change in direction of a wave at a boundary or fixed end destructive interference when two identical waves arrive at refraction bending of a wave as it passes from one medium the same point exactly out of phase that is precisely aligned crest to trough inversion vertical flipping of a wave after reflection from a fixed end longitudinal wave wave in which the disturbance is parallel to another medium with a different density standing wave wave made by the superposition of two waves of the same amplitude and wavelength moving in opposite directions and which appears to vibrate in place superposition phenomenon that occurs when two or more to the direction of propagation waves arrive at the same point mechanical wave wave that requires a medium through which it can travel transverse wave wave in which the disturbance is perpendicular to the direction of propagation medium solid, liquid, or gas material through which a wave disturbance that moves from its source and carries wave propagates energy nodes points where the string does not move; more generally, points where the wave disturbance is zero in a standing wave periodic wave wave that repeats the same oscillation for several cycles and is associated with simple harmonic SECTION SUMMARY 13.1 Types of Waves • A wave is a disturbance that moves from the point of creation and carries energy but not mass. • Mechanical waves must travel through a medium. • Sound waves, water waves, and earthquake waves are all examples of mechanical waves. • Light is not a mechanical wave since it can travel through a vacuum. • A periodic wave is a wave that repeats for several cycles, whereas a pulse wave has only one crest or a few crests and is associated with a sudden disturbance. • Periodic waves are associated with simple harmonic motion. • A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period • A wave is a disturbance that moves from the point of creation at a wave velocity vw. •
A wave has a wavelength , which is the distance between adjacent identical parts of the wave. • The wave velocity and the wavelength are related to the wave’s frequency and period by or Access for free at openstax.org. wave velocity speed at which the disturbance moves; also called the propagation velocity or propagation speed wavelength distance between adjacent identical parts of a wave • The time for one complete wave cycle is the period T. • The number of waves per unit time is the frequency ƒ. • The wave frequency and the period are inversely related to one another. 13.3 Wave Interaction: Superposition and Interference • Superposition is the combination of two waves at the same location. • Constructive interference occurs when two identical waves are superimposed exactly in phase. • Destructive interference occurs when two identical waves are superimposed exactly out of phase. • A standing wave is a wave produced by the superposition of two waves. It varies in amplitude but does not propagate. • The nodes are the points where there is no motion in standing waves. • An antinode is the location of maximum amplitude of a standing wave. Inversion occurs when a wave reflects from a fixed end. • Reflection causes a wave to change direction. • • Refraction causes a wave’s path to bend and occurs when a wave passes from one medium into another medium with a different density. KEY EQUATIONS 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period wave velocity or CHAPTER REVIEW Concept Items 13.1 Types of Waves 1. Do water waves push water from one place to another? Explain. a. No, water waves transfer only energy from one place to another. b. Yes, water waves transfer water from one place to another. 2. With reference to waves, what is a trough? a. b. c. d. the lowermost position of a wave the uppermost position of a wave the final position of a wave the initial position of the wave 3. Give an example of longitudinal waves. light waves a. b. water waves in a lake sound waves in air c. seismic waves in Earth’s surface d. 4. What does the speed of a mechanical wave depend on? a. b. c. d. the properties of the material through which it travels the shape of the material through which it travels the size of the material through which it travels the color of the material through which it travels 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 5. Which characteristic of a transverse wave is measured along the direction of propagation? a. The amplitude of a transverse wave is measured along the direction of propagation. b. The amplitude and the wavelength of a transverse wave are measured along the direction of propagation. c. The wavelength of a transverse wave is measured along the direction of propagation. d. The displacement of the particles of the medium in a transverse wave is measured along the direction of propagation. 6. Which kind of seismic waves cannot travel through Chapter 13 • Key Equations 407 period frequency liquid? a. b. P-waves c. d. S-waves compressional waves longitudinal waves 7. What is the period of a wave? a. b. c. d. the time that a wave takes to complete a half cycle the time that a wave takes to complete one cycle the time that a wave takes to complete two cycles the time that a wave takes to complete four cycles 8. When the period of a wave increases, what happens to its frequency? a. b. c. Its frequency decreases. Its frequency increases. Its frequency remains the same. 13.3 Wave Interaction: Superposition and Interference 9. Is this statement true or false? The amplitudes of waves add up only if they are propagating in the same line. a. True b. False 10. Why is sound from a stereo louder in one part of the room and softer in another? a. Sound is louder in parts of the room where the density is greatest. Sound is softer in parts of the room where density is smallest. b. Sound is louder in parts of the room where the density is smallest. Sound is softer in parts of the room where density is greatest. c. Sound is louder in parts of the room where constructive interference occurs and softer in parts where destructive interference occurs. d. Sound is louder in parts of the room where destructive interference occurs and softer in parts where constructive interference occurs. 11. In standing waves on a string, what does the frequency depend on? a. The frequency depends on the propagation speed and the density of the string. 408 Chapter 13 • Chapter Review b. The frequency depends on the propagation speed a. Refraction is the phenomenon in which waves and the length of the string. c. The frequency depends on the density and the length of the string. d. The frequency depends on the propagation speed, the density, and the length of the string. 12. Is the following statement true or false? Refraction is useful in fiber optic cables for transmitting signals. a. False b. True 13. What is refraction? Critical Thinking Items 13.1 Types of Waves 14. Give an example of a wave that propagates only through a solid. a. Light wave b. Sound wave c. Seismic wave d. Surface wave 15. Can mechanical waves be periodic waves? a. No, mechanical waves cannot be periodic waves. b. Yes, mechanical waves can be periodic. 16. In a sound wave, which parameter of the medium varies with every cycle? a. The density of the medium varies with every cycle. b. The mass of the medium varies with every cycle. c. The resistivity of the medium varies with every cycle. d. The volume of the medium varies with every cycle. 17. What is a transverse wave in an earthquake called? a. L-wave b. P-wave c. S-wave d. R-wave 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 18. If the horizontal distance, that is, the distance in the direction of propagation, between a crest and the adjacent trough of a sine wave is 1 m, what is the wavelength of the wave? a. 0.5 m b. 1 m c. 2 m d. 4 m 19. How is the distance to the epicenter of an earthquake determined? Access for free at openstax.org. change their path of propagation at the interface of two media with different densities. b. Refraction is the phenomenon in which waves change their path of propagation at the interface of two media with the same density. c. Refraction is the phenomenon in which waves become non-periodic at the boundary of two media with different densities. d. Refraction is the phenomenon in which waves become non-periodic at the boundary of two media with the same density. a. The wavelength difference between P-waves and S- waves is used to measure the distance to the epicenter. b. The time difference between P-waves and S-waves is used to measure the distance to the epicenter. c. The frequency difference between P-waves and Swaves is used to measure the distance to the epicenter. d. The phase difference between P-waves and S-waves is used to measure the distance to the epicenter. 20. Two identical waves superimpose in pure constructive interference. What is the height of the resultant wave if the amplitude of each of the waves is 1 m? a. 1 m b. 2 m 3 m c. d. 4 m 13.3 Wave Interaction: Superposition and Interference 21. Two identical waves with an amplitude superimpose in a way that pure constructive interference occurs. What is the amplitude of the resultant wave? a. b. c. d. 22. In which kind of wave is the amplitude at each point constant? a. Seismic waves b. Pulse wave c. Standing waves d. Electromagnetic waves 23. Which property of a medium causes refraction? a. Conductivity b. Opacity c. Ductility d. Density 24. What is added together when two waves superimpose? a. Amplitudes Problems 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 25. If a seagull sitting in water bobs up and down once every 2 seconds and the distance between two crests of the water wave is 3 m, what is the velocity of the wave? 1.5 m/s a. 3 m/s b. c. 6 m/s Performance Task 13.3 Wave Interaction: Superposition and Interference 27. Ocean waves repeatedly crash against beaches and coasts. Their energy can lead to erosion and collapse of land. Scientists and engineers need to study how waves interact with beaches in order to assess threats to coastal communities and construct breakwater systems. In this task, you will construct a wave tank and fill it with water. Simulate a beach by placing sand at one end. Create waves by moving a piece of wood or plastic up and down in the water. Measure or estimate the TEST PREP Multiple Choice 13.1 Types of Waves 28. What kind of waves are sound waves? a. Mechanical waves b. Electromagnetic waves Chapter 13 • Test Prep 409 b. Wavelengths c. Velocities d. 12 m/s 26. A boat in the trough of a wave takes 3 seconds to reach the highest point of the wave. The velocity of the wave is 5 m/s. What is its wavelength? a. 0.83 m 15 m b. 30 m c. 180 m d. wavelength, period, frequency, and amplitude of the wave, and observe the effect of the wave on the sand. Produce waves of different amplitudes and frequencies, and record your observations each time. Use mathematical representations to demonstrate the relationships between different wave properties. Change the position of the sand to create a steeper beach and record your observations. Give a qualitative analysis of the effects of the waves on the beach. What kind of wave causes the most damage? At what height, wavelength, and frequency do waves break? How does the steepness of the beach affect the waves? that creates a wave. It refers to the wavelength of the wave. It refers to the speed of the wave. c. d. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 29. What kind of a wave does a tuning fork create? 32. Which of these is not a characteristic of a wave? a. Pulse wave b. Periodic wave c. Electromagnetic wave 30. What kind of waves are electromagnetic waves? a. amplitude b. period c. mass d. velocity a. Longitudinal waves b. Transverse waves c. Mechanical waves d. P-waves a. 31. With reference to waves, what is a disturbance? It refers to the resistance produced by some
particles of a material. It refers to an oscillation produced by some energy b. 33. If you are in a boat at a resting position, how much will your height change when you are hit by the peak of a wave with a height of 2 m? a. 0 m 1 m b. c. 2 m d. 4 m 34. What is the period of a wave with a frequency of 0.5 Hz? 410 Chapter 13 • Test Prep a. 0.5 s 1 s b. c. 2 s 3 s d. 35. What is the relation between the amplitude of a wave and its speed? a. The amplitude of a wave is independent of its speed. b. The amplitude of a wave is directly proportional to its speed. c. The amplitude of a wave is directly proportional to the square of the inverse of its speed. d. The amplitude of a wave is directly proportional to the inverse of its speed. 36. What does the speed of seismic waves depend on? a. The speed of seismic waves depends on the size of the medium. b. The speed of seismic waves depends on the shape of the medium. c. The speed of seismic waves depends on the rigidity of the medium. 13.3 Wave Interaction: Superposition and Interference 37. What is added together when two waves superimpose? a. amplitudes b. wavelengths c. velocities 38. Pure constructive interference occurs between two waves when they have the same _____. Short Answer 13.1 Types of Waves frequency and are in phase frequency and are out of phase a. b. c. amplitude and are in phase d. amplitude and are out of phase 39. What kind(s) of interference can occur between two identical waves moving in opposite directions? a. Constructive interference only b. Destructive interference only c. Both constructive and destructive interference d. Neither constructive nor destructive interference 40. What term refers to the bending of light at the junction interference of two media? a. b. diffraction scattering c. refraction d. 41. Which parameter of a wave gets affected after superposition? a. wavelength b. direction c. amplitude frequency d. 42. When do the amplitudes of two waves get added? a. When their amplitudes are the same b. When their amplitudes are different c. When they propagate in perpendicular directions d. When they are propagating along the same line in opposite directions d. The cone of a speaker vibrates to create small changes in the resistance of the air. 43. Give an example of a non-mechanical wave. a. A radio wave is an example of a nonmechanical wave. b. A sound wave is an example of a nonmechanical wave. 45. What kind of wave is thunder? a. Transverse wave b. Pulse wave c. R-wave d. P-wave c. A surface wave is an example of a nonmechanical 46. Are all ocean waves perfectly sinusoidal? wave. d. A seismic wave is an example of a nonmechanical wave. 44. How is sound produced by an electronic speaker? a. The cone of a speaker vibrates to create small changes in the temperature of the air. b. The cone of a speaker vibrates to create small changes in the pressure of the air. c. The cone of a speaker vibrates to create small changes in the volume of the air. a. No, all ocean waves are not perfectly sinusoidal. b. Yes, all ocean waves are perfectly sinusoidal. 47. What are orbital progressive waves? a. Waves that force the particles of the medium to follow a linear path from the crest to the trough b. Waves that force the particles of the medium to follow a circular path from the crest to the trough c. Waves that force the particles of the medium to follow a zigzag path from the crest to the trough d. Waves that force the particles of the medium to Access for free at openstax.org. follow a random path from the crest to the trough 48. Give an example of orbital progressive waves. a. light waves b. ocean waves sound waves c. seismic waves d. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 49. What is the relation between the amplitude and height of a transverse wave? a. The height of a wave is half of its amplitude. b. The height of a wave is equal to its amplitude. c. The height of a wave is twice its amplitude. d. The height of a wave is four times its amplitude. 50. If the amplitude of a water wave is 0.2 m and its frequency is 2 Hz, how much distance would a bird sitting on the water’s surface move with every wave? How many times will it do this every second? a. The bird will go up and down a distance of 0.4 m. It will do this twice per second. b. The bird will go up and down a distance of 0.2 m. It will do this twice per second. c. The bird will go up and down a distance of 0.4 m. It will do this once per second. d. The bird will go up and down a distance of 0.2 m. It will do this once per second. 51. What is the relation between the amplitude and the frequency of a wave? a. The amplitude and the frequency of a wave are independent of each other. b. The amplitude and the frequency of a wave are equal. c. The amplitude decreases with an increase in the frequency of a wave. d. The amplitude increases with an increase in the frequency of a wave. 52. What is the relation between a wave’s energy and its amplitude? a. There is no relation between the energy and the amplitude of a wave. b. The magnitude of the energy is equal to the magnitude of the amplitude of a wave. c. The energy of a wave increases with an increase in the amplitude of the wave. d. The energy of a wave decreases with an increase in the amplitude of a wave. 53. A wave travels every 2 cycles. What is its wavelength? a. Chapter 13 • Test Prep 411 b. c. d. 54. A water wave propagates in a river at 6 m/s. If the river moves in the opposite direction at 3 m/s, what is the effective velocity of the wave? a. 3 m/s b. 6 m/s c. 9 m/s 18 m/s d. 13.3 Wave Interaction: Superposition and Interference 55. Is this statement true or false? Spherical waves can superimpose. a. True b. False 56. Is this statement true or false? Waves can superimpose if their frequencies are different. a. True b. False 57. When does pure destructive interference occur? a. When two waves with equal frequencies that are perfectly in phase and propagating along the same line superimpose. b. When two waves with unequal frequencies that are perfectly in phase and propagating along the same line superimpose. c. When two waves with unequal frequencies that are perfectly out of phase and propagating along the same line superimpose. d. When two waves with equal frequencies that are perfectly out of phase and propagating along the same line superimpose. 58. Is this statement true or false? The amplitude of one wave is affected by the amplitude of another wave only when they are precisely aligned. a. True b. False 59. Why does a standing wave form on a guitar string? a. due to superposition with the reflected waves from the ends of the string b. due to superposition with the reflected waves from the walls of the room c. due to superposition with waves generated from the body of the guitar 60. Is the following statement true or false? A standing wave is a superposition of two identical waves that are in phase and propagating in the same direction. 412 Chapter 13 • Test Prep a. True b. False 61. Why do water waves traveling from the deep end to the shallow end of a swimming pool experience refraction? a. Because the pressure of water at the two ends of the c. Because the density of water at the two ends of the pool is same d. Because the density of water at the two ends of the pool is different 62. Is the statement true or false? Waves propagate faster in pool is same b. Because the pressures of water at the two ends of the pool are different a less dense medium if the stiffness is the same. a. True b. False Extended Response 13.1 Types of Waves 63. Why can light travel through outer space while sound cannot? a. Sound waves are mechanical waves and require a medium to propagate. Light waves can travel through a vacuum. b. Sound waves are electromagnetic waves and require a medium to propagate. Light waves can travel through a vacuum. c. Light waves are mechanical waves and do not require a medium to propagate; sound waves require a medium to propagate. d. Light waves are longitudinal waves and do not require a medium to propagate; sound waves require a medium to propagate. 64. Do periodic waves require a medium to travel through? a. No, the requirement of a medium for propagation does not depend on whether the waves are pulse waves or periodic waves. b. Yes, the requirement of a medium for propagation depends on whether the waves are pulse waves or periodic waves. 65. How is the propagation of sound in solids different from that in air? a. Sound waves in solids are transverse, whereas in and moves with the wave in its direction. b. The gull experiences mostly side-to-side motion but does not move with the wave in its direction. c. The gull experiences mostly up-and-down motion and moves with the wave in its direction. d. The gull experiences mostly up-and-down motion but does not move in the direction of the wave. 67. Why does a good-quality speaker have a woofer and a tweeter? a. In a good-quality speaker, sounds with high frequencies or short wavelengths are reproduced accurately by woofers, while sounds with low frequencies or long wavelengths are reproduced accurately by tweeters. b. Sounds with high frequencies or short wavelengths are reproduced more accurately by tweeters, while sounds with low frequencies or long wavelengths are reproduced more accurately by woofers. 68. The time difference between a 2 km/s S-wave and a 6 km/s P-wave recorded at a certain point is 10 seconds. How far is the epicenter of the earthquake from that point? a. b. c. d. 15 m 30 m 15 km 30 km air, they are longitudinal. b. Sound waves in solids are longitudinal, whereas in 13.3 Wave Interaction: Superposition and Interference air, they are transverse. c. Sound waves in solids can be both longitudinal and transverse, whereas in air, they are longitudinal. d. Sound waves in solids are longitudinal, whereas in air, they can be both longitudinal and transverse. 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period 66. A seagull is s
itting in the water surface and a simple water wave passes under it. What sort of motion does the gull experience? Why? a. The gull experiences mostly side-to-side motion 69. Why do water waves sometimes appear like a complex criss-cross pattern? a. The crests and the troughs of waves traveling in the same direction combine to form a criss-cross pattern. b. The crests and the troughs of waves traveling in different directions combine to form a criss-cross pattern. 70. What happens when two dissimilar waves interfere? a. pure constructive interference b. pure destructive interference c. a combination of constructive and destructive Access for free at openstax.org. interference 71. Occasionally, during earthquakes, areas near the epicenter are not damaged while those farther away are damaged. Why could this occur? a. Destructive interference results in waves with greater amplitudes being formed in places farther away from the epicenter. b. Constructive interference results in waves with greater amplitudes being formed in places farther away from the epicenter. c. The standing waves of great amplitudes are formed in places farther away from the epicenter. d. The pulse waves of great amplitude are formed in places farther away from the epicenter. 72. Why does an object appear to be distorted when you Chapter 13 • Test Prep 413 view it through a glass of water? a. The glass and the water reflect the light in different directions. Hence, the object appears to be distorted. b. The glass and the water absorb the light by different amounts. Hence, the object appears to be distorted. c. Water, air, and glass are media with different densities. Light rays refract and bend when they pass from one medium to another. Hence, the object appears to be distorted. d. The glass and the water disperse the light into its components. Hence, the object appears to be distorted. 414 Chapter 13 • Test Prep Access for free at openstax.org. CHAPTER 14 Sound Figure 14.1 This tree fell some time ago. When it fell, particles in the air were disturbed by the energy of the tree hitting the ground. This disturbance of matter, which our ears have evolved to detect, is called sound. (B.A. Bowen Photography) Chapter Outline 14.1 Speed of Sound, Frequency, and Wavelength 14.2 Sound Intensity and Sound Level 14.3 Doppler Effect and Sonic Booms 14.4 Sound Interference and Resonance INTRODUCTION If a tree falls in a forest (see Figure 14.1) and no one is there to hear it, does it make a sound? The answer to this old philosophical question depends on how you define sound. If sound only exists when someone is around to perceive it, then the falling tree produced no sound. However, in physics, we know that colliding objects can disturb the air, water or other matter surrounding them. As a result of the collision, the surrounding particles of matter began vibrating in a wave-like fashion. This is a sound wave. Consequently, if a tree collided with another object in space, no one would hear it, because no sound would be produced. This is because, in space, there is no air, water or other matter to be disturbed and produce sound waves. In this chapter, we’ll learn more about the wave properties of sound, and explore hearing, as well as some special uses for sound. 416 Chapter 14 • Sound 14.1 Speed of Sound, Frequency, and Wavelength Section Learning Objectives By the end of this section, you will be able to do the following: • Relate the characteristics of waves to properties of sound waves • Describe the speed of sound and how it changes in various media • Relate the speed of sound to frequency and wavelength of a sound wave Section Key Terms rarefaction sound Properties of Sound Waves Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion. A vibrating string produces a sound wave as illustrated in Figure 14.2, Figure 14.3, and Figure 14.4. As the string oscillates back and forth, part of the string’s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions. The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves. Figure 14.2 A vibrating string moving to the right compresses the air in front of it and expands the air behind it. Figure 14.3 As the string moves to the left, it creates another compression and rarefaction as the particles on the right move away from the string. Access for free at openstax.org. 14.1 • Speed of Sound, Frequency, and Wavelength 417 Figure 14.4 After many vibrations, there is a series of compressions and rarefactions that have been transmitted from the string as a sound wave. The graph shows gauge pressure (Pgauge) versus distance xfrom the source. Gauge pressure is the pressure relative to atmospheric pressure; it is positive for pressures above atmospheric pressure, and negative for pressures below it. For ordinary, everyday sounds, pressures vary only slightly from average atmospheric pressure. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But some of the energy is also absorbed by objects, such as the eardrum in Figure 14.5, and some of the energy is converted to thermal energy in the air. Figure 14.4 shows a graph of gauge pressure versus distance from the vibrating string. From this figure, you can see that the compression of a longitudinal wave is analogous to the peak of a transverse wave, and the rarefaction of a longitudinal wave is analogous to the trough of a transverse wave. Just as a transverse wave alternates between peaks and troughs, a longitudinal wave alternates between compression and rarefaction. Figure 14.5 Sound wave compressions and rarefactions travel up the ear canal and force the eardrum to vibrate. There is a net force on the eardrum, since the sound wave pressures differ from the atmospheric pressure found behind the eardrum. A complicated mechanism converts the vibrations to nerve impulses, which are then interpreted by the brain. The Speed of Sound The speed of sound varies greatly depending upon the medium it is traveling through. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. The greater the density of a medium, the slower the speed of sound. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases. Table 14.1 shows the speed of sound in various media. Since temperature affects density, the speed of sound varies with the temperature of the medium through which it’s traveling to some extent, especially for gases. Medium vw (m/s) Gases at 0 °C Air Carbon dioxide Oxygen Helium 331 259 316 965 Hydrogen 1290 Liquids at 20 °C Ethanol Mercury Water, fresh Sea water 1160 1450 1480 1540 Human tissue 1540 Solids (longitudinal or bulk) Vulcanized rubber 54 Polyethylene 920 Marble Glass, Pyrex Lead Aluminum Steel 3810 5640 1960 5120 5960 Table 14.1 Speed of Sound in Various Media 418 Chapter 14 • Sound Access for free at openstax.org. The Relationship Between the Speed of Sound and the Frequency and Wavelength of a Sound Wave 14.1 • Speed of Sound, Frequency, and Wavelength 419 Figure 14.6 When fireworks explode in the sky, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (Dominic Alves, Flickr) Sound, like all waves, travels at certain speeds through different media and has the properties of frequency and wavelength. Sound travels much slower than light—you can observe this while watching a fireworks display (see Figure 14.6), since the flash of an explosion is seen before its sound is heard. The relationship between the speed of sound, its frequency, and wavelength is the same as for all waves: where vis the speed of sound (in units of m/s), fis its frequency (in units of hertz), and is its wavelength (in units of meters). Recall that wavelength is defined as the distance between adjacent identical parts of a wave. The wavelength of a sound, therefore, is the distance between adjacent identical parts of a sound wave. Just as the distance between adjacent crests in a transverse wave is one wavelength, the distance between adjacent compressions in a sound wave is also one wavelength, as shown in Figure 14.7. The frequency of a sound wave is the same as that of the source. For example, a tuning fork vibrating at a given frequency would produce sound waves that oscillate at that same frequency. The frequency of a sound is the number of waves that pass a point per unit time. 14.1 Figure 14.7 A sound wave emanates from a source vibrating at a frequency f, propagates at v, and has a wavelength . One of the more important properties of sound is that its speed is nearly independent of frequency. If this were not the case, and high-frequency sounds traveled faster, for example, then the farther you were from a band in a football stadium, the more the sound from the low-pitch instruments would lag behind the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies
must travel at nearly the same speed. Recall that between fand is inverse: The higher the frequency, the shorter the wavelength of a sound wave. , and in a given medium under fixed temperature and humidity, vis constant. Therefore, the relationship The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If vchanges and fremains the same, then the wavelength must change. Since frequency. , the higher the speed of a sound, the greater its wavelength for a given Virtual Physics Sound Click to view content (https://www.openstax.org/l/28sound) 420 Chapter 14 • Sound This simulation lets you see sound waves. Adjust the frequency or amplitude (volume) and you can see and hear how the wave changes. Move the listener around and hear what she hears. Switch to the Two Source Interference tab or the Interference by Reflection tab to experiment with interference and reflection. TIPS FOR SUCCESS Make sure to have audio enabled and set to Listener rather than Speaker, or else the sound will not vary as you move the listener around. GRASP CHECK In the first tab, Listen to a Single Source, move the listener as far away from the speaker as possible, and then change the frequency of the sound wave. You may have noticed that there is a delay between the time when you change the setting and the time when you hear the sound get lower or higher in pitch. Why is this? a. Because, intensity of the sound wave changes with the frequency. b. Because, the speed of the sound wave changes when the frequency is changed. c. Because, loudness of the sound wave takes time to adjust after a change in frequency. d. Because it takes time for sound to reach the listener, so the listener perceives the new frequency of sound wave after a delay. Is there a difference in the amount of delay depending on whether you make the frequency higher or lower? Why? a. Yes, the speed of propagation depends only on the frequency of the wave. b. Yes, the speed of propagation depends upon the wavelength of the wave, and wavelength changes as the frequency changes. c. No, the speed of propagation depends only on the wavelength of the wave. d. No, the speed of propagation is constant in a given medium; only the wavelength changes as the frequency changes. Snap Lab Voice as a Sound Wave In this lab you will observe the effects of blowing and speaking into a piece of paper in order to compare and contrast different sound waves. • • • sheet of paper tape table Instructions Procedure 1. Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table, for example. 2. Gently blow air near the edge of the bottom of the sheet and note how the sheet moves. 3. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Interpret the results. 4. GRASP CHECK Which sound wave property increases when you are speaking more loudly than softly? a. amplitude of the wave frequency of the wave b. speed of the wave c. Access for free at openstax.org. 14.1 • Speed of Sound, Frequency, and Wavelength 421 d. wavelength of the wave WORKED EXAMPLE What Are the Wavelengths of Audible Sounds? Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in conditions where sound travels at 348.7 m/s. STRATEGY To find wavelength from frequency, we can use . Solution (1) Identify the knowns. The values for vand fare given. (2) Solve the relationship between speed, frequency and wavelength for . (3) Enter the speed and the minimum frequency to give the maximum wavelength. (4) Enter the speed and the maximum frequency to give the minimum wavelength. 14.2 14.3 14.4 Discussion Because the product of fmultiplied by be, and vice versa. Note that you can also easily rearrange the same formula to find frequency or velocity. equals a constant velocity in unchanging conditions, the smaller fis, the larger must Practice Problems 1. What is the speed of a sound wave with frequency and wavelength ? a. b. c. d. 2. Dogs can hear frequencies of up to . What is the wavelength of a sound wave with this frequency traveling in air at ? a. b. c. d. 422 Chapter 14 • Sound LINKS TO PHYSICS Echolocation Figure 14.8 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance. Echolocation is the use of reflected sound waves to locate and identify objects. It is used by animals such as bats, dolphins and whales, and is also imitated by humans in SONAR—Sound Navigation and Ranging—and echolocation technology. Bats, dolphins and whales use echolocation to navigate and find food in their environment. They locate an object (or obstacle) by emitting a sound and then sensing the reflected sound waves. Since the speed of sound in air is constant, the time it takes for the sound to travel to the object and back gives the animal a sense of the distance between itself and the object. This is called ranging. Figure 14.8 shows a bat using echolocation to sense distances. Echolocating animals identify an object by comparing the relative intensity of the sound waves returning to each ear to figure out the angle at which the sound waves were reflected. This gives information about the direction, size and shape of the object. Since there is a slight distance in position between the two ears of an animal, the sound may return to one of the ears with a bit of a delay, which also provides information about the position of the object. For example, if a bear is directly to the right of a bat, the echo will return to the bat’s left ear later than to its right ear. If, however, the bear is directly ahead of the bat, the echo would return to both ears at the same time. For an animal without a sense of sight such as a bat, it is important to know whereother animals are as well as whatthey are; their survival depends on it. Principles of echolocation have been used to develop a variety of useful sensing technologies. SONAR, is used by submarines to detect objects underwater and measure water depth. Unlike animal echolocation, which relies on only one transmitter (a mouth) and two receivers (ears), manmade SONAR uses many transmitters and beams to get a more accurate reading of the environment. Radar technologies use the echo of radio waves to locate clouds and storm systems in weather forecasting, and to locate aircraft for air traffic control. Some new cars use echolocation technology to sense obstacles around the car, and warn the driver who may be about to hit something (or even to automatically parallel park). Echolocation technologies and training systems are being developed to help visually impaired people navigate their everyday environments. GRASP CHECK If a predator is directly to the left of a bat, how will the bat know? a. The echo would return to the left ear first. b. The echo would return to the right ear first. Check Your Understanding 3. What is a rarefaction? a. Rarefaction is the high-pressure region created in a medium when a longitudinal wave passes through it. b. Rarefaction is the low-pressure region created in a medium when a longitudinal wave passes through it. c. Rarefaction is the highest point of amplitude of a sound wave. d. Rarefaction is the lowest point of amplitude of a sound wave. 4. What sort of motion do the particles of a medium experience when a sound wave passes through it? a. Simple harmonic motion Access for free at openstax.org. 14.2 • Sound Intensity and Sound Level 423 b. Circular motion c. Random motion d. Translational motion 5. What does the speed of sound depend on? a. The wavelength of the wave b. The size of the medium c. The frequency of the wave d. The properties of the medium 6. What property of a gas would affect the speed of sound traveling through it? a. The volume of the gas b. The flammability of the gas c. The mass of the gas d. The compressibility of the gas 14.2 Sound Intensity and Sound Level Section Learning Objectives By the end of this section, you will be able to do the following: • Relate amplitude of a wave to loudness and energy of a sound wave • Describe the decibel scale for measuring sound intensity • Solve problems involving the intensity of a sound wave • Describe how humans produce and hear sounds Section Key Terms amplitude decibel hearing loudness pitch sound intensity sound intensity level Amplitude, Loudness and Energy of a Sound Wave Figure 14.9 Noise on crowded roadways like this one in Delhi makes it hard to hear others unless they shout. (Lingaraj G J, Flickr) In a quiet forest, you can sometimes hear a single leaf fall to the ground. But in a traffic jam filled with honking cars, you may have to shout just so the person next to you can hear Figure 14.9.The loudness of a sound is related to how energetically its source is vibrating. In cartoons showing a screaming person, the cartoonist often shows an open mouth with a vibrating uvula (the hanging tissue at the back of the mouth) to represent a loud sound coming from the throat. Figure 14.10 shows such a cartoon depiction of a bird loudly expressing its opinion. A useful quantity for describing the loudness of sounds is called sound intensity. In general, the intensity of a wave is the power per unit area carried by the wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity Iis where Pis the power through an area A. The SI unit for Iis W/m2. The intensity of a sound depends upon its pressure amplitude. 14.5 424 Chapter 14 • Sound The relationship between the intensity of a sound wave and its pressure amplitude (or pressure variation Δp) is 14.6 where ρis the density of the material in which the sound wave travels, in units of kg/m3, and vis the speed
of sound in the medium, in units of m/s. Pressure amplitude has units of pascals (Pa) or N/m2. Note that Δpis half the difference between the maximum and minimum pressure in the sound wave. We can see from the equation that the intensity of a sound is proportional to its amplitude squared. The pressure variation is proportional to the amplitude of the oscillation, and so Ivaries as (Δp)2. This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed during the vibration. Because the power of a sound wave is the rate at which energy is transferred, the energy of a sound wave is also proportional to its amplitude squared. TIPS FOR SUCCESS Pressure is usually denoted by capital P, but we are using a lowercase pfor pressure in this case to distinguish it from power Pabove. Figure 14.10 Graphs of the pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations and has greater pressure maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters. The Decibel Scale You may have noticed that when people talk about the loudness of a sound, they describe it in units of decibels rather than watts per meter squared. While sound intensity (in W/m2) is the SI unit, the sound intensity level in decibels (dB) is more relevant for how humans perceive sounds. The way our ears perceive sound can be more accurately described by the logarithm of the intensity of a sound rather than the intensity of a sound directly. The sound intensity level βis defined to be 14.7 where Iis sound intensity in watts per meter squared, and I0 = 10–12 W/m2 is a reference intensity. I0 is chosen as the reference point because it is the lowest intensity of sound a person with normal hearing can perceive. The decibel level of a sound having an intensity of 10–12 W/m2 is β= 0 dB, because log10 1 = 0. That is, the threshold of human hearing is 0 decibels. Each factor of 10 in intensity corresponds to 10 dB. For example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, 103 times) as intense. Another example is that if one sound is 107 as intense as another, it is 70 dB higher. Since βis defined in terms of a ratio, it is unit-less. The unit called decibel (dB) is used to indicate that this ratio is multiplied by 10. The sound intensity level is not the same as sound intensity—it tells you the levelof the sound relative to a reference intensity rather than the actual intensity. Access for free at openstax.org. 14.2 • Sound Intensity and Sound Level 425 Snap Lab Feeling Sound In this lab, you will play music with a heavy beat to literally feel the vibrations and explore what happens when the volume changes. • CD player or portable electronic device connected to speakers • • a lightweight table rock or rap music CD or mp3 Procedure 1. Place the speakers on a light table, and start playing the CD or mp3. 2. Place your hand gently on the table next to the speakers. 3. 4. Increase the volume and note the level when the table just begins to vibrate as the music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations? GRASP CHECK Do you think that when you double the volume of a sound wave you are doubling the sound intensity level (in dB) or the sound intensity (in a. The sound intensity in b. The sound intensity level in c. The sound intensity in d. The sound intensity level in , because it is a closer measure of how humans perceive sound. because it is a closer measure of how humans perceive sound. because it is the only unit to express the intensity of sound. because it is the only unit to express the intensity of sound. )? Why? Solving Sound Wave Intensity Problems WORKED EXAMPLE Calculating Sound Intensity Levels: Sound Waves Calculate the sound intensity level in decibels for a sound wave traveling in air at 0 ºC and having a pressure amplitude of 0.656 Pa. STRATEGY We are given Δp, so we can calculate Iusing the equation . Using I, we can calculate βstraight from its definition in . Solution (1) Identify knowns: Sound travels at 331 m/s in air at 0 °C. Air has a density of 1.29 kg/m3 at atmospheric pressure and 0ºC. (2) Enter these values and the pressure amplitude into . (3) Enter the value for Iand the known value for I0 into decibels. . Calculate to find the sound intensity level in 426 Chapter 14 • Sound Discussion This 87.0 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any intensities differing by a factor of five. WORKED EXAMPLE Change Intensity Levels of a Sound: What Happens to the Decibel Level? Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher. STRATEGY You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms. Solution (1) Identify knowns: The ratio of the two intensities is 2 to 1, or: We want to show that the difference in sound levels is about 3 dB. That is, we want to show Note that (2) Use the definition of βto get Therefore, 14.8 14.9 14.10 Discussion This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio I2/I1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so on. Practice Problems 7. Calculate the intensity of a wave if the power transferred is 10 W and the area through which the wave is transferred is 5 square meters. a. 200 W / m2 50 W / m2 b. c. 0.5 W / m2 d. 2 W / m2 8. Calculate the sound intensity for a sound wave traveling in air at and having a pressure amplitude of . a. b. c. d. Hearing and Voice People create sounds by pushing air up through their lungs and through elastic folds in the throat called vocal cords. These folds Access for free at openstax.org. 14.2 • Sound Intensity and Sound Level 427 open and close rhythmically, creating a pressure buildup. As air travels up and past the vocal cords, it causes them to vibrate. This vibration escapes the mouth along with puffs of air as sound. A voice changes in pitch when the muscles of the larynx relax or tighten, changing the tension on the vocal chords. A voice becomes louder when air flow from the lungs increases, making the amplitude of the sound pressure wave greater. Hearing is the perception of sound. It can give us plenty of information—such as pitch, loudness, and direction. Humans can normally hear frequencies ranging from approximately 20 to 20,000 Hz. Other animals have hearing ranges different from that of humans. Dogs can hear sounds as high as 45,000 Hz, whereas bats and dolphins can hear up to 110,000 Hz sounds. You may have noticed that dogs respond to the sound of a dog whistle which produces sound out of the range of human hearing. Sounds below 20 Hz are called infrasound, whereas those above 20,000 Hz are ultrasound. The perception of frequency is called pitch, and the perception of intensity is called loudness. The way we hear involves some interesting physics. The sound wave that hits our ear is a pressure wave. The ear converts sound waves into electrical nerve impulses, similar to a microphone. Figure 14.11 shows the anatomy of the ear with its division into three parts: the outer ear or ear canal; the middle ear, which runs from the eardrum to the cochlea; and the inner ear, which is the cochlea itself. The body part normally referred to as the ear is technically called the pinna. Figure 14.11 The illustration shows the anatomy of the human ear. The outer ear, or ear canal, carries sound to the eardrum protected inside of the ear. The middle ear converts sound into mechanical vibrations and applies these vibrations to the cochlea. The lever system of the middle ear takes the force exerted on the eardrum by sound pressure variations, amplifies it and transmits it to the inner ear via the oval window. Two muscles in the middle ear protect the inner ear from very intense sounds. They react to intense sound in a few milliseconds and reduce the force transmitted to the cochlea. This protective reaction can also be triggered by your own voice, so that humming during a fireworks display, for example, can reduce noise damage. Figure 14.12 shows the middle and inner ear in greater detail. As the middle ear bones vibrate, they vibrate the cochlea, which contains fluid. This creates pressure waves in the fluid that cause the tectorial membrane to vibrate. The motion of the tectorial membrane stimulates tiny cilia on specialized cells called hair cells. These hair cells, and their attached neurons, transform the motion of the tectorial membrane into electrical signals that are sent to the brain. The tectorial membrane vibrates at different positions based on the frequency of the incoming sound. This allows us to detect the pitch of sound. Additional processing in the brain also allows us to determine which direction the sound is coming from (based on comparison of the sound’s arrival time and intensity between our two ears). 428 Chapter 14 • Sound Figure 14.12 The inner ear, or cochlea, is a coiled tube about 3 mm in diameter and 3 cm in length when uncoiled. As the stapes vibrates against the oval window, it creates pressure waves that travel through fluid in the cochlea. These waves vibrate the tectorial membrane, which bends the cilia and stimulates nerves in the organ of Corti. These nerves then send information about the sound to the br
ain. FUN IN PHYSICS Musical Instruments Figure 14.13 Playing music, also known as “rocking out”, involves creating vibrations using musical instruments. (John Norton) Yet another way that people make sounds is through playing musical instruments (see the previous figure). Recall that the perception of frequency is called pitch. You may have noticed that the pitch range produced by an instrument tends to depend upon its size. Small instruments, such as a piccolo, typically make high-pitch sounds, while larger instruments, such as a tuba, typically make low-pitch sounds. High-pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds, just as a large instrument creates long-wavelength sounds. Most of us have excellent relative pitch, which means that we can tell whether one sound has a different frequency from another. We can usually distinguish one sound from another if the frequencies of the two sounds differ by as little as 1 Hz. For example, 500.0 and 501.5 Hz are noticeably different. Musical notes are particular sounds that can be produced by most instruments, and are the building blocks of a song. In Western music, musical notes have particular names, such as A-sharp, C, or E-flat. Some people can identify musical notes just by listening to them. This rare ability is called perfect, or absolute, pitch. When a violin plays middle C, there is no mistaking it for a piano playing the same note. The reason is that each instrument produces a distinctive set of frequencies and intensities. We call our perception of these combinations of frequencies and intensities the timbreof the sound. It is more difficult to quantify timbre than loudness or pitch. Timbre is more subjective. Evocative adjectives such as dull, brilliant, warm, cold, pure, and rich are used to describe the timbre of a sound rather than Access for free at openstax.org. 14.2 • Sound Intensity and Sound Level 429 quantities with units, which makes for a difficult topic to dissect with physics. So the consideration of timbre takes us into the realm of perceptual psychology, where higher-level processes in the brain are dominant. This is also true for other perceptions of sound, such as music and noise. But as a teenager, you are likely already aware that one person’s music may be another person’s noise. GRASP CHECK If you turn up the volume of your stereo, will the pitch change? Why or why not? a. No, because pitch does not depend on intensity. b. Yes, because pitch is directly related to intensity. Check Your Understanding 9. What is sound intensity? a. b. c. d. Intensity is the energy per unit area carried by a wave. Intensity is the energy per unit volume carried by a wave. Intensity is the power per unit area carried by a wave. Intensity is the power per unit volume carried by a wave. 10. How is power defined with reference to a sound wave? a. Power is the rate at which energy is transferred by a sound wave. b. Power is the rate at which mass is transferred by a sound wave. c. Power is the rate at which amplitude of a sound wave changes. d. Power is the rate at which wavelength of a sound wave changes. 11. What word or phrase is used to describe the loudness of sound? frequency or oscillation intensity level or decibel timbre a. b. c. d. pitch 12. What is the mathematical expression for sound intensity level ? a. b. c. d. 13. What is the range frequencies that humans are capable of hearing? a. b. c. d. to to to 14. How do humans change the pitch of their voice? a. Relaxing or tightening their glottis b. Relaxing or tightening their uvula c. Relaxing or tightening their tongue d. Relaxing or tightening their larynx References Nave, R. Vocal sound production—HyperPhysics. Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/music/voice.html 430 Chapter 14 • Sound 14.3 Doppler Effect and Sonic Booms Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the Doppler effect of sound waves • Explain a sonic boom • Calculate the frequency shift of sound from a moving object by the Doppler shift formula, and calculate the speed of an object by the Doppler shift formula Section Key Terms Doppler effect sonic boom The Doppler Effect of Sound Waves The Doppler effect is a change in the observed pitch of a sound, due to relative motion between the source and the observer. An example of the Doppler effect due to the motion of a source occurs when you are standing still, and the sound of a siren coming from an ambulance shifts from high-pitch to low-pitch as it passes by. The closer the ambulance is to you, the more sudden the shift. The faster the ambulance moves, the greater the shift. We also hear this shift in frequency for passing race cars, airplanes, and trains. An example of the Doppler effect with a stationary source and moving observer is if you ride a train past a stationary warning bell, you will hear the bell’s frequency shift from high to low as you pass by. What causes the Doppler effect? Let’s compare three different scenarios: Sound waves emitted by a stationary source (Figure 14.14), sound waves emitted by a moving source (Figure 14.15), and sound waves emitted by a stationary source but heard by moving observers (Figure 14.16). In each case, the sound spreads out from the point where it was emitted. If the source and observers are stationary, then observers on either side see the same wavelength and frequency as emitted by the source. But if the source is moving and continues to emit sound as it travels, then the air compressions (crests) become closer together in the direction in which it’s traveling and farther apart in the direction it’s traveling away from. Therefore, the wavelength is shorter in the direction the source is moving (on the right in Figure 14.15), and longer in the opposite direction (on the left in Figure 14.15). Finally, if the observers move, as in Figure 14.16, the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency (and therefore shorter wavelength), and the person moving away from the source receives them at a lower frequency (and therefore longer wavelength). Figure 14.14 Sounds emitted by a source spread out in spherical waves. Because the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. Figure 14.15 Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is Access for free at openstax.org. reduced and, consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitch sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. 14.3 • Doppler Effect and Sonic Booms 431 Figure 14.16 The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary. We know that wavelength and frequency are related by medium and has the same speed vin that medium whether the source is moving or not. Therefore, fmultiplied by constant. Because the observer on the right in Figure 14.15 receives a shorter wavelength, the frequency she perceives must be higher. Similarly, the observer on the left receives a longer wavelength and therefore perceives a lower frequency. where vis the fixed speed of sound. The sound moves in a is a The same thing happens in Figure 14.16. A higher frequency is perceived by the observer moving toward the source, and a lower frequency is perceived by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the perceived frequency. Relative motion apart decreases the perceived frequency. The greater the relative speed is, the greater the effect. WATCH PHYSICS Introduction to the Doppler Effect This video explains the Doppler effect visually. Click to view content (https://www.openstax.org/l/28doppler) GRASP CHECK If you are standing on the sidewalk facing the street and an ambulance drives by with its siren blaring, at what point will the frequency that you observe most closely match the actual frequency of the siren? a. when it is coming toward you b. when it is going away from you c. when it is in front of you For a stationary observer and a moving source of sound, the frequency (fobs) of sound perceived by the observer is 14.11 where fs is the frequency of sound from a source, vs is the speed of the source along a line joining the source and observer, and vw is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away from the observer. TIPS FOR SUCCESS Rather than just memorizing rules, which are easy to forget, it is better to think about the rules of an equation intuitively. Using a minus sign in will decrease the denominator and increase the observed frequency, which is consistent with the expected outcome of the Doppler effect when the source is moving toward the observer. Using a plus sign will increase the denominator and decrease the observed frequency, consistent with what you would expect for the source 432 Chapter 14 • Sound moving away from the observer. This may be more helpful to keep in mind rather than memorizing the fact that “the minus sign is used for motion toward the observer and the plus sign for motion away from the observer.” Note that the greater the speed of the source, the greater the Doppler effect. Similarly, for a stationary source and moving observer, the frequency perceived by
the observer fobs is given by 14.12 where vobs is the speed of the observer along a line joining the source and observer. Here the plus sign is for motion toward the source, and the minus sign is for motion away from the source. Sonic Booms What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency fs. The greater the plane’s speed, vs, the greater the Doppler shift and the greater the value of fobs. Now, as vs approaches the speed of sound, vw, fobs approaches infinity, because the denominator in approaches zero. This result means that at the speed of sound, in front of the source, each wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is theoretically infinite. If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the source when it was approaching are stacked up with those from it when receding, creating a sonic boom. A sonic boom is a constructive interference of sound created by an object moving faster than sound. An aircraft creates two sonic booms, one from its nose and one from its tail (see Figure 14.17). During television coverage of space shuttle landings, two distinct booms could often be heard. These were separated by exactly the time it would take the shuttle to pass by a point. Observers on the ground often do not observe the aircraft creating the sonic boom, because it has passed by before the shock wave reaches them. If the aircraft flies close by at low altitude, pressures in the sonic boom can be destructive enough to break windows. Because of this, supersonic flights are banned over populated areas of the United States. Figure 14.17 Two sonic booms, created by the nose and tail of an aircraft, are observed on the ground after the plane has passed by. Solving Problems Using the Doppler Shift Formula WATCH PHYSICS Doppler Effect Formula for Observed Frequency This video explains the Doppler effect formula for cases when the source is moving toward the observer. Click to view content (https://www.openstax.org/l/28dopplerform) GRASP CHECK Let’s say that you have a rare phobia where you are afraid of the Doppler effect. If you see an ambulance coming your way, what would be the best strategy to minimize the Doppler effect and soothe your Doppleraphobia? Access for free at openstax.org. 14.3 • Doppler Effect and Sonic Booms 433 a. Stop moving and become stationary till it passes by. b. Run toward the ambulance. c. Run alongside the ambulance. WATCH PHYSICS Doppler Effect Formula When Source is Moving Away This video explains the Doppler effect formula for cases when the source is moving away from the observer. Click to view content (https://www.openstax.org/l/28doppleraway) GRASP CHECK Sal uses two different formulas for the Doppler effect-one for when the source is moving toward the observer and another for when the source is moving away. However, in this textbook we use only one formula. Explain. a. The combined formula that can be used is, Use ( source is moving away from the observer. b. The combined formula that can be used is, ) when the source is moving toward the observer and ( ) when the source is moving away from the ) when the . Use ( observer and ( ) when the source is moving toward the observer. c. The combined formula that can be used is, . Use ( ) when the source is moving toward the observer and ( ) when the source is moving away from the observer. d. The combined formula that can be used is, . Use ( ) when the source is moving away from the observer and ( ) when the source is moving toward the observer. WORKED EXAMPLE Calculate Doppler Shift: A Train Horn Suppose a train that has a 150 Hz horn is moving at 35 m/s in still air on a day when the speed of sound is 340 m/s. What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes? Strategy To find the observed frequency, must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. Solution (1) Enter known values into to calculate the frequency observed by a stationary person as the train approaches: (2) Use the same equation but with the plus sign to find the frequency heard by a stationary person as the train recedes. Discussion The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining the train and the observer. In both cases, the shift is significant and easily noticed. Note that the shift is approximately 20 Hz for motion toward and approximately 10 Hz for motion away. The shifts are not symmetric. 434 Chapter 14 • Sound Practice Problems 15. What is the observed frequency when the source having frequency is moving towards the observer at a speed of and the speed of sound is ? a. b. c. d. 16. A train is moving away from you at a speed of . If you are standing still and hear the whistle at a frequency of , what is the actual frequency of the produced whistle? (Assume speed of sound to be .) a. b. c. d. Check Your Understanding 17. What is the Doppler effect? a. The Doppler effect is a change in the observed speed of a sound due to the relative motion between the source and the observer. b. The Doppler effect is a change in the observed frequency of a sound due to the relative motion between the source and the observer. c. The Doppler effect is a change in the observed intensity of a sound due to the relative motion between the source and the observer. d. The Doppler effect is a change in the observed timbre of a sound, due to the relative motion between the source and the observer. 18. Give an example of the Doppler effect caused by motion of the source. a. The sound of a vehicle horn shifts from low-pitch to high-pitch as we move towards it. b. The sound of a vehicle horn shifts from low-pitch to high-pitch as we move away from it. c. The sound of a vehicle horn shifts from low-pitch to high-pitch as it passes by. d. The sound of a vehicle horn shifts from high-pitch to low-pitch as it passes by. 19. What is a sonic boom? a. b. c. d. It is a destructive interference of sound created by an object moving faster than sound. It is a constructive interference of sound created by an object moving faster than sound. It is a destructive interference of sound created by an object moving slower than sound. It is a constructive interference of sound created by an object moving slower than sound. 20. What is the relation between speed of source and value of observed frequency when the source is moving towards the observer? a. They are independent of each other. b. The greater the speed, the greater the value of observed frequency. c. The greater the speed, the smaller the value of observed frequency. d. The speed of the sound is directly proportional to the square of the frequency observed. 14.4 Sound Interference and Resonance Section Learning Objectives By the end of this section, you will be able to do the following: • Describe resonance and beats • Define fundamental frequency and harmonic series • Contrast an open-pipe and closed-pipe resonator • Solve problems involving harmonic series and beat frequency Access for free at openstax.org. 14.4 • Sound Interference and Resonance 435 Section Key Terms beat beat frequency damping fundamental harmonics natural frequency overtones resonance resonate Resonance and Beats Sit in front of a piano sometime and sing a loud brief note at it while pushing down on the sustain pedal. It will sing the same note back at you—the strings that have the same frequencies as your voice, are resonating in response to the forces from the sound waves that you sent to them. This is a good example of the fact that objects—in this case, piano strings—can be forced to oscillate but oscillate best at their natural frequency. A driving force (such as your voice in the example) puts energy into a system at a certain frequency, which is not necessarily the same as the natural frequency of the system. Over time the energy dissipates, and the amplitude gradually reduces to zero- this is called damping. The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. Most of us have played with toys where an object bobs up and down on an elastic band, something like the paddle ball suspended from a finger in Figure 14.18. At first you hold your finger steady, and the ball bounces up and down with a small amount of damping. If you move your finger up and down slowly, the ball will follow along without bouncing much on its own. As you increase the frequency at which you move your finger up and down, the ball will respond by oscillating with increasing amplitude. When you drive the ball at its natural frequency, the ball’s oscillations increase in amplitude with each oscillation for as long as you drive it. As the driving frequency gets progressively higher than the resonant or natural frequency, the amplitude of the oscillations becomes smaller, until the oscillations nearly disappear and your finger simply moves up and down with little effect on the ball. Figure 14.18 The paddle ball on its rubber band moves in response to the finger supporting it. If the finger moves with the natural frequency of the ball on the rubber band, then a resonance is achieved, and the amplitude of the ball’s oscillations increases dramatically. At higher and lower driving frequencies, energy is transferred to the ball less efficiently, and it responds with lower-amplitude oscil
lations. Another example is that when you tune a radio, you adjust its resonant frequency so that it oscillates only at the desired station’s broadcast (driving) frequency. Also, a child on a swing is driven (pushed) by a parent at the swing’s natural frequency to reach the maximum amplitude (height). In all of these cases, the efficiency of energy transfer from the driving force into the oscillator is best at resonance. 436 Chapter 14 • Sound Figure 14.19 Some types of headphones use the phenomena of constructive and destructive interference to cancel out outside noises. All sound resonances are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle to the recognizability of a great singer’s voice, resonance and standing waves play a vital role in sound. Interference happens to all types of waves, including sound waves. In fact, one way to support that something is a waveis to observe interference effects. Figure 14.19 shows a set of headphones that employs a clever use of sound interference to cancel noise. To get destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. In addition to resonance, superposition of waves can also create beats. Beats are produced by the superposition of two waves with slightly different frequencies but the same amplitude. The waves alternate in time between constructive interference and destructive interference, giving the resultant wave an amplitude that varies over time. (See the resultant wave in Figure 14.20). This wave fluctuates in amplitude, or beats, with a frequency called the beat frequency. The equation for beat frequency is where f1 and f2 are the frequencies of the two original waves. If the two frequencies of sound waves are similar, then what we hear is an average frequency that gets louder and softer at the beat frequency. TIPS FOR SUCCESS Don’t confuse the beat frequency with the regular frequency of a wave resulting from superposition. While the beat frequency is given by the formula above, and describes the frequency of the beats, the actual frequency of the wave resulting from superposition is the average of the frequencies of the two original waves. 14.13 Figure 14.20 Beats are produced by the superposition of two waves of slightly different frequencies but identical amplitudes. The waves alternate in time between constructive interference and destructive interference, giving the resulting wave a time-varying amplitude. Virtual Physics Wave Interference Click to view content (https://www.openstax.org/l/28interference) Access for free at openstax.org. 14.4 • Sound Interference and Resonance 437 For this activity, switch to the Sound tab. Turn on the Sound option, and experiment with changing the frequency and amplitude, and adding in a second speaker and a barrier. GRASP CHECK According to the graph, what happens to the amplitude of pressure over time. What is this phenomenon called, and what causes it ? a. The amplitude decreases over time. This phenomenon is called damping. It is caused by the dissipation of energy. b. The amplitude increases over time. This phenomenon is called feedback. It is caused by the gathering of energy. c. The amplitude oscillates over time. This phenomenon is called echoing. It is caused by fluctuations in energy. Fundamental Frequency and Harmonics Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 14.21, Figure 14.22, and Figure 14.23. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This means that the air column has only certain natural frequencies. The figures show how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown. Figure 14.21 Resonance of air in a tube closed at one end, caused by a tuning fork. A disturbance moves down the tube. Figure 14.22 Resonance of air in a tube closed at one end, caused by a tuning fork. The disturbance reflects from the closed end of the tube. Figure 14.23 Resonance of air in a tube closed at one end, caused by a tuning fork. If the length of the tube Lis just right, the disturbance gets back to the tuning fork half a cycle later and interferes constructively with the continuing sound from the tuning fork. This interference forms a standing wave, and the air column resonates. The standing wave formed in the tube has its maximum air displacement (an antinode) at the open end, and no displacement (a 438 Chapter 14 • Sound node) at the closed end. Recall from the last chapter on waves that motion is unconstrained at the antinode, and halted at the node. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; therefore, . This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 14.24. Figure 14.24 The same standing wave is created in the tube by a vibration introduced near its closed end. Since maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube see Figure 14.25). Here the standing wave has three-fourths of its wavelength in the tube, or , so that . There is a whole series of shorter-wavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. All resonant frequencies are multiples of the fundamental, and are called harmonics. The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 14.26 shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end. Figure 14.25 Another resonance for a tube closed at one end. This has maximum air displacements at the open end, and none at the closed end. The wavelength is shorter, with three-fourths equaling the length of the tube, so that . This higher-frequency vibration is the first overtone. Figure 14.26 The fundamental and three lowest overtones for a tube closed at one end. All have maximum air displacements at the open end and none at the closed end. The fundamental and overtones can be present at the same time in a variety of combinations. For example, the note middle C on a trumpet sounds very different from middle C on a clarinet, even though both instruments are basically modified versions of a Access for free at openstax.org. 14.4 • Sound Interference and Resonance 439 tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different. This mix is what gives musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Open-Pipe and Closed-Pipe Resonators The resonant frequencies of a tube closed at one end (known as a closed-pipe resonator) are where f1 is the fundamental, f3 is the first overtone, and so on. Note that the resonant frequencies depend on the speed of sound vand on the length of the tube L. Another type of tube is one that is openat both ends (known as an open-pipe resonator). Examples are some organ pipes, flutes, and oboes. The air columns in tubes open at both ends have maximum air displacements at both ends. (See Figure 14.27). Standing waves form as shown. Figure 14.27 The resonant frequencies of a tube open at both ends are shown, including the fundamental and the first three overtones. In all cases the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end. The resonant frequencies of an open-pipe resonator are where f1 is the fundamental, f2 is the first overtone, f3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube since it has more overtones. An open-pipe resonator has more overtones than a closed-pipe resonator because it has even multiples of the fundamental as well as odd, whereas a closed tube has only odd multiples. In this section we have covered resonance and standing waves for wind instruments, but vibrating strings on stringed instruments also resonate and have fundamentals and overtones similar to those for wind instruments. Solving Problems Involving Harmonic Series and Beat Frequency WORKED EXAMPLE Finding the Length of a Tube for a Closed-Pipe Resonator If sound travels through the air at a speed of 344 m/s, what should be the length of a tube closed at one end to have a fundamental frequency of 128 Hz? 440 Chapter 14 • Sound Strategy The length Lcan be found by rearranging the equation . Solution (1) Ident
ify knowns. • The fundamental frequency is 128 Hz. • The speed of sound is 344 m/s. (2) Use to find the fundamental frequency (n= 1). (3) Solve this equation for length. (4) Enter the values of the speed of sound and frequency into the expression for L. 14.14 14.15 14.16 Discussion Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and therefore, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into loops. WORKED EXAMPLE Finding the Third Overtone in an Open-Pipe Resonator If a tube that’s open at both ends has a fundamental frequency of 120 Hz, what is the frequency of its third overtone? Strategy Since we already know the value of the fundamental frequency (n = 1), we can solve for the third overtone (n = 4) using the equation . Solution Since fundamental frequency (n = 1) is and 14.17 14.18 Discussion To solve this problem, it wasn’t necessary to know the length of the tube or the speed of the air because of the relationship between the fundamental and the third overtone. This example was of an open-pipe resonator; note that for a closed-pipe resonator, the third overtone has a value of n = 7 (not n = 4). WORKED EXAMPLE Using Beat Frequency to Tune a Piano Piano tuners use beats routinely in their work. When comparing a note with a tuning fork, they listen for beats and adjust the string until the beats go away (to zero frequency). If a piano tuner hears two beats per second, and the tuning fork has a Access for free at openstax.org. 14.4 • Sound Interference and Resonance 441 frequency of 256 Hz, what are the possible frequencies of the piano? Strategy Since we already know that the beat frequency fBis 2, and one of the frequencies (let’s say f2) is 256 Hz, we can use the equation to solve for the frequency of the piano f1. Solution Since , we know that either or . Solving for f1, Substituting in values, So, 14.19 14.20 14.21 Discussion The piano tuner might not initially be able to tell simply by listening whether the frequency of the piano is too high or too low and must tune it by trial and error, making an adjustment and then testing it again. If there are even more beats after the adjustment, then the tuner knows that he went in the wrong direction. Practice Problems 21. Two sound waves have frequencies and . What is the beat frequency produced by their superposition? a. b. c. d. 22. What is the length of a pipe closed at one end with fundamental frequency ? (Assume the speed of sound in air is .) a. b. c. d. Check Your Understanding 23. What is damping? a. Over time the energy increases and the amplitude gradually reduces to zero. This is called damping. b. Over time the energy dissipates and the amplitude gradually increases. This is called damping. c. Over time the energy increases and the amplitude gradually increases. This is called damping. d. Over time the energy dissipates and the amplitude gradually reduces to zero. This is called damping. 24. What is resonance? When can you say that the system is resonating? a. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. b. The phenomenon of driving a system with a frequency higher than its natural frequency is called resonance, and a system being driven at its natural frequency does not resonate. c. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance, and a system being driven at its natural frequency does not resonate. d. The phenomenon of driving a system with a frequency higher than its natural frequency is called resonance, and a system being driven at its natural frequency is said to resonate. 442 Chapter 14 • Sound 25. In the tuning fork and tube experiment, in case a standing wave is formed, at what point on the tube is the maximum disturbance from the tuning fork observed? Recall that the tube has one open end and one closed end. a. At the midpoint of the tube b. Both ends of the tube c. At the closed end of the tube d. At the open end of the tube 26. In the tuning fork and tube experiment, when will the air column produce the loudest sound? a. b. c. d. If the tuning fork vibrates at a frequency twice that of the natural frequency of the air column. If the tuning fork vibrates at a frequency lower than the natural frequency of the air column. If the tuning fork vibrates at a frequency higher than the natural frequency of the air column. If the tuning fork vibrates at a frequency equal to the natural frequency of the air column. 27. What is a closed-pipe resonator? a. A pipe or cylindrical air column closed at both ends b. A pipe with an antinode at the closed end c. A pipe with a node at the open end d. A pipe or cylindrical air column closed at one end 28. Give two examples of open-pipe resonators. a. piano, violin b. drum, tabla c. d. rlectric guitar, acoustic guitar flute, oboe Access for free at openstax.org. KEY TERMS amplitude the amount that matter is disrupted during a sound wave, as measured by the difference in height between the crests and troughs of the sound wave. beat a phenomenon produced by the superposition of two waves with slightly different frequencies but the same amplitude beat frequency the frequency of the amplitude fluctuations of a wave damping the reduction in amplitude over time as the energy of an oscillation dissipates decibel a unit used to describe sound intensity levels Doppler effect an alteration in the observed frequency of a sound due to relative motion between the source and the observer fundamental harmonics the term used to refer to the fundamental and the lowest-frequency resonance its overtones hearing the perception of sound SECTION SUMMARY 14.1 Speed of Sound, Frequency, and Wavelength • Sound is one type of wave. • Sound is a disturbance of matter that is transmitted from its source outward in the form of longitudinal waves. • The relationship of the speed of sound v, its frequency f, and its wavelength is given by same relationship given for all waves. , which is the • The speed of sound depends upon the medium through • which the sound wave is travelling. In a given medium at a specific temperature (or density), the speed of sound vis the same for all frequencies and wavelengths. 14.2 Sound Intensity and Sound Level • The intensity of a sound is proportional to its amplitude squared. • The energy of a sound wave is also proportional to its amplitude squared. • Sound intensity level in decibels (dB) is more relevant for how humans perceive sounds than sound intensity (in W/m2), even though sound intensity is the SI unit. • Sound intensity level is not the same as sound intensity—it tells you the levelof the sound relative to a reference intensity rather than the actual intensity. • Hearing is the perception of sound and involves that transformation of sound waves into vibrations of parts within the ear. These vibrations are then transformed Chapter 14 • Key Terms 443 loudness the perception of sound intensity natural frequency the frequency at which a system would oscillate if there were no driving and no damping forces overtones all resonant frequencies higher than the fundamental pitch the perception of the frequency of a sound rarefaction a low-pressure region in a sound wave resonance the phenomenon of driving a system with a frequency equal to the system's natural frequency resonate to drive a system at its natural frequency sonic boom a constructive interference of sound created by an object moving faster than sound sound a disturbance of matter that is transmitted from its source outward by longitudinal waves sound intensity the power per unit area carried by a sound wave sound intensity level the level of sound relative to a fixed standard related to human hearing into neural signals that are interpreted by the brain. • People create sounds by pushing air up through their lungs and through elastic folds in the throat called vocal cords. 14.3 Doppler Effect and Sonic Booms • The Doppler effect is a shift in the observed frequency of a sound due to motion of either the source or the observer. • The observed frequency is greater than the actual source’s frequency when the source and the observer are moving closer together, either by the source moving toward the observer or the observer moving toward the source. • A sonic boom is constructive interference of sound created by an object moving faster than sound. 14.4 Sound Interference and Resonance • A system’s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces. • A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate. • Beats occur when waves of slightly different frequencies • are superimposed. In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are 444 Chapter 14 • Key Equations called harmonics. • The resonant frequencies of a tube closed at one end are , where f1is the fundamental and L is the length of the tube. • The resonant frequencies of a tube open at both ends are KEY EQUATIONS 14.1 Speed of Sound, Frequency, and Wavelength speed of sound 14.2 Sound Intensity and Sound Level intensity sound intensity sound intensity level CHAPTER REVIEW Concept Items 14.1 Speed of Sound, Frequency, and Wavelength 1. What is the amplitude of a sound wave perceived by the loudness human ear? a. b. pitch c. d. intensity timbre 2. The compressibility of air and hydrogen is almost the same. Which factor is the reason that sound travels faster in hydrogen than in air? a. Hydrogen is more dense than air. b. Hydrogen is less dense than air. c. Hydrogen atoms are heavier than air mol
ecules. d. Hydrogen atoms are lighter than air molecules. 14.3 Doppler Effect and Sonic Booms Doppler effect observed frequency (moving source) Doppler effect observed frequency (moving observer) 14.4 Sound Interference and Resonance beat frequency resonant frequencies of a closed-pipe resonator resonant frequencies of an open-pipe resonator a. b. c. d. 4. How does the "decibel" get its name? a. The meaning of deci is “hundred” and the number of decibels is one-hundredth of the logarithm to base 10 of the ratio of two sound intensities. b. The meaning of deci is "ten" and the number of decibels is one-tenth of the logarithm to base 10 of the ratio of two sound intensities. c. The meaning of deci is “one-hundredth” and the number of decibels is hundred times the logarithm to base 10 of the ratio of two sound intensities. d. The meaning of deci is “one-tenth” and the number of decibels is ten times the logarithm to base 10 of the ratio of two sound intensities. 14.2 Sound Intensity and Sound Level 5. What is “timbre” of sound? a. Timbre is the quality of the sound that distinguishes 3. What is the mathematical relationship between intensity, it from other sound power, and area? Access for free at openstax.org. Chapter 14 • Chapter Review 445 b. Timbre is the loudness of the sound that distinguishes it from other sound. c. Timbre is the pitch of the sound that distinguishes it from other sound. down is greater than the amplitude of the yo-yo b. when the amplitude of the finger moving up and down is less than the amplitude of the yo-yo c. when the frequency of the finger moving up and d. Timbre is the wavelength of the sound that down is equal to the resonant frequency of the yo-yo distinguishes it from other sound. d. when the frequency of the finger moving up and 14.3 Doppler Effect and Sonic Booms 6. Two sources of sound producing the same frequency are moving towards you at different speeds. Which one would sound more high-pitched? the one moving slower a. the one moving faster b. 7. When the speed of the source matches the speed of sound, what happens to the amplitude of the sound wave? Why? a. It approaches zero. This is because all wave crests are superimposed on one another through constructive interference. It approaches infinity. This is because all wave crests are superimposed on one another through constructive interference. It approaches zero, because all wave crests are superimposed on one another through destructive interference. It approaches infinity, because all wave crests are superimposed on one another through destructive interference. b. c. d. 8. What is the mathematical expression for the frequency perceived by the observer in the case of a stationary observer and a moving source? a. b. c. d. 14.4 Sound Interference and Resonance 9. When does a yo-yo travel the farthest from the finger? a. when the amplitude of the finger moving up and down is different from the resonant frequency of the yo-yo 10. What is the difference between harmonics and overtones? a. Harmonics are all multiples of the fundamental frequency. The first overtone is actually the first harmonic. b. Harmonics are all multiples of the fundamental frequency. The first overtone is actually the second harmonic. c. Harmonics are all multiples of the fundamental frequency. The second overtone is actually the first harmonic. d. Harmonics are all multiples of the fundamental frequency. The third overtone is actually the second harmonic. 11. What kind of waves form in pipe resonators? a. damped waves b. propagating waves c. high-frequency waves d. standing waves 12. What is the natural frequency of a system? a. The natural frequency is the frequency at which a system oscillates when it undergoes forced vibration. b. The natural frequency is the frequency at which a system oscillates when it undergoes damped oscillation. c. The natural frequency is the frequency at which a system oscillates when it undergoes free vibration without a driving force or damping. d. The natural frequency is the frequency at which a system oscillates when it undergoes forced vibration with damping. Critical Thinking Items d. It remains constant. 14.1 Speed of Sound, Frequency, and Wavelength 13. What can be said about the frequency of a monotonous sound? a. b. c. It decreases with time. It decreases with distance. It increases with distance. 14. A scientist notices that a sound travels faster through a solid material than through the air. Which of the following can explain this? a. Solid materials are denser than air. b. Solid materials are less dense than air. c. A solid is more rigid than air. d. A solid is easier to compress than air. 446 Chapter 14 • Chapter Review 14.2 Sound Intensity and Sound Level 15. Which property of the wave is related to its intensity? How? a. The frequency of the wave is related to the intensity of the sound. The larger-frequency oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. b. The wavelength of the wave is related to the intensity of the sound. The longer-wavelength oscillations indicate greater pressure maxima and minima, and the pressure is higher in greaterintensity sound. c. The amplitude of the wave is related to the intensity of the sound. The larger-amplitude oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. d. The speed of the wave is related to the intensity of the sound. The higher-speed oscillations indicate greater pressure maxima and minima, and the pressure is higher in greater-intensity sound. 16. Why is decibel (dB) used to describe loudness of sound? a. Because, human ears have an inverse response to the amplitude of sound. b. Because, human ears have an inverse response to the intensity of sound. c. Because, the way our ears perceive sound can be more accurately described by the amplitude of a sound rather than the intensity of a sound directly. d. Because, the way our ears perceive sound can be more accurately described by the logarithm of the intensity of a sound rather than the intensity of a sound directly. 17. How can humming while shooting a gun reduce ear damage? a. Humming can trigger those two muscles in the outer ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. b. Humming can trigger those three muscles in the outer ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. c. Humming can trigger those two muscles in the middle ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. d. Humming can trigger those three muscles in the middle ear that react to intense sound produced while shooting and reduce the force transmitted to the cochlea. 18. A particular sound, S1, has an intensity times that of Access for free at openstax.org. another sound, S2. What is the difference in sound intensity levels measured in decibels? a. b. c. d. 14.3 Doppler Effect and Sonic Booms 19. When the source of sound is moving through the air, does the speed of sound change with respect to a stationary person standing nearby? a. Yes b. No 20. Why is no sound heard by the observer when an object approaches him at a speed faster than that of sound? a. If the source exceeds the speed of sound, then destructive interference occurs and no sound is heard by the observer when an object approaches him. If the source exceeds the speed of sound, the frequency of sound produced is beyond the audible range of sound. If the source exceeds the speed of sound, all the sound waves produced approach minimum intensity and no sound is heard by the observer when an object approaches him. If the source exceeds the speed of sound, all the sound waves produced are behind the source. Hence, the observer hears the sound only after the source has passed. b. c. d. 21. Does the Doppler effect occur when the source and observer are both moving towards each other? If so, how would this affect the perceived frequency? a. Yes, the perceived frequency will be even lower in this case than if only one of the two were moving. b. No, the Doppler effect occurs only when an observer is moving towards a source. c. No, the Doppler effect occurs only when a source is moving towards an observer. d. Yes, the perceived frequency will be even higher in this case than if only one of the two were moving. 14.4 Sound Interference and Resonance 22. When does the amplitude of an oscillating system become maximum? a. When two sound waves interfere destructively. b. When the driving force produces a transverse wave in the system. c. When the driving force of the oscillator to the oscillating system is at a maximum amplitude. d. When the frequency of the oscillator equals the natural frequency of the oscillating system. 23. How can a standing wave be formed with the help of a tuning fork and a closed-end tube of appropriate length? a. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes destructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly one b. c. Problems 14.1 Speed of Sound, Frequency, and Wavelength 25. A bat produces a sound at and wavelength . What is the speed of the sound? a. b. c. d. 26. A sound wave with frequency of is traveling . By how much will its wavelength through air at change when it enters aluminum? a. b. c. d. 14.2 Sound Intensity and Sound Level 27. Calculate the sound intensity for a sound wave traveling through air at 15° C and having a pressure amplitude of 0.80 Pa. (Hint—Speed o
f sound in air at 15° C is 340 m/s .) a. 9.6×10−3 W / m2 7.7×10−3 W / m2 b. c. 9.6×10−4 W / m2 7.7×10−4 W / m2 d. Chapter 14 • Chapter Review 447 d. full cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly one full cycle later, and it interferes destructively with the continuing sound produced by the tuning fork. 24. A tube open at both ends has a fundamental frequency . What will the frequency be if one end is of closed? a. b. c. d. 14.3 Doppler Effect and Sonic Booms 29. An ambulance is moving away from you. You are standing still and you hear its siren at a frequency of . You know that the actual frequency of the siren . What is the speed of the ambulance? .) is (Assume the speed of sound to be a. b. c. d. 30. An ambulance passes you at a speed of . If its siren has a frequency of , what is difference in the frequencies you perceive before and after it passes you? (Assume the speed of sound in air is a. b. c. d. .) 14.4 Sound Interference and Resonance 31. What is the length of an open-pipe resonator with a ? (Assume the .) fundamental frequency of speed of sound is a. b. c. d. 28. The sound level in dB of a sound traveling through air at is . Calculate its pressure amplitude. 32. An open-pipe resonator has a fundamental frequency of . By how much would its length have to be a. b. c. d. changed to get a fundamental frequency of (Assume the speed of sound is a. b. c. d. .) ? 448 Chapter 14 • Test Prep Performance Task 14.4 Sound Interference and Resonance 33. Design and make an open air resonator capable of playing at least three different pitches (frequencies) of sound using a selection of bamboo of varying widths and lengths, which can be obtained at a local hardware store. Choose a piece of bamboo for creating a musical TEST PREP Multiple Choice 14.1 Speed of Sound, Frequency, and Wavelength 34. What properties does a loud, shrill whistle have? a. high amplitude, high frequency b. high amplitude, low frequency low amplitude, high frequency c. low amplitude, low frequency d. 35. What is the speed of sound in fresh water at degrees Celsius? a. b. c. d. 36. A tuning fork oscillates at a frequency of , creating sound waves. How many waves will reach the eardrum of a person near that fork in seconds? a. b. c. d. 37. Why does the amplitude of a sound wave decrease with distance from its source? a. The amplitude of a sound wave decreases with distance from its source, because the frequency of the sound wave decreases. b. The amplitude of a sound wave decreases with distance from its source, because the speed of the sound wave decreases. c. The amplitude of a sound wave decreases with distance from its source, because the wavelength of the sound wave increases. d. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. 38. Does the elasticity of the medium affect the speed of sound? How? a. No, there is no relationship that exists between the speed of sound and elasticity of the medium. Access for free at openstax.org. pipe. Calculate the length required for a certain frequency to resonate and then mark the locations where holes should be placed in the pipe to achieve their desired pitches. Use a simple hand drill or ask your wood shop department for help drilling holes. Use tuning forks to test and calibrate your instrument. Demonstrate your pipe for the class. b. Yes. When particles are more easily compressed in a medium, sound does not travel as quickly through the medium. c. Yes. When the particles in a medium do not compress much, sound does not travel as quickly through the medium. d. No, the elasticity of a medium affects frequency and wavelength, not wave speed. 14.2 Sound Intensity and Sound Level 39. Which of the following terms is a useful quantity to intensity frequency describe the loudness of a sound? a. b. c. pitch d. wavelength 40. What is the unit of sound intensity level? a. decibels b. hertz c. watts 41. If a particular sound S1 is times more intense than another sound S2, then what is the difference in sound intensity levels in dB for these two sounds? a. b. c. 42. By what minimum amount should frequencies vary for humans to be able to distinguish two separate sounds? a. b. c. d. 43. Why is I0chosen as the reference for sound intensity? a. Because, it is the highest intensity of sound a person with normal hearing can perceive at a frequency of 100 Hz. b. Because, it is the lowest intensity of sound a person with normal hearing can perceive at a frequency of 100 Hz. c. Because, it is the highest intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. d. Because, it is the lowest intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. 14.3 Doppler Effect and Sonic Booms 44. In which of the following situations is the Doppler effect absent? a. The source and the observer are moving towards each other. b. The observer is moving toward the source. c. The source is moving away from the observer. d. Neither the source nor the observer is moving relative to one another. 45. What does the occurrence of the sonic boom depend on? speed of the source a. b. frequency of source c. amplitude of source d. distance of observer from the source 46. What is the observed frequency when the observer is ? The source moving away from the source at frequency is . and the speed of sound is a. b. c. d. 47. How will your perceived frequency change if the source is moving towards you? Short Answer 14.1 Speed of Sound, Frequency, and Wavelength 52. What component of a longitudinal sound wave is analogous to a trough of a transverse wave? a. b. c. node d. antinode compression rarefaction 53. What is the frequency of a sound wave as perceived by the human ear? timbre a. loudness b. intensity c. d. pitch 54. What properties of a solid determine the speed of sound traveling through it? Chapter 14 • Test Prep 449 a. The frequency will become lower. b. The frequency will become higher. 14.4 Sound Interference and Resonance 48. Observation of which phenomenon can be considered interference proof that something is a wave? a. b. noise c. d. reflection conduction 49. Which of the resonant frequencies has the greatest amplitude? a. The first harmonic b. The second harmonic c. The first overtone d. The second overtone 50. What is the fundamental frequency of an open-pipe resonator? a. b. c. d. ? 51. What is the beat frequency produced by the superposition of two waves with frequencies and a. b. c. d. a. mass and density b. rigidity and density c. volume and density d. shape and rigidity 55. Does the density of a medium affect the speed of sound? a. No b. Yes 56. Does a bat make use of the properties of sound waves to locate its prey? a. No b. Yes 57. Do the properties of a sound wave change when it travels from one medium to another? a. No b. Yes 450 Chapter 14 • Test Prep 14.2 Sound Intensity and Sound Level meter squared. 58. When a passing driver has his stereo turned up, you 61. Why is the reference intensity cannot even hear what the person next to you is saying. Why is this so? a. The sound from the passing car’s stereo has a higher amplitude and hence higher intensity compared to the intensity of the sound coming from the person next to you. The higher intensity corresponds to greater loudness, so the first sound dominates the second. b. The sound from the passing car’s stereo has a higher amplitude and hence lower intensity compared to the intensity of the sound coming from the person next to you. The lower intensity corresponds to greater loudness, so the first sound dominates the second. c. The sound from the passing car’s stereo has a higher frequency and hence higher intensity compared to the intensity of the sound coming from the person next to you. The higher frequency corresponds to greater loudness so the first sound dominates the second. d. The sound from the passing car’s stereo has a lower frequency and hence higher intensity compared to the intensity of the sound coming from the person next to you. The lower frequency corresponds to greater loudness, so the first sound dominates the second. 59. For a constant area, what is the relationship between intensity of a sound wave and power? a. The intensity is inversely proportional to the power transmitted by the wave, for a constant area. b. The intensity is inversely proportional to the square of the power transmitted by the wave, for a constant area. c. The intensity is directly proportional to the square of the power transmitted by the wave, for a constant area. d. The intensity is directly proportional to the power transmitted by the wave, for a constant area. ? decibels, a. The upper limit of human hearing is , i.e. . For . b. The lower threshold of human hearing is decibels, i.e. . For , c. The upper limit of human hearing is decibels, i.e. . For , d. The lower threshold of human hearing is decibels, i.e., . For , 62. Given that the sound intensity level of a particular wave , what will be the sound intensity for that wave? is a. b. c. d. 63. For a sound wave with intensity , calculate the pressure amplitude given that the sound . travels through air at a. b. c. d. 64. Which nerve carries auditory information to the brain? a. buccal nerve b. peroneal nerve c. cochlear nerve d. mandibular nerve 65. Why do some smaller instruments, such as piccolos, produce higher-pitched sounds than larger instruments, such as tubas? a. Smaller instruments produce sounds with shorter wavelengths, and thus higher frequencies. b. Smaller instruments produce longer wavelength, and thus higher amplitude, sounds. c. Smaller instruments produce lower amplitude, and thus longer wavelength sounds. 60. What does stand for in the equation d. Smaller instruments produce higher amplitude, ? What is i
ts unit? and thus lower frequency, sounds. a. Yes, is the sound intensity in watts per meter squared in the equation, . 14.3 Doppler Effect and Sonic Booms is the sound illuminance and its unit is lumen 66. How will your perceived frequency change if you move b. c. per meter squared. is the sound intensity and its unit is watts per meter cubed. d. is the sound intensity and its unit is watts per Access for free at openstax.org. away from a stationary source of sound? a. The frequency will become lower. b. The frequency will be doubled. c. The frequency will be tripled. d. The frequency will become higher. 67. True or false—The Doppler effect also occurs with waves other than sound waves. a. False b. True 68. A source of sound is moving towards you. How will what you hear change if the speed of the source increases? a. The sound will become more high-pitched. b. The sound will become more low-pitched. c. The pitch of the sound will not change. 69. Do sonic booms continue to be created when an object is traveling at supersonic speeds? a. No, a sonic boom is created only when the source exceeds the speed of sound. b. Yes, sonic booms continue to be created when an Chapter 14 • Test Prep 451 a. Human speech is produced by shaping the cavity formed by the throat and mouth, the vibration of vocal cords, and using the tongue to adjust the fundamental frequency and combination of overtones. b. Human speech is produced by shaping the cavity formed by the throat and mouth into a closed pipe and using tongue to adjust the fundamental frequency and combination of overtones. c. Human speech is produced only by the vibrations of the tongue. d. Human speech is produced by elongating the vocal cords. 75. What is the possible number of nodes and antinodes along one full wavelength of a standing wave? a. nodes and antinodes or antinodes and object is traveling at supersonic speeds. nodes. 70. Suppose you are driving at a speed of and you . ? hear the sound of a bell at a frequency of What is the actual frequency of the bell if the speed of sound is a. b. c. d. 71. What is the frequency of a stationary sound source if you hear it at 1200.0 Hz while moving towards it at a speed of 50.0 m/s? (Assume speed of sound to be 331 m/s.) a. b. c. d. 1410 Hz 1380 Hz 1020 Hz 1042 Hz 14.4 Sound Interference and Resonance 72. What is the actual frequency of the wave produced as a result of superposition of two waves? a. It is the average of the frequencies of the two original waves that were superimposed. It is the difference between the frequencies of the two original waves that were superimposed. It is the product of the frequencies of the two original waves that were superimposed. It is the sum of the frequencies of the two original waves that were superimposed. b. c. d. 73. Can beats be produced through a phenomenon different from resonance? How? a. No, beats can be produced only by resonance. b. Yes, beats can be produced by superimposition of any two waves having slightly different frequencies. 74. How is human speech produced? b. c. nodes and antinodes or antinodes and nodes. nodes and antinodes or antinodes and nodes. d. nodes and antinodes or antinodes and nodes. 76. In a pipe resonator, which frequency will be the least second overtone frequency intense of those given below? a. b. first overtone frequency fundamental frequency c. third overtone frequency d. 77. A flute is an open-pipe resonator. If a flute is long, what is the longest wavelength it can produce? a. b. c. d. 78. What is the frequency of the second overtone of a ? closed-pipe resonator with a length of (Assume the speed of sound is a. b. c. d. .) when the speed of sound is 79. An open-pipe resonator has a fundamental frequency of . What will its fundamental frequency be when the speed of sound is a. b. c. d. ? 452 Chapter 14 • Test Prep Extended Response eardrum. 14.1 Speed of Sound, Frequency, and Wavelength 80. How is a human able to hear sounds? a. Sound waves cause the eardrum to vibrate. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person as sound. b. Sound waves cause the ear canal to vibrate. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person as sound. c. Sound waves transfer electrical impulses to the eardrum. A complicated mechanism converts the electrical impulses to sound. d. Sound waves transfer mechanical vibrations to the ear canal, and the eardrum converts them to electrical impulses. 81. Why does sound travel faster in iron than in air even though iron is denser than air? a. The density of iron is greater than that of air. However, the rigidity of iron is much greater than that of air. Hence, sound travels faster in it. b. The density of iron is greater than that of air. However, the rigidity of iron is much less than that of air. Hence, sound travels faster in it. c. The density of iron is greater than that of air. However, the rigidity of iron is equal to that of air. Hence, sound travels faster in it. d. The mass of iron is much less than that of air and the rigidity of iron is much greater than that of air. Hence, sound travels faster in it. 82. Is the speed of sound dependent on its frequency? a. No b. Yes 14.2 Sound Intensity and Sound Level 83. Why is the sound from a tire burst louder than that from a finger snap? a. The sound from the tire burst has higher pressure amplitudes, hence it can exert smaller force on the eardrum. b. The sound from the tire burst has lower pressure amplitudes, hence it can exert smaller force on the eardrum. c. The sound from the tire burst has lower pressure amplitudes, hence it can exert larger force on the ear drum. d. The sound from the tire burst has higher pressure amplitudes, hence it can exert larger force on the Access for free at openstax.org. 84. Sound A is times more intense than Sound B. What will be the difference in decibels in their sound intensity levels? a. b. c. d. 85. The ratio of the pressure amplitudes of two sound waves . What will be the is traveling through water at difference in their sound intensity levels in dB? a. b. c. d. 86. Which of the following most closely models how sound is produced by the vocal cords? a. A person plucks a string. b. A person blows over the mouth of a half-filled glass bottle. c. A person strikes a hammer against a hard surface. d. A person blows through a small slit in a wide, stretched rubber band. 14.3 Doppler Effect and Sonic Booms 87. True or false—The Doppler effect occurs only when the sound source is moving. a. False b. True 88. True or false—The observed frequency becomes infinite when the source is moving at the speed of sound. a. False b. True 89. You are driving alongside a train. You hear its horn at a pitch that is lower than the actual frequency. What should you do to match the speed of the train? Why? a. In order to match the speed of the train, one would need to increase or decrease the speed of his/her car because a lower pitch means that either the train (the source) is moving away or that you (the observer) are moving away. In order to match the speed of the train, one would need to drive at a constant speed because a lower pitch means that the train and the car are at the same speed. b. 14.4 Sound Interference and Resonance 90. How are the beat frequency and the regular frequency of a wave resulting from superposition of two waves different? a. Beat frequency is the sum of two frequencies and regular frequency is the difference between frequencies of two original waves. overtone so resonance will occur. c. The frequency formed is a harmonic and third overtone so resonance will occur. b. Beat frequency is the difference between the d. The frequency formed is a harmonic and fourth Chapter 14 • Test Prep 453 constituent frequencies, but the regular frequency is the average of the frequencies of the two original waves. c. Beat frequency is the sum of two frequencies and regular frequency is the average of frequencies of two original waves. d. Beat frequency is the average of frequencies of two original waves and regular frequency is the sum of two original frequencies. 91. In the tuning fork and tube experiment, if resonance is is the length of the tube formed for , where and is the wavelength of the sound wave, can resonance also be formed for a wavelength Why? a. The frequency formed is a harmonic and first ? overtone so resonance will occur. b. The frequency formed is a harmonic and second overtone so resonance will occur. 92. True or false—An open-pipe resonator has more overtones than a closed-pipe resonator. a. False b. True 93. A flute has finger holes for changing the length of the resonating air column, and therefore, the frequency of the note played. How far apart are two holes that, when closed, play two frequencies that are apart, if the first hole is the flute? a. b. c. d. away from the mouthpiece of 454 Chapter 14 • Test Prep Access for free at openstax.org. CHAPTER 15 Light Figure 15.1 Human eyes detect these orange sea goldiefish swimming over a coral reef in the blue waters of the Gulf of Eilat, in the Red Sea, using visible light. (credit: David Darom, Wikimedia Commons) Chapter Outline 15.1 The Electromagnetic Spectrum 15.2 The Behavior of Electromagnetic Radiation INTRODUCTION The beauty of a coral reef, the warm radiance of sunshine, the sting of sunburn, the X-ray revealing a broken bone, even microwave popcorn—all are brought to us by electromagnetic waves. The list of the various types of electromagnetic waves, ranging from radio transmission waves to nuclear gamma-ray (γ-ray) emissions, is interesting in itself. Even more intriguing is that all of these different phenomena are manifestations of the same thing—electromagnetic waves (see Figure 15.1). What are electromagnetic waves? How are they created, and how do they travel? How can we understand their widely varying properties? What is the relationship between electric and
magnetic effects? These and other questions will be explored. 15.1 The Electromagnetic Spectrum Section Learning Objectives By the end of this section, you will be able to do the following: • Define the electromagnetic spectrum, and describe it in terms of frequencies and wavelengths • Describe and explain the differences and similarities of each section of the electromagnetic spectrum and the applications of radiation from those sections Section Key Terms electric field electromagnetic radiation (EMR) magnetic field Maxwell’s equations 456 Chapter 15 • Light The Electromagnetic Spectrum We generally take light for granted, but it is a truly amazing and mysterious form of energy. Think about it: Light travels to Earth across millions of kilometers of empty space. When it reaches us, it interacts with matter in various ways to generate almost all the energy needed to support life, provide heat, and cause weather patterns. Light is a form of electromagnetic radiation (EMR). The term lightusually refers to visible light, but this is not the only form of EMR. As we will see, visible light occupies a narrow band in a broad range of types of electromagnetic radiation. Electromagnetic radiation is generated by a moving electric charge, that is, by an electric current. As you will see when you study electricity, an electric current generates both an electric field, E, and a magnetic field, B. These fields are perpendicular to each other. When the moving charge oscillates, as in an alternating current, an EM wave is propagated. Figure 15.2 shows how an electromagnetic wave moves away from the source—indicated by the ~ symbol. WATCH PHYSICS Electromagnetic Waves and the Electromagnetic Spectrum This video, link below, is closely related to the following figure. If you have questions about EM wave properties, the EM spectrum, how waves propagate, or definitions of any of the related terms, the answers can be found in this video (http://www.openstax.org/l/28EMWaves) . Click to view content (https://www.openstax.org/l/28EMWaves) GRASP CHECK In an electromagnetic wave, how are the magnetic field, the electric field, and the direction of propagation oriented to each other? a. All three are parallel to each other and are along the x-axis. b. All three are mutually perpendicular to each other. c. The electric field and magnetic fields are parallel to each other and perpendicular to the direction of propagation. d. The magnetic field and direction of propagation are parallel to each other along the y-axis and perpendicular to the electric field. Virtual Physics Radio Waves and Electromagnetic Fields Click to view content (https://www.openstax.org/l/28Radiowaves) This simulation demonstrates wave propagation. The EM wave is propagated from the broadcast tower on the left, just as in Figure 15.2. You can make the wave yourself or allow the animation to send it. When the wave reaches the antenna on the right, it causes an oscillating current. This is how radio and television signals are transmitted and received. GRASP CHECK Where do radio waves fall on the electromagnetic spectrum? a. Radio waves have the same wavelengths as visible light. b. Radio waves fall on the high-frequency side of visible light. c. Radio waves fall on the short-wavelength side of visible light. d. Radio waves fall on the low-frequency side of visible light. Access for free at openstax.org. 15.1 • The Electromagnetic Spectrum 457 Figure 15.2 A part of the electromagnetic wave sent out from an oscillating charge at one instant in time. The electric and magnetic fields (E and B) are in phase, and they are perpendicular to each other and to the direction of propagation. For clarity, the waves are shown only along one direction, but they propagate out in other directions too. From your study of sound waves, recall these features that apply to all types of waves: • Wavelength—The distance between two wave crests or two wave troughs, expressed in various metric measures of distance • Frequency—The number of wave crests that pass a point per second, expressed in hertz (Hz or s–1) • Amplitude: The height of the crest above the null point As mentioned, electromagnetic radiation takes several forms. These forms are characterized by a range of frequencies. Because frequency is inversely proportional to wavelength, any form of EMR can also be represented by its range of wavelengths. Figure 15.3 shows the frequency and wavelength ranges of various types of EMR. With how many of these types are you familiar? Figure 15.3 The electromagnetic spectrum, showing the major categories of electromagnetic waves. The range of frequencies and wavelengths is remarkable. The dividing line between some categories is distinct, whereas other categories overlap. Take a few minutes to study the positions of the various types of radiation on the EM spectrum, above. Sometimes all radiation with frequencies lower than those of visible light are referred to as infrared (IR) radiation. This includes radio waves, which overlap with the frequencies used for media broadcasts of TV and radio signals. The microwave radiation that you see on the diagram is the same radiation that is used in a microwave oven. What we feel as radiant heat is also a form of low-frequency EMR. All the high-frequency radiation to the right of visible light is sometimes referred to as ultraviolet (UV) radiation. This includes X-rays and gamma (γ) rays. The narrow band that is visible light extends from lower-frequency red light to higher-frequency violet light, thus the terms are infrared(below red) and ultraviolet(beyond violet). BOUNDLESS PHYSICS Maxwell’s Equations The Scottish physicist James Clerk Maxwell (1831–1879) is regarded widely to have been the greatest theoretical physicist of the 458 Chapter 15 • Light nineteenth century. Although he died young, Maxwell not only formulated a complete electromagnetic theory, represented by Maxwell’s equations, he also developed the kinetic theory of gases, and made significant contributions to the understanding of color vision and the nature of Saturn’s rings. Maxwell brought together all the work that had been done by brilliant physicists, such as Ørsted, Coulomb, Ampere, Gauss, and Faraday, and added his own insights to develop the overarching theory of electromagnetism. Maxwell’s equations are paraphrased here in words because their mathematical content is beyond the level of this text. However, the equations illustrate how apparently simple mathematical statements can elegantly unite and express a multitude of concepts—why mathematics is the language of science. Maxwell’s Equations 1. Electric field lines originate on positive charges and terminate on negative charges. The electric field is defined as the force per unit charge on a test charge, and the strength of the force is related to the electric constant, ε0. 2. Magnetic field lines are continuous, having no beginning or end. No magnetic monopoles are known to exist. The strength of the magnetic force is related to the magnetic constant, μ0. 3. A changing magnetic field induces an electromotive force (emf) and, hence, an electric field. The direction of the emf opposes the change, changing direction of the magnetic field. 4. Magnetic fields are generated by moving charges or by changing electric fields. Maxwell’s complete theory shows that electric and magnetic forces are not separate, but different manifestations of the same thing—the electromagnetic force. This classical unification of forces is one motivation for current attempts to unify the four basic forces in nature—the gravitational, electromagnetic, strong nuclear, and weak nuclear forces. The weak nuclear and electromagnetic forces have been unified, and further unification with the strong nuclear force is expected; but, the unification of the gravitational force with the other three has proven to be a real head-scratcher. One final accomplishment of Maxwell was his development in 1855 of a process that could produce color photographic images. In 1861, he and photographer Thomas Sutton worked together on this process. The color image was achieved by projecting red, blue, and green light through black-and-white photographs of a tartan ribbon, each photo itself exposed in different-colored light. The final image was projected onto a screen (see Figure 15.4). Figure 15.4 Maxwell and Sutton’s photograph of a colored ribbon. This was the first durable color photograph. The plaid tartan of the Scots made a colorfulphotographic subject. GRASP CHECK Describe electromagnetic force as explained by Maxwell’s equations. a. According to Maxwell’s equations, electromagnetic force gives rise to electric force and magnetic force. b. According to Maxwell’s equations, electric force and magnetic force are different manifestations of electromagnetic force. c. According to Maxwell’s equations, electric force is the cause of electromagnetic force. d. According to Maxwell’s equations, magnetic force is the cause of electromagnetic force. Characteristics of Electromagnetic Radiation All the EM waves mentioned above are basically the same form of radiation. They can all travel across empty space, and they all Access for free at openstax.org. 15.1 • The Electromagnetic Spectrum 459 travel at the speed of light in a vacuum. The basic difference between types of radiation is their differing frequencies. Each frequency has an associated wavelength. As frequency increases across the spectrum, wavelength decreases. Energy also increases with frequency. Because of this, higher frequencies penetrate matter more readily. Some of the properties and uses of the various EM spectrum bands are listed in Table 15.1. Types of EM Waves Radio and TV Production Applications Life Sciences Aspect Issues Accelerating charges Communications, remote controls MRI Requires controls for band use Microwaves Accelerating charges & thermal agitation Communications, microwave ovens, radar Deep heating Cell p
hone use Infrared Thermal agitation & electronic transitions Thermal imaging, heating Visible Light Thermal agitation & electronic transitions All pervasive Greenhouse effect Absorption by atmosphere Photosynthesis, human vision Ultraviolet Thermal agitation & electronic transitions Sterilization, slowing abnormal growth of cells Vitamin D production Ozone depletion, causes cell damage X-rays Gamma Rays Inner electronic transitions & fast collisions Medical, security Medical diagnosis, cancer therapy Causes cell damage Nuclear decay Nuclear medicine, security Medical diagnosis, cancer therapy Causes cell damage, radiation damage Table 15.1 Electromagnetic Waves This table shows how each type of EM radiation is produced, how it is applied, as well as environmental and health issues associated with it. The narrow band of visible light is a combination of the colors of the rainbow. Figure 15.5 shows the section of the EM spectrum that includes visible light. The frequencies corresponding to these wavelengths are at the red end to at the violet end. This is a very narrow range, considering that the EM spectrum spans about 20 orders of magnitude. Figure 15.5 A small part of the electromagnetic spectrum that includes its visible components. The divisions between infrared, visible, and ultraviolet are not perfectly distinct, nor are the divisions between the seven rainbow colors TIPS FOR SUCCESS Wavelengths of visible light are often given in nanometers, nm. One nm equals wavelength of about 600 nm, or m. m. For example, yellow light has a 460 Chapter 15 • Light As a child, you probably learned the color wheel, shown on the left in Figure 15.6. It helps if you know what color results when you mix different colors of paint together. Mixing two of the primary pigmentcolors—magenta, yellow, or cyan—together results in a secondary color. For example, mixing cyan and yellow makes green. This is called subtractivecolor mixing. Mixing different colors of lighttogether is quite different. The diagram on the right shows additivecolor mixing. In this case, the primary colors are red, green, and blue, and the secondary colors are cyan, magenta, and yellow. Mixing pigments and mixing light are different because materials absorb light by a different set of rules than does the perception of light by the eye. Notice that, when all colors are subtracted, the result is no color, or black. When all colors are added, the result is white light. We see the reverse of this when white sunlight is separated into the visible spectrum by a prism or by raindrops when a rainbow appears in the sky. Figure 15.6 Mixing colored pigments follows the subtractive color wheel, and mixing colored light follows the additive color wheel. Virtual Physics Color Vision Click to view content (https://www.openstax.org/l/28Colorvision) This video demonstrates additive color and color filters. Try all the settings except Photons. GRASP CHECK Explain why only light from a blue bulb passes through the blue filter. a. A blue filter absorbs blue light. b. A blue filter reflects blue light. c. A blue filter absorbs all visible light other than blue light. d. A blue filter reflects all of the other colors of light and absorbs blue light. LINKS TO PHYSICS Animal Color Perception The physics of color perception has interesting links to zoology. Other animals have very different views of the world than humans, especially with respect to which colors can be seen. Color is detected by cells in the eye called cones. Humans have three cones that are sensitive to three different ranges of electromagnetic wavelengths. They are called red, blue, and green cones, although these colors do not correspond exactly to the centers of the three ranges. The ranges of wavelengths that each cone detects are red, 500 to 700 nm; green, 450 to 630 nm; and blue, 400 to 500 nm. Most primates also have three kinds of cones and see the world much as we do. Most mammals other than primates only have two cones and have a less colorful view of things. Dogs, for example see blue and yellow, but are color blind to red and green. You might think that simplerspecies, such as fish and insects, would have less sophisticated vision, but this is not the case. Many birds, reptiles, amphibians, and insects have four or five different cones in their eyes. These species don’t have a wider range of perceived colors, but they see more hues, or combinations of colors. Also, some animals, such as bees or rattlesnakes, see a Access for free at openstax.org. 15.1 • The Electromagnetic Spectrum 461 range of colors that is as broad as ours, but shifted into the ultraviolet or infrared. These differences in color perception are generally adaptations that help the animals survive. Colorful tropical birds and fish display some colors that are too subtle for us to see. These colors are believed to play a role in the species mating rituals. Figure 15.7 shows the colors visible and the color range of vision in humans, bees, and dogs. Figure 15.7 Humans, bees, and dogs see colors differently. Dogs see fewer colors than humans, and bees see a different range of colors. GRASP CHECK The belief that bulls are enraged by seeing the color red is a misconception. What did you read in this Links to Physics that shows why this belief is incorrect? a. Bulls are color-blind to every color in the spectrum of colors. b. Bulls are color-blind to the blue colors in the spectrum of colors. c. Bulls are color-blind to the red colors in the spectrum of colors. d. Bulls are color-blind to the green colors in the spectrum of colors. Humans have found uses for every part of the electromagnetic spectrum. We will take a look at the uses of each range of frequencies, beginning with visible light. Most of our uses of visible light are obvious; without it our interaction with our surroundings would be much different. We might forget that nearly all of our food depends on the photosynthesis process in plants, and that the energy for this process comes from the visible part of the spectrum. Without photosynthesis, we would also have almost no oxygen in the atmosphere. The low-frequency, infrared region of the spectrum has many applications in media broadcasting. Television, radio, cell phone, and remote-control devices all broadcast and/or receive signals with these wavelengths. AM and FM radio signals are both lowfrequency radiation. They are in different regions of the spectrum, but that is not their basic difference. AM and FM are abbreviations for amplitude modulationand frequency modulation. Information in AM signals has the form of changes in amplitudeof the radio waves; information in FM signals has the form of changes in wave frequency. Another application of long-wavelength radiation is found in microwave ovens. These appliances cook or warm food by irradiating it with EM radiation in the microwave frequency range. Most kitchen microwaves use a frequency of Hz. These waves have the right amount of energy to cause polar molecules, such as water, to rotate faster. Polar molecules are those that have a partial charge separation. The rotational energy of these molecules is given up to surrounding matter as heat. The first microwave ovens were called Radarangesbecause they were based on radar technology developed during World War II. Radar uses radiation with wavelengths similar to those of microwaves to detect the location and speed of distant objects, such as airplanes, weather formations, and motor vehicles. Radar information is obtained by receiving and analyzing the echoes of microwaves reflected by an object. The speed of the object can be measured using the Doppler shift of the returning waves. This is the same effect you learned about when you studied sound waves. Like sound waves, EM waves are shifted to higher frequencies by an object moving toward an observer, and to lower frequencies by an object moving away from the observer. Astronomers use this same Doppler effect to measure the speed at which distant galaxies are moving away from us. In this case, the shift in frequency is called the red shift, because visible frequencies are shifted toward the lower-frequency, red end of the spectrum. 462 Chapter 15 • Light Exposure to any radiation with frequencies greater than those of visible light carries some health hazards. All types of radiation in this range are known to cause cell damage. The danger is related to the high energy and penetrating ability of these EM waves. The likelihood of being harmed by any of this radiation depends largely on the amount of exposure. Most people try to reduce exposure to UV radiation from sunlight by using sunscreen and protective clothing. Physicians still use X-rays to diagnose medical problems, but the intensity of the radiation used is extremely low. Figure 15.8 shows an X-ray image of a patient’s chest cavity. One medical-imaging technique that involves no danger of exposure is magnetic resonance imaging (MRI). MRI is an important imaging and research tool in medicine, producing highly detailed two- and three-dimensional images. Radio waves are broadcast, absorbed, and reemitted in a resonance process that is sensitive to the density of nuclei, usually hydrogen nuclei—protons. Figure 15.8 This shadow X-ray image shows many interesting features, such as artificial heart valves, a pacemaker, and wires used to close the sternum. (credit: P.P. Urone) Check Your Understanding 1. Identify the fields produced by a moving charged particle. a. Both an electric field and a magnetic field will be produced. b. Neither a magnetic field nor an electric field will be produced. c. A magnetic field, but no electric field will be produced. d. Only the electric field, but no magnetic field will be produced. 2. X-rays carry more energy than visible light. Compare the frequencies and wavelengths of these two types of EM radiation. a. Visible light has higher frequencies and shorter wavelengths than X-rays. b. Visible li
ght has lower frequencies and shorter wavelengths than X-rays. c. Visible light has higher frequencies and longer wavelengths than X-rays. d. Visible light has lower frequencies and longer wavelengths than X-rays. 3. How does wavelength change as frequency increases across the EM spectrum? a. The wavelength increases. b. The wavelength first increases and then decreases. c. The wavelength first decreases and then increases. d. The wavelength decreases. 4. Why are X-rays used in imaging of broken bones, rather than radio waves? a. X-rays have higher penetrating energy than radio waves. b. X-rays have lower penetrating energy than radio waves. c. X-rays have a lower frequency range than radio waves. d. X-rays have longer wavelengths than radio waves. 5. Identify the fields that make up an electromagnetic wave. a. both an electric field and a magnetic field b. neither a magnetic field nor an electric field Access for free at openstax.org. 15.2 • The Behavior of Electromagnetic Radiation 463 c. only a magnetic field, but no electric field d. only an electric field, but no magnetic field 15.2 The Behavior of Electromagnetic Radiation Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the behavior of electromagnetic radiation • Solve quantitative problems involving the behavior of electromagnetic radiation Section Key Terms illuminance interference lumens luminous flux lux polarized light Types of Electromagnetic Wave Behavior In a vacuum, all electromagnetic radiation travels at the same incredible speed of 3.00 × 108 m/s, which is equal to 671 million miles per hour. This is one of the fundamental physical constants. It is referred to as the speed of light and is given the symbol c. The space between celestial bodies is a near vacuum, so the light we see from the Sun, stars, and other planets has traveled here at the speed of light. Keep in mind that all EM radiation travels at this speed. All the different wavelengths of radiation that leave the Sun make the trip to Earth in the same amount of time. That trip takes 8.3 minutes. Light from the nearest star, besides the Sun, takes 4.2 years to reach Earth, and light from the nearest galaxy—a dwarf galaxy that orbits the Milky Way—travels 25,000 years on its way to Earth. You can see why we call very long distances astronomical. When light travels through a physical medium, its speed is always less than the speed of light. For example, light travels in water at three-fourths the value of c. In air, light has a speed that is just slightly slower than in empty space: 99.97 percent of c. Diamond slows light down to just 41 percent of c. When light changes speeds at a boundary between media, it also changes direction. The greater the difference in speeds, the more the path of light bends. In other chapters, we look at this bending, called refraction, in greater detail. We introduce refraction here to help explain a phenomenon called thin-film interference. Have you ever wondered about the rainbow colors you often see on soap bubbles, oil slicks, and compact discs? This occurs when light is both refracted by and reflected from a very thin film. The diagram shows the path of light through such a thin film. The symbols n1, n2, and n3 indicate that light travels at different speeds in each of the three materials. Learn more about this topic in the chapter on diffraction and interference. Figure 15.9 shows the result of thin film interference on the surface of soap bubbles. Because ray 2 travels a greater distance, the two rays become out of phase. That is, the crests of the two emerging waves are no longer moving together. This causes interference, which reinforces the intensity of the wavelengths of light that create the bands of color. The color bands are separated because each color has a different wavelength. Also, the thickness of the film is not uniform, and different thicknesses cause colors of different wavelengths to interfere in different places. Note that the film must be very, very thin—somewhere in the vicinity of the wavelengths of visible light. 464 Chapter 15 • Light Figure 15.9 Light striking a thin film is partially reflected (ray 1) and partially refracted at the top surface. The refracted ray is partially reflected at the bottom surface and emerges as ray 2. These rays will interfere in a way that depends on the thickness of the film and the indices of refraction of the various media. You have probably experienced how polarized sunglasses reduce glare from the surface of water or snow. The effect is caused by the wave nature of light. Looking back at , we see that the electric field moves in only one direction perpendicular to the direction of propagation. Light from most sources vibrates in all directions perpendicular to propagation. Light with an electric field that vibrates in only one direction is called polarized. A diagram of polarized light would look like . Polarized glasses are an example of a polarizing filter. These glasses absorb most of the horizontal light waves and transmit the vertical waves. This cuts down glare, which is caused by horizontal waves. Figure 15.10 shows how waves traveling along a rope can be used as a model of how a polarizing filter works. The oscillations in one rope are in a vertical plane and are said to be vertically polarized. Those in the other rope are in a horizontal plane and are horizontally polarized. If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field oscillation is analogous to the disturbances on the ropes. Figure 15.10 The transverse oscillations in one rope are in a vertical plane, and those in the other rope are in a horizontal plane. The first is said to be vertically polarized, and the other is said to be horizontally polarized. Vertical slits pass vertically polarized waves and block horizontally polarized waves. Light can also be polarized by reflection. Most of the light reflected from water, glass, or any highly reflective surface is polarized horizontally. Figure 15.11 shows the effect of a polarizing lens on light reflected from the surface of water. Access for free at openstax.org. 15.2 • The Behavior of Electromagnetic Radiation 465 Figure 15.11 These two photographs of a river show the effect of a polarizing filter in reducing glare in light reflected from the surface of water. Part (b) of this figure was taken with a polarizing filter and part (a) was taken without. As a result, the reflection of clouds and sky observed in part (a) is not observed in part (b). Polarizing sunglasses are particularly useful on snow and water. WATCH PHYSICS Polarization of Light, Linear and Circular This video explains the polarization of light in great detail. Before viewing the video, look back at the drawing of an electromagnetic wave from the previous section. Try to visualize the two-dimensional drawing in three dimensions. Click to view content (https://www.openstax.org/l/28Polarization) GRASP CHECK How do polarized glasses reduce glare reflected from the ocean? a. They block horizontally polarized and vertically polarized light. b. They are transparent to horizontally polarized and vertically polarized light. c. They block horizontally polarized rays and are transparent to vertically polarized rays. d. They are transparent to horizontally polarized light and block vertically polarized light. Snap Lab Polarized Glasses • EYE SAFETY—Looking at the Sun directly can cause permanent eye damage. Avoid looking directly at the Sun. • • two pairs of polarized sunglasses OR two lenses from one pair of polarized sunglasses Procedure 1. Look through both or either polarized lens and record your observations. 2. Hold the lenses, one in front of the other. Hold one lens stationary while you slowly rotate the other lens. Record your observations, including the relative angles of the lenses when you make each observation. 3. Find a reflective surface on which the Sun is shining. It could be water, glass, a mirror, or any other similar smooth surface. The results will be more dramatic if the sunlight strikes the surface at a sharp angle. 4. Observe the appearance of the surface with your naked eye and through one of the polarized lenses. 5. Observe any changes as you slowly rotate the lens, and note the angles at which you see changes. 466 Chapter 15 • Light GRASP CHECK If you buy sunglasses in a store, how can you be sure that they are polarized? a. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to refraction of light. b. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to reflection of light. c. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the polarization of light. d. When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the bending of light waves. Quantitative Treatment of Electromagnetic Waves We can use the speed of light, c, to carry out several simple but interesting calculations. If we know the distance to a celestial object, we can calculate how long it takes its light to reach us. Of course, we can also make the reverse calculation if we know the time it takes for the light to travel to us. For an object at a very great distance from Earth, it takes many years for its light to reach us. This means that we are looking at the object as it existed in the distant past. The object may, in fact, no longer exist. Very large distance
s in the universe are measured in light years. One light year is the distance that light travels in one year, which is kilometers or miles (…and 1012 is a trillion!). A useful equation involving cis where fis frequency in Hz, and is wavelength in meters. WORKED EXAMPLE Frequency and Wavelength Calculation For example, you can calculate the frequency of yellow light with a wavelength of STRATEGY Rearrange the equation to solve for frequency. Solution Substitute the values for the speed of light and wavelength into the equation. m. 15.2 15.1 15.3 Discussion Manipulating exponents of 10 in a fraction can be tricky. Be sure you keep track of the + and – exponents correctly. Checking back to the diagram of the electromagnetic spectrum in the previous section shows that 1014 is a reasonable order of magnitude for the frequency of yellow light. The frequency of a wave is proportional to the energy the wave carries. The actual proportionality constant will be discussed in a later chapter. Since frequency is inversely proportional to wavelength, we also know that wavelength is inversely proportional to energy. Keep these relationships in mind as general rules. The rate at which light is radiated from a source is called luminous flux, P, and it is measured in lumens (lm). Energy-saving light bulbs, which provide more luminous flux for a given use of electricity, are now available. One of these bulbs is called a compact fluorescent lamp; another is an LED(light-emitting diode) bulb. If you wanted to replace an old incandescent bulb with an energy saving bulb, you would want the new bulb to have the same brightness as the old one. To compare bulbs accurately, you would need to compare the lumens each one puts out. Comparing wattage—that is, the electric power used—would be Access for free at openstax.org. 15.2 • The Behavior of Electromagnetic Radiation 467 misleading. Both wattage and lumens are stated on the packaging. The luminous flux of a bulb might be 2,000 lm. That accounts for all the light radiated in all directions. However, what we really need to know is how much light falls on an object, such as a book, at a specific distance. The number of lumens per square meter is called illuminance, and is given in units of lux (lx). Picture a light bulb in the middle of a sphere with a 1-m radius. The total surface of the sphere equals 4πr2 m2. The illuminance then is given by What happens if the radius of the sphere is increased 2 m? The illuminance is now only one-fourth as great, because the r2 term in the denominator is 4 instead of 1. Figure 15.12 shows how illuminance decreases with the inverse square of the distance. 15.4 Figure 15.12 The diagram shows why the illuminance varies inversely with the square of the distance from a source of light. WORKED EXAMPLE Calculating Illuminance A woman puts a new bulb in a floor lamp beside an easy chair. If the luminous flux of the bulb is rated at 2,000 lm, what is the illuminance on a book held 2.00 m from the bulb? STRATEGY Choose the equation and list the knowns. Equation: P= 2,000 lm π = 3.14 r= 2.00 m Solution Substitute the known values into the equation. Discussion Try some other distances to illustrate how greatly light fades with distance from its source. For example, at 3 m the illuminance is only 17.7 lux. Parents often scold children for reading in light that is too dim. Instead of shouting, “You’ll ruin your eyes!” it might be better to explain the inverse square law of illuminance to the child. 468 Chapter 15 • Light Practice Problems 6. Red light has a wavelength of 7.0 × 10−7 m and a frequency of 4.3 × 1014 Hz. Use these values to calculate the speed of light in a vacuum. a. b. c. d. 3 × 1020 m/s 3 × 1015 m/s 3 × 1014 m/s 3 × 108 m/s 7. A light bulb has a luminous flux of 942 lumens. What is the illuminance on a surface from the bulb when it is lit? a. b. c. d. Check Your Understanding 8. Give an example of a place where light travels at the speed of 3.00 × 108 m/s. a. outer space b. water c. Earth’s atmosphere d. quartz glass 9. Explain in terms of distances and the speed of light why it is currently very unlikely that humans will visit planets that circle stars other than our Sun. a. The spacecrafts used for travel are very heavy and thus very slow. b. Spacecrafts do not have a constant source of energy to run them. c. If a spacecraft could attain a maximum speed equal to that of light, it would still be too slow to cover astronomical distances. d. Spacecrafts can attain a maximum speed equal to that of light, but it is difficult to locate planets around stars. Access for free at openstax.org. Chapter 15 • Key Terms 469 KEY TERMS electric field a field that tells us the force per unit charge at all locations in space around a charge distribution electromagnetic radiation (EMR) radiant energy that consists of oscillating electric and magnetic fields lux unit of measure for illuminance magnetic field the directional lines around a magnetic material that indicates the direction and magnitude of the magnetic force illuminance number of lumens per square meter, given in Maxwell’s equations equations that describe the units of lux (lx) interference increased or decreased light intensity caused by the phase differences between waves lumens unit of measure for luminous flux luminous flux rate at which light is radiated from a source interrelationship between electric and magnetic fields, and how these fields combine to form electromagnetic radiation polarized light light whose electric field component vibrates in a specific plane SECTION SUMMARY 15.1 The Electromagnetic Spectrum • The electromagnetic spectrum is made up of a broad range of frequencies of electromagnetic radiation. • All frequencies of EM radiation travel at the same speed in a vacuum and consist of an electric field and a magnetic field. The types of EM radiation have different frequencies and wavelengths, and different energies and penetrating ability. 15.2 The Behavior of Electromagnetic Radiation • EM radiation travels at different speeds in different media, produces colors on thin films, and can be polarized to oscillate in only one direction. • Calculations can be based on the relationship among the speed, frequency, and wavelength of light, and on the relationship among luminous flux, illuminance, and distance. KEY EQUATIONS 15.2 The Behavior of Electromagnetic Radiation frequency and wavelength illuminance CHAPTER REVIEW Concept Items 15.1 The Electromagnetic Spectrum 1. Use the concepts on which Maxwell’s equations are based to explain why a compass needle is deflected when the compass is brought near a wire that is carrying an electric current. a. The charges in the compass needle and the charges in the electric current have interacting electric fields, causing the needle to deflect. b. The electric field from the moving charges in the current interacts with the magnetic field of the compass needle, causing the needle to deflect. c. The magnetic field from the moving charges in the current interacts with the electric field of the compass needle, causing the needle to deflect. d. The moving charges in the current produce a magnetic field that interacts with the compass needle’s magnetic field, causing the needle to deflect. 2. Consider these colors of light: yellow, blue, and red. Part A. Put these light waves in order according to wavelength, from shortest wavelength to longest wavelength. Part B. Put these light waves in order according to frequency, from lowest frequency to highest frequency. a. wavelength: blue, yellow, red frequency: blue, yellow, red b. wavelength: red, yellow, blue frequency: red, yellow, blue c. wavelength: red, yellow, blue frequency: blue, yellow, red d. wavelength: blue, yellow, red frequency: red, yellow, blue 3. Describe the location of gamma rays on the electromagnetic spectrum. 470 Chapter 15 • Chapter Review a. At the high-frequency and long-wavelength end of the spectrum b. At the high-frequency and short-wavelength end of thickness of the wall of a soap bubble? Explain your answer. a. The thickness of the bubble wall is ten times that of the spectrum the wavelength of light. c. At the low-frequency and long-wavelength end of b. The thickness of the bubble wall is similar to that of the spectrum the wavelength of light. d. At the low-frequency and short-wavelength end of c. The thickness of the bubble wall is half the the spectrum wavelength of light. 4. In which region of the electromagnetic spectrum would you find radiation that is invisible to the human eye and has low energy? a. Long-wavelength and high-frequency region b. Long-wavelength and low-frequency region c. Short-wavelength and high-frequency region d. Short-wavelength and low-frequency region 15.2 The Behavior of Electromagnetic Radiation 5. Light travels at different speeds in different media. Put these media in order, from the slowest light speed to the fastest light speed: air, diamond, vacuum, water. a. diamond, water, air, vacuum b. vacuum, diamond, air, water c. diamond, air, water, vacuum d. air, diamond, water, vacuum 6. Visible light has wavelengths in the range of about 400 to 800 nm. What does this indicate about the approximate Critical Thinking Items 15.1 The Electromagnetic Spectrum 8. Standing in front of a fire, we can sense both its heat and its light. How are the light and heat radiated by the fire the same, and how are they different? a. Both travel as waves, but only light waves are a form of electromagnetic radiation. b. Heat and light are both forms of electromagnetic radiation, but light waves have higher frequencies. c. Heat and light are both forms of electromagnetic radiation, but heat waves have higher frequencies. d. Heat and light are both forms of electromagnetic radiation, but light waves have higher wavelengths. 9. Light shines on a picture of the subtractive color wheel. The light is a mixture of red, blue, and green light. Part A—Which part of the color wheel
will look blue? Explain in terms of absorbed and reflected light. Part B—Which part of the color wheel will look yellow? Explain in terms of absorbed and reflected light. a. A. The yellow section of the wheel will look blue because it will reflect blue light and absorb red Access for free at openstax.org. d. The thickness of the bubble wall equals the cube of the wavelength of light. 7. Bright sunlight is reflected from an icy pond. You look at the glare of the reflected light through polarized glasses. When you take the glasses off, rotate them 90°, and look through one of the lenses again, the light you see becomes brighter. Explain why the light you see changes. a. The glass blocks horizontally polarized light, and the light reflected from the icy pond is, in part, polarized horizontally. b. The glass blocks vertically polarized light, and the light reflected from the icy pond is, in part, polarized vertically. c. The glass allows horizontally polarized light to pass, and the light reflected from the icy pond is, in part, polarized vertically. d. The glass allows horizontally polarized light to pass, and the light reflected from the icy pond is, in part, polarized horizontally. and green. B. The blue section of the wheel will look yellow because it will reflect red and green light and absorb blue. b. A. The blue section of the wheel will look blue because it will absorb blue light and reflect red and green. B. The yellow section of the wheel will look yellow because it will absorb red and green light and reflect blue. c. A. The yellow section of the wheel will look blue because it will absorb blue light and reflect red and green. B. The blue section of the wheel will look yellow because it will absorb red and green light and reflect blue. d. A. The blue section of the wheel will look blue because it will reflect blue light and absorb red and green. B. The yellow section of the wheel will look yellow because it will reflect red and green light and absorb blue. 10. Part A. When you stand in front of an open fire, you can sense light waves with your eyes. You sense another type of electromagnetic radiation as heat. What is this other type of radiation? Part B. How is this other type of radiation different front light waves? a. A. X-rays B. The X-rays have higher frequencies and shorter wavelengths than the light waves. b. A. X-rays B. The X-rays have lower frequencies and longer wavelengths than the light waves. c. A. infrared rays B. The infrared rays have higher frequencies and shorter wavelengths than the light waves. d. A. infrared rays B. The infrared rays have lower frequencies and longer wavelengths than the light waves. 11. Overexposure to this range of EM radiation is dangerous, and yet it is used by doctors to diagnose medical problems. Part A—Identify the type of radiation. Part B—Locate the position of this radiation on the EM spectrum by comparing its frequency and wavelength to visible light. Part C—Explain why this radiation is both dangerous and therapeutic in terms of its energy, based on your answer to Part B. a. A. X-rays B. X-rays have shorter wavelengths (1 × 10–8 – 5 × 10–12 m) and higher frequencies (3 × 1016 – 6 × 1019 Hz) than visible light (7.5 × 10–7 – 4.0 × 10–7 m; 4.0 × 1014 – 7.5 × 1014 Hz). C. X-rays have low energies because of their high frequencies, and so can penetrate matter to greater depths. b. A. X-rays B. X-rays have shorter wavelengths (1 × 10–8 – 5 × 10–12 m) and higher frequencies (3 × 1010 – 6 × 1013 Hz) than visible light (7.5 × 10–7 – 4.0 × 10–7 m; 4.0 × 1014 – 7.5 × 1014 Hz). C. X-rays have low energies because of their low frequencies, and so can penetrate matter to greater depths. c. A. X-rays B. X-rays have longer wavelengths (1 × 10–6 – 5 × 10–7 m) and higher frequencies (3 × 1015 – 6 × 1015 Hz) than visible light (7.5 × 10–7 – 4.0 × 10–7 m; 4.0 × 1014 – 7.5 × 1014 Hz). C. X-rays have high energies because of their high Chapter 15 • Chapter Review 471 frequencies, and therefore can penetrate matter to greater depths. d. A. X-rays B. X-rays have shorter wavelengths (1 × 10–8 – 5 × 10–12 m) and higher frequencies (3 × 1016 – 6 × 1019 Hz) than visible light (7.5 × 10–7 – 4.0 × 10–7 m; 4.0 × 1014 – 7.5 × 1014 Hz). C. X-rays have high energies because of their high frequencies, and so can penetrate matter to greater depths. 15.2 The Behavior of Electromagnetic Radiation 12. Explain how thin-film interference occurs. Discuss in terms of the meaning of interference and the pathways of light waves. a. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely in phase, and so undergoes constructive interference. b. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner surfaces is completely in phase, and so undergoes destructive interference. c. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely out of phase, and so undergoes constructive interference. d. For a particular thickness of film, light of a given wavelength that reflects from the outer and inner film surfaces is completely out of phase, and so undergoes no interference. 13. When you move a rope up and down, waves are created. If the waves pass through a slot, they will be affected differently, depending on the orientation of the slot. Using the rope waves and the slot as a model, explain how polarizing glasses affect light waves. a. If the wave—electric field—is vertical and slit—polarizing molecules in the glass—is horizontal, the wave will pass. If the wave—electric field— is vertical and slit—polarizing molecules in the glass—is vertical, the wave will not pass. If the wave—electric field—is horizontal and slit—polarizing molecules in the glass—is horizontal, the wave will pass. If the wave—electric field—is horizontal and slit—polarizing molecules in the glass—is horizontal, the wave will not pass. b. c. d. 472 Chapter 15 • Test Prep Problems 15.2 The Behavior of Electromagnetic Radiation 14. Visible light has a range of wavelengths from about 400 nm to 800 nm . What is the range of frequencies for visible light? a. b. c. d. 3.75 × 106 Hz to 7.50 × 106 Hz 3.75 Hz to 7.50 Hz 3.75 × 10−7 Hz to 7.50 × 10−7 Hz 3.75 × 1014 Hz to 7.50 × 1014 Hz Performance Task 15.2 The Behavior of Electromagnetic Radiation 16. Design an experiment to observe the phenomenon of thin-film interference. Observe colors of visible light, and relate each color to its corresponding wavelength. Comparison with the magnitudes of visible light wavelength will give an appreciation of just how very thin a thin film is. Thin-film interference has a number of practical applications, such as anti-reflection coatings and optical filters. Thin films used in filters can be designed to reflect or transmit specific wavelengths of light. This is done by depositing a film one molecular layer at a time from a vapor, thus allowing the thickness of the film to be exactly controlled. • EYE SAFETY—Chemicals in this lab are poisonous if ingested. If chemicals are ingested, inform your teacher immediately. • FUMES—Certain chemicals or chemical reactions in this lab create a vapor that is harmful if inhaled. Follow your teacher's instructions for the use of fume hoods and other safety apparatus designed to prevent fume inhalation. Never smell or otherwise breath in any chemicals or vapors in the lab. • FLAMMABLE—Chemicals in this lab are highly flammable and can ignite, especially if exposed to a spark or open flame. Follow your teacher's TEST PREP Multiple Choice 15.1 The Electromagnetic Spectrum 17. Which type of EM radiation has the shortest wavelengths? a. gamma rays b. c. blue light d. microwaves infrared waves Access for free at openstax.org. 15. Light travels through the wall of a soap bubble that is 600 nm thick and is reflected from the inner surface back into the air. Assume the bubble wall is mostly water and that light travels in water at 75 percent of the speed of light in vacuum. How many seconds behind will the light reflected from the inner surface arrive compared to the light that was reflected from the outer surface? a. 4.0 × 10–8 s 5.3 × 10–6 s b. c. 2.65 × 10–15 s 5.3 × 10–15 s d. instructions carefully on how to handle flammable chemicals. Do not expose any chemical to a flame or other heat source unless specifically instructed by your teacher. • HAND WASHING—Some materials may be hazardous if in extended contact with the skin. Be sure to wash your hands with soap after handling and disposing of these materials during the lab. • WASTE—Some things in this lab are hazardous and need to be disposed of properly. Follow your teacher's instructions for disposal of all items. • A large flat tray with raised sides, such as a baking tray • Small volumes of motor oil, lighter fluid or a penetrating oil of the type used to loosen rusty bolts, and cooking oil • Water • A camera a. Thin-film interference causes colors to appear on the surface of a thin transparent layer. Do you expect to see a pattern to the colors? b. How could you make a permanent record of your observations? c. What data would you need to look up to help explain any patterns that you see? d. What could explain colors failing to appear under some conditions? 18. Which form of EM radiation has the most penetrating ability? a. red light b. microwaves c. gamma rays d. infrared radiation 19. Why are high-frequency gamma rays more dangerous to humans than visible light? a. Gamma rays have a lower frequency range than visible light. 22. What is the wavelength of red light with a frequency of Chapter 15 • Test Prep 473 b. Gamma rays have a longer wavelength range than visible light. c. Gamma rays have greater energy than visible light for penetrating matter. d. Gamma rays have less energy than visible light for penetrating matter. 20. A dog would have a hard time stalking and catching a
red bird hiding in a field of green grass. Explain this in terms of cone cells and color perception. a. Dogs are red-green color-blind because they can see only blue and yellow through two kinds of cone cells present in their eyes. b. Dogs are only red color-blind because they can see only blue and yellow through two kinds of cones cells present in their eyes. c. Dogs are only green color-blind because they can see only blue and yellow through two kinds of cones cells present in their eyes. d. Dogs are color-blind because they have only rods and no cone cells present in their eyes. 15.2 The Behavior of Electromagnetic Radiation 21. To compare the brightness of light bulbs for sale in a frequency store, you should look on the labels to see how they are rated in terms of ____. a. b. watts c. amps d. lumens 4.00 × 1014 Hz? a. 2.50 × 1014 m b. 4.00 × 1015 m c. 2.50 × 106 m d. 4.00 × 10-7 m 23. What is the distance of one light year in kilometers? a. 2.59 × 1010 km 1.58 × 1011 km b. c. 2.63 × 109 km d. 9.46 × 1012 km 24. How does the illuminance of light change when the distance from the light source is tripled? Cite the relevant equation and explain how it supports your answer. a. if distance is tripled, then the illuminance increases by 19 times. b. c. d. if distance is tripled, then the illuminance decreases by 13 times. then the illuminance decreases by 9 times. if distance is tripled, if distance tripled, then the illuminance increases by 3 times. 25. A light bulb has an illuminance of 19.9 lx at a distance of 2 m . What is the luminous flux of the bulb? 500 lm a. b. 320 lm c. 250 lm d. 1,000 lm Short Answer 15.1 The Electromagnetic Spectrum 26. Describe one way in which heat waves—infrared radiation—are different from sound waves. a. Sound waves are transverse waves, whereas heat waves—infrared radiation—are longitudinal waves. b. Sound waves have shorter wavelengths than heat waves. c. Sound waves require a medium, whereas heat waves—infrared radiation—do not. d. Sound waves have higher frequencies than heat waves. 27. Describe the electric and magnetic fields that make up an electromagnetic wave in terms of their orientation relative to each other and their phases. a. They are perpendicular to and out of phase with each other. b. They are perpendicular to and in phase with each other. c. They are parallel to and out of phase with each other. d. They are parallel to and in phase with each other. 28. Explain how X-radiation can be harmful and how it can be a useful diagnostic tool. a. Overexposure to X-rays can cause HIV, though normal levels of X-rays can be used for sterilizing needles. b. Overexposure to X-rays can cause cancer, though in limited doses X-rays can be used for imaging internal body parts. c. Overexposure to X-rays causes diabetes, though normal levels of X-rays can be used for imaging internal body parts. d. Overexposure to X-rays causes cancer, though normal levels of X-rays can be used for reducing 474 Chapter 15 • Test Prep cholesterol in the blood. 31. What is it about the nature of light reflected from snow 29. Explain how sunlight is the original source of the energy in the food we eat. a. Sunlight is converted into chemical energy by plants; this energy is released when we digest food. b. Sunlight is converted into chemical energy by animals; this energy is released when we digest food. c. Sunlight is converted into chemical energy by fish; this energy is released when we digest food. that causes skiers to wear polarized sunglasses? a. The reflected light is polarized in the vertical direction. b. The reflected light is polarized in the horizontal direction. c. The reflected light has less intensity than the incident light. d. The reflected light has triple the intensity of the incident light. d. Sunlight is converted into chemical energy by 32. How many lumens are radiated from a candle which has humans; this energy is released when we digest food. 15.2 The Behavior of Electromagnetic Radiation 30. Describe what happens to the path of light when the light slows down as it passes from one medium to another? a. The path of the light remains the same. b. The path of the light becomes circular. c. The path of the light becomes curved. d. The path of the light changes. an illuminance of 3.98 lx at a distance of 2.00 m? a. 400 lm 100 lm b. c. 50 lm d. 200 lm 33. Saturn is 1.43×1012 m from the Sun. How many minutes does it take the Sun’s light to reach Saturn? a. b. c. d. 7.94 × 109 minutes 3.4 × 104 minutes 3.4 × 10–6 minutes 79.4 minutes Extended Response C. ultraviolet radiation 15.1 The Electromagnetic Spectrum 35. A mixture of red and green light is shone on each of the 34. A frequency of red light has a wavelength of 700 nm. Part A—Compare the wavelength and frequency of violet light to red light. Part B—Identify a type of radiation that has lower frequencies than red light. Part C—Identify a type of radiation that has shorter wavelengths than violet light. a. A. Violet light has a lower frequency and longer wavelength than red light. B. ultraviolet radiation infrared radiation C. b. A. Violet light has a lower frequency and longer wavelength than red light. B. infrared radiation C. ultraviolet radiation c. A. Violet light has a higher frequency and shorter wavelength than red light. B. ultraviolet radiation infrared radiation C. d. A. Violet light has a higher frequency and shorter wavelength than red light. infrared radiation B. Access for free at openstax.org. subtractive colors. Part A—Which of these colors of light are reflected from magenta? Part B—Which of these colors of light are reflected from yellow? Part C—Which these colors of light are reflected from cyan? a. Part A. red and green Part B. green Part C. red b. Part A. red and green Part B. red Part C. green c. Part A. green Part B. red and green Part C. red d. Part A. red Part B. red and green Part C. green 15.2 The Behavior of Electromagnetic Radiation 36. Explain why we see the colorful effects of thin-film interference on the surface of soap bubbles and oil slicks, but not on the surface of a window pane or clear plastic bag. a. The thickness of a window pane or plastic bag is more than the wavelength of light, and interference occurs for thicknesses smaller than the wavelength of light. b. The thickness of a window pane or plastic bag is less than the wavelength of light, and interference occurs for thicknesses similar to the wavelength of light. c. The thickness of a window pane or plastic bag is more than the wavelength of light, and interference occurs for thicknesses similar to the wavelength of light. d. The thickness of a window pane or plastic bag is Chapter 15 • Test Prep 475 less than the wavelength of light, and interference occurs for thicknesses larger than the wavelength of light. 37. The Occupational Safety and Health Administration (OSHA) recommends an illuminance of for desktop lighting. An office space has lighting hung . above desktop level that provides only To what height would the lighting fixtures have to be lowered to provide a. b. c. d. on desktops? 476 Chapter 15 • Test Prep Access for free at openstax.org. CHAPTER 16 Mirrors and Lenses Figure 16.1 Flat, smooth surfaces reflect light to form mirror images. (credit: NASA Goddard Photo and Video, via Flickr) Chapter Outline 16.1 Reflection 16.2 Refraction 16.3 Lenses INTRODUCTION “In another moment Alice was through the glass, and had jumped lightly down into the Looking-glass room.” —Through the Looking Glass by Lewis Carol Through the Looking Glasstells of the adventures of Alice after she steps from the real world, through a mirror, and into the virtual world. In this chapter we examine the optical meanings of real and virtual, as well as other concepts that make up the field of optics. The light from this page or screen is formed into an image by the lens of your eyes, much as the lens of the camera that made the photograph at the beginning of this chapter. Mirrors, like lenses, can also form images, which in turn are captured by your eyes. Optics is the branch of physics that deals with the behavior of visible light and other electromagnetic waves. For now, we concentrate on the propagation of light and its interaction with matter. It is convenient to divide optics into two major parts based on the size of objects that light encounters. When light interacts with an object that is several times as large as the light’s wavelength, its observable behavior is similar to a ray; it does not display its 478 Chapter 16 • Mirrors and Lenses wave characteristics prominently. We call this part of optics geometric optics. This chapter focuses on situations for which geometric optics is suited. 16.1 Reflection Section Learning Objectives By the end of this section, you will be able to do the following: • Explain reflection from mirrors, describe image formation as a consequence of reflection from mirrors, apply ray diagrams to predict and interpret image and object locations, and describe applications of mirrors • Perform calculations based on the law of reflection and the equations for curved mirrors Section Key Terms angle of incidence angle of reflection central axis concave mirror convex mirror diffused focal length focal point geometric optics law of reflection law of refraction ray real image specular virtual image Characteristics of Mirrors There are three ways, as shown in Figure 16.2, in which light can travel from a source to another location. It can come directly from the source through empty space, such as from the Sun to Earth. Light can travel to an object through various media, such as air and glass. Light can also arrive at an object after being reflected, such as by a mirror. In all these cases, light is modeled as traveling in a straight line, called a ray. Light may change direction when it encounters the surface of a different material (such as a mirror) or when it passes from one material to another (such as when passing from air i
nto glass). It then continues in a straight line—that is, as a ray. The word raycomes from mathematics. Here it means a straight line that originates from some point. It is acceptable to visualize light rays as laser rays (or even science fiction depictions of ray guns). Figure 16.2 Three methods for light to travel from a source to another location are shown. (a) Light reaches the upper atmosphere of Earth by traveling through empty space directly from the source (the Sun). (b) This light can reach a person in one of two ways. It can travel through a medium, such as air or glass, and typically travels from one medium to another. It can also reflect from an object, such as a mirror. Because light moves in straight lines, that is, as rays, and changes directions when it interacts with matter, it can be described through geometry and trigonometry. This part of optics, described by straight lines and angles, is therefore called geometric optics. There are two laws that govern how light changes direction when it interacts with matter: the law of reflection, for situations in which light bounces off matter; and the law of refraction, for situations in which light passes through matter. In this section, we consider the geometric optics of reflection. Whenever we look into a mirror or squint at sunlight glinting from a lake, we are seeing a reflection. How does the reflected light travel from the object to your eyes? The law of reflection states: The angle of reflection, , equals the angle of incidence, Access for free at openstax.org. 16.1 • Reflection 479 .This law governs the behavior of all waves when they interact with a smooth surface, and therefore describe the behavior of light waves as well. The reflection of light is simplified when light is treated as a ray. This concept is illustrated in Figure 16.3, which also shows how the angles are measured relative to the line perpendicular to the surface at the point where the light ray strikes it. This perpendicular line is also called the normal line, or just the normal. Light reflected in this way is referred to as specular (from the Latin word for mirror: speculum). We expect to see reflections from smooth surfaces, but Figure 16.4, illustrates how a rough surface reflects light. Because the light is reflected from different parts of the surface at different angles, the rays go in many different directions, so the reflected light is diffused. Diffused light allows you to read a printed page from almost any angle because some of the rays go in different directions. Many objects, such as people, clothing, leaves, and walls, have rough surfaces and can be seen from many angles. A mirror, on the other hand, has a smooth surface and reflects light at specific angles. Figure 16.3 The law of reflection states that the angle of reflection, θr, equals the angle of incidence, θi. The angles are measured relative to the line perpendicular to the surface at the point where the ray strikes the surface. The incident and reflected rays, along with the normal, lie in the same plane. Figure 16.4 Light is diffused when it reflects from a rough surface. Here, many parallel rays are incident, but they are reflected at many different angles because the surface is rough. When we see ourselves in a mirror, it appears that our image is actually behind the mirror. We see the light coming from a direction determined by the law of reflection. The angles are such that our image is exactly the same distance behind the mirror, di, as the distance we stand away from the mirror, do. Although these mirror images make objects appear to be where they cannot be (such as behind a solid wall), the images are not figments of our imagination. Mirror images can be photographed and videotaped by instruments and look just as they do to our eyes, which are themselves optical instruments. An image in a mirror is said to be a virtual image, as opposed to a real image. A virtual image is formed when light rays appear to diverge from a point without actually doing so. Figure 16.5 helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and reflecting into the observer’s eye. The rays can diverge slightly, and both still enter the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, allowing us to locate the image. The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror. Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, as defined earlier. 480 Chapter 16 • Mirrors and Lenses Figure 16.5 When two sets of rays from common points on an object are reflected by a flat mirror into the eye of an observer, the reflected rays seem to originate from behind the mirror, which determines the position of the virtual image. FUN IN PHYSICS Mirror Mazes Figure 16.6 is a chase scene from an old silent film called The Circus, starring Charlie Chaplin. The chase scene takes place in a mirror maze. You may have seen such a maze at an amusement park or carnival. Finding your way through the maze can be very difficult. Keep in mind that only one image in the picture is real—the others are virtual. Figure 16.6 Charlie Chaplin is in a mirror maze. Which image is real? One of the earliest uses of mirrors for creating the illusion of space is seen in the Palace of Versailles, the former home of French royalty. Construction of the Hall of Mirrors (Figure 16.7) began in 1678. It is still one of the most popular tourist attractions at Versailles. Figure 16.7 Tourists love to wander in the Hall of Mirrors at the Palace of Versailles. (credit: Michal Osmenda, Flickr) GRASP CHECK Only one Charlie in this image (Figure 16.8) is real. The others are all virtual images of him. Can you tell which is real? Hint—His hat is tilted to one side. Access for free at openstax.org. 16.1 • Reflection 481 Figure 16.8 a. The virtual images have their hats tilted to the right. b. The virtual images have their hats tilted to the left. c. The real images have their hats tilted to the right. d. The real images have their hats tilted to the left. WATCH PHYSICS Virtual Image This video explains the creation of virtual images in a mirror. It shows the location and orientation of the images using ray diagrams, and relates the perception to the human eye. Click to view content (https://openstax.org/l/28Virtualimage) Compare the distance of an object from a mirror to the apparent distance of its virtual image behind the mirror. a. The distances of the image and the object from the mirror are the same. b. The distances of the image and the object from the mirror are always different. c. The image is formed at infinity if the object is placed near the mirror. d. The image is formed near the mirror if the object is placed at infinity. Some mirrors are curved instead of flat. A mirror that curves inward is called a concave mirror, whereas one that curves outward is called a convex mirror. Pick up a well-polished metal spoon and you can see an example of each type of curvature. The side of the spoon that holds the food is a concave mirror; the back of the spoon is a convex mirror. Observe your image on both sides of the spoon. TIPS FOR SUCCESS You can remember the difference between concave and convex by thinking, Concave means caved in. Ray diagrams can be used to find the point where reflected rays converge or appear to converge, or the point from which rays appear to diverge. This is called the focal point, F. The distance from F to the mirror along the central axis (the line perpendicular to the center of the mirror’s surface) is called the focal length, f. Figure 16.9 shows the focal points of concave and convex mirrors. 482 Chapter 16 • Mirrors and Lenses Figure 16.9 (a, b) The focal length for the concave mirror in (a), formed by converging rays, is in front of the mirror, and has a positive value. The focal length for the convex mirror in (b), formed by diverging rays, appears to be behind the mirror, and has a negative value. Images formed by a concave mirror vary, depending on which side of the focal point the object is placed. For any object placed on the far side of the focal point with respect to the mirror, the rays converge in front of the mirror to form a real image, which can be projected onto a surface, such as a screen or sheet of paper However, for an object located inside the focal point with respect to the concave mirror, the image is virtual. For a convex mirror the image is always virtual—that is, it appears to be behind the mirror. The ray diagrams in Figure 16.10 show how to determine the nature of the image formed by concave and convex mirrors. Figure 16.10 (a) The image of an object placed outside the focal point of a concave mirror is inverted and real. (b) The image of an object Access for free at openstax.org. placed inside the focal point of a concave mirror is erect and virtual. (c) The image of an object formed by a convex mirror is erect and virtual. The information in Figure 16.10 is summarized in Table 16.1. Type of Mirror Object to Mirror Distance, do Image Characteristics 16.1 • Reflection 483 Concave Concave Convex Real and inverted Virtual and erect Virtual and erect Table 16.1 Curved Mirror Images This table details the type and orientation of images formed by concave and convex mirrors. Snap Lab Concave and Convex Mirrors • Silver spoon and silver polish, or a new spoon made of any shiny metal Instructions Procedure 1. Choose any small object with a top and a bottom, such as a short nail or tack, or a coin, such as a quarter. Observe the object’s reflection on the back of the spoon. 2. Observe the reflection of the object on the front (bowl side) of the spoon when held away from the spoon at a distance of several inches.
3. Observe the image while slowly moving the small object toward the bowl of the spoon. Continue until the object is all the way inside the bowl of the spoon. 4. You should see one point where the object disappears and then reappears. This is the focal point. WATCH PHYSICS Parabolic Mirrors and Real Images This video uses ray diagrams to show the special feature of parabolic mirrors that makes them ideal for either projecting light energy in parallel rays, with the source being at the focal point of the parabola, or for collecting at the focal point light energy from a distant source. Click to view content (https://www.openstax.org/l/28Parabolic) Explain why using a parabolic mirror for a car headlight throws much more light on the highway than a flat mirror. a. The rays do not polarize after reflection. b. The rays are dispersed after reflection. c. The rays are polarized after reflection. d. The rays become parallel after reflection. You should be able to notice everyday applications of curved mirrors. One common example is the use of security mirrors in stores, as shown in Figure 16.11. 484 Chapter 16 • Mirrors and Lenses Figure 16.11 Security mirrors are convex, producing a smaller, upright image. Because the image is smaller, a larger area is imaged compared with what would be observed for a flat mirror; hence, security is improved. (credit: Laura D’Alessandro, Flickr) Some telescopes also use curved mirrors and no lenses (except in the eyepieces) both to magnify images and to change the path of light. Figure 16.12 shows a Schmidt-Cassegrain telescope. This design uses a spherical primary concave mirror and a convex secondary mirror. The image is projected onto the focal plane by light passing through the perforated primary mirror. The effective focal length of such a telescope is the focal length of the primary mirror multiplied by the magnification of the secondary mirror. The result is a telescope with a focal length much greater than the length of the telescope itself. Figure 16.12 This diagram shows the design of a Schmidt–Cassegrain telescope. A parabolic concave mirror has the very useful property that all light from a distant source, on reflection by the mirror surface, is directed to the focal point. Likewise, a light source placed at the focal point directs all the light it emits in parallel lines away from the mirror. This case is illustrated by the ray diagram in Figure 16.13. The light source in a car headlight, for example, is located at the focal point of a parabolic mirror. Figure 16.13 The bulb in this ray diagram of a car headlight is located at the focal point of a parabolic mirror. Parabolic mirrors are also used to collect sunlight and direct it to a focal point, where it is transformed into heat, which in turn can be used to generate electricity. This application is shown in Figure 16.14. Figure 16.14 Parabolic trough collectors are used to generate electricity in southern California. (credit: kjkolb, Wikimedia Commons) Access for free at openstax.org. 16.1 • Reflection 485 Using a concave mirror, you look at the reflection of a faraway object. The image size changes if you move the object closer to the mirror. Why does the image disappear entirely when the object is at the mirror's focal point? a. The height of the image became infinite. b. The height of the object became zero. c. The intensity of intersecting light rays became zero. d. The intensity of intersecting light rays increased. The Application of the Curved Mirror Equations Curved mirrors and the images they create involve a fairly small number of variables: the mirror’s radius of curvature, R; the focal length, f; the distances of the object and image from the mirror, doand di, respectively; and the heights of the object and image, hoand hi, respectively. The signs of these values indicate whether the image is inverted, erect (upright), real, or virtual. We now look at the equations that relate these variables and apply them to everyday problems. Figure 16.15 shows the meanings of most of the variables we will use for calculations involving curved mirrors. The basic equation that describes both lenses and mirrors is the lens/mirror equation Figure 16.15 Look for the variables, do, di, ho, hi,and fin this figure. This equation can be rearranged several ways. For example, it may be written to solve for focal length. Magnification, m, is the ratio of the size of the image, hi, to the size of the object, ho. The value of mcan be calculated in two ways. This relationship can be written to solve for any of the variables involved. For example, the height of the image is given by We saved the simplest equation for last. The radius of curvature of a curved mirror, R, is simply twice the focal length. We can learn important information from the algebraic sign of the result of a calculation using the previous equations: • A negative diindicates a virtual image; a positive value indicates a real image • A negative hiindicates an inverted image; a positive value indicates an erect image • For concave mirrors, fis positive; for convex mirrors, fis negative Now let’s apply these equations to solve some problems. 486 Chapter 16 • Mirrors and Lenses WORKED EXAMPLE Calculating Focal Length A person standing 6.0 m from a convex security mirror forms a virtual image that appears to be 1.0 m behind the mirror. What is the focal length of the mirror? STRATEGY The person is the object, so do= 6.0 m. We know that, for this situation, dois positive. The image is virtual, so the value for the image distance is negative, so di= –1.0 m. Now, use the appropriate version of the lens/mirror equation to solve for focal length by substituting the known values. Solution Discussion The negative result is expected for a convex mirror. This indicates the focal point is behind the mirror. WORKED EXAMPLE Calculating Object Distance Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR radiation follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m in front of the mirror, where are the coils with respect to the mirror? STRATEGY We are told that the concave mirror projects a real image of the coils at an image distance di= 3.00 m. The coils are the object, and we are asked to find their location—that is, to find the object distance do. We are also given the radius of curvature of the mirror, so that its focal length is f= R/2 = 25.0 cm (a positive value, because the mirror is concave, or converging). We can use the lens/mirror equation to solve this problem. Solution Because diand fare known, the lens/mirror equation can be used to find do. Rearranging to solve for do, we have Entering the known quantities gives us 16.1 16.2 16.3 Discussion Note that the object (the coil filament) is farther from the mirror than the mirror’s focal length. This is a case 1image (do > fand f positive), consistent with the fact that a real image is formed. You get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. In general, this is not desirable because it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of the mirror. Note that the filament here is not much farther from the mirror than the focal length, and that the image produced is considerably farther away. Access for free at openstax.org. 16.2 • Refraction 487 Practice Problems 1. A concave mirror has a radius of curvature of . What is the focal length of the mirror? a. b. c. d. 2. What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person’s face is 12.0 cm away? Construct a ray diagram using paper, a pencil and a ruler to confirm your calculation. a. –36.0 cm b. –7.20 cm c. d. 7.20 cm 36.0 cm Check Your Understanding 3. How does the object distance, do, compare with the focal length, f, for a concave mirror that produces an image that is real and inverted? a. do > f, where do and f are object distance and focal length, respectively. b. do < f, where do and f are object distance and focal length, respectively. c. do = f, where do and f are object distance and focal length, respectively. d. do = 0, where do is the object distance. 4. Use the law of reflection to explain why it is not a good idea to polish a mirror with sandpaper. a. The surface becomes smooth, and a smooth surface produces a sharp image. b. The surface becomes irregular, and an irregular surface produces a sharp image. c. The surface becomes smooth, and a smooth surface transmits light, but does not reflect it. d. The surface becomes irregular, and an irregular surface produces a blurred image. 5. An object is placed in front of a concave mirror at a distance that is greater than the focal length of the mirror. Will the image produced by the mirror be real or virtual? Will it be erect or inverted? a. b. c. d. It is real and erect. It is real and inverted. It is virtual and inverted. It is virtual and erect. 16.2 Refraction Section Learning Objectives By the end of this section, you will be able to do the following: • Explain refraction at media boundaries, predict the path of light after passing through a boundary (Snell’s law), describe the index of refraction of materials, explain total internal reflection, and describe applications of refraction and total internal reflection • Perform calculations based on the law of refraction, Snell’s law, and the conditions for total internal reflection Section Key Terms angle of refraction corner reflector critical angle dispersion incident ray index of refraction refracted ray Snell’s law total internal reflection The Law of Refraction You may have noticed some odd optical phenomena when looking into a fish tank. For example, you may see the same fish appear to be in two different places (Figure 16.16). T
his is because light coming to you from the fish changes direction when it 488 Chapter 16 • Mirrors and Lenses leaves the tank and, in this case, light rays traveling along two different paths both reach our eyes. The changing of a light ray’s direction (loosely called bending) when it passes a boundary between materials of different composition, or between layers in single material where there are changes in temperature and density, is called refraction. Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibers. Figure 16.16 Looking at the fish tank as shown, we can see the same fish in two different locations, because light changes directions when it passes from water to air. In this case, light rays traveling on two different paths change direction as they travel from water to air, and so reach the observer. Consequently, the fish appears to be in two different places. This bending of light is called refractionand is responsible for many optical phenomena. Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. This behavior is typical of all waves and is especially easy to apply to light because light waves have very small wavelengths, and so they can be treated as rays. Before we study the law of refraction, it is useful to discuss the speed of light and how it varies between different media. The speed of light is now known to great precision. In fact, the speed of light in a vacuum, c, is so important, and is so precisely known, that it is accepted as one of the basic physical quantities, and has the fixed value where the approximate value of 3.00 matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, given that its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction, n, of a material to be 108 m/s is used whenever three-digit precision is sufficient. The speed of light through 16.4 where vis the observed speed of light in the material. Because the speed of light is always less than cin matter and equals conly in a vacuum, the index of refraction (plural: indices of refraction) is always greater than or equal to one. Table 16.2 lists the indices of refraction in various common materials. Medium n Gases at 0 °C and 1 atm Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. Access for free at openstax.org. 16.2 • Refraction 489 Medium n Air 1.000293 Carbon dioxide 1.00045 Hydrogen Oxygen Liquids at 20 °C Benzene Carbon disulfide 1.000139 1.000271 1.501 1.628 Carbon tetrachloride 1.461 Ethanol Glycerin Water, fresh Solids at 20 °C Diamond Fluorite Glass, crown Glass, flint Ice at 0 °C Plexiglas Polystyrene 1.361 1.473 1.333 2.419 1.434 1.52 1.66 1.309 1.51 1.49 Quartz, crystalline 1.544 Quartz, fused Sodium chloride 1.458 1.544 Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. 490 Chapter 16 • Mirrors and Lenses Medium n Zircon 1.923 Table 16.2 Indices of Refraction The table lists the indices of refraction for various materials that are transparent to light. Note, that light travels the slowest in the materials with the greatest indices of refraction. Figure 16.17 provides an analogy for and a description of how a ray of light changes direction when it passes from one medium to another. As in the previous section, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure 16.17, medium 2 has a greater index of refraction than medium 1. This difference in index of refraction means that the speed of light is less in medium 2 than in medium 1. Note that, in Figure 16.17(a), the path of the ray moves closer to the perpendicular when the ray slows down. Conversely, in Figure 16.17(b), the path of the ray moves away from the perpendicular when the ray speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the right front wheel is slowed and pulled to the side as shown. This is the same change in direction for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the left front wheel moves faster than the others, and the mower changes direction as shown. This, too, is the same change in direction as light going from slow to fast. Figure 16.17 The change in direction of a light ray depends on how the speed of light changes when it crosses from one medium to another. For the situations shown here, the speed of light is greater in medium 1 than in medium 2. (a) A ray of light moves closer to the perpendicular when it slows down. This is analogous to what happens when a lawnmower goes from a footpath (medium 1) to grass (medium 2). (b) A ray of light moves away from the perpendicular when it speeds up. This is analogous to what happens when a lawnmower goes from grass (medium 2) to the footpath (medium 1). The paths are exactly reversible. Snap Lab Bent Pencil A classic observation of refraction occurs when a pencil is placed in a glass filled halfway with water. Do this and observe the shape of the pencil when you look at it sideways through air, glass, and water. • A full-length pencil • A glass half full of water Instructions Procedure 1. Place the pencil in the glass of water. 2. Observe the pencil from the side. Access for free at openstax.org. 16.2 • Refraction 491 3. Explain your observations. Virtual Physics Bending Light Click to view content (https://www.openstax.org/l/28Bendinglight) The Bending Light simulation in allows you to show light refracting as it crosses the boundaries between various media (download animation first to view). It also shows the reflected ray. You can move the protractor to the point where the light meets the boundary and measure the angle of incidence, the angle of refraction, and the angle of reflection. You can also insert a prism into the beam to view the spreading, or dispersion, of white light into colors, as discussed later in this section. Use the ray option at the upper left. A light ray moving upward strikes a horizontal boundary at an acute angle relative to the perpendicular and enters the medium above the boundary. What must be true for the light to bend away from the perpendicular? a. The medium below the boundary must have a greater index of refraction than the medium above. b. The medium below the boundary must have a lower index of refraction than the medium above. c. The medium below the boundary must have an index of refraction of zero. d. The medium above the boundary must have an infinite index of refraction. The amount that a light ray changes direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in the angle of refraction. The exact mathematical relationship is the law of refraction, or Snell’s law, which is stated in equation form as In terms of speeds, Snell’s law becomes Here, n1 and n2 are the indices of refraction for media 1 and 2, respectively, and θ1 and θ2 are the angles between the rays and the perpendicular in the respective media 1 and 2, as shown in Figure 16.17. The incoming ray is called the incident rayand the outgoing ray is called the refracted ray. The associated angles are called the angle of incidenceand the angle of refraction. Later, we apply Snell’s law to some practical situations. Dispersion is defined as the spreading of white light into the wavelengths of which it is composed. This happens because the index of refraction varies slightly with wavelength. Figure 16.18 shows how a prism disperses white light into the colors of the rainbow. 16.5 492 Chapter 16 • Mirrors and Lenses Figure 16.18 (a) A pure wavelength of light ( ) falls onto a prism and is refracted at both surfaces. (b) White light is dispersed by the prism (spread of light exaggerated). Because the index of refraction varies with wavelength, the angles of refraction vary with wavelength. A sequence of red to violet is produced, because the index of refraction increases steadily with decreasing wavelength. Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you turn your back to the Sun. Light enters a drop of water and is reflected from the back of the drop, as shown in Figure 16.19. The light is refracted both as it enters and as it leaves the drop. Because the index of refraction of water varies with wavelength, the light is dispersed and a rainbow is observed. Figure 16.19 Part of the light falling on this water drop enters and is reflected from the back of the drop. This light is refracted and dispersed both as it enters and as it leaves the drop. WATCH PHYSICS Dispersion This video explains how refraction disperses white light into its composite colors. Click to view content (https://www.openstax.org/l/28Raindrop) Which colors of the rainbow bend most when refracted? a. Colors with a longer wavelength and higher frequency bend most when refracted. b. Colors with a shorter wavelength and higher frequency bend most when r
efracted. c. Colors with a shorter wavelength and lower frequency bend most when refracted. d. Colors with a longer wavelength and a lower frequency bend most when refracted. Access for free at openstax.org. 16.2 • Refraction 493 A good-quality mirror reflects more than 90 percent of the light that falls on it; the mirror absorbs the rest. But, it would be useful to have a mirror that reflects all the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction. Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 16.20(a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than the first, the ray bends away from the perpendicular. Because n1 > n2, the angle of refraction is greater than the angle of incidence—that is, increased. This causes The critical angle, refraction of 90°. That is, as shown in Figure 16.20(c), then all the light is reflected back into medium 1, a condition called total internal reflection. , which produces an angle of , is greater than the critical angle, , for a combination of two materials is defined to be the incident angle, is the incident angle for which =90°. If the incident angle, to increase as well. The largest the angle of refraction, . Now, imagine what happens as the incident angle is , can be is 90°, as shown in Figure 16.20(b). > Figure 16.20 (a) A ray of light crosses a boundary where the speed of light increases and the index of refraction decreases—that is, n2 < n1. The refracted ray bends away from the perpendicular. (b) The critical angle, , is the one for which the angle of refraction is 90°. (c) Total internal reflection occurs when the incident angle is greater than the critical angle. Recall that Snell’s law states the relationship between angles and indices of refraction. It is given by 16.6 When the incident angle equals the critical angle ( Snell’s law in this case becomes = ), the angle of refraction is 90° ( = 90°). Noting that sin 90° = 1, 494 Chapter 16 • Mirrors and Lenses The critical angle, , for a given combination of materials is thus for n1 > n2. Total internal reflection occurs for any incident angle greater than the critical angle, medium has an index of refraction less than the first. Note that the previous equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in Figure 16.20. , and it can only occur when the second There are several important applications of total internal reflection. Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4°; so, when light enters a diamond, it has trouble getting back out (Figure 16.21). Although light freely enters the diamond at different angles, it can exit only if it makes an angle less than 24.4° with the normal to a given surface. Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Diamonds with very few impurities are very clear, so the light makes many internal reflections and is concentrated at the few places it can exit—hence the sparkle. Figure 16.21 Light cannot escape a diamond easily because its critical angle with air is so small. Most reflections are total and the facets are placed so that light can exit only in particular ways, thus concentrating the light and making the diamond sparkle. A light ray that strikes an object that consists of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came. This parallel reflection is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. Such an object is called a corner reflectorbecause the light bounces from its inside corner. Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to return light in the direction from which it originates. Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45°. One use of these perfect mirrorsis in binoculars, as shown in Figure 16.22. Another application is for periscopes used in submarines. Figure 16.22 These binoculars use corner reflectors with total internal reflection to get light to the observer’s eyes. Fiber optics are one common application of total internal reflection. In communications, fiber optics are used to transmit telephone, internet, and cable TV signals, and they use the transmission of light down fibers of plastic or glass. Because the Access for free at openstax.org. fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (Figure 16.23). The index of refraction outside the fiber must be smaller than inside, a condition that is satisfied easily by coating the outside of the fiber with a material that has an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal reflection. Rays are reflected around corners as shown in the figure, making the fibers into tiny light pipes. 16.2 • Refraction 495 Figure 16.23 (a) Fibers in bundles are clad by a material that has a lower index of refraction than the core to ensure total internal reflection, even when fibers are in contact with one another. A single fiber with its cladding is shown. (b) Light entering a thin fiber may strike the inside surface at large, or grazing, angles, and is completely reflected if these angles exceed the critical angle. Such rays continue down the fiber, even following it around corners, because the angles of reflection and incidence remain large. LINKS TO PHYSICS Medicine: Endoscopes A medical device called an endoscopeis shown in Figure 16.24. Figure 16.24 Endoscopes, such as the one drawn here, send light down a flexible fiber optic tube, which sends images back to a doctor in charge of performing a medical procedure. The word endoscope means looking inside. Doctors use endoscopes to look inside hollow organs in the human body and inside body cavities. These devices are used to diagnose internal physical problems. Images may be transmitted to an eyepiece or sent to a video screen. Another channel is sometimes included to allow the use of small surgical instruments. Such surgical procedures include collecting biopsies for later testing, and removing polyps and other growths. Identify the process that allows light and images to travel through a tube that is not straight. a. The process is refraction of light. b. The process is dispersion of light. c. The process is total internal reflection of light. d. The process is polarization of light. Calculations with the Law of Refraction The calculation problems that follow require application of the following equations: 16.8 496 Chapter 16 • Mirrors and Lenses and These are the equations for refractive index, the mathematical statement of the law of refraction (Snell’s law), and the equation for the critical angle. WATCH PHYSICS Snell’s Law Example 1 This video leads you through calculations based on the application of the equation that represents Snell’s law. Click to view content (https://www.openstax.org/l/28Snellslaw) Which two types of variables are included in Snell’s law? a. The two types of variables are density of a material and the angle made by the light ray with the normal. b. The two types of variables are density of a material and the thickness of a material. c. The two types of variables are refractive index and thickness of each material. d. The two types of variables are refractive index of a material and the angle made by a light ray with the normal. WORKED EXAMPLE Calculating Index of Refraction from Speed Calculate the index of refraction for a solid medium in which the speed of light is 2.012 substance, based on the previous table of indicies of refraction. STRATEGY We know the speed of light, c, is 3.00 of refraction, n. 108 m/s, and we are given v. We can simply plug these values into the equation for index 108 m/s, and identify the most likely Solution This value matches that of polystyrene exactly, according to the table of indices of refraction (Table 16.2). Discussion The three-digit approximation for cis used, which in this case is all that is needed. Many values in the table are only given to three significant figures. Note that the units for speed cancel to yield a dimensionless answer, which is correct. 16.9 WORKED EXAMPLE Calculating Index of Refraction from Angles Suppose you have an unknown, clear solid substance immersed in water and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.00°, and you observe the angle of refraction to be 40.30°. What are the index of refraction of the substance and its likely identity? STRATEGY We must use the mathematical expression for the law of refraction to solve this problem because we are given angle data, not speed data. The subscripts 1 and 2 refer to values for water and the unknown, respectively, where 1 represents the medium from which the 16.10 Access for free at openstax.org. light is coming and 2 is the new medium it is entering. We are given the angle values, and the table of indicies of refraction gives us nfor water as 1.333. All we have to do before solving the problem is rearrange the equation 16.2 • Refraction 497 Solution 16.11 16.12 The best match from Table 16.2 is fused quartz, with n= 1.458. Discussion Note the relative sizes of the variables involved. For example, a larger angle has a larger sine value. This
checks out for the two indicates the ray has bent towardnormal. This result is to angles involved. Note that the smaller value of compared with be expected if the unknown substance has a greater nvalue than that of water. The result shows that this is the case. WORKED EXAMPLE Calculating Critical Angle Verify that the critical angle for light going from water to air is 48.6°. (See Table 16.2, the table of indices of refraction.) STRATEGY First, choose the equation for critical angle Then, look up the nvalues for water, n1, and air, n2. Find the value of and it compare with the given angle of 48.6°. . Last, find the angle that has a sine equal to this value Solution For water, n1 = 1.333; for air, n2 = 1.0003. So, 16.13 16.14 Discussion Remember, when we try to find a critical angle, we look for the angle at which light can no longer escape past a medium boundary by refraction. It is logical, then, to think of subscript 1 as referring to the medium the light is trying to leave, and subscript 2 as where it is trying (unsuccessfully) to go. So water is 1 and air is 2. Practice Problems 6. The refractive index of ethanol is 1.36. What is the speed of light in ethanol? a. 2.25×108 m/s b. 2.21×107 m/s c. 2.25×109 m/s d. 2.21×108 m/s 7. The refractive index of air is and the refractive index of crystalline quartz is . What is the critical angle for a ray of light going from crystalline quartz into air? a. b. c. d. 498 Chapter 16 • Mirrors and Lenses Check Your Understanding 8. Which law is expressed by the equation ? a. This is Ohm’s law. b. This is Wien’s displacement law. c. This is Snell’s law. d. This is Newton’s law. 9. Explain why the index of refraction is always greater than or equal to one. a. The formula for index of refraction, , of a material is where , so is always greater than one. b. The formula for index of refraction, , of a material is where , so is always greater than one. c. The formula for index of refraction, , of a material is than one. d. The formula for refractive index, , of a material is , so is always greater than one. 10. Write an equation that expresses the law of refraction. a. b. c. d. where , , so is always greater where 16.3 Lenses Section Learning Objectives By the end of this section, you will be able to do the following: • Describe and predict image formation and magnification as a consequence of refraction through convex and concave lenses, use ray diagrams to confirm image formation, and discuss how these properties of lenses determine their applications • Explain how the human eye works in terms of geometric optics • Perform calculations, based on the thin-lens equation, to determine image and object distances, focal length, and image magnification, and use these calculations to confirm values determined from ray diagrams Section Key Terms aberration chromatic aberration concave lens converging lens convex lens diverging lens eyepiece objective ocular parfocal Characteristics of Lenses Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera’s zoom lens. In this section, we use the law of refraction to explore the properties of lenses and how they form images. Some of what we learned in the earlier discussion of curved mirrors also applies to the study of lenses. Concave, convex, focal point F, and focal length fhave the same meanings as before, except each measurement is made from the center of the lens instead of the surface of the mirror. The convex lens shown in Figure 16.25 has been shaped so that all light rays that enter it parallel to its central axis cross one another at a single point on the opposite side of the lens. The central axis, or axis, is defined to be a line normal to the lens at its center. Such a lens is called a converging lens because of the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown in Figure 16.25 to illustrate how the ray changes Access for free at openstax.org. direction both as it enters and as it leaves the lens. Because the index of refraction of the lens is greater than that of air, the ray moves toward the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) As a result of the shape of the lens, light is thus bent toward the axis at both surfaces. 16.3 • Lenses 499 Figure 16.25 Rays of light entering a convex, or converging, lens parallel to its axis converge at its focal point, F. Ray 2 lies on the axis of the lens. The distance from the center of the lens to the focal point is the focal length, ƒ, of the lens. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. Note that rays from a light source placed at the focal point of a converging lens emerge parallel from the other side of the lens. You may have heard of the trick of using a converging lens to focus rays of sunlight to a point. Such a concentration of light energy can produce enough heat to ignite paper. Figure 16.26 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the figure is the axis of the lens). The concave lens is a diverging lens because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length, or “ƒ,” of the lens. Note that the focal length of a diverging lens is defined to be negative. An expanded view of the path of one ray through the lens is shown in Figure 16.26 to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and diverge. Figure 16.26 Rays of light enter a concave, or diverging, lens parallel to its axis diverge and thus appear to originate from its focal point, F. The dashed lines are not rays; they indicate the directions from which the rays appear to come. The focal length, ƒ, of a diverging lens is negative. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. The power, P, of a lens is very easy to calculate. It is simply the reciprocal of the focal length, expressed in meters 16.15 The units of power are diopters, D, which are expressed in reciprocal meters. If the focal length is negative, as it is for the diverging lens in Figure 16.26, then the power is also negative. In some circumstances, a lens forms an image at an obvious location, such as when a movie projector casts an image onto a screen. In other cases, the image location is less obvious. Where, for example, is the image formed by eyeglasses? We use ray 500 Chapter 16 • Mirrors and Lenses tracing for thin lenses to illustrate how they form images, and we develop equations to describe the image-formation quantitatively. These are the rules for ray tracing: 1. A ray entering a converging lens parallel to its axis passes through the focal point, F, of the lens on the other side 2. A ray entering a diverging lens parallel to its axis seems to come from the focal point, F, on the side of the entering ray 3. A ray passing through the center of either a converging or a diverging lens does not change direction 4. A ray entering a converging lens through its focal point exits parallel to its axis 5. A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis Consider an object some distance away from a converging lens, as shown in Figure 16.27. To find the location and size of the image formed, we trace the paths of select light rays originating from one point on the object. In this example, the originating point is the top of a woman’s head. Figure 16.27 shows three rays from the top of the object that can be traced using the raytracing rules just listed. Rays leave this point traveling in many directions, but we concentrate on only a few, which have paths that are easy to trace. The first ray is one that enters the lens parallel to its axis and passes through the focal point on the other side (rule 1). The second ray passes through the center of the lens without changing direction (rule 3). The third ray passes through the nearer focal point on its way into the lens and leaves the lens parallel to its axis (rule 4). All rays that come from the same point on the top of the person’s head are refracted in such a way as to cross at the same point on the other side of the lens. The image of the top of the person’s head is located at this point. Rays from another point on the object, such as the belt buckle, also cross at another common point, forming a complete image, as shown. Although three rays are traced in Figure 16.27, only two are necessary to locate the image. It is best to trace rays for which there are simple ray-tracing rules. Before applying ray tracing to other situations, let us consider the example shown in Figure 16.27 in more detail. Access for free at openstax.org. 16.3 • Lenses 501 Figure 16.27 Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced. The three chosen rays each follow one of the rules for ray tracing, so their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed. The image formed in Figure 16.27 is a real image—meaning, it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye. In Figure 16.27, the object distance, do, is greater than f.Now we consider a ray
diagram for a convex lens where do<f, and another diagram for a concave lens. 502 Chapter 16 • Mirrors and Lenses Virtual Physics Geometric Optics Click to view content (https://www.openstax.org/l/28Geometric) This animation shows you how the image formed by a convex lens changes as you change object distance, curvature radius, refractive index, and diameter of the lens. To begin, choose Principal Rays in the upper left menu and then try varying some of the parameters indicated at the upper center. Show Help supplies a few helpful labels. , change with an increasing radius of curvature? How does change with an increasing How does the focal length, refractive index? a. The focal length increases in both cases: when the radius of curvature and the refractive index increase. b. The focal length decreases in both cases: when the radius of curvature and the refractive index increase. c. The focal length increases when the radius of curvature increases; it decreases when the refractive index increases. d. The focal length decreases when the radius of curvature increases; it increases in when the refractive index increases. Type Formed When Image Type di M Case 1 fpositive, do> f Real Positive Negative m>, <, or = ‒1 Case 2 fpositive, do< f Virtual Negative Positive m> 1 Case 3 fnegative Virtual Negative Positive m< 1 Table 16.3 Three Types of Images Formed by Lenses The examples in Figure 16.27 and Figure 16.28 represent the three possible cases—case 1, case 2, and case 3—summarized in Table 16.3. In the table, mis magnification; the other symbols have the same meaning as they did for curved mirrors. Figure 16.28 (a) The image is virtual and larger than the object. (b) The image is virtual and smaller than the object. Access for free at openstax.org. 16.3 • Lenses 503 Snap Lab Focal Length • Temperature extremes—Very hot or very cold temperatures are encountered in this lab that can cause burns. Use protective mitts, eyewear, and clothing when handling very hot or very cold objects. Notify your teacher immediately of any burns. • EYE SAFETY—Looking at the Sun directly can cause permanent eye damage. Do not look at the Sun through any lens. • Several lenses • A sheet of white paper • A ruler or tape measure Instructions Procedure 1. Find several lenses and determine whether they are converging or diverging. In general, those that are thicker near the edges are diverging and those that are thicker near the center are converging. 2. On a bright, sunny day take the converging lenses outside and try focusing the sunlight onto a sheet of white paper. 3. Determine the focal lengths of the lenses. Have one partner slowly move the lens toward and away from the paper until you find the distance at which the light spot is at its brightest. Have the other partner measure the distance from the lens to the bright spot. Be careful, because the paper may start to burn, depending on the type of lens. True or false—The bright spot that appears in focus on the paper is an image of the Sun. a. True b. False Image formation by lenses can also be calculated from simple equations. We learn how these calculations are carried out near the end of this section. Some common applications of lenses with which we are all familiar are magnifying glasses, eyeglasses, cameras, microscopes, and telescopes. We take a look at the latter two examples, which are the most complex. We have already seen the design of a telescope that uses only mirrors in . Figure 16.29 shows the design of a telescope that uses two lenses. Part (a) of the figure shows the design of the telescope used by Galileo. It produces an upright image, which is more convenient for many applications. Part (b) shows an arrangement of lenses used in many astronomical telescopes. This design produces an inverted image, which is less of a problem when viewing celestial objects. 504 Chapter 16 • Mirrors and Lenses Figure 16.29 (a) Galileo made telescopes with a convex objective and a concave eyepiece. They produce an upright image and are used in spyglasses. (b) Most simple telescopes have two convex lenses. The objective forms a case 1 image, which is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified. Figure 16.30 shows the path of light through a typical microscope. Microscopes were first developed during the early 1600s by eyeglass makers in the Netherlands and Denmark. The simplest compound microscope is constructed from two convex lenses, as shown schematically in Figure 16.30. The first lens is called the objective lens; it has typical magnification values from 5 to 100 . In standard microscopes, the objectives are mounted such that when you switch between them, the sample remains in focus. Objectives arranged in this way are described as parfocal. The second lens, the eyepiece, also referred to as the ocular, has several lenses that slide inside a cylindrical barrel. The focusing ability is provided by the movement of both the objective lens and the eyepiece. The purpose of a microscope is to magnify small objects, and both lenses contribute to the final magnification. In addition, the final enlarged image is produced in a location far enough from the observer to be viewed easily because the eye cannot focus on objects or images that are too close. Access for free at openstax.org. 16.3 • Lenses 505 Figure 16.30 A compound microscope composed of two lenses, an objective and an eyepiece. The objective forms a case 1 image that is larger than the object. This first image is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified even further. Real lenses behave somewhat differently from how they are modeled using rays diagrams or the thin-lens equations. Real lenses produce aberrations. An aberration is a distortion in an image. There are a variety of aberrations that result from lens size, material, thickness, and the position of the object. One common type of aberration is chromatic aberration, which is related to color. Because the index of refraction of lenses depends on color, or wavelength, images are produced at different places and with different magnifications for different colors. The law of reflection is independent of wavelength, so mirrors do not have this problem. This result is another advantage for the use of mirrors in optical systems such as telescopes. Figure 16.31(a) shows chromatic aberration for a single convex lens, and its partial correction with a two-lens system. The index of refraction of the lens increases with decreasing wavelength, so violet rays are refracted more than red rays, and are thus focused closer to the lens. The diverging lens corrects this in part, although it is usually not possible to do so completely. Lenses made of different materials and with different dispersions may be used. For example, an achromatic doublet consisting of a converging lens made of crown glass in contact with a diverging lens made of flint glass can reduce chromatic aberration dramatically (Figure 16.31(b)). Figure 16.31 (a) Chromatic aberration is caused by the dependence of a lens’s index of refraction on color (wavelength). The lens is more powerful for violet (V) than for red (R), producing images with different colors, locations, and magnifications. (b) Multiple-lens systems can correct chromatic aberrations in part, but they may require lenses of different materials and add to the expense of optical systems such as cameras. 506 Chapter 16 • Mirrors and Lenses Physics of the Eye The eye is perhaps the most interesting of all optical instruments. It is remarkable in how it forms images and in the richness of detail and color they eye can detect. However, our eyes commonly need some correction to reach what is called normalvision, but should be called idealvision instead. Image formation by our eyes and common vision correction are easy to analyze using geometric optics. Figure 16.32 shows the basic anatomy of the eye. The cornea and lens form a system that, to a good approximation, acts as a single thin lens. For clear vision, a real image must be projected onto the light-sensitive retina, which lies at a fixed distance from the lens. The lens of the eye adjusts its power to produce an image on the retina for objects at different distances. The center of the image falls on the fovea, which has the greatest density of light receptors and the greatest acuity (sharpness) in the visual field. There are no receptors at the place where the optic nerve meets the eye, which is called the blind spot. An image falling on this spot cannot be seen. The variable opening (or pupil) of the eye along with chemical adaptation allows the eye to detect light intensities from the lowest observable to 1010 times greater (without damage). Ten orders of magnitude is an incredible range of detection. Our eyes perform a vast number of functions, such as sense direction, movement, sophisticated colors, and distance. Processing of visual nerve impulses begins with interconnections in the retina and continues in the brain. The optic nerve conveys signals received by the eye to the brain. Figure 16.32 The cornea and lens of an eye act together to form a real image on the light-sensing retina, which has its densest concentration of receptors in the fovea, and a blind spot over the optic nerve. The power of the lens of an eye is adjustable to provide an image on the retina for varying object distances. Refractive indices are crucial to image formation using lenses. Table 16.4 shows refractive indices relevant to the eye. The biggest change in the refractive index—and the one that causes the greatest bending of rays—occurs at the cornea rather than the lens. The ray diagram in Figure 16.33 shows image formation by the cornea and lens of the eye. The rays bend according to the refractive indices provided in Table 16.4. The cornea provides about two-thirds of the magnification of the eye because the speed of light changes considerab
ly while traveling from air into the cornea. The lens provides the remaining magnification needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, although the light rays pass through several layers of material (such as the cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens. This result is a case 1 image. Images formed in the eye are inverted, but the brain inverts them once more to make them seem upright. Material Index of Refraction Water Air Cornea Aqueous humor 1.33 1.00 1.38 1.34 The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Access for free at openstax.org. 16.3 • Lenses 507 Material Index of Refraction Lens 1.41 average Vitreous humor 1.34 *The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Figure 16.33 An image is formed on the retina, with light rays converging most at the cornea and on entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale. As noted, the image must fall precisely on the retina to produce clear vision—that is, the image distance, di, must equal the lensto-retina distance. Because the lens-to-retina distance does not change, di must be the same for objects at all distances. The eye manages to vary the distance by varying the power (and focal length) of the lens to accommodate for objects at various distances. In Figure 16.33, you can see the small ciliary muscles above and below the lens that change the shape of the lens and, thus, the focal length. The need for some type of vision correction is very common. Common vision defects are easy to understand, and some are simple to correct. Figure 16.34 illustrates two common vision defects. Nearsightedness, or myopia, is the inability to see distant objects clearly while close objects are in focus. The nearsighted eye overconvergesthe nearly parallel rays from a distant object, and the rays cross in front of the retina. More divergent rays from a close object are converged on the retina, producing a clear image. Farsightedness, or hyperopia, is the inability to see close objects clearly whereas distant objects may be in focus. A farsighted eye does not converge rays from a close object sufficiently to make the rays meet on the retina. Less divergent rays from a distant object can be converged for a clear image. Figure 16.34 (a) The nearsighted (myopic) eye converges rays from a distant object in front of the retina; thus, they are diverging when they strike the retina, and produce a blurry image. This divergence can be caused by the lens of the eye being too powerful (in other words, too short a focal length) or the length of the eye being too great. (b) The farsighted (hyperopic) eye is unable to converge the rays from a close object by the time they strike the retina and produce ... blurry close vision. This poor convergence can be caused by insufficient power (in 508 Chapter 16 • Mirrors and Lenses other words, too long a focal length) in the lens or by the eye being too short. Because the nearsighted eye overconverges light rays, the correction for nearsightedness involves placing a diverging spectacle lens in front of the eye. This lens reduces the power of an eye that has too short a focal length (Figure 16.35(a)). Because the farsighted eye underconvergeslight rays, the correction for farsightedness is to place a converging spectacle lens in front of the eye. This lens increases the power of an eye that has too long a focal length (Figure 16.35(b)). Figure 16.35 (a) Correction of nearsightedness requires a diverging lens that compensates for the overconvergence by the eye. The diverging lens produces an image closer to the eye than the object so that the nearsighted person can see it clearly. (b) Correction of farsightedness uses a converging lens that compensates for the underconvergence by the eye. The converging lens produces an image farther from the eye than the object so that the farsighted person can see it clearly. In both (a) and (b), the rays that meet at the retina represent corrected vision, and the other rays represent blurred vision without corrective lenses. Calculations Using Lens Equations As promised, there are no new equations to memorize. We can use equations already presented for solving problems involving curved mirrors. Careful analysis allows you to apply these equations to lenses. Here are the equations you need where Pis power, expressed in reciprocal meters (m–1) rather than diopters (D), and fis focal length, expressed in meters (m). You also need where, as before, do and di are object distance and image distance, respectively. Remember, this equation is usually more useful if rearranged to solve for one of the variables. For example, The equations for magnification, m, are also the same as for mirrors where hi and ho are the image height and object height, respectively. Remember, also, that a negative di value indicates a virtual image and a negative hi value indicates an inverted image. These are the steps to follow when solving a lens problem: • Step 1. Examine the situation to determine that image formation by a lens is involved. • Step 2. Determine whether ray tracing, the thin-lens equations, or both should be used. A sketch is very helpful even if ray tracing is not specifically required by the problem. Write useful symbols and values on the sketch. • Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). • Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. Although these are just names for types of images, they have certain characteristics (given in Table 16.3) that can be of great use in solving problems. Access for free at openstax.org. 16.3 • Lenses 509 • Step 5. If ray tracing is required, use the ray-tracing rules listed earlier in this section. • Step 6. Most quantitative problems require the use of the thin-lens equations. These equations are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples were included earlier and can serve as guides. • Step 7. Check whether the answer is reasonable. Does it make sense?If you identified the type of image (case 1, 2, or 3) correctly, you should assess whether your answer is consistent with the type of image, magnification, and so on. All problems will be solved by one or more of the equations just presented, with ray tracing used only for general analysis of the problem. The steps then simplify to the following: Identify the unknown. Identify the knowns. 1. 2. 3. Choose an equation, plug in the knowns, and solve for the unknown. Here are some worked examples: WORKED EXAMPLE The Power of a Magnifying Glass Strategy The Sun is so far away that its rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus, the focal length of the lens is the distance from the lens to the spot, and its power, in diopters (D), is the inverse of this distance (in reciprocal meters). Solution The focal length of the lens is the distance from the center of the lens to the spot, which we know to be 8.00 cm. Thus, 16.16 To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. Discussion This result demonstrates a relatively powerful lens. Remember that the power of a lens in diopters should not be confused with the familiar concept of power in watts. 16.17 WORKED EXAMPLE Image Formation by a Convex Lens A clear glass light bulb is placed 0.75 m from a convex lens with a 0.50 m focal length, as shown in Figure 16.36. Use ray tracing to get an approximate location for the image. Then, use the mirror/lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin-lens and magnification equations produce consistent results. Figure 16.36 A light bulb placed 0.75 m from a lens with a 0.50 m focal length produces a real image on a poster board, as discussed in the previous example. Ray tracing predicts the image location and size. 510 Chapter 16 • Mirrors and Lenses Strategy Because the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to the one illustrated in the previous figure of a series of drawings showing a woman standing to the left of a lens. Ray tracing to scale should produce similar results for di. Numerical solutions for di and mcan be obtained using the thin-lens and magnification equations, noting that do = 0.75 m and f= 0.50 m. Solution The ray tracing to scale in Figure 16.36 shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus, the image distance, di, is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of two, and the image is inverted. Thus, mis about –2. The minus sign indicates the image is inverted. The lens equation can be rearranged to solve for di from the given information. Now, we use to find m. 16.18 16.19 Discussion Note that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the lens equation produce consistent results. The thin-lens equation gives the most precise results, and is limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you draw, b
ut it is highly useful both conceptually and visually. WORKED EXAMPLE Image Formation by a Concave Lens Suppose an object, such as a book page, is held 6.50 cm from a concave lens with a focal length of –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced? Strategy This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is therefore the same, but the results are different in important ways. Solution 16.20 Now the magnification equation can be used to find the magnification, m, because both di and do are known. Entering their values gives 16.21 Discussion A number of results in this example are true of all case 3 images. Magnification is positive (as calculated), meaning the image is upright. The magnification is also less than one, meaning the image is smaller than the object—in this case, a little more than half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. The image is virtual. The image is closer to the lens than the object, because the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, because the image is smaller than the object, you may think it is farther away; however, the image is closer than the object—a fact that is useful in correcting nearsightedness. Access for free at openstax.org. 16.3 • Lenses 511 WATCH PHYSICS The Lens Equation and Problem Solving The video shows calculations for both concave and convex lenses. It also explains real versus virtual images, erect versus inverted images, and the significance of negative and positive signs for the involved variables. Click to view content (https://www.openstax.org/l/28Lenses) If a lens has a magnification of a. The image is erect and is half as tall as the object. b. The image is erect and twice as tall as the object. c. The image is inverted and is half as tall as the object. d. The image is inverted and is twice as tall as the object. , how does the image compare with the object in height and orientation? Practice Problems 11. A lens has a focal length of . What is the power of the lens? a. The power of the lens is b. The power of the lens is c. The power of the lens is d. The power of the lens is . . . . 12. If a lens produces a 5.00 -cm tall image of an 8.00 -cm -high object when placed 10.0 cm from the lens, what is the apparent image distance? Construct a ray diagram using paper, a pencil, and a ruler to confirm your calculation. a. −3.12 cm b. −6.25 cm 3.12 cm c. d. 6.25 cm Check Your Understanding 13. A lens has a magnification that is negative. What is the orientation of the image? a. Negative magnification means the image is erect and real. b. Negative magnification means the image is erect and virtual. c. Negative magnification means the image is inverted and virtual. d. Negative magnification means the image is inverted and real. 14. Which part of the eye controls the amount of light that enters? a. b. c. d. the pupil the iris the cornea the retina 15. An object is placed between the focal point and a convex lens. Describe the image that is formed in terms of its orientation, and whether the image is real or virtual. a. The image is real and erect. b. The image is real and inverted. c. The image is virtual and erect. d. The image is virtual and inverted. 16. A farsighted person buys a pair of glasses to correct her farsightedness. Describe the main symptom of farsightedness and the type of lens that corrects it. a. Farsighted people cannot focus on objects that are far away, but they can see nearby objects easily. A convex lens is used to correct this. b. Farsighted people cannot focus on objects that are close up, but they can see far-off objects easily. A concave lens is used to correct this. 512 Chapter 16 • Mirrors and Lenses c. Farsighted people cannot focus on objects that are close up, but they can see distant objects easily. A convex lens is used to correct this. d. Farsighted people cannot focus on objects that are either close up or far away. A concave lens is used to correct this. Access for free at openstax.org. Chapter 16 • Key Terms 513 KEY TERMS aberration a distortion in an image produced by a lens angle of incidence the angle, with respect to the normal, at which a ray meets a boundary between media or a reflective surface focal point the point at which rays converge or appear to converge incident ray the incoming ray toward a medium boundary or a reflective surface angle of reflection the angle, with respect to the normal, at index of refraction the speed of light in a vacuum divided which a ray leaves a reflective surface by the speed of light in a given material angle of refraction the angle between the normal and the law of reflection the law that indicates the angle of refracted ray reflection equals the angle of incidence central axis a line perpendicular to the center of a lens or law of refraction the law that describes the relationship mirror extending in both directions chromatic aberration an aberration related to color concave lens a lens that causes light rays to diverge from the central axis concave mirror a mirror with a reflective side that is curved inward converging lens a convex lens convex lens a lens that causes light rays to converge toward the central axis between refractive indices of materials on both sides of a boundary and the change in the path of light crossing the boundary, as given by the equation n1 sin = n2 sin ray light traveling in a straight line real image an optical image formed when light rays converge and pass through the image, producing an image that can be projected onto a screen refracted ray the light ray after it has been refracted Snell’s law the law of refraction expressed mathematically convex mirror a mirror with a reflective side that is curved as outward critical angle an incident angle that produces an angle of refraction of 90° dispersion separation of white light into its component wavelengths diverging lens a concave lens focal length the distance from the focal point to the mirror SECTION SUMMARY 16.1 Reflection • The angle of reflection equals the angle of incidence. • Plane mirrors and convex mirrors reflect virtual, erect images. Concave mirrors reflect light to form real, inverted images or virtual, erect images, depending on the location of the object. Image distance, height, and other characteristics can be calculated using the lens/mirror equation and the magnification equation. • 16.2 Refraction • The index of refraction for a material is given by the speed of light in a vacuum divided by the speed of light in that material. • Snell’s law states the relationship between indices of KEY EQUATIONS 16.1 Reflection lens/mirror equation (reciprocal version) total internal reflection reflection of light traveling through a medium with a large refractive index at a boundary of a medium with a low refractive index under conditions such that refraction cannot occur virtual image the point from which light rays appear to diverge without actually doing so refraction, the incident angle, and the angle of refraction. • The critical angle, , determines whether total internal refraction can take place, and can be calculated according to . 16.3 Lenses • The characteristics of images formed by concave and convex lenses can be predicted using ray tracing. Characteristics include real versus virtual, inverted versus upright, and size. • The human eye and corrective lenses can be explained using geometric optics. • Characteristics of images formed by lenses can be calculated using the mirror/lens equation. lens/mirror equation (solved version) 514 Chapter 16 • Chapter Review magnification equation critical angle radius/focal length equation R= 2f 16.2 Refraction index of Refraction Snell’s law >Snell’s law in terms of speed CHAPTER REVIEW Concept Items 16.1 Reflection 1. Part A. Can you see a virtual image? Part B. Can you photograph one? Explain your answers. a. A. yes; B. No, an image from a flat mirror cannot be photographed. b. A. no; B. Yes, an image from a flat mirror can be photographed. c. A. yes; B. Yes, an image from a flat mirror can be photographed. d. A. no; B. No, an image from a flat mirror cannot be photographed. 2. State the law of reflection. a. b. c. d. , where is the angle of incidence. , where is the angle of incidence. , where is the angle of incidence. , where is the angle of reflection and is the angle of reflection and is the angle of reflection and is the angle of reflection. 16.2 Refraction 3. Does light change direction toward or away from the normal when it goes from air to water? Explain. a. The light bends away from the normal because the index of refraction of water is greater than that of air. b. The light bends away from the normal because the index of refraction of air is greater than that of water. Access for free at openstax.org. 16.3 Lenses power and focal length mirror/lens (or thin-lens) equation rearranged mirror/lens equation magnification equation c. The light bends toward the normal because the index of refraction of water is greater than that of air. d. The light bends toward the normal because the index of refraction of air is greater than that of water. 16.3 Lenses 4. An object is positioned in front of a lens with its base resting on the principal axis. Describe two rays that could be traced from the top of the object and through the lens that would locate the top of an image. a. A ray perpendicular to the axis and a ray through the center of the lens b. A ray parallel to the axis and a ray that does not pass through the center of the lens c. A ray parallel to the axis and a ray through the center of the lens d. A ray parallel to the axis and a ray that does not pass through the focal point 5. A person timing the
moonrise looks at her watch and then at the rising moon. Describe what happened inside her eyes that allowed her to see her watch clearly one second and then see the moon clearly. a. The shape of the lens was changed by the sclera, and thus its focal length was also changed, so that each of the images focused on the retina. b. The shape of the lens was changed by the choroid, and thus its focal length was also changed, so that each of the images focused on the retina. c. The shape of the lens was changed by the iris, and thus its focal length was also changed, so that each of the images focused on the retina. d. The shape of the lens was changed by the muscles, and thus its focal length was also changed, so that each of the images focused on the retina. 6. For a concave lens, if the image distance, di, is negative, where does the image appear to be with respect to the object? a. The image always appears on the same side of the Critical Thinking Items 16.1 Reflection 7. Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one? a. It gives a wide range of view. The image appears to be closer than the actual object. It gives a narrow range of view. The image appears to be farther than the actual object. It gives a narrow range of view. The image appears to be closer than the actual object. It gives a wide range of view. The image appears to be farther than the actual object. b. c. d. 16.2 Refraction 8. A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light. a. Diamond and air have a small difference in their refractive indices that results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. b. Diamond and air have a small difference in their refractive indices that results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. c. Diamond has a high index of refraction with respect to air, which results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. d. Diamond has a high index of refraction with respect to air, which results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. 9. The most common type of mirage is an illusion in which light from far-away objects is reflected by a pool of water that is not really there. Mirages are generally observed in Chapter 16 • Chapter Review 515 lens. b. The image appears on the opposite side of the lens. c. The image appears on the opposite side of the lens only if the object distance is greater than the focal length. d. The image appears on the same side of the lens only if the object distance is less than the focal length. deserts, where there is a hot layer of air near the ground. Given that the refractive index of air is less for air at higher temperatures, explain how mirages can be formed. a. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. b. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in refractive index is large, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. c. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a small critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. d. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in the refractive index is large, which results in a small critical angle. The light rays coming from the horizontal strike the hot air at large angles, so they are reflected as they would be from water. 16.3 Lenses 10. When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be kept at a fixed distance from the film for both near and distant objects? a. To focus on a distant object, you need to increase the image distance. b. To focus on a distant object, you need to increase the focal length of the lens. c. To focus on a distant object, you need to decrease the focal length of the lens. d. To focus on a distant object, you may need to increase or decrease the focal length of the lens. 11. Part A—How do the refractive indices of the cornea, 516 Chapter 16 • Chapter Review aqueous humor, and the lens of the eye compare with the refractive index of air? Part B—How do the comparisons in part A explain how images are focused on the retina? a. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted away from the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted away from b. Problems 16.1 Reflection 12. Some telephoto cameras use a mirror rather than a lens. What radius of curvature is needed for a concave mirror to replace a 0.800 -m focal-length telephoto lens? a. 0.400 m 1.60 m b. c. 4.00 m 16.0 m d. 13. What is the focal length of a makeup mirror that produces a magnification of 2.00 when a person’s face is 8.00 cm away? a. –16 cm b. –5.3 cm c. d. 5.3 cm 16 cm c. d. the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. Calculate the amount the ray is displaced by the glass (Δx), given that the incident angle is 40.0° and the glass is 1.00 cm thick. a. 0.839 cm b. 0.619 cm c. 0.466 cm d. 0.373 cm 16.2 Refraction 14. An optical fiber uses flint glass (n= 1.66) clad with crown 16.3 Lenses glass (n= 1.52) . What is the critical angle? a. 33.2° b. 23.7° c. 0.92 rad 1.16 rad d. 15. Suppose this figure represents a ray of light going from air (n= 1.0003) through crown glass (n= 1.52) into water, similar to a beam of light going into a fish tank. 16. A camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm . What is its range of powers? a. The lowest power is 0.05 D and the highest power is 0.125 D. b. The lowest power is 0.08 D and the highest power is 0.20 D. c. The lowest power is 5.00 D and the highest power is 12.5 D. d. The lowest power is 80 D and the highest power is 200 D. 17. Suppose a telephoto lens with a focal length of 200 mm is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1,000-m-high cliff on one of the mountains? a. (a) The image is 0.200 m on the same side of the lens. (b) The height of the image is – 2.00 cm. (a) The image is 0.200 m on the opposite side of the Access for free at openstax.org. b. Chapter 16 • Test Prep 517 c. lens. (b) The height of the image is – 2.00 cm. (a) The image is 0.200 m on the opposite side of the lens. (b) The height of the image is +2.00 cm. d. (a) The image is 0.100 m on the same side of the lens. (b) The height of the image is +2.00 cm. Performance Task 16.3 Lenses 18. In this performance task, you will investigate the lens- like properties of a clear bottle. • a water bottle or glass with a round cross-section and smooth, vertical sides • enough water to fill the bottle • a meter stick or tape measure • a bright light source with a small bulb, such as a pen light • a small bright object, such as a silver spoon. Instructions TEST PREP Multiple Choice 16.1 Reflection 19. In geometric optics, a straight line emerging from a point is called a (an) ________. a. b. c. d. object distance ray focal point image 20. An image of a 2.0 -cm object reflected from a mirror is 5.0 cm tall. What is the magnification of the mirror? a. 0.4 b. 2.5 3 c. 10 d. 21. Can a virtual image be projected onto a screen with additional lenses or mirrors? Explain your answer. a. Yes, the rays actually meet behind the lens or mirror. b. No, the image is formed by rays that converge to a point in front of the mirror or lens. Procedure 1. Look through a clear glass or plastic bottle and describe what you see. 2. Next, fill the bottle with water and describe what you see. 3. Use the water bottle as a lens to produce the image of a bright object. 4. Estimate the focal length of the water bottle lens. a. How can you find the focal length of the lens using the light and a blank wall? b. How can you find the focal length of the lens using the bright object? c. Why did the water change the lens properties of the bottle? b. c. d. the refractive index the speed of light in a vacuum the speed of light in a transparent material 23. What is the term for the minimum angle at which a light ray is reflected back into a material and cannot pass into the surrounding med
ium? critical angle a. b. incident angle c. angle of refraction d. angle of reflection 24. Consider these indices of refraction: glass: 1.52, air: 1.0003, water: 1.333. Put these materials in order from the one in which the speed of light is fastest to the one in which it is slowest. a. The speed of light in water > the speed of light in air > the speed of light in glass. b. The speed of light in glass > the speed of light in water > the speed of light in air. c. The speed of light in air > the speed of light in water > the speed of light in glass. d. The speed of light in glass > the speed of light in air c. Yes, any image that can be seen can be manipulated > the speed of light in water. so that it can be projected onto a screen. d. No, the image can only be perceived as being behind the lens or mirror. 16.2 Refraction 22. What does crepresent in the equation ? a. the critical angle 25. Explain why an object in water always appears to be at a depth that is more shallow than it actually is. a. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to 518 Chapter 16 • Test Prep be at a depth that is more shallow than the actual depth. b. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. c. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. d. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. 16.3 Lenses 26. For a given lens, what is the height of the image divided by the height of the object ( ) equal to? a. power focal length b. c. magnification d. radius of curvature Short Answer 16.1 Reflection 30. Distinguish between reflection and refraction in terms of how a light ray changes when it meets the interface between two media. a. Reflected light penetrates the surface whereas refracted light is bentas it travels from one medium to the other. b. Reflected light penetrates the surface whereas refracted light travels along a curved path. c. Reflected light bouncesfrom the surface whereas refracted light travels along a curved path. d. Reflected light bouncesfrom the surface whereas refracted light is bentas it travels from one medium to the other. 31. Sometimes light may be both reflected and refracted as it meets the surface of a different medium. Identify a material with a surface that when light travels through Access for free at openstax.org. 27. Which part of the eye has the greatest density of light receptors? a. b. c. d. the lens the fovea the optic nerve the vitreous humor 28. What is the power of a lens with a focal length of 10 cm? a. b. c. d. 10 m–1, or 10 D 10 cm–1, or 10 D 10 m, or 10 D 10 cm, or 10 D 29. Describe the cause of chromatic aberration. a. Chromatic aberration results from the dependence of the frequency of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. b. Chromatic aberration results from the dispersion of different wavelengths of light by a curved mirror so that each color has a different focal point. c. Chromatic aberration results from the dependence of the reflection angle at a spherical mirror’s surface on the distance of light rays from the principal axis so that different colors have different focal points. d. Chromatic aberration results from the dependence of the wavelength of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. the air it is both reflected and refracted. Explain how this is possible. a. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the transparent glass. b. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface, and it is refracted while passing into the opaque glass. c. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the opaque glass. d. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface; it is refracted while passing into the transparent glass. 32. A concave mirror has a focal length of 5.00 cm. What is the image distance of an object placed 7.00 cm from the center of the mirror? a. −17.5 cm b. −2.92 cm c. 2.92 cm 17.5 cm d. 33. An 8.0 -cm tall object is placed 6.0 cm from a concave mirror with a magnification of –2.0. What are the image height and the image distance? a. hi = – 16 cm, di = – 12 cm b. hi = – 16 cm, di = 12 cm c. hi = 16 cm, di = – 12 cm d. hi = 16 cm, di = 12 cm 16.2 Refraction 34. At what minimum angle does total internal reflection of toward light occur if it travels from water ice ? a. b. c. d. 35. Water floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0° and an angle of refraction of 40.1°. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride? a. b. c. d. 1.60 1.49 1.21 1.46 36. Describe what happens to a light ray when it is refracted. Include in your explanation comparison of angles, comparison of refractive indices, and the term normal. a. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of interference. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is greater in the medium with the greater refractive index. b. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the greater refractive index. Chapter 16 • Test Prep 519 another medium with a different refractive index, the ray does not change its path. The angle between the ray and the normal (the line parallel to the surfaces of the two media) is the same in both media. d. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index. 16.3 Lenses 37. What are two equivalent terms for a lens that always causes light rays to bend away from the principal axis? a. a diverging lens or a convex lens b. a diverging lens or a concave lens c. a converging lens or a concave lens d. a converging lens or a convex lens 38. Define the term virtual image. a. A virtual image is an image that cannot be projected onto a screen. b. A virtual image is an image that can be projected onto a screen. c. A virtual image is an image that is formed on the opposite side of the lens from where the object is placed. d. A virtual image is an image that is always bigger than the object. 39. Compare nearsightedness (myopia) and farsightedness (hyperopia) in terms of focal point. a. The eyes of a nearsighted person have focal points beyond the retina. A farsighted person has eyes with focal points between the lens and the retina. b. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points beyond the retina. c. A nearsighted person has eyes with focal points between the lens and the choroid. A farsighted person has eyes with focal points beyond the choroid. d. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points on the retina. 40. Explain how a converging lens corrects farsightedness. a. A converging lens disperses the rays so they focus on the retina. b. A converging lens bends the rays closer together so they do not focus on the retina. c. A converging lens bends the rays closer together so c. When a ray of light goes from one medium to they focus on the retina. 520 Chapter 16 • Test Prep d. A converging lens disperses the rays so they do not 42. What is the magnification of a lens if it produces a focus on the retina. 41. Solve the equation for in such a way that it is not expressed as a reciprocal. 12-cm-high image of a 4 -cm -high object? The image is virtual and erect. a. b. a. b. c. d. c. d. Extended Response 16.1 Reflection 43. The diagram shows a lightbulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. What angle does the ray make from the instructor’s face with the normal to the water (n= 1.33) at the point where the ray enters? Assume n= 1.00 for air. a. 68° b. 25° 19° c. 34° d. 46. Describe total internal reflection. Include a definition of the crit
ical angle and how it is related to total internal reflection. Also, compare the indices of refraction of the interior material and the surrounding material. a. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. b. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. c. When the interior material has the same index of refraction as the surrounding material, the incident ray approaches the boundary at an angle (called the critical angle) such that the refraction Where is the light source in relation to the focal point or radius of curvature of each mirror? Explain your answer. a. The bulb is at the center of curvature of the small mirror and at the focal point of the large mirror. b. The bulb is at the focal point of the small mirror and at the focal point of the large mirror. c. The bulb is at the center of curvature of the small mirror and at the center of curvature of the large mirror. d. The bulb is at the focal point of the small mirror and at the center of curvature of the large mirror. in front of a mirror that has . What is the radius of curvature 44. An object is placed a magnification of of the mirror? a. b. c. d. 16.2 Refraction 45. A scuba diver training in a pool looks at his instructor, as shown in this figure. The angle between the ray in the water and the normal to the water is 25°. Access for free at openstax.org. angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. d. When the interior material has a greater index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. 16.3 Lenses 47. The muscles that change the shape of the lens in the eyes have become weak, causing vision problems for a person. In particular, the muscles cannot pull hard enough on the edges of the lens to make it less convex. Part A—What condition does inability cause? Part B—Where are images focused with respect to the retina? Part C—Which type of lens corrects this person’s problem? Explain. a. Part A—This condition causes hyperopia. Part B—Images are focused between the lens and the retina. Chapter 16 • Test Prep 521 Part C—A converging lens gathers the rays slightly so they focus onto the retina. b. Part A—This condition causes myopia. Part B—Images are focused between the lens and the retina. Part C—A converging lens gathers the rays slightly so they focus onto the retina. c. Part A—This condition causes hyperopia. Part B—Images are focused between the lens and the retina. Part C—A diverging lens spreads the rays slightly so they focus onto the retina. d. Part A—This condition causes myopia. Part B—Images are focused between the lens and the retina. Part C—A diverging lens spreads the rays slightly so they focus onto the retina. 48. If the lens-to-retina distance is , what is the power of the eye when viewing an object a. b. c. d. away? 522 Chapter 16 • Test Prep Access for free at openstax.org. CHAPTER 17 Diffraction and Interference Figure 17.1 The colors reflected by this compact disc vary with angle and are not caused by pigments. Colors such as these are direct evidence of the wave character of light. (credit: Reggie Mathalone) Chapter Outline 17.1 Understanding Diffraction and Interference 17.2 Applications of Diffraction, Interference, and Coherence Examine a compact disc under white light, noting the colors observed and their locations on the disc. Using INTRODUCTION the CD, explore the spectra of a few light sources, such as a candle flame, an incandescent bulb, and fluorescent light. If you have ever looked at the reds, blues, and greens in a sunlit soap bubble and wondered how straw-colored soapy water could produce them, you have hit upon one of the many phenomena that can only be explained by the wave character of light. That and other interesting phenomena, such as the dispersion of white light into a rainbow of colors when passed through a narrow slit, cannot be explained fully by geometric optics. In such cases, light interacts with small objects and exhibits its wave characteristics. The topic of this chapter is the branch of optics that considers the behavior of light when it exhibits wave characteristics. 17.1 Understanding Diffraction and Interference Section Learning Objectives By the end of this section, you will be able to do the following: • Explain wave behavior of light, including diffraction and interference, including the role of constructive and destructive interference in Young’s single-slit and double-slit experiments • Perform calculations involving diffraction and interference, in particular the wavelength of light using data from a two-slit interference pattern Section Key Terms diffraction Huygens’s principle monochromatic wavefront 524 Chapter 17 • Diffraction and Interference Diffraction and Interference We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation m/s is the speed of light in vacuum, fis the frequency of the electromagnetic wave in Hz (or s–1), and where is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave. In Figure 17.2, both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves. Figure 17.2 (a) The light beam emitted by a laser at the Paranal Observatory (part of the European Southern Observatory in Chile) acts like a ray, traveling in a straight line. (credit: Yuri Beletsky, European Southern Observatory) (b) A laser beam passing through a grid of vertical slits produces an interference pattern—characteristic of a wave. (credit: Shim’on and Slava Rybka, Wikimedia Commons) That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves. Figure 17.3 shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern. Figure 17.3 Incoming waves (at the top of the picture) pass through the gaps in the rocks and create an interference pattern (in the foreground). Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f, remains the same. The speed of light in a medium is where nis its index of refraction. If you divide both sides of the equation is the wavelength in a medium, and by n, you get . Therefore, , where , Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 525 where is the wavelength in vacuum and nis the medium’s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n= 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or not, since colors are associated with frequency. 285–570 nm. Although wavelengths change while traveling from one medium to another, colors do The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts, which are the points on a wave’s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens’s principle states, “Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.” Figure 17.4 shows how Huygens’s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn later at a time, t, so that they have moved a distance . The new wavefront is a line tangent to the wavelets and is where the wave is located at time t. Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere. Figure 17.4 Huygens’s principle applied to a straight wavefront. Each point on the wavefront emits a semicircular wav
elet that moves a distance . The new wavefront is a line tangent to the wavelets. What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners. If light passes through smaller openings, often called slits, you can use Huygens’s principle to show that light bends as sound does (see Figure 17.5). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a 526 Chapter 17 • Diffraction and Interference wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in Figure 17.2 is evidence that light has the properties of a wave. Figure 17.5 Huygens’s principle applied to a straight wavefront striking an opening. The edges of the wavefront bend after passing through the opening, a process called diffraction. The amount of bending is more extreme for a small opening, consistent with the fact that wave characteristics are most noticeable for interactions with objects about the same size as the wavelength. Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in Figure 17.6. Figure 17.6 Ocean waves pass through an opening in a reef, resulting in a diffraction pattern. Diffraction occurs because the opening is similar in width to the wavelength of the waves. WATCH PHYSICS Single-Slit Interference This video works through the math needed to predict diffraction patterns that are caused by single-slit interference. Click to view content (https://www.openstax.org/l/28slit) Which values of mdenote the location of destructive interference in a single-slit diffraction pattern? a. whole integers, excluding zero b. whole integers c. d. real numbers excluding zero real numbers The fact that Huygens’s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light’s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773–1829) did his now-classic double-slit experiment (see Figure 17.7). Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 527 Figure 17.7 Young’s double-slit experiment. Here, light of a single wavelength passes through a pair of vertical slits and produces a diffraction pattern on the screen—numerous vertical light and dark lines that are spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen. When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 17.8 (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in Figure 17.8 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below. Figure 17.8 Double slits produce two sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. The waves overlap and interfere constructively (bright lines) and destructively (dark regions). You can only see the effect if the light falls onto a screen and is scattered into your eyes. (b) The double-slit interference pattern for water waves is nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. Virtual Physics Wave Interference Click to view content (https://www.openstax.org/l/28interference) This simulation demonstrates most of the wave phenomena discussed in this section. First, observe interference between two sources of electromagnetic radiation without adding slits. See how water waves, sound, and light all show interference patterns. Stay with light waves and use only one source. Create diffraction patterns with one slit and then with two. You may have to adjust slit width to see the pattern. 528 Chapter 17 • Diffraction and Interference Visually compare the slit width to the wavelength. When do you get the best-defined diffraction pattern? a. when the slit width is larger than the wavelength b. when the slit width is smaller than the wavelength c. when the slit width is comparable to the wavelength d. when the slit width is infinite Calculations Involving Diffraction and Interference The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young’s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More , then generally, if the paths taken by the two waves differ by any half-integral number of wavelengths destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths , then constructive interference occurs. Figure 17.9 shows how to determine the path-length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference, , where dis the distance between the slits, , to be To obtain constructive interference for a double slit, the path-length difference must be an integral multiple of the wavelength, or Similarly, to obtain destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength, or The number mis the orderof the interference. For example, m= 4 is fourth-order interference. Figure 17.9 The paths from each slit to a common point on the screen differ by an amount , assuming the distance to the screen is much greater than the distance between the slits (not to scale here). Figure 17.10 shows how the intensity of the bands of constructive interference decreases with increasing angle. Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 529 Figure 17.10 The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit. Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits. Figure 17.11 shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side. Figure 17.11 (a) Single-slit diffraction pattern. Monochromatic light passing through a single slit produces a central maximum and many smaller and dimmer maxima on either side. The central maximum is six times higher than shown. (b) The drawing shows the bright central maximum and dimmer and thinner maxima on either side. (c) The location of the minima are shown in terms of and D. The analysis of single-slit diffraction is illustrated in Figure 17.12. Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference. or relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or farther than the ray from the top edge of the slit, they arrive out When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle out of phase. Thus, a ray from the center travels a distance of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray. farther to the common point on the screen, and so 530 Chapter 17 • Diffraction and Interference Figure 17.12 Equations for a single

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