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b) Which planet’s semi-major axis is the closest in length to its semi-minor axis? 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 337 12. Three identical coins 24. A slingshot is used to propel a stone vertically C B A are placed on a rotating platter as shown. As the frequency of rotation increases, identify which coin will begin to slide off first. Explain your answer. 13. Briefly explain how Neptune was discovered. Use the terms orbital perturbation, force of gravity, and orbital velocity in your explanation. 14. Your friend argues that Neptune would not have been discovered as soon as it was if Neptune were a much smaller planet and Uranus much bigger. Is she right? Defend your answer. 15. What difficulties do astronomers face when searching for extrasolar planets that might have life as we know it? CHAPTER 6 16. Express a joule in terms of kilograms, metres, and seconds. 17. If work is a scalar quantity, why is it affected by the directions of the force and displacement? 18. What happens to an object’s gravitational potential energy when it is in free fall? 19. Explain why doubling the speed of an object does not result in a doubling of its kinetic energy. 20. A large mass and a small mass with the same kinetic energy are sliding on a horizontal frictionless surface. If forces of equal magnitude act on each of these bodies to bring them to rest, which one will stop in the shorter distance? Explain. 21. Describe the energy changes of a roller coaster car from the time when it is just coming over the crest of one hill until it arrives at the crest of the next hill. 22. A cart is pulled up an inclined plane by a force that is just large enough to keep the cart moving without a change in its speed. Is this an isolated system? Explain why or why not. Is the force used to move the cart up the incline a conservative force? 23. A cart at the top of an inclined plane is allowed to roll down the plane. Under what conditions can this system be considered isolated? If the conditions that make this an isolated system do not exist, is the force that moves the cart down the plane still considered a conservative force? Explain. upward. Describe the energy changes that are involved from the time the stone is placed in the slingshot until the stone reaches its maximum height. 25. If a
force that acts on an object results in a change in the object’s kinetic energy, what can be said about the nature of this force? 26. How do you calculate work from a force- displacement graph? 27. According to the work-energy theorem, how much work is done on an isolated system? 28. Does power affect the amount of work you are able to do? Explain why or why not. Applications 29. Electrons in an electric (AC) circuit vibrate at 60 Hz. What is their period? 30. A cell phone is set to vibrate when it rings. It vibrates with a period of 0.0160 s. What is the frequency of the ring? 31. A toy top spins at 300.0 rpm. What is the frequency (in Hz) and the period of the top? 32. The Moon orbits Earth once every 27.3 days at a mean orbital radius of 3.844 105 km. What is its speed? 33. A child sits in a pretend airplane on a ride at an amusement park. The airplane is at the end of a long arm that moves in a circular path with a radius of 4.0 m at a speed of 1.57 m/s. What is the period of the ride? 34. A person sliding down a water slide at a speed of 5.56 m/s encounters a turn with a radius of 10.0 m. Determine the acceleration that he experiences in the turn. 35. A pilot of a jet airplane makes a sharp turn to create an acceleration of 4.00 times the acceleration of gravity. If the turn has a radius of 500.0 m, what is the speed of the plane? 36. A cork (m 2.88 g) is caught in a small whirlpool created in the basin of a sink. What is the centripetal force acting on the cork, if its speed is 0.314 m/s at a radius of 4.00 cm? 37. When braking to a stop, the maximum force that friction exerts on a 1250-kg auto is 3200 N. (a) If the original speed of the auto is 12.0 m/s, what is the minimum stopping distance? If the speed of the car were twice as great, how would that affect the minimum stopping distance? (b) Unit III Circular Motion, Work, and Energy 337 06-PearsonPhys20-Chap06 7/24/08 12
:56 PM Page 338 38. A force of 250 N [up] is applied to a mass of 15.0 kg over a displacement of 9.60 m [up]. (a) How much work does the force do on the mass? (b) What is the change in gravitational (c) potential energy? If it is an isolated system, explain the difference between the answers for (a) and (b). 39. A car with a mass of 2.00 103 kg is travelling at a velocity of 15.0 m/s [0] on a horizontal stretch of highway. The driver presses on the accelerator so that the force propelling the car forward is increased to 3.30 103 N [0]. The force acts over a displacement of 55.0 m [0] during which force of friction on the car is 5.00 102 N in magnitude. (a) Draw a free-body diagram to analyze the forces acting on the car. What is the net force on the car? (b) What is the work done by net force over the displacement? (c) What is the final kinetic energy of the car? (d) What is the final speed of the car? 40. A block with a mass of 0.800 kg is initially at rest on a frictionless inclined plane. A force of 5.00 N, applied parallel to the inclined plane, moves the block a distance of 4.50 m up the plane. If, at the end of the effort, the block has a speed of 6.00 m/s up the incline, what is the change in height through which it moved? 41. A varying force acts on a 25.0-kg mass over a displacement as shown in the graph below. The mass has an initial velocity of 12.0 m/s [0]. Recall that the area of a force-displacement graph is equivalent to the work done by the force. (a) What is the ) ] ˚ 0 [ 60 45 N (a) What is the change in gravitational potential energy for block A? (b) What was the change in height through which block A moved? B m 0 5 1. 1.50 m A 43. For question 42, draw the graph that shows the potential, kinetic, and mechanical energies for the system as a function of the displacement assuming (a) there is no friction, (b) there is friction, but block A still accelerates up the hill. 44. A pend
ulum bob with a mass of 0.750 kg is initially at rest at its equilibrium position. You give the bob a push so that when it is at a height of 0.150 m above its equilibrium position it is moving at a speed of 2.00 m/s. (a) How much work did you do on the bob? If you pushed on the bob with a force of (b) 40.0 N parallel to the displacement, how far did you push it? 45. A billiard ball with a speed of 2.00 m/s strikes a second ball initially at rest. After the collision, the first ball is moving at a speed of 1.50 m/s. If this is an elastic collision (i.e., energy is conserved), what is the speed of the second ball after the collision? Assume that the balls have identical masses of 0.200 kg. 46. A cannon ball (m 3.00 kg) is fired at a velocity of 280 m/s at an angle of 20 above the horizontal. The cannon is on a cliff that is 450 m above the ocean. (a) What is the mechanical energy of the work that the force did on the mass? ( F e c r o F 30 15 (b) What is the final speed of the mass? 0 10 20 Displacement d (m [0˚]) 30 cannon ball relative to the base of the cliff? (b) What is the greatest height above the ocean 40 that the cannon ball reaches? (c) What is the speed of the cannon ball when it lands in the ocean? 42. In the following diagram, block A is at rest on a frictionless inclined plane. It is attached to block B by a light cord over a frictionless pulley. Block A has a mass of 4.50 kg and B has a mass of 5.50 kg. When they are released, block A moves up the incline so that after it has moved a distance of 1.50 m along the incline it has a speed of 3.00 m/s. 47. A Styrofoam™ ball is dropped from a height of 5.00 m. The mass of the ball is 0.200 kg. When the ball hits the ground it has a speed of 3.00 m/s. (a) What change in mechanical energy does the ball undergo while it falls? (b) What is the average force that air friction exerted on the falling ball? 338
Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 339 48. What is the average power output if an engine lifts a 250-kg mass a distance of 30.0 m in 20.0 s? 49. What is the effective power required to maintain a constant speed of 108 km/h if the force opposing the motion is 540 N in magnitude? 50. An airplane engine has an effective power output of 150 kW. What will be the speed of the plane if the drag (air friction opposing the motion of the plane) exerts a force of 2.50 103 N? Extensions 51. A 90.9-kg gymnast swings around a horizontal bar in a vertical circle. When he is directly over top of the bar, his arms experience a tug of 108.18 N. What is the speed of his body in this position? (Assume that the gymnast’s mass is centred 1.20 m from the bar.) 52. Suppose a solar system, with three planets circling a star, exists in a galaxy far far away. One of the planets orbits the star with the period of 6.31 107 s and a radius of 3.00 1011 m. It has a moon that orbits it with a period of 1.73 106 s at a radius of 6.00 108 m. (a) What is the mass of the planet’s star? (b) What is the mass of the planet? (c) What is the speed of the planet’s moon? 53. A soil-moving machine called a bucket wheel loader has a large metallic wheel with a radius of 3.05 m that has many scoops attached to it. The scoops are designed to dig into the ground and lift soil out as the wheel turns around. If the wheel turns with a frequency of 0.270 Hz, will the soil fall out of a scoop when it gets to the top of the wheel? 54. Three blocks (A, B, and C), with masses 6.00 kg, 4.00 kg, and 2.00 kg, respectively, are initially at rest on a horizontal frictionless surface as shown in diagram (a) on the right. A force of 48.0 N [90] acts on block A over a displacement of 7.50 m [90]. Block A is connected to block B by a string that is 1.00 m long and block
B is connected to block C by a string that is 1.50 m long. Initially, the three blocks are touching each other. As the blocks move and the strings become taut, they end up as shown in diagram (b). Is this an isolated or a non-isolated system? Explain. (a) What is the speed of the blocks after the force has acted for the full 7.50 m? (b) What is the speed of block A when the force has acted over a displacement of 2.00 m? Hint: Find the work done by the force. (a) 1.50 m string connecting B to C 1.00 m string connecting A to B force meter (b) Mass A Mass B Mass C 0 m 1. 0 force meter 0 m 1. 5 F F Skills Practice 55. Your cousin doesn’t understand how a satellite can stay in orbit without falling toward Earth. Using the knowledge you have gained from this unit, provide a short explanation. 56. Figure 6.21 on page 303 shows the impact crater for a meteor that landed in Arizona. How widespread is the evidence of meteors striking Earth? Where is the impact crater closest to where you live? Do Internet research to identify locations of impact craters in Alberta and Canada. On a map of Alberta, show the location of meteor impacts. What clues on maps show meteor landings? For each crater, identify how much kinetic energy the meteor would have had when it struck Earth. 57. Explain how you would experiment to determine the quantity of external work done on a system for a cart accelerating down an inclined plane. Could you confirm this quantity by measuring forces? Self-assessment 58. Describe to a classmate one misconception you had about circular motion before studying this unit. Explain what you know about this concept now. 59. Describe to a classmate the relationship between the roles of science and technology in the development of new energy resources. 60. Can science provide solutions to all of the problems associated with the impact of energy consumption on the environment? Give reasons for your answer. e TEST To check your understanding of circular motion, work, and energy, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit III Circular Motion, Work, and Energy 339 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 340 U N I T IV Oscillatory Oscillatory Motion and Motion and Mechanical Mechanical Waves Waves
An earthquake more than two thousand kilometres away sent this tsunami speeding across the Indian Ocean. Waves, a form of simple harmonic oscillations, can efficiently transport incredible amounts of energy over great distances. How does a wave move through its medium? How does understanding simple harmonic motion help us understand how waves transport energy? 340 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 341 Unit at a Glance C H A P T E R 7 Oscillatory motion requires a set of conditions. 7.1 Period and Frequency 7.2 Simple Harmonic Motion 7.3 Position, Velocity, Acceleration, and Time Relationships 7.4 Applications of Simple Harmonic Motion C H A P T E R 8 Mechanical waves transmit energy in a variety of ways. 8.1 The Properties of Waves 8.2 Transverse and Longitudinal Waves 8.3 Superposition and Interference 8.4 The Doppler Effect Unit Themes and Emphases • Change, Energy, and Matter • Scientific Inquiry • Nature of Science Focussing Questions As you study this unit, consider these questions: • Where do we observe oscillatory motion? • How do mechanical waves transmit energy? • How can an understanding of the natural world improve how society, technology, and the environment interact? Unit Project • By the time you complete this unit, you will have the knowledge and skill to research earthquakes, the nature of earthquake shock waves, and the use of the Richter scale to rate earthquake intensity. On completion of your research, you will demonstrate the operation of a seismograph. e WEB To learn more about earthquakes and their environmental effects, follow the links at www.pearsoned.ca/school/physicssource. Unit IV Oscillatory Motion and Mechanical Waves 341 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 342 Oscillatory motion requires a set of conditions. On October 15, 1997, NASA launched the Cassini-Huygens space probe toward Saturn — a distance of 1 500 000 000 km from Earth. The probe’s flight path took it by Venus twice, then Earth, and finally past Jupiter on its way to Saturn. This route was planned so that, as the probe approached each planet, it would be accelerated by the planet’s gravitational force. Each time, it picked up more speed, allowing it to get to Saturn more quickly (Figure 7.1). Recall from Chapter
4 that increasing the probe’s speed this way is referred to as gravity assist. The entire journey of 3 500 000 000 km took seven years. For this incredible feat to succeed, scientists had to know where the planets would be seven years in the future. How could they do this? They relied on the fact that planets follow predictable paths around the Sun. Nature is full of examples of repetitive, predictable motion. Water waves, a plucked guitar string, the orbits of planets, and even a bumblebee flapping its wings are just a few. In this chapter, you will examine oscillatory motion. Oscillatory motion is a slightly different form of motion from the circular motion you studied in Chapter 5. You will better understand why bungee jumpers experience the greatest pull of the bungee cord when they are at the bottom of a fall and why trees sway in the wind. You may notice the physics of many objects that move with oscillatory motion and gain a new insight into the wonders of the natural world. C H A P T E R 7 Key Concepts In this chapter, you will learn about: ■ oscillatory motion ■ simple harmonic motion ■ restoring force ■ oscillating spring and pendulum ■ mechanical resonance Learning Outcomes When you have completed this chapter, you will be able to: Knowledge ■ describe oscillatory motion in terms of period and frequency ■ define simple harmonic motion as being due to a restoring force that is directly proportional and opposite to the displacement of an object from an equilibrium position ■ explain the quantitative relationships among displacement, acceleration, velocity, and time for simple harmonic motion ■ define mechanical resonance Science, Technology, and Society ■ explain that the goal of science is knowledge about the natural world Figure 7.1 The Cassini-Huygens probe successfully orbits Saturn after a seven-year journey through the solar system. 342 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 343 7-1 QuickLab 7-1 QuickLab Oscillatory Motion of Toys Problem What is the time taken by one complete back-and-forth motion of a toy? Procedure 1 Fully wind the spring mechanism of one of the toys. 2 Release the winding knob and start your stopwatch. Materials several different wind-up toys, such as: • swimming frog • hopping toy • monkey with cymbals • walking toy yo-yo stopwatch (or wristwatch with a second hand) Figure 7.2 Think About It 3 Count the number of complete back-and
-forth movements the toy makes in 10 s. These back-andforth movements are called oscillations. 4 Record the number of oscillations made by the toy. 5 Repeat steps 1 to 4 for each toy. For the yo-yo, first achieve a steady up-and-down rhythm. Then do steps 3 and 4, counting the up-and-down motions. This type of repetitive movement is also an oscillation. Questions 1. In what ways are the motions of all the toys similar? 2. Divide the number of oscillations of each toy by the time. Be sure to retain the units in your answer. What does this number represent? (Hint: Look at the units of the answer.) 3. Which toy had the most oscillations per second? Which had the least? 4. Divide the time (10 s) by the number of oscillations for each toy. Be sure to retain the units in your answer. How long is the interval of oscillation of each toy? 5. Which toy had the longest time for one oscillation? Which had the shortest time? 1. How are the oscillations per second and the time for one oscillation related? 2. What do you think influences the number of oscillations per second of the toys you tested? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 7 Oscillatory motion requires a set of conditions. 343 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 344 7.1 Period and Frequency On a warm summer day in your backyard, you can probably hear bees buzzing around, even if they are a few metres away. That distinctive sound is caused by the very fast, repetitive up-and-down motion of the bees’ wings (Figure 7.3). Take a closer look at the bumblebee. The motion of a bee’s wings repeats at regular intervals. Imagine that you can examine the bee flying through the air. If you start your observation when its wings are at the highest point (Figure 7.4(a)), you see them move down to their lowest point (Figure 7.4(c)), then back up again. When the wings are in the same position as when they started (Figure 7.4(e)), one complete oscillation has occurred. An oscillation is a repetitive backand-forth motion.
One complete oscillation is called a cycle. Figure 7.3 The wings of a bee in flight make a droning sound because of their motion. oscillatory motion: motion in which the period of each cycle is constant info BIT Earthquake waves can have periods of up to several hundred seconds. (a) (b) (c) (d) (e) Figure 7.4 The bee’s wings make one full cycle from (a) to (e). The time for this motion is called the period. The time required for the wings to make one complete oscillation is the period (T). If the period of each cycle remains constant, then the wings are moving up and down with oscillatory motion. Recall from Chapter 5 that the number of oscillations per second is the frequency (f), measured in hertz (Hz). The equation that relates frequency and period is: f 1 T (1) Table 7.1 shows the period of a bee’s wings as it hovers, along with other examples of periods. ▼ Table 7.1 Periods of Common Items Object Bumblebee wings Hummingbird wings Medical ultrasound technology Middle C on a piano Electrical current in a house Period 0.00500 s 0.0128 s 1 106 to 5 108 s 0.0040 s 0.0167 s 344 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 345 A piston in the engine of a car also undergoes oscillatory motion if it is moving up and down in equal intervals of time. The piston shown in Figure 7.5 moves from position (a) (its highest point) through (b) to position (c), where it is at its lowest point. It begins moving back up again through (d) to (e), where it returns to its highest position. The range of movement from (a) to (e) is one cycle. A single piston in a Formula 1 racecar can achieve a frequency of 300 cycles/second or 300 Hz (18 000 rpm). The piston makes 300 complete cycles in only 1 s. Conversely, the period of the piston, which is the time for one complete cycle, is the inverse of the frequency. It is a mere 0.003 s or about 100 times faster than the blink of an eye! (a) (b) (c) (d) (e) e MATH Using a graphing calculator,
plot period as a function of frequency. Use the following values of frequency: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. If possible, print this graph, label it, and add it to your notes. What is the shape of this graph? info BIT A Formula 1 racecar has 10 cylinders but the engine size is limited to 3.0 L, no bigger than many engines in medium-sized cars. The fuel efficiency of F1 cars is approximately 1.3 km/L. Figure 7.5 The piston makes one complete cycle from positions (a) to (e). The time it takes to do this is its period. The number of times it does this in 1 s is its frequency. Example 7.1 What is the frequency of an automobile engine in which the pistons oscillate with a period of 0.0625 s? Analysis and Solution f 1 T 1 0.0625 s 16.0 Hz The frequency of the engine is 16.0 Hz. Practice Problems 1. Earthquake waves that travel along Earth’s surface can have periods of up to 5.00 minutes. What is their frequency? 2. A hummingbird can hover when it flaps its wings with a frequency of 78 Hz. What is the period of the wing’s motion? Answers 1. 3.33 103 Hz 2. 0.013 s M I N D S O N Examples of Oscillatory Motion Working with a partner or group, make a list of three or four natural or humanmade objects that move with the fastest oscillatory motion that you can think of. Make a similar list of objects that have very long periods of oscillatory motion. Beside each object estimate its period. The lists must not include the examples already mentioned. Be prepared to share your lists with the class. Chapter 7 Oscillatory motion requires a set of conditions. 345 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 346 7-2 Inquiry Lab 7-2 Inquiry Lab Relating Period and Frequency Your teacher may want to do this Inquiry Lab in the gym instead of the science lab. Question What is the relationship between the period and the frequency of a bouncing ball? Hypothesis Write a hypothesis that relates the period of the ball’s bounce to its frequency. Remember to use an “if/then” statement. Variables The variables in this lab are the height of the bounce, period, and frequency
. Read the procedure, then determine and label the controlled, manipulated, and responding variables. Materials and Equipment stopwatch chair basketball metre-stick tape Procedure 1 Copy Table 7.2 into your notes. ▼ Table 7.2 Bounce, Period, and Frequency Bounce Height (cm) Time for 20 Bounces (s) Period (s/bounce) Frequency (bounces/s) Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Find a convenient place to bounce the basketball near the wall. Using the metre-stick, place tape at heights of 50, 75, 100, 125, and 150 cm. Mark the heights on the tape. 3 Using just a flick of the wrist, begin bouncing the ball at the 50-cm mark. The top of the ball should just make it to this height at the top of its bounce. The ball should bounce with a steady rhythm. 4 While one person is bouncing the ball, another person uses the stopwatch to record the time taken for 20 complete bounces. Record this time in Table 7.2. 5 Reset the stopwatch and begin bouncing the ball to the next height up. Record the time for 20 complete bounces as you did in step 4. To achieve the proper height you may have to stand on a chair. Analysis 1. Using the data for 20 bounces, determine the period and frequency for each height. Record the values in the table. 2. Draw a graph of frequency versus period. What type of relationship is this? 3. What effect did increasing the bounce height have on the period of a bounce? 50 75 100 125 150 346 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 347 In 7-2 Inquiry Lab, the ball will make many more bounces in a certain length of time if it travels a shorter distance than a longer one. Its frequency will be high. By necessity, the amount of time it takes to make one bounce (its period) will be small. The next section explores oscillatory motion by going one step further. You will examine a type of oscillatory motion in which the range of motion is related to the applied force. 7.1 Check and Reflect 7.1 Check and Reflect Knowledge 1. What conditions describe oscillatory motion? 2. Which unit is equivalent to cycles/s? 3. Define period and frequency. 4. How are period and frequency related? 5. Is
it possible to increase the period of an oscillatory motion without increasing the frequency? Explain. 6. Give three examples of oscillatory motion that you have observed. Applications 7. What is the frequency of a swimming water toy that makes 20.0 kicking motions per second? 8. Do the oars on a rowboat move with 12. A dog, happy to see its owner, wags its tail 2.50 times a second. (a) What is the period of the dog’s wagging tail? (b) How many wags of its tail will the dog make in 1 min? Extensions 13. Use your library or the Internet to research the frequency of four to six different types of insect wings. Rank these insect wings from highest to lowest frequency. 14. Which of these motions is oscillatory? Explain. (a) a figure skater moving with a constant speed, performing a figure eight (b) a racecar racing on an oval track (c) your heartbeat oscillatory motion? Explain. 15. Many objects exhibit oscillatory motion. 9. Determine the frequency of a guitar string that oscillates with a period of 0.004 00 s. 10. A dragonfly beats its wings with the frequency of 38 Hz. What is the period of the wings? 11. A red-capped manakin is a bird that can flap its wings faster than a hummingbird, at 4800 beats per minute. What is the period of its flapping wings? Use your library or the Internet to find the frequency or range of frequencies of the objects below. (a) fluorescent light bulbs (b) overhead power lines (c) human voice range (d) FM radio range (e) lowest note on a bass guitar e TEST To check your understanding of period and frequency, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 347 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 348 info BIT A human eardrum can oscillate back and forth up to 20 000 times a second. 7.2 Simple Harmonic Motion Children on swings can rise to heights that make their parents nervous. But to the children, the sensation of flying is thrilling. At what positions on a swing does a child move fastest? When does the child’s motion stop? Many objects that move with oscillatory motion exhibit the same properties that a child
on a swing does. A piston moves up and down in the cylinder of an engine. At the extreme of its motion, it stops for a brief instant as it reverses direction and begins to accelerate downward until it reaches the bottom of its stroke. There it stops again and accelerates back, just as the swing does. In order for the piston or a child on a swing to experience acceleration, it must experience a non-zero net force. This section explores how the net force affects an object’s motion in a special type of oscillatory motion called simple harmonic motion. 7-3 QuickLab 7-3 QuickLab Determining the Stiffness of a Spring Problem How does the force applied to a spring affect its displacement? Materials spring (with loops at each end) spring scale metre-stick tape Procedure 1 Make a two-column table in your notebook. Write “Displacement (cm)” as the heading of the left column, and “Force (N)” as the heading of the right column. 2 Determine the maximum length the spring can be pulled without permanently deforming it. If you are unsure, ask your instructor what the maximum displacement of the spring is. Divide this length by 5. This will give you an idea of the even increments through which to pull your spring. 3 Lie the spring flat on the surface of the table so that it lies in a straight line. Leave room for it to be stretched. Do not pull on the spring. 4 Fix one end of the spring to the desk by holding the loop at the end of the spring with your hand. Do not let this end of the spring move. 5 Attach the spring scale to the free end of the spring but do not pull on it yet. 6 Align the 0-cm mark of the metre-stick with the other end of the spring at exactly where the spring scale is attached. Tape the stick to the desk (Figure 7.6). spring scale Figure 7.6 7 Pull the spring, using the spring scale, by the incremental distance determined in step 2. Record the values of the displacement and force in your table. 8 Repeat step 7, until you have five values each for displacement and force in your table. 9 Gently release the tension from the spring. Clean up and put away the equipment at your lab station. 348 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page
349 Questions 1. Determine the controlled, manipulated, and responding 4. Determine the slope of the line. What are the units of the slope? variables. 2. Plot a graph of force versus displacement. Be sure to use a scale that allows you to use the full graph paper. Draw a line of best fit. 3. What kind of relationship does the line of best fit represent? 5. Extrapolate where the line intercepts the horizontal axis. Why does it intercept there? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Hooke’s Law Robert Hooke (1635–1703) was a British scientist best remembered for using a microscope to discover plant cells, but his talents extended into many areas (Figure 7.7). He is credited with inventing the universal joint, which is used today on many mechanical devices (including cars); the iris diaphragm used to adjust the amount of light that enters a camera lens; the respirator to help people breathe; and the compound microscope, just to name a few of his inventions. In the field of oscillatory motion, he is acknowledged for his work with elastic materials and the laws that apply to them (Figure 7.8). In 1676, Hooke recognized that the more stress (force) is applied to an object, the more strain (deformation) it undergoes. The stress can be applied in many ways. For example, an object can be squeezed, pulled, or twisted. Elastic materials will usually return to their original state after the stress has been removed. This will not occur if too much force is applied or if the force is applied for too long a time. In those cases, the object will become permanently deformed because it was strained beyond the material’s ability to withstand the deformation. The point at which the material cannot be stressed farther, without permanent deformation, is called the elastic limit. A spring is designed to be a very elastic device, and the deformation of a spring (whether it is stretched or compressed) is directly related to the force applied. The deformation of a spring is referred to as its displacement. From his observations, Hooke determined that the deformation (displacement) is proportional to the applied force. This can also be stated as “force varies directly with displacement.” It can be written mathematically as: F x This relationship is known as Hooke’
s law, which states: The deformation of an object is proportional to the force causing the deformation. Figure 7.7 Robert Hooke lived at the same time as Sir Isaac Newton. Figure 7.8 Hooke’s notes show the simple devices he used to derive his law. Chapter 7 Oscillatory motion requires a set of conditions. 349 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 350 Hooke’s law: the deformation of an object is proportional to the force causing it Figure 7.9(a) shows a spring that conforms to Hooke’s law. It experiences a displacement that is proportional to the applied force. When no mass is applied to the spring, it is not compressed, and it is in its equilibrium position. As mass is added in increments of 10 g, the displacement (deformation) of the spring increases proportionally in increments of 5 cm, as shown in Figures 7.9(b), (c), and (d). spring constant: amount of stiffness of a spring (a) x 5 cm (b) 10 g (c) (d) x 0 x 10 cm 20 g x 15 cm 30 g Figure 7.9 The spring pictured above conforms to Hooke’s law. If the mass is doubled, the displacement will also double, as seen in (b) and (c). If the force (mass) is tripled, the displacement will triple, as seen in (b) and (d). Each spring is different, so the force required to deform it will change from spring to spring. The stiffness of the spring, or spring constant, is represented by the letter k. Using k, you can write the equation for Hooke’s law as: kx F where F is the applied force that extends or compresses the spring, k is the spring constant, and x is the displacement from the equilibrium position. Graphing Hooke’s Law Figure 7.10 shows how you can use a force meter to measure the applied force required to pull the spring from its equilibrium position to successively farther displacements. For simplicity, we’ll assume that all the springs used in this text are “ideal” springs, meaning that they have no mass. FA FA x 0 Figure 7.10 A force meter is attached to a spring that has not been stretched. The spring is then pulled through several displacements. Each time
, the force required for the displacement is recorded. 350 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 351 Figure 7.11 Graph of data from Table 7.3 Table 7.3 shows the data collected for this spring and the results plotted on the graph shown in Figure 7.11. ▼ Table 7.3 Data for Figure 7.11 Force vs. Displacement of a Spring Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.0 5.0 10.0 15.0 20.0 20 ) N ( e c r o F 10 0.0 0.1 0.2 0.3 0.4 Displacement (m) Notice that the relationship is linear (a straight line), which means force is proportional to the displacement. The slope of the line can be determined by the following calculations: slope F x (20.0 N 0.0 N) (0.40 m 0.00 m) 50 N/m This slope represents the spring constant k. The variables F and x are vectors but here we are calculating their scalar quantities so no vector arrows are used. In this example, an applied force of 50 N is needed to stretch (or compress) this spring 1 m. Therefore, the units for the spring constant are newtons per metre (N/m). By plotting the force-displacement graph of a spring and finding its slope, you can determine the spring constant of any ideal spring or spring that obeys Hooke’s law. Of the many objects that display elastic properties, springs are arguably the best to examine because they obey Hooke’s law over large displacements. Steel cables are also elastic when stretched through relatively small displacements. Even concrete displays elastic properties and obeys Hooke’s law through very small displacements. In simpler terms, the property of elasticity gives a material the ability to absorb stress without breaking. This property is vital to consider when structural engineers and designers build load-bearing structures such as bridges and buildings (Figure 7.12). You will learn more about the factors that must be considered in bridge and building design later in this chapter. Figure 7.12 The Jin Mao Tower in Shanghai, China, is 88 storeys high. Skyscrapers are built with elastic materials so they can sway in high winds and withstand the
shaking of an earthquake. The main building materials for the Jin Mao Tower are concrete and steel. Chapter 7 Oscillatory motion requires a set of conditions. 351 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 352 Example 7.2 Practice Problems 1. A spring is stretched through several displacements and the force required is recorded. The data are shown below. Determine the spring constant of this spring by plotting a graph and finding the slope. Displacement Force (m) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 (N) 0.0 20.0 50.0 80.0 95.0 130.0 150.0 2. Determine the spring constant of a spring that has the forcedisplacement graph shown in Figure 7.15. Force vs. Displacement 40 30 20 10 ) Displacement (m) Figure 7.15 Answers 1. 2.5 102 N/m 2. 15 N/m To determine the spring constant of a spring, a student attaches a force meter to one end of the spring, and the other end to a wall as shown in Figure 7.13. She pulls the spring incrementally to successive displacements, and records the values of displacement and force in Table 7.4. Plot the values on a graph of force as a function of displacement. Using a line of best fit, determine the spring constant of the spring. x 0 Figure 7.13 Given ▼ Table 7.4 Data for Figure 7.14 Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.25 0.35 0.55 0.85 Required spring constant (k) Analysis and Solution Using the values from Table 7.4, plot the graph and draw a line of best fit. Force vs. Displacement of a Spring 1..5 (x2, y2) (0.30, 0.60) (x1, y1) (0.10, 0.20) 0.0 0.1 0.2 0.3 Displacement (m) 0.4 Figure 7.14 352 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 353 The slope of the line gives the spring constant (k). Pick two points from the
line and solve for the slope using the equation below. Note the points used in the equation are not data points. slope k F x point 1 (0.10, 0.20) point 2 (0.30, 0.60) (0.60 N 0.20 N) (0.30 m 0.10 m) k 2.0 N/m Paraphrase The spring constant of the spring is 2.0 N/m. The Restoring Force Imagine that you have applied a force to pull a spring to a positive dis- placement (x ) as shown in Figure 7.16. While you hold it there, the spring exerts an equal and opposite force in your hand, as described by Newton’s third law in Chapter 3. However, this force is to the left, in the negative direction, and attempts to restore the spring to its equilibrium position. This force is called the restoring force. left right FA x 0 FR x Figure 7.16 The spring system is pulled from its equilibrium position to displacement xx. The displacement is positive, but the restoring force is negative. The restoring force always acts in a direction opposite to the displacement. Therefore, Hooke’s law is properly written with a negative sign when representing the restoring force. kx F (2) In this case, while the spring is held in this position, the applied force and the restoring force have equal magnitudes but opposite directions, so the net force on the system is zero. In the next section, a mass will be attached to the spring and it will slide on a frictionless horizontal surface. The restoring force will be the only force in the system and will give rise to a repetitive back-and-forth motion called simple harmonic motion. restoring force: a force acting opposite to the displacement to move the object back to its equilibrium position Chapter 7 Oscillatory motion requires a set of conditions. 353 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 354 Example 7.3 A spring has a spring constant of 30.0 N/m. This spring is pulled to a distance of 1.50 m from equilibrium as shown in Figure 7.17. What is the restoring force? Practice Problems 1. Determine the restoring force of a spring displaced 55.0 cm. The spring constant is 48.0 N/m. 2. A spring requires a force of 100.0 N to compress it a displacement of 4.0 cm. What
is its spring constant? Answers 1. 26.4 N 2. 2.5 103 N/m Analysis and Solution Draw a diagram to represent the stretched spring. Displacement to the right is positive, so the restoring force is negative because it is to the left, according to Hooke’s law. kx F 30.0 45.0 N (1.50 m) N m The restoring force is 45.0 N [left]. left right FR FA x 1.50 m equilibrium Figure 7.17 Simple Harmonic Motion of Horizontal Mass-spring Systems Figure 7.18 shows a mass attached to an ideal spring on a horizontal frictionless surface. This simple apparatus can help you understand the relationship between the oscillating motion of an object and the effect the restoring force has on it. x 0 left right Fnet 0 m Figure 7.18 The mass is in its equilibrium position (x 0) and is at rest. There is no net force acting on it. Any displacement of the mass to the right is positive, and to the left, negative. e SIM Observe a simulation of simple harmonic motion in a horizontal mass-spring system. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. The position of the mass is represented by the variable x and is measured in metres. In Figure 7.18, there is no tension on the spring nor restoring force acting on the mass, because the mass is in its equilibrium position. Figure 7.19 shows how the restoring force affects the acceleration, displacement, and velocity of the mass when the mass is pulled to a positive displacement and released. 354 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 355 The mass has been pulled to its maximum displacement, Figure 7.19(a) called its amplitude (symbol A). When the mass is released, it begins oscillating with a displacement that never exceeds this distance. The greater the amplitude, the more energy a system has. In this diagram, x A. At maximum displacement, the restoring force is at a maximum value, and therefore, so is the acceleration of the mass, as explained by Newton’s second law (F ma). When the mass is released, it accelerates from rest (v 0) toward its equilibrium position. As the mass approaches this position, its velocity is increasing. But the restoring force is decreasing because the spring is not
stretched as much. Remember that force varies directly with displacement. Figure 7.19(b) As the mass returns to its equilibrium position (x 0), it achieves its maximum velocity. It is moving toward the left (the negative direction), but the restoring force acting on it is zero because its displacement is zero. The mass continues to move through the equilibrium position and begins to compress the spring. As it compresses the spring, the restoring force acts on the mass toward the right (the positive direction) to return it to its equilibrium position. This causes the mass to slow down, and its velocity approaches zero. Figure 7.19(c) After passing through the equilibrium position, the mass experiences a restoring force that opposes its motion and brings it to a stop at the point of maximum compression. Its amplitude here is equal, but opposite to its amplitude when it started. At maximum displacement, the velocity is zero. The restoring force has reached its maximum value again. The restoring force is positive, and the displacement is negative. The restoring force again accelerates the mass toward equilibrium. Figure 7.19(d) The mass has accelerated on its way to the equilibrium position where it is now. The restoring force and acceleration are again zero, and the velocity has achieved the maximum value toward the right. At equilibrium, the mass is moving to the right. It has attained the same velocity as in diagram (b), but in the opposite direction. Figure 7.19(e) The mass has returned to the exact position where it was released. Again the restoring force and acceleration are negative and the velocity is zero. The oscillation will repeat again as it did in diagram (a). (a) (b) (c) (d) (e) F max a max v 0 m x is positive F 0 a 0 v max m x 0 F max a max v 0 m x is negative F 0 a 0 v max m x 0 F max a max v 0 m x is positive In Figure 7.19(e), the mass has returned to the position where it started, and one full oscillation has occurred. Throughout its entire motion, the mass-spring system obeys Hooke’s law. In other words, at any instant, the restoring force is proportional to the displacement of the mass. Any object that obeys Hooke’s law undergoes simple harmonic motion (SHM). SHM is oscillatory motion where the restoring force is proportional to the displacement of the mass. An object that moves with
SHM is called a simple harmonic oscillator. simple harmonic motion: oscillatory motion where the restoring force is proportional to the displacement of the mass simple harmonic oscillator: an object that moves with simple harmonic motion Chapter 7 Oscillatory motion requires a set of conditions. 355 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 356 (a) (b) Fspring equilibrium Fnet 0 Fg Figure 7.20 The spring in (a) has no mass attached. In (b), the spring stretches until the force exerted by the mass is equal and opposite to the force of gravity, and equilibrium is reached. The net force (or restoring force) is the vector sum of the force of gravity and the tension of the spring. In this case, it is zero. Figure 7.21 The net force (the restoring force) is the vector sum of the upward force exerted by the spring and the downward force of gravity. The force of gravity is always negative and constant, but the force exerted by the spring varies according to the displacement, so the net force changes as the position of the mass changes from (a) to (e). The values of F and v are identical to the horizontal mass-spring system. net, a, PHYSICS INSIGHT For any frictionless simple harmonic motion, the restoring force is equal to the net force. Simple Harmonic Motion of Vertical Mass-spring Systems Figure 7.20(a) shows a spring without a mass attached, anchored to a ceiling. Assume that the spring itself is massless, so it will not experience any displacement. When a mass is attached, the spring is pulled down and deforms as predicted by Hooke’s law. The mass will come to rest when the downward force of gravity is equal to the upward pull (tension) of the spring (Figure 7.20(b)). The displacement of the spring depends on its spring constant. A weak spring has a small spring constant. It will stretch farther than a spring with a large spring constant. In Figure 7.20(b), the net force (or restoring force) acting on the mass is zero. It is the result of the upward tension exerted by the spring balancing the downward force of gravity. This position is considered the equilibrium position and the displacement is zero. If the mass is lifted to the position shown in Figure 7.21(a) and released, it will begin oscillating with simple harmonic motion. Its amplitude will equal
its initial displacement. Regardless of the position of the mass, the force of gravity remains constant but the tension of the spring varies. In the position shown in Figures 7.21(b) and (d), the net (restoring) force is zero. This is where the spring’s tension is equal and opposite to the force of gravity. In the position shown in Figure 7.21(c), the displacement of the spring is equal to the amplitude, and the tension exerted by the spring is at its maximum. The mass experiences the greatest restoring force, which acts upward. up down x 0 (a) (b) (c) (d) (e) Fs Fg Fnet max Fnet a max v 0 Fg Fs Fnet Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fnet Fg Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fg Fnet When the mass is below the equilibrium position, the upward force exerted by the tension of the spring is greater than the gravitational force. So the net force — and therefore the restoring force — is upward. Above the equilibrium position, the downward force of gravity exceeds the upward tension of the spring, and the restoring force is downward. The values of velocity, acceleration, and restoring force change in exactly the same way that they do in a horizontal massspring system. 356 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 357 Example 7.4 A spring is hung from a hook on a ceiling. When a mass of 510.0 g is attached to the spring, the spring stretches a distance of 0.500 m. What is the spring constant? Practice Problems 1. Five people with a combined mass of 275.0 kg get into a car. The car’s four springs are each compressed a distance of 5.00 cm. Determine the spring constant of the springs. Assume the mass is distributed evenly to each spring. 2. Two springs are hooked together and one end is attached to a ceiling. Spring A has a spring constant (k) of 25 N/m, and spring B has a spring constant (k) of 60 N/m. A mass weighing 40.0 N
is attached to the free end of the spring system to pull it downward from the ceiling. What is the total displacement of the mass? Answers 1. 1.35 104 N/m 2. 2.3 m s and F g Given x 0.500 m m 510.0 g 0.5100 kg g 9.81 m/s2 Required spring constant (k) Analysis and Solution Draw a diagram to show the mass-spring system and the forces acting on the mass. up down x 0.500 m equilibrium Fs Fg Figure 7.22 The mass is not moving so the net force on the mass is zero. F are therefore equal in magnitude. kx mg mg x m (0.5100 kg )9.81 s2 0.500 m k 10.0 N/m Paraphrase The spring constant is 10.0 N/m. Chapter 7 Oscillatory motion requires a set of conditions. 357 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 358 Examples of Simple Harmonic Motion Provided the cord doesn’t go slack, a person making a bungee jump will bob up and down with SHM, as shown in Figures 7.23 and 7.24. The cord acts as the spring and the person is the mass. up down x 0 Fcord Fg Fcord Fg Fnet Fcord Fg Fnet Fcord Fnet Fg Fcord Fg Fnet 0 Fnet Fcord Fg Figure 7.24 The bungee jumper bouncing up and down on the cord after a jump in (a) is a vertical mass-spring system. The cord acts as a spring and the jumper is the mass. The restoring (net) force acting on the bungee jumper is the same as it was for the vertical mass-spring system. When the oscillating finally stops, the jumper will come to a stop in the equilibrium position. The reeds of woodwind instruments, such as the clarinet, behave as simple harmonic oscillators. As the musician blows through the mouthpiece, the reed vibrates as predicted by the laws of SHM. Once a simple harmonic oscillator is set in motion, it will slowly come to rest because of friction unless a force is continually applied. We will examine these conditions in section 7.4 on resonance. SHM is repetitive and predictable, so we can state the following: • The restoring force acts in the opposite
direction to the displacement. • At the extremes of SHM, the displacement is at its maximum and is referred to as the amplitude. At this point, force and acceleration are also at their maximum, and the velocity of the object is zero. • At the equilibrium position, the force and acceleration are zero, and the velocity of the object is at its maximum. Figure 7.23 A bungee jumper experiences SHM as long as the cord does not go slack. e WEB After the first few oscillations following the jump, the bungee jumper oscillates with simple harmonic motion. To learn more about simple harmonic motion in the vertical direction, follow the links at www.pearsoned.ca/school/ physicssource. 358 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 359 Concept Check 1. Must the line on a graph of force versus displacement for 2. a spring always intercept the origin? Explain. In what situation might the line on a graph of force as a function of displacement for a spring become non-linear? 3. A student wants to take a picture of a vertical mass-spring system as it oscillates up and down. At what point in the mass’s motion would you suggest that she press the button to take the clearest picture? Instead of plotting a force-displacement graph for a spring, a student plots a restoring force-displacement graph. Sketch what this graph might look like. 4. 5. How would you write the equation for Hooke’s law to reflect the shape of the graph above? Simple Harmonic Motion of a Pendulum The Cassini-Huygens space probe featured at the beginning of this chapter is named in honour of two distinguished scientists. Among many other notable accomplishments, the Italian astronomer Giovanni Cassini (1625–1712) observed the planets Mars and Jupiter and measured their periods of rotation. Christiaan Huygens (1629–1695), a Dutch mathematician and astronomer, invented the first accurate clock. It used a swinging pendulum and was a revolution in clock making (Figure 7.25). For small displacements, a swinging pendulum exhibits SHM. Since SHM is oscillatory, a clock mechanism that uses a pendulum to keep time could be very accurate. Up until Huygens’s time, clocks were very inaccurate. Even the best clocks could be out by as
much as 15 minutes a day. They used a series of special gears and weights that didn’t always produce a uniform rate of rotation — a necessity for an accurate mechanical clock. Huygens recognized that if he could take advantage of the uniform oscillations of a pendulum, he could produce a much better clock. When completed, his pendulum clock was accurate to within one minute a day. This may not be very accurate by today’s standards, but was easily the best of its time. Pendulum clocks became the standard in time keeping for the next 300 years. Let’s examine cases where an ideal pendulum swings through a small angle, as explained in Figure 7.26 (a) to (e). In this book, all pendulums are considered ideal. That is, we assume that the system is frictionless, and the entire mass of the pendulum is concentrated in the weight. While this is not possible in reality, it is reasonable to make these assumptions here because they provide reasonably accurate results. Figure 7.25 A replica of Huygens’s pendulum clock Chapter 7 Oscillatory motion requires a set of conditions. 359 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 360 (a) left right v 0 max FR a max Fg Fg FR (b) left right v max 0 FR a 0 (c) left right v 0 max FR a max Fg Fg FR (d) left right v max 0 FR a 0 (e) left right v 0 max FR a max Figure 7.26(a) The mass (called a “bob”) is attached to the string and has been pulled from its equilibrium (rest) position through a displacement angle of. It has a mass m. When the bob’s displacement is farthest to the right, the restoring force is a maximum negative value and velocity is zero. When the pendulum is released, gravity becomes the restoring force. Given the direction of the force of gravity, the acceleration due to gravity is straight down. However, the motion of the pendulum is an arc. A component of gravity acts along this arc to pull the bob back toward equilibrium and is, by definition, the restoring force (F R). We can express FR in terms of Fg with the following equation: FR Fg(sin ) Figure 7.26(b) As the bob accelerates downward, its velocity begins to
increase and the restoring force (F R) becomes less and less. When it reaches the equilibrium position, no component of gravity is acting parallel to the motion of the bob, so the restoring force is zero, but the velocity has reached its maximum value. Figure 7.26(c) restoring force has also reached a maximum value but it acts toward the right. The bob’s velocity is zero again. The bob has reached its maximum displacement to the left. The The bob passes through the equilibrium position and begins to move upward. As it does so, the restoring force becomes larger as the displacement of the bob increases. But just like the mass-spring system, the restoring force is acting in a direction opposite to the bob’s displacement. At the other extreme of the bob’s displacement, the restoring force has slowed the bob to an instantaneous stop. In this position, the displacement and restoring force are a maximum, and the bob’s velocity is zero. Figure 7.26(d) The bob has achieved its maximum velocity, but this time it is to the right. The bob’s displacement is again zero, and so is the restoring force. On its back swing, the bob moves through the equilibrium position again, as shown here. The velocity is a maximum value, just as it was in Figure 7.26(b), but now it’s in the opposite direction. The restoring force once more brings the bob’s motion to a stop for an Figure 7.26(e) instant at the position farthest to the right. The restoring force is a maximum negative value, and the bob’s velocity is zero. The pendulum has made one complete oscillation as shown in diagrams (a) to (e). Note that the motion of the pendulum bob and the mass-spring systems are similar. Figure 7.19(a–e) on page 355 and Figure 7.26(a–e) on this page are comparable because both systems undergo the same changes to force, velocity, and acceleration at the same displacements. 360 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 361 Motion with Large Amplitudes Earlier in this chapter, you read that a pendulum acts as a simple harmonic oscillator for small angles. Why is that? How is its motion different from SHM at larger angles? The best way to answer these questions is to plot
a graph of force versus displacement like those done for springs earlier in this chapter (e.g., Figure 7.11 on page 351). The displacement of the pendulum can be measured by its angle from the vertical. If the graph is linear, then the restoring force is proportional to the displacement, and the pendulum has moved in SHM, as described by Hooke’s law. To create this graph, use the equation for restoring force that you saw in the explanation of Figure 7.26(a) on the previous page: FR Fg(sin ) (3) From this equation, we can plot the values for angles up to 90 for a bob with a mass of 1.0 kg. As the graph in Figure 7.27 shows, the line is not linear, so the restoring force does not vary proportionally with the displacement. Strictly speaking, a pendulum is not a true simple harmonic oscillator. However, the line is almost linear up to about 20. At angles of less than 15, the deviation from a straight line is so small that, for all practical purposes, it is linear. ▼ Table 7.5 Data for Figure 7.27 Angle (°) Restoring Force (N) 0 10 20 30 40 50 60 70 80 90 0 1.70 3.36 4.91 6.31 7.51 8.50 9.22 9.66 9.81 Force vs. Displacement of a Pendulum 10 10 20 40 50 60 30 Displacement (°) 70 80 90 Figure 7.27 For the pendulum to be a true simple harmonic oscillator, its graph of restoring force versus displacement should be linear, as the dotted line suggests. After 15, its line departs from the straight line, and its motion can no longer be considered SHM. Chapter 7 Oscillatory motion requires a set of conditions. 361 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 362 Example 7.5 Practice Problems 1. Determine the restoring force of a pendulum that is pulled to an angle of 12.0 left of the vertical. The mass of the bob is 300.0 g. 2. At what angle must a pendulum be displaced to create a restoring force of 4.00 N [left] on a bob with a mass of 500.0 g? Answers 1. 0.612 N [right] 2. 54.6 Determine the magnitude of the restoring
force for a pendulum bob of mass 100.0 g that has been pulled to an angle of 10.0 from the vertical. Given g 9.81 m/s2 m 100.0 g 0.1000 kg Required restoring force (FR) Analysis and Solution Draw a diagram of the pendulum in its displaced position to show the forces acting on the bob. left right 10.0° FT Fg Fg FR Figure 7.28 The restoring force F arc path of the pendulum. R is the component of F g that is tangential to the FR Fg(sin ) mg(sin ) (0.1000 kg)9.81 0.170 N m s2 (sin 10.0) Paraphrase The magnitude of the restoring force acting on the pendulum is 0.170 N. When Christiaan Huygens designed the first pendulum clock, his primary concern was to have the clock operate with a very consistent period. For a uniform period, he could use gear ratios in the mechanism to translate the motion of the pendulum to meaningful units of time, such as minutes and hours. Which factors influence the period of a pendulum, and which do not? To discover how a pendulum’s mass, amplitude, and length influence its period, do 7-4 Inquiry Lab. 362 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 363 Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Attach the thermometer clamp to the top of the retort stand. Attach the thread to the thermometer clamp. Make sure the thread is a little shorter than the height of the clamp from the table. 3 Squeeze one end of the thread in the clamp and use a slip knot on the other end to attach the first mass. The mass should hang freely above the table. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this as the length of the pendulum at the top of Table 7.6. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in your table. 7 Remove the mass and replace it with the next
mass. Loosen the clamp and adjust the length of the thread so that it is the same length as for the previous mass. (Remember to measure length to the middle of the mass.) Repeat steps 5 to 7 until all the masses are used. Analysis 1. Determine the frequency and period of each mass. Record the numbers in your table. 2. Plot a graph of period versus mass. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the mass and the period of a pendulum? Explain your answer and show any relevant calculations. 7-4 Inquiry Lab 7-4 Inquiry Lab A Pendulum and Simple Harmonic Motion Question What is the relationship between the period of a pendulum and its mass, amplitude, and length? Materials and Equipment thermometer clamp retort stand 1.00-m thread (e.g., dental floss) 4 masses: 50 g, 100 g, 150 g, 200 g ruler (30 cm) or metre-stick protractor stopwatch or watch with a second hand Hypothesis Before you begin parts A, B, and C, state a suitable hypothesis for each part of the lab. Remember to write your hypotheses as “if/then” statements. Variables The variables are the length of the pendulum, the mass of the pendulum, elapsed time, and the amplitude of the pendulum. Read the procedure for each part and identify the controlled, manipulated, and responding variables each time. Part A: Mass and Period Procedure 1 Copy Table 7.6 into your notebook. ▼ Table 7.6 Mass and Period Length of Pendulum Mass (g) No. of Cycles/20 s Frequency (Hz) Period (s) 50 100 150 200 Chapter 7 Oscillatory motion requires a set of conditions. 363 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 364 Part B: Amplitude and Period Part C: Length and Period Procedure Procedure 1 Copy Table 7.7 into your notebook. 1 Copy Table 7.8 into your notebook. ▼ Table 7.7 Amplitude and Period ▼ Table 7.8 Length and Period Length of Pendulum Amplitude () No. of Cycles/20 s Frequency (Hz) Period (s) Length (m) No. of Cycles/20 s Frequency (Hz) Period (s) 5 10 15 20 2 Use the same apparatus as in part A. 3 Attach a 200
-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length at the top of Table 7.7. 5 Pull the mass on the thread back until it makes an angle of 5 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.7. 7 Repeat steps 5 and 6, each time increasing the amplitude by 5. Analysis 1. Determine the frequency and period for each amplitude and record the numbers in the appropriate columns in the table. 2. Plot a graph of period versus amplitude. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the amplitude and the period of a pendulum? Show any relevant calculations. 2 Use the same apparatus as in part A. Start with a pendulum length of 1.00 m. 3 Attach a 200-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length in Table 7.8. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.8. 7 Repeat steps 4 to 6, but each time decrease the length of the pendulum by half. Analysis 1. Determine the frequency and period for each length and record the values in the appropriate column in the table. 2. Plot a graph of period versus length. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the length and the period of a pendulum? Show any relevant calculations. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Pendulums and mass-spring systems are not the only simple harmonic oscillators. There are many other examples: a plucked guitar string, molecules vibrating within a solid, and water waves are just a few. In section 7.4, you will explore some human-made examples of SHM and learn about an interesting property called resonance. 364 Unit IV O
scillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 365 7.2 Check and Reflect 7.2 Check and Reflect Knowledge 7. Two students are given the task of 1. The restoring force of a vertical mass- spring system is determined by the mass attached to the spring and the spring constant k. What two factors determine the restoring force of a pendulum? 2. Copy the following tables into your notes. Then fill in the blanks by using the words, “zero” or “maximum.” determining the spring constant of a spring as accurately as possible. To do this, they attach a force meter to a spring that is lying on a desk and is anchored at the other end. One student pulls the spring through several displacements, while the other records the force applied, as shown in the table below. Using this table, plot a graph of force versus displacement. Find the spring constant by determining the slope of the line of best fit. Pendulum Displace- Accelerment ation System max x max a max v min F Velocity Restoring Force Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.15 0.33 0.42 0.60 Displace- Acceler- ment ation Velocity Restoring Force Massspring System max x max a max v min F 8. Determine the restoring force for a pendulum bob with a mass of 0.400 kg that is pulled to an angle of 5.0 from the vertical. 9. A toy car, with a wind-up spring motor, on a horizontal table is pulled back to a displacement of 20.0 cm to the left and released. If the 10.0-g car initially accelerates at 0.55 m/s2 to the right, what is the spring constant of the car’s spring? (Hint: The restoring force is F ma.) 3. Explain why a pendulum is not a true simple harmonic oscillator. Applications Extension 4. A mass of 2.0 kg is attached to a spring with a spring constant of 40.0 N/m on a horizontal frictionless surface. Determine the restoring force acting on the mass when the spring is compressed to a displacement of 0.15 m. 5. A spring hangs vertically from a ceiling and has a spring constant of 25.0 N/m. How far will
the spring be stretched when a 4.0-kg mass is attached to its free end? 6. An applied force of 25.0 N is required to compress a spring 0.20 m. What force will pull it to a displacement of 0.15 m? 10. Obtain three different types of rulers: plastic, metal, and wooden. Fix one end of each ruler to the side of a desk so the ruler juts out horizontally a distance of 25 cm from the edge. Hang enough weight on the end that sticks out to make the ruler bend downward by 2 to 3 cm. Record the deflection of the ruler and the mass used in each case. (Note: The deflection does not have to be the same for each ruler.) Use these data to determine the spring constant for each ruler. Rank the rulers from highest spring constant to lowest. e TEST To check your understanding of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 365 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 366 info BIT An accelerometer is a device used to measure acceleration. It is designed like a mass-spring system. The force exerted by the accelerating object causes the mass to compress the spring. The displacement of the mass is used to determine the positive or negative acceleration of the object. Accelerometers are commonly used in airbag systems in cars (see Chapter 3). If the car slows down too quickly, the displacement of the mass is large and it triggers the airbag to deploy. PHYSICS INSIGHT An object does not have to be moving to experience acceleration. 7.3 Position, Velocity, Acceleration, and Time Relationships One way for ball players to practise their timing is by attempting to throw a ball through a tire swinging on a rope. Someone just beginning this kind of practice might throw too early or too late, missing the tire altogether. Part of the difficulty has to do with the continually changing velocity of the tire. Choosing the best time to throw the ball is an exercise in physics. With practice, the human brain can learn to calculate the proper time to throw without even being aware that it is doing so. Throwing the ball through the tire is much more difficult than it sounds because the tire is a simple harmonic oscillator for small amplitudes. Not only is the velocity continually changing, but so is the restoring force
and the acceleration. The only constant for a swinging tire is its period. In this section, you will mathematically analyze acceleration, velocity, and period for SHM in a mass-spring system, and then determine the period of a pendulum. Both mass-spring systems and pendulums are simple harmonic oscillators, as described in section 7.2, but they are different from each other. The mass-spring system has a spring constant k, but the pendulum does not. For this reason, we will look at each separately, starting with the mass-spring system. Acceleration of a Mass-spring System In section 7.2, you learned that two equations can be used to describe force in the mass-spring system: Newton’s second law and Hooke’s law. They can be written mathematically as: • Newton’s second law: F • Hooke’s law: F kx ma (2) net Since both equations refer to the restoring force, you can equate them: F F net ma kx x a k m (4) where aa is the acceleration in metres per second squared; k is the spring constant in newtons per metre; x is the displacement in metres; and m is the mass of the oscillator in kilograms. 366 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 367 e MATH To see how the spring constant, mass, position and acceleration are related graphically, visit www.pearsoned.ca/school/ physicssource. Figure 7.29 The acceleration of a simple harmonic oscillator depends on its position. In position (a), the oscillator moves from its maximum displacement and maximum positive acceleration through to position (b), where the displacement and acceleration are zero. It then moves to position (c), where the oscillator again experiences a maximum acceleration and displacement in the other direction. maximum height v 0 a 9.81 m/s2 The acceleration of a horizontal mass-spring simple harmonic oscillator can be determined by its spring constant, displacement, and mass. It’s logical that the acceleration of the mass depends on how stiff the spring is and how far it is stretched from its equilibrium position. It is also reasonable to assume that, if the mass is large, then the acceleration will be small. This assumption is based on Newton’s second law. The acceleration depends on the displacement of
the mass, so the acceleration changes throughout the entire motion as shown in Figure 7.29. Since acceleration of a simple harmonic oscillator is not uniform, only the instantaneous acceleration of the mass can be determined by equation 4. () () (b) (c) () Acceleration (a) (a) () (b) () Displacement () (c) x 0 The Relationship Between Acceleration and Velocity of a Mass-spring System The acceleration of a simple harmonic oscillator is continually changing, so it should come as no surprise that the velocity changes too. As we have just seen, the maximum acceleration occurs when the oscillator is at its maximum displacement. At this position, it is tempting to think that the velocity will be at its maximum as well, but we know that this is not the case. Remember, the acceleration is at its greatest magnitude at the extremes of the motion, yet the oscillator has actually stopped in these positions! In some ways, a ball thrown vertically into the air is similar (Figure 7.30). The acceleration of gravity acts on the ball to return it to the ground. When the ball reaches its maximum height, it comes to a stop for an instant, just like the mass-spring system you studied earlier in this chapter. Figure 7.30 At its maximum height, the ball stops for a brief instant, yet the acceleration of gravity acting on it is not zero. This is similar to the mass-spring system. Chapter 7 Oscillatory motion requires a set of conditions. 367 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 368 () () (a) (b) (c) () Velocity (b) (a) (c) Displacement () x 0 Figure 7.31 The velocity of a simple harmonic oscillator is not uniform. The mass experiences its greatest acceleration at the extremes of its motion where the velocity is zero. Only after the mass accelerates from position (a) to (b) does its velocity reach its maximum value. The mass then decelerates from (b) to (c) where it comes to a stop again. Since the acceleration of the oscillator decreases as it approaches the equilibrium position (Figure 7.31(a)), the velocity does not increase at a uniform rate. The velocity-displacement graph looks like Figure 7.31(b). Figure 7.32 shows a diagram of a simple harmonic oscillator (a mass-spring system) as it moves through one-
half of a complete oscillation from (a) to (b) to (c). Below the diagram are the acceleration-displacement and velocity-displacement graphs. The diagram of the oscillator and the graphs are vertically aligned so the graphs show you what is happening as the massspring system moves. In the diagram at the top of Figure 7.32, you can see the oscillator in position (a). It is at its farthest displacement to the left and the spring is compressed. The velocity-displacement graph shows that the oscillator’s velocity in this position is zero (graph 2). You can also see from the accelerationdisplacement graph that the acceleration at that moment is positive and a maximum, but the displacement is negative (graph 1). () () Acceleration and Displacement — Always Opposite Graph 1: Acceleration vs. Displacement (a) (a) (b) (c) () Acceleration () (b) () Displacement () (c) () Velocity (b) Graph 2: Velocity vs. Displacement (a) (c) Displacement () x 0 Figure 7.32 The mass-spring system experiences a changing acceleration and velocity as it makes one-half of a full oscillation from position (a) to position (c). In fact, if you look closely at graph 1, you might notice how the acceleration and displacement are always opposite to one another, regardless of the position of the mass. The acceleration is positive while the displacement is negative, and vice versa. This isn’t surprising, however, because it is what the negative sign in the equation for Hooke’s law illustrates: kx. F Look again at Figure 7.32 and follow the mass as it moves from position (a) to positions (b) and (c). As the oscillator accelerates from position (a) to the right, it picks up speed. The velocity-displacement graph (graph 2) shows that the velocity is positive and increasing as the oscillator approaches position (b), yet the acceleration is decreasing, as shown in graph 1. The oscillator goes through the equilibrium position with a maximum positive velocity, but now the acceleration becomes negative as the spring tries to pull the oscillator back (graph 1). This is why the oscillator slows down and the velocity-displacement graph returns to zero in position (c). 368 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-
Chap07.qxd 7/24/08 1:09 PM Page 369 Consider a vertical mass-spring system. A bungee jumper will experience a positive acceleration when she is below the equilibrium position and a negative acceleration when above it (Figures 7.33 and 7.34). (a) (b) (c) equilibrium x max x 0 x max a v a v a v Displacement Figure 7.33 Displacement Figure 7.34 After a jump, the bungee jumper is shown in three positions: At the lowest point (a), in the equilibrium position (b), and at her maximum displacement (c). In each case the circled region on the graphs indicates her acceleration and velocity. Maximum Speed of a Mass-spring System Now you know that a simple harmonic oscillator will experience its greatest speed at the equilibrium position. What factors influence this speed, and how can we calculate it? In our examples, the mass-spring system is frictionless, and no external forces act on it. This is referred to as an isolated system and the law of conservation of energy applies. We will use this concept to derive the equation for the maximum speed. Recall from Chapter 6 that the total mechanical energy in an isolated system remains constant. That means that the kinetic and potential energy of the system may vary, but their sum is always the same. Chapter 7 Oscillatory motion requires a set of conditions. 369 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 370 In other words, at any position in the motion of a mass-spring system, the sum of kinetic and potential energies must be equal to the total energy of the system. Recall that kinetic energy is expressed as: Ek 1 mv2 2 Recall that elastic potential energy is expressed as: Ep 1 kx2 2 Let’s begin by looking at the energy of the system in two positions: • When the mass is at the maximum displacement (Figure 7.35(a)). • When the mass is at the minimum displacement (Figure 7.35(b)). v 0 Ek 0 Ep max m x max The oscillator at its maximum displacement. Figure 7.35(a) Potential energy has reached a maximum value (Ep oscillator’s displacement is a maximum (x A). The kinetic energy is zero (v 0). ) because the max v max Ek max Ep 0 m x 0 Figure 7.35(b) The oscillator at its minimum displacement (x 0
). Kinetic energy has reached a maximum value (Ekmax has a maximum velocity. The potential energy is zero (x 0). ) because the oscillator Remember that the total energy of the system remains constant regardless of the oscillator’s position. The equation for the total energy is: ET Ep Ek The kinetic energy of the oscillator at its maximum displacement is zero so the total energy of the oscillator at that position must be: ET Epmax The potential energy of the oscillator at its minimum displacement is zero so the total energy of the oscillator at that position must be: ET Ekmax Because the total energy is always the same, we can write: Ekmax Epmax or 1 2 mv 2 max 1 kx 2 2 max 370 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 371 max mv2 1 kA2 2 If we use A to represent xmax, we can write: 1 2 We can then simplify this equation to solve for vmax: 1 2 1 kA2 2 mv 2 max mv 2 max kA2 v 2 max kA2 m Then we take the square root of each side: vmax kA2 m or vmax A k m (5) e WEB To find out how these factors are taken into account in bungee jumping, follow the links at www.pearsoned.ca/school/ physicssource. Factors That Influence the Maximum Speed of a Mass-spring System Three factors influence the maximum speed of a mass-spring system: • the amplitude of the oscillations: If the oscillator moves through a large amplitude, the restoring force increases in proportion to the amplitude. As the restoring force increases, so does the acceleration, and the oscillator will achieve a greater velocity by the time it reaches the equilibrium position. • the stiffness of the spring: A stiffer spring with a higher spring constant exerts a stronger restoring force and creates a greater maximum velocity for the same reasons that increasing the amplitude does. • the mass of the oscillator: Changing the mass of an oscillator has a different effect. If the mass increases, the velocity of the oscillations decreases. This is because the oscillator has more inertia. A larger mass is harder to accelerate so it won’t achieve as great a speed as a similar mass-spring system with less mass. Concept Check 1. When acceleration is negative, displacement
is positive and vice versa. Why? 2. Why is the velocity-time graph of a simple harmonic oscillator a curved line? 3. The acceleration-displacement graph and velocity-displacement graph are shown in Figure 7.32 on page 368 for half of an oscillation only. Sketch three more acceleration-displacement and velocitydisplacement graphs for the second half of the oscillation. 4. Suppose the amplitude of an object’s oscillation is doubled. How would this affect the object’s maximum velocity? Chapter 7 Oscillatory motion requires a set of conditions. 371 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 372 PHYSICS INSIGHT When a mass is hanging from a vertical spring at rest in the equilibrium position, the downward g is force of gravity F equal and opposite to the upward force exerted s, so the by the spring F restoring force is zero. The force of gravity acting on the mass doesn’t change. If the spring is displaced from the equilibrium position, the restoring force will just be the force of the spring due to its displacement x. This is kx. expressed as F R Example 7.6 A 100.0-g mass hangs motionless from a spring attached to the ceiling. The spring constant (k) is 1.014 N/m. The instructor pulls the mass through a displacement of 40.0 cm [down] and releases it. Determine: (a) the acceleration when the mass is at a displacement of 15.0 cm [up], and (b) the maximum speed of the mass. up v down a x 15.0 cm equilibrium x 40.0 cm FR Figure 7.36 Given m 100.0 g 0.1000 kg k 1.014 N/m x 40.0 cm [down] 0.400 m [down] Required (a) acceleration (a) when x 15.0 cm [up] 0.150 m [up] (b) maximum speed (vmax) Practice Problems 1. A 0.724-kg mass is oscillating on a horizontal frictionless surface attached to a spring (k 8.21 N/m). What is the mass’s displacement when its instantaneous acceleration is 4.11 m/s2 [left]? 2. A 50.0-g mass is attached to a spring with a spring constant (k) of 4.00 N/m. The mass
oscillates with an amplitude of 1.12 m. What is its maximum speed? 3. An instructor sets up an oscillating vertical mass-spring system (k 6.05 N/m). The maximum displacement is 81.7 cm and the maximum speed is 2.05 m/s. What is the mass of the oscillator? Answers 1. 0.362 m [right] 2. 10.0 m/s 3. 0.961 kg Analysis and Solution (a) The mass will begin to oscillate when released. Acceleration R net is a vector quantity so direction is important. F F ma kx kx a m N (0.150 m) 1.014 m 0.1000 kg 1.52 m/s2 (b) The maximum speed occurs when the mass is in the equilibrium position, whether it is moving up or down. The displacement of the mass before it is released is the amplitude (A) of the mass’s oscillation. vmax A k m N 1.014 m 0.1000 kg 0.400 m 1.27 m/s Paraphrase (a) The mass has an acceleration of 1.52 m/s2 [down] when it is 15.0 cm above the equilibrium position. (b) The maximum speed of the mass is 1.27 m/s. 372 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 373 Period of a Mass-spring System The next time you are travelling in a vehicle at night, watch for bicycles moving in the same direction as your vehicle. Notice the peculiar motion of the pedals as they reflect the light from your headlights (Figure 7.37). From a distance, these reflectors don’t appear to be moving in a circular path, but seem to be moving up and down. The apparent up-and-down motion of the pedals is the same kind of motion as a mass-spring system oscillating back and forth, so it is simple harmonic motion. This observation proves useful because it is an example of how circular motion can be used to describe simple harmonic motion. The next few pages show how to derive equations for the period and maximum speed of a simple harmonic oscillator. Two conditions are necessary if circular motion is to be used to replicate simple harmonic motion: 1. The period of both the circular motion and the simple harmonic motion must be the same. 2. The radius of
the circular motion must match the amplitude of the oscillator. For example, look at Figure 7.38, where a mass moving in a circular path with a radius r is synchronized with a mass-spring simple harmonic oscillator. This illustration demonstrates how circular motion can be used to describe SHM. Figure 7.37 From a distance, the reflectors on the bicycle pedals would appear to be moving up and down instead of in a circle. (a) (b) (c) (d) (e Figure 7.38 A mass moving in a circle is a simple harmonic oscillator that corresponds to the mass-spring oscillator shown below it. One-half of a complete cycle is shown here. For our purposes, the following conditions are true: • The radius of the circular motion is equal to the amplitude of the oscillator (r A, as shown in Figure 7.38(a)). • The mass in circular motion moves at a constant speed. • The periods of the mass in circular motion and the oscillator in the mass-spring system are the same. Chapter 7 Oscillatory motion requires a set of conditions. 373 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 374 Deriving the Equation for the Period of a Mass-spring System Recall that the maximum velocity of a simple harmonic oscillator occurs when it is in its equilibrium position, which is position (c) in Figure 7.38. At the exact moment that the mass in circular motion is in position (c), its velocity is in the same direction as the velocity of the mass-spring system, and they are both moving at the same speed. But if the mass moving in a circular path is moving at a constant speed, then it must always be moving at the maximum speed of the mass-spring oscillator! Therefore, the maximum speed (vmax) of the mass-spring system is equal to the speed (v) of the circular mass system. The speed of an object moving in a cir- cular path was derived in Chapter 5. It is: v 2r T (6) Figure 7.39 The strings of a piano all have different masses. Even if they vibrate with the same amplitude they will have a different period of vibration because each string has a different mass. A heavy string will vibrate with a longer period (and lower frequency) than a lighter string. The speed of the circular motion (v) matches the maximum speed on the mass-spring
oscillator (vmax), and the radius of the circle matches its amplitude. Therefore, we can customize the equation for the mass-spring oscillator: 2A T vmax (7) If we equate equation 5 and equation 7, we get: A k m 2A T We can then solve for T: A k m 2A T 2π k T T 2 m m k (8) This equation describes the period of a simple harmonic oscillator, where T is the period of the oscillator in seconds; k is the spring constant in newtons per metre; and m is the mass of the oscillator in kilograms. Figure 7.39 is an example of an application of this equation. PHYSICS INSIGHT The period of a simple harmonic oscillator does not depend on displacement. 374 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 375 Factors Affecting the Period of an Oscillating Mass The larger the oscillating mass is, the longer its period of oscillation is. This seems reasonable since a large mass takes longer to speed up or slow down. It would also seem reasonable that the period should be inversely related to the spring constant, as the equation suggests. After all, the stiffer the spring, the more force it exerts over smaller displacements. Therefore, you could expect the mass to oscillate more quickly and have a smaller period. What is interesting about this equation is not what influences the period but what does not. It may seem odd that the displacement of the mass has no influence on the period of oscillation. This means that if you were to pull a mass-spring system to a displacement x and then let go, it would have the same period of oscillation as it would if you pulled it to a displacement of 2x and released it! The two identical mass-spring systems in Figure 7.40 have different amplitudes but the same period. k1 10.0 N/m k2 10.0 N/m amplitude 20 cm m1 x 0 amplitude 10 cm m2 x 0 1 10° θ 2 5° θ Figure 7.40 Two identical mass-spring systems have the same spring constant and mass, but different amplitudes. Which has the longest period? They have the same period because displacement doesn’t affect period. Figure 7.41 Two identical pendulums have the same mass and length, but different amplitudes.
Which one has the longest period? They have the same period because displacement doesn’t affect the period of a simple harmonic oscillator. This relationship is true for any simple harmonic oscillator, including a pendulum with a small amplitude. It is easy enough to test. Take two pendulums with the same mass and length (Figure 7.41). Pull both bobs back to different displacements. Remember to keep the displacements small so the pendulums oscillate with SHM. Release them at the same time. You will discover that both make one full oscillation in unison. This means they return to the point of release at exactly the same time. The pendulum that begins with the larger displacement has farther to travel but experiences a larger restoring force that compensates for this. Chapter 7 Oscillatory motion requires a set of conditions. 375 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 376 Example 7.7 What is the period of oscillation of a mass-spring system that is oscillating with an amplitude of 12.25 cm and has a maximum speed of 5.13 m/s? The spring constant (k) is 5.03 N/m. Practice Problems 1. A mass of 2.50 kg is attached to a horizontal spring and oscillates with an amplitude of 0.800 m. The spring constant is 40.0 N/m. Determine: a) the acceleration of the mass when it is at a displacement of 0.300 m the maximum speed the period b) c) 2. A 2.60-g mass experiences an acceleration of 20.0 m/s2 at a displacement of 0.700 m on a spring. What is k for this spring? 3. What is the mass of a vertical mass-spring system if it oscillates with a period of 2.0 s and has a spring constant of 20.0 N/m? 4. What is the period of a vertical mass-spring system that has an amplitude of 71.3 cm and maximum speed of 7.02 m/s? The spring constant is 12.07 N/m. Answers 1. a) 4.80 m/s2 b) 3.20 m/s c) 1.57 s 2. 0.0743 N/m 3. 2.0 kg 4. 0.638 s A 12.25 cm left right Given A 12.25 cm 0.1225 m k 5.
03 N/m vmax 5.13 m/s Required period of the oscillations (T ) equilibrium Analysis and Solution To determine the period of the oscillator, you need to know the oscillator’s mass. Use the maximum speed equation (equation 5) to find the mass: Figure 7.42 EPmax Ekmax A2 2 k mv max 2 2 mv 2 max kA2 0.1225 m)2 5.03 m m 2 5.13 s 2.868 103 kg Then use equation 8 to determine the period: T 2 m k 2 2.868 103 kg N 5.03 m 0.150 s Paraphrase The period of the mass-spring oscillator is 0.150 s. 376 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 377 Concept Check 2. 1. What effect does doubling the displacement have on the period of oscillation of a simple harmonic oscillator? Explain your answer. In order to compare circular motion to the motion of a simple harmonic oscillator, what two conditions must be satisfied? If the mass and spring constant of a mass-spring oscillator were doubled, what effect would this have on the period of the oscillations? 3. 4. Two mass-spring systems with identical masses are set oscillating side by side. Compare the spring constants of the two systems if the period of one system is twice the other. The Period of a Pendulum Christiaan Huygens recognized that a pendulum was ideally suited for measuring time because its period isn’t affected by as many of the factors that influence a mass-spring system. A pendulum doesn’t have a spring constant, k, like the mass-spring system does, and unlike the mass-spring system, the mass of the pendulum does not affect its period. Because of these factors, a new equation for a pendulum’s period must be derived. In doing so, you will discover why its mass is irrelevant and what factors play a role in its period of oscillation. Take a closer look at the pendulum when it is at a small displacement of 15 or less, as shown in Figure 7.43. For a small angle (), the displacement of the bob can be taken as x. The sine of angle is expressed as: x l sin Recall that the restoring force for a pendulum is FR above expression for
sin in this equation: Fg sin. Use the Fg FR x l θ l x Recall also that in a mass-spring system, the restoring force is F kx. We want to solve for the period (T ), which is a scalar quantity, so the negative sign in Hooke’s law can be omitted. The two equations for restoring force can then be equated: Figure 7.43 For a pendulum with a small displacement of 15 or less, the displacement is x. kx Fg x l Fg mg kx (mg) x l k mg l Chapter 7 Oscillatory motion requires a set of conditions. 377 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 378 None of the values of mass, gravitational field strength, or length change for a pendulum. They are constants, which are represented by k. Substitute them into equation 8 (page 374): T 2 m k g) m (m 2 T 2 l g l (9) PHYSICS INSIGHT The period of a pendulum does not depend on its mass or amplitude. e SIM Learn more about the motion of a pendulum and the factors that affect it. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. where l is the length of the pendulum string in metres; and g is the gravitational field strength in newtons per kilogram. Recall that the length of the pendulum is always measured from the point where it is attached at the top, to the centre of mass of the bob, not the point at which the string or wire is attached to the bob. Also recall that the period of the pendulum’s swing does not depend on the mass of the pendulum bob. This may not seem logical but it is indeed the case — just as the acceleration of an object in free fall doesn’t depend on the mass of the object. The Pendulum and Gravitational Field Strength Equation 9 is useful when it is manipulated to solve for g, the gravitational field strength. As you learned in section 4.3 of Chapter 4, the gravitational field strength varies with altitude and latitude. The magnitude of the gravitational field is 9.81 N/kg at any place on Earth’s surface that corresponds to the average radius of Earth. However, very few places on the surface of Earth are at exactly the average radius. To determine the exact value of g at any
point, you can use a pendulum. If you manipulate equation 9 and solve for g, you get: g 42l T 2 (10) Due to the changing nature of Earth’s gravity, Christiaan Huygens’s pendulum clock (introduced in section 7.2) was only accurate if it was manufactured for a specific place. For example, pendulum clocks designed to operate in London could not be sold in Paris because the accuracy could not be maintained. The difference in gravitational field strength between London and Paris meant that the period of oscillation would be slightly different. The difference in g between two locations could be quite small, but the cumulative effect on a pendulum clock would be significant. An extreme example of the varying value of g at different geographic locations can be illustrated by using a pendulum to determine the gravitational field strength at the top of Mount Everest. 378 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 379 Example 7.8 What is the gravitational field strength at the top of Mount Everest at an altitude of 8954.0 m, if a pendulum with a length of 1.00 m has a period of 2.01 s? Analysis and Solution Use equation 10 to determine g. Note that no vector arrow is required with the symbol g because you are calculating a scalar quantity. g 42l T 2 (42)(1.00 m) (2.01 s)2 9.77 m/s2 9.77 N/kg T 2.01 s 8954.0 m Figure 7.44 Mount Everest g 9.81 N/kg The gravitational field strength at the top of Mount Everest is 9.77 N/kg, which is very close to the accepted value of 9.81 N/kg. The extra height of Mount Everest adds very little to the radius of Earth. Practice Problems 1. What is the gravitational field strength on Mercury, if a 0.500-m pendulum swings with a period of 2.30 s? 2. A pendulum swings with a period of 5.00 s on the Moon, where the gravitational field strength is 1.62 N/kg [down]. What is the pendulum’s length? 3. What period would a 30.0-cm pendulum have on Mars, where the gravitational field strength is 3.71 N/kg [down]? Answers 1. 3.73 N
/kg [down] 2. 1.03 m 3. 1.79 s At the top of Mount Everest, a pendulum will swing with a slightly different period than at sea level. So a pendulum clock on Mount Everest, oscillating with a longer period than one at sea level, will report a different time. Huygens’s clocks also suffered from another problem: the pendulum arm would expand or contract in hot or cold weather. Since the length of the arm also determines the period of oscillation, these clocks would speed up or slow down depending on the ambient temperature. Given their limitations, pendulum clocks were not considered the final solution to accurate timekeeping. Further innovations followed that you may want to research on your own. e WEB To learn more about pendulum clocks and the evolution of timekeeping, create a timeline of the evolution of clock design. In your timeline include what the innovation was, who invented it, and the year it was introduced. Begin your search at www.pearsoned.ca/school/ physicssource. Concept Check 1. An archer is doing target practice with his bow and arrow. He ties an apple to a string and sets it oscillating left to right, down range. In what position of the apple should he aim so that he increases his chances of hitting it? Explain your answer. 2. What factors affect the accuracy of pendulum clocks? Why? Chapter 7 Oscillatory motion requires a set of conditions. 379 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 380 7.3 Check and Reflect 7.3 Check and Reflect Knowledge 1. Explain the effect that changing each of the following factors has on the period of a mass-spring system: (a) amplitude (b) spring constant (c) mass 2. Explain what effect changing each of the following factors has on the period of a pendulum: (a) amplitude (b) gravitational field strength (c) mass 9. A pendulum bob (m 250.0 g) experiences a restoring force of 0.468 N. Through what angle is it displaced? 10. A 50.0-cm pendulum is placed on the Moon, where g is 1.62 N/kg. What is the period of the pendulum? Extensions 11. A horizontal mass-spring system oscillates with an amplitude of 1.50 m. The spring constant is 10.00 N/m. Another mass moving in a
circular path with a radius of 1.50 m at a constant speed of 5.00 m/s is synchronized with the mass-spring system. Determine the mass-spring system’s: 3. Describe the positions that a mass-spring system and pendulum are in when: (a) period (b) mass (a) acceleration is a maximum (b) velocity is a maximum (c) restoring force is maximum 4. Why is the acceleration of a simple harmonic oscillator not uniform? 5. A mass-spring system has a negative displacement and a positive restoring force. What is the direction of acceleration? Applications 6. What length of pendulum would oscillate with a period of 4.0 s on the surface of Mars (g 3.71 N/kg)? 7. A mass of 3.08 kg oscillates on the end of a horizontal spring with a period of 0.323 s. What acceleration does the mass experience when its displacement is 2.85 m to the right? 8. A 50.0-kg girl bounces up and down on a pogo stick. The girl has an instantaneous acceleration of 2.0 m/s2 when the displacement is 8.0 cm. What is the spring constant of the pogo stick’s spring? (c) maximum acceleration 12. A quartz crystal (m 0.200 g) oscillates with simple harmonic motion at a frequency of 10.0 kHz and has an amplitude of 0.0500 mm. What is its maximum speed? 13. A horizontal mass-spring system has a mass of 0.200 kg, a maximum speed of 0.803 m/s, and an amplitude of 0.120 m. What is the mass’s position when its acceleration is 3.58 m/s2 to the west? 14. Suppose an inquisitive student brings a pendulum aboard a jet plane. The plane is in level flight at an altitude of 12.31 km. What period do you expect for a 20.0-cm pendulum? (Hint: First determine the gravitational field strength as shown in Chapter 4.) e TEST To check your understanding of position, velocity, acceleration, and time relationships in mass-spring systems and pendulums, follow the eTest links at www.pearsoned.ca/school/physicssource. 380 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM
Page 381 info BIT When you walk with a drink in your hand at the right speed, your motion creates resonance in the liquid. This makes waves that splash over the edge of the cup. To prevent this, people walk slowly so resonance doesn’t occur, often without knowing why this works. resonant frequency: the natural frequency of vibration of an object 7.4 Applications of Simple Harmonic Motion People’s arms swing as they walk. An annoying rattle can develop in a car when it reaches a certain speed. A child can make large waves in the bathtub by sliding back and forth. Many things can be made to vibrate, and when they do, they seem to do it with a period of motion that is unique to them. After all, how often do you think about your arms swinging as you walk? You don’t — they seem to swing of their own accord and at their own frequency. The water in the bathtub will form very large waves when the child makes a back-and-forth motion at just the right rate. Any other rate won’t create the waves that splash over the edge and soak the floor, which, of course, is the goal. In all these cases, the object vibrates at a natural frequency. Resonant frequency is the natural frequency of vibration of an object. In other words, objects that are caused to vibrate do so at a natural frequency that depends on the physical properties of the object. All objects that can vibrate have a resonant frequency, including a pendulum. Maintaining a Pendulum’s Resonant Frequency A pendulum swings back and forth at its resonant frequency. Since the acceleration of gravity does not change if we stay in the same place, the only factor that affects the resonant frequency is the pendulum’s length. All pendulums of the same length oscillate with the same natural (resonant) frequency. Huygens made use of this fact when he designed his pendulum clock (Figure 7.45). He knew that all pendulum clocks would keep the same time as long as the length of the pendulum arms was the same. Their resonant frequencies would be identical. However, Huygens faced some challenges in making a pendulum clock. The arm of the pendulum would expand or contract with temperature, affecting its period. But this was a relatively minor issue compared to another difficulty that had to be overcome — friction. Unless something was done,
friction would very quickly stop the pendulum from swinging. To compensate for the effects of friction, he designed his clocks so that the pendulum was given a small push at just the right moment in its swing. The timing of these pushes coincided with the resonant frequency of the pendulum. By doing this, Huygens could make the pendulum swing for as long as the periodic force was applied. Figure 7.45 The interior of Huygens’s clock Chapter 7 Oscillatory motion requires a set of conditions. 381 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 382 forced frequency: the frequency at which an external force is applied to an oscillating object PHYSICS INSIGHT The forced frequency that is the same as the resonant frequency will increase the amplitude of the SHM, but will not change the resonant frequency. mechanical resonance: the increase in amplitude of oscillation of a system as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the system Forced Frequency To visualize how this works, imagine a child on a swing. A swing is an example of a pendulum, with the child as the bob. The swing moves back and forth at its natural frequency, which depends only on its length. To keep the swing going with the same amplitude, all the parent has to do is push at just the right moment. The timing of the pushes must match the frequency of the swing. As anyone who has pushed a swing can attest, it takes very little energy to keep a swing swinging to the same height. The frequency at which the force is applied to keep the swing moving is called the forced frequency. If the forced frequency matches or is close to the resonant frequency of the object, then very little force is required to keep the object moving. The resonant frequency won’t change though, because it depends only on the length of the pendulum. If the parent decides to push a little harder each time the swing returns, then the swing’s amplitude will increase, but not its frequency. A larger force than is needed to overcome friction will create a larger amplitude of motion. If the forced frequency isn’t close to the resonant frequency, then the object will not vibrate very much and will have a small amplitude. Imagine trying to increase the frequency of a pendulum by increasing the forced frequency. Much of the force won’t be transferred to the pend
ulum because the pendulum won’t be in the right position when the force is applied. The pendulum will bounce around but there will be no increase in its amplitude of vibration, and its motion will become harder to predict. The flowchart in Figure 7.46 on the next page summarizes the relationship between forced frequency and resonant frequency. Mechanical Resonance A forced frequency that matches the resonant frequency is capable of creating very large amplitudes of oscillation. This is referred to as mechanical resonance. This can be a good or bad thing. The larger the amplitude, the more energy the system has. Huygens’s pendulum clock didn’t need to have large oscillations, so a very small force could keep the pendulum swinging. A small weight-driven mechanism was used to provide the force needed. The force simply had to be applied with the same frequency as the pendulum. Huygens managed to do this without much difficulty. His pendulum clocks were a great success but weren’t completely practical since they had to be placed on solid ground. A pendulum clock would not work aboard a ship because sailing ships of the time were buffeted by the waves more than today’s large ocean-going vessels are. The motion of the ship on the waves would disturb a pendulum’s SHM, so sailors could not take advantage of the increased accuracy these clocks provided. 382 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 383 The key to successfully navigating across an ocean (where there are no landmarks) was to use an accurate clock on the ship. This clock could be synchronized to a clock in Greenwich, England, which is situated on the prime meridian of 0° longitude. As the ship travelled east or west, the sailors could compare their local time, using the Sun and a sundial, to the ship’s clock, which was still synchronized to the time on the prime meridian. The difference in time between the two clocks could be used to compute their longitudinal position. However, it wasn’t until the 1700s that a brilliant clockmaker, John Harrison, successfully made a marine chronometer (ship’s clock) that was immune to the buffeting of waves and temperature. It contained several ingenious innovations and, for better or worse, made the pendulum obsolete in navigation. NO The amplitude of vibration is not increased
or may be decreased. Is the forced frequency close to the resonant frequency? YES A small force is required to keep the object vibrating with the same amplitude. The force is increased. The amplitude of vibration increases. Figure 7.46 Flowchart of the effect of forced frequency on resonant amplitude Chapter 7 Oscillatory motion requires a set of conditions. 383 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 384 7-5 QuickLab 7-5 QuickLab Investigating Mechanical Resonance Problem How can we cause a pendulum to begin oscillating at its resonant frequency using a forced frequency? 8 Repeat step 7 three more times. Each time, lengthen the thread of mass 2 by 10 cm. For your last trial, the thread of mass 2 should be 40.0 cm long. Materials retort stands string thread 2 identical masses (200 g each) string thread mass 2 retort stand mass 1 two equal masses Figure 7.47 Procedure Part A 1 Read the questions in the next column before doing the lab. 2 Set up the two retort stands about 75 cm apart. 3 Tie the string to both retort stands at the same height (50.0 cm) on each stand, as shown in Figure 7.47. Clamp the ends of the string to the retort stands so they don’t slip. The string should be taut. 4 Pull the two retort stands farther apart if you need to remove slack from the string. 5 Attach the thread to one mass and tie the other end to the string so that the distance from the mass to the string is 30 cm. This is mass 1. 6 Repeat step 5 for the second mass (mass 2) so that it has a length of 10 cm and is attached to the string about 15 cm from the first mass. 7 Make sure neither mass is moving, then pull mass 2 back a small distance and release it. Observe the motion of mass 1 as mass 2 oscillates. Make a note of the maximum amplitude that mass 1 achieves. Part B 9 Adjust the thread length of mass 2 so that it is as close to the thread length of mass 1 as possible. 10 Make sure both masses are motionless. Pull back and release mass 2. Note the amplitude of vibration that mass 1 achieves. 11 Pull the retort stands farther apart and hold them there so the tension in the string is increased, and the string is almost horizontal. 12 Make sure both masses are motion
less, then pull back mass 2 and release it. Note the amplitude of vibration that mass 1 achieves. Questions Part A 1. At what thread length did mass 2 create the maximum oscillation of mass 1? Explain why this happened, in terms of frequency. 2. At what thread length did mass 2 create the minimum oscillation of mass 1? Explain why this happened, in terms of frequency. 3. Why did mass 1 have a large amplitude of vibration in only one case? Part B 4. What effect did increasing the tension on the string have on the amplitude achieved by mass 1? 5. Why did increasing the tension alter the maximum oscillation of mass 1? 6. Write a sentence describing the effect that increasing the tension had on the resonant amplitude of mass 1. Use the terms forced frequency and resonant amplitude in your answer. e LAB For a probeware lab, go to www.pearsoned.ca/school/physicssource. 384 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 385 Resonance Effects on Buildings and Bridges A forced frequency that matches the resonant frequency can create problems for designers of bridges and skyscrapers. A bridge has a resonant frequency that can be amplified by the effect of wind. Air flows around the top and bottom of a bridge and can cause it to vibrate. The bridge will vibrate at its resonant frequency, with a large amplitude, even though the force applied by the wind may be relatively small. As the bridge vibrates, it may flex more than it is designed to and could conceivably vibrate to pieces. A skyscraper is also susceptible to forced vibrations caused by the wind. Most skyscrapers have a huge surface area and catch a lot of wind. Even though a building is a rigid structure, the force of the wind can make it sway back and forth. The wind causes a phenomenon called “vortex shedding.” It can create a forced vibration that matches the natural frequency of the building’s back-and-forth vibration. The unfortunate result is to increase the sway (amplitude) of the building. The occupants on the top floors of the skyscraper will feel the effects the most. Over time, the continual large sway could weaken the building’s structural supports and reduce its lifespan. Reducing Resonance Effects To counter resonance effects on bridges and buildings, engineers build them in such a way as
to reduce the amplitude of resonance. Bridge designers make bridges more streamlined so that the wind passes over without imparting much energy. They also make bridges stiff, so a larger force is needed to create a large amplitude. The second-largest bridge in the world, the Great Belt East Bridge of Denmark is built with a smooth underside, like an airplane wing, that greatly increases its streamlined shape (Figure 7.48). It is not likely that a forced vibration would cause it to resonate. Skyscraper designers employ many strategies to lessen resonant vibration. One very effective approach is to use a large mass at the top of the building, called a “tuned mass damper,” which is free to oscillate back and forth (Figure 7.49). Controlled by computers, it can be made to vibrate at the resonant frequency of the building. When the building sways left, the mass moves right, and when the building sways right, the mass moves left. This has the effect of cancelling the vibration of the building. Any process that lessens the amplitude of an object’s oscillations is referred to as “damping.” Figure 7.48 The Great Belt East Bridge of Denmark is 6.8 km long and is constructed with a smooth underside. This allows air to flow by without inducing a resonant frequency. Chapter 7 Oscillatory motion requires a set of conditions. 385 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 386 building movement cables damper movement Figure 7.49 The Taipei 101 building in Taiwan was completed in 2004 and stands 101 stories high. The inset shows a tuned mass damper in the building designed by Motioneering Inc. of Guelph, Ontario. It has a huge mass and vibrates opposite to the direction of the building, cancelling much of the amplitude of the resonant vibration. THEN, NOW, AND FUTURE Stressed-out Airplanes Ask any mechanical engineers, and they will tell you the importance of designing equipment to minimize vibration. Vibration causes excess wear on parts and stress on materials. Nowhere is this more evident than on an airplane. Yancey Corden knows this better than most people. Yancey is an aircraft maintenance engineer, and one of his jobs is to inspect aircraft for excess metal fatigue. Yancey was born on the Pavilion Reserve in south central British Columbia but grew up north of Williams Lake. His father maintained their car
, boat, and other equipment around the home. Yancey watched and helped his father, and during this time, his interest in mechanics grew. Not long after finishing high school, Yancey enrolled in the aircraft maintenance engineer program at the British Columbia Institute of Technology located in Burnaby. He is now qualified with an M1 certification, which allows him to work on planes under 12 500 kg, and an M2 certification, which allows him to work on larger planes. He received specialized training in structural maintenance. This shift of force. This happens because the airplane springs upward due to the lighter load, and as a result, the wings tend to flutter up and down. This vibration causes stress fractures on the wing, and it is Yancey’s job to find them. If a problem is found, Yancey designs the solution. This could involve fabricating a new part or simply fixing the existing one. He enjoys his job because each day is different and brings new challenges. He is thankful that he had the foresight to maintain good marks when he went to high school because the physics and science courses he took directly applied to his training. He is very proud of his heritage but he also believes it is important to focus on who you are and where you are going. Questions 1. What factors contribute to metal fatigue on a firefighting airplane? 2. What steps must be taken to gain a licence as an aircraft maintenance engineer? 3. To what factors does Yancey attribute his success? Figure 7.50 Yancey Corden involves an in-depth knowledge of the skin and frame of the plane. In 2003 Yancey moved to Alberta where he works in Red Deer for a company called Air Spray. Air Spray maintains and repairs Lockheed L-188 airplanes. These planes were originally manufactured as passenger craft in the 1950s, but because of their rugged design, they have been converted to firefighting aircraft today. They carry over 10 000 kg of water and fire retardant and are capable of dumping the entire amount in three seconds. When a dump of water and fire retardant occurs, the wings and airframe of the plane undergo a huge 386 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 387 Quartz Clocks The technology of clock design and manufacture has taken huge leaps since the 1600s when Huygens built his first pendulum clock. Today, quartz clocks are the most accurate timepieces commercially
available. They have an accuracy of about 1/2000 of a second a day. A quartz clock works on the principle of resonance. Inside each quartz clock is a tiny crystal of quartz. Quartz is a mineral that naturally forms into crystals. It also has a property unique to just a handful of materials: it will bend when a voltage is applied to it. If a pulse of voltage is applied to it, the crystal will begin to vibrate at its resonant frequency, just as a cymbal vibrates when hit by a drumstick. Once the quartz crystal is set vibrating, the circuitry of the clock times successive voltage pulses to synchronize with the frequency of the crystal. The synchronized voltage provides the forced frequency to keep the crystal oscillating just as the pendulums of Huygens’s clocks needed a synchronized forced frequency to keep them from running down. The difference is that the pendulum clock receives the forced frequency through mechanical means, while the quartz crystal clocks get the forced vibration from electrical means. Resonant Frequency of a Quartz Crystal The crystal’s resonant frequency depends on its size and shape and is not affected significantly by temperature. This makes it ideal for keeping time. As the crystal gets larger, more voltage is required to make it oscillate, and its resonant frequency decreases. A piece of quartz could be cut to oscillate once every second, but it would be far too large for a wristwatch and would require a large voltage to operate. If the crystal size is decreased, less voltage is required to make it oscillate. info BIT A substance that deforms with an applied voltage is called a piezoelectric material. A piezoelectric material will also create a voltage if stressed. Some butane lighters use a piezoelectric material to create a flame. As the lighter trigger is pressed, butane gas is released and the piezoelectric material undergoes stress. The piezoelectric material creates a voltage that causes a spark to jump a very small gap at the end of the lighter, igniting the butane. e WEB Atomic clocks keep time extremely precisely. Do they use a principle of resonance to keep such accurate time? Begin your search at www.pearsoned.ca/school/ physicssource. Quartz crystals are cut to a size and shape small enough to fit into a watch and use a small voltage (Figure 7.51). In most of today’s quartz watches, the crystal vibr
ates with a resonant frequency of about 30 kHz and operates at 1.5 V. A small microprocessor in the watch combines these oscillations to make one oscillation per second so the watch can display time in a meaningful way. The topic of resonant frequencies is large and can’t possibly be fully covered in this unit. You will learn more about resonance in musical instruments in Chapter 8. Figure 7.51 The quartz crystal in a wristwatch is enclosed in the small metal cylinder (lower right). Chapter 7 Oscillatory motion requires a set of conditions. 387 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 388 12. What factors affect the resonance of a quartz crystal? 13. (a) What are two advantages of a quartz clock over a pendulum clock? (b) Are there any disadvantages of a quartz clock compared with a pendulum clock? Extensions 14. Use the knowledge you have gained about the design of a pendulum clock and the equation for its period in section 7.3 to answer the following question. What would the length of the pendulum’s arm have to be so that it would oscillate with a resonant frequency of 1.00 Hz in Alberta (g 9.81 N/kg)? Under what conditions would it be most accurate? 15. Use your local library or the Internet to find out what automobile manufacturers do to reduce resonant frequencies in cars. 16. Investigate other methods not mentioned in the text that bridge designers use to lessen resonant vibrations. 17. Tuned mass dampers are not just used on buildings; cruise ships also have them. Explain why a cruise ship might have them and how they would be used. 18. Use your local library or the Internet to explore orbital resonance. In one or two paragraphs, explain how it applies to Saturn’s rings. e TEST To check your understanding of applications of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 7.4 Check and Reflect 7.4 Check and Reflect Knowledge 1. What provides the force necessary to start a building or bridge oscillating? 2. What is forced frequency? 3. Explain what engineers use to reduce resonant vibrations of buildings and how these devices or structures work. 4. Explain the effect of applying a force to a vibrating object with the same frequency. 5. Identify two limitations of Huygens’s pendulum clock.
6. Can a pendulum clock built to operate at the equator have the same accuracy at the North Pole? Explain. 7. What is damping? Use an example in your explanation. Applications 8. How could a person walking across a rope bridge prevent resonant vibration from building up in the bridge? 9. An opera singer can shatter a champagne glass by sustaining the right musical note. Explain how this happens. 10. Tuning forks are Y-shaped metal bars not much bigger than a regular fork. They can be made to vibrate at a specific frequency when struck with a rubber hammer. A piano tuner uses tuning forks to tune a piano. Explain, in terms of resonance, how this might be done. 11. Students are asked to find ways to dampen or change the resonant frequency of a pendulum. Here is a list of their suggestions. Identify the ones that would work and those that would not. In each case, justify your answer. (a) Apply a forced frequency that is different from the resonant frequency. (b) Place the pendulum in water. (c) Increase the mass of the pendulum bob. (d) Move the pendulum to a higher altitude. 388 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 389 CHAPTER 7 SUMMARY Key Terms and Concepts period frequency oscillation cycle oscillatory motion Hooke’s law spring constant restoring force simple harmonic motion simple harmonic oscillator resonant frequency amplitude forced frequency mechanical resonance Key Equations kx F vmax A k m T 2 m k T 2l g Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. where the time for each interval is the Oscillatory Motion can produce which is the inverse of frequency measured in can take the form of simple harmonic motion mechanical resonance that obeys whereas written mathematically as forced frequency can cause an object to cycles/s F kx where F is the where x is the with a large displacement amplitude Figure 7.52 Chapter 7 Oscillatory motion requires a set of conditions. 389 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 390 CHAPTER 7 REVIEW Knowledge 1. (7.1) What is oscillatory motion? Use an example 16. Determine the spring constant from the following graph
: in your answer. 2. (7.1) Under what conditions must a ball be bounced so it has oscillatory motion? 3. (7.2) What is the defining property of an elastic material? 4. (7.2) What force, or forces, act on an isolated, frictionless simple harmonic oscillator? 5. (7.2) State the directional relationship that exists between the restoring force and displacement of a simple harmonic oscillator. 6. (7.2) What quantity does the slope of a force- displacement graph represent? 7. (7.2) What can be said about a pendulum’s position if the restoring force is a non-zero value? 8. (7.3) Why isn’t acceleration uniform for a simple harmonic oscillator? 9. (7.3) Why is it acceptable to consider a pendulum a simple harmonic oscillator for small displacements, but not for large displacements? 10. (7.4) If the forced frequency and the resonant frequency are similar, what effect does this have on an oscillator? Applications 11. Determine the restoring force acting on a 1.0-kg pendulum bob when it is displaced: (a) 15 (b) 5 12. Determine the frequency of a guitar string that oscillates with a period of 0.0040 s. 13. What is the period of a ball with a frequency of 0.67 Hz? 14. After a diver jumps off, a diving board vibrates with a period of 0.100 s. What is its frequency? 15. What is the restoring force on a 2.0-kg pendulum bob displaced 15.0? Force vs. Displacement 160 140 120 100 80 60 40 20 ) N ( e c r o F 0 0.2 0.4 0.6 0.8 Displacement (m) 17. A spring hangs from the ceiling in a physics lab. The bottom of the spring is 1.80 m from the floor. When the teacher hangs a mass of 100 g from the bottom of the spring, the spring stretches 50.0 cm. (a) What is its spring constant? (b) What force must a person apply to pull the 100.0-g mass on the bottom of the spring down through a displacement of 20.0 cm? (c) The 100.0-g mass is removed and a 300.0-g mass is attached. What is the
distance of the mass above the floor? 18. Two different springs, A and B, are attached together at one end. Spring A is fixed to the wall as shown. The spring constant of A is 100.0 N/m and B is 50.0 N/m. What is the combined stretch of the two springs when a force of 25.0 N [right] is applied to the free end of spring B? A B wall 390 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 391 19. Students stretch an elastic band attached to a force meter through several displacements and gather the following data. Use a graphing calculator or another acceptable method to plot the graph of this data and determine if the elastic band moves as predicted by Hooke’s law. Displacement (cm) 0.00 10.0 20.0 30.0 40.0 50.0 Force (N) 0.00 3.80 15.2 34.2 60.8 95.0 20. How long must the arm of a pendulum clock be to swing with a period of 1.00 s, where the gravitational field strength is 9.81 N/kg? 21. What is the period of a 10.0-kg mass attached to a spring with a spring constant of 44.0 N/m? 22. Determine the maximum velocity of a 2.00-t crate suspended from a steel cable (k 2000.0 N/m) that is oscillating up and down with an amplitude of 12.0 cm. 23. A 0.480-g mass is oscillating vertically on the end of a thread with a maximum displacement of 0.040 m and a maximum speed of 0.100 m/s. What acceleration does the mass have if it is displaced 0.0200 m upwards from the equilibrium position? 24. Determine the period of oscillation of a pendulum that has a length of 25.85 cm. 25. An astronaut who has just landed on Pluto wants to determine the gravitational field strength. She uses a pendulum that is 0.50 m long and discovers it has a frequency of vibration of 0.182 Hz. What value will she determine for Pluto’s gravity? 26. A student is given the relationship for a pendulum: T 2X (a) What does X represent? (b) The student records the period of the pendulum
and finds it is 1.79 s. What is the pendulum’s length? Extensions 27. A spring (k 10.0 N/m) is suspended from the ceiling and a mass of 250.0 g is hanging from the end at rest. The mass is pulled to a displacement of 20.0 cm and released. (a) What is the maximum velocity of the mass? (b) What is the period of oscillation of the mass if it is displaced 15.0 cm and released? 28. A horizontal mass-spring system has a mass M attached to a spring that oscillates back and forth at a frequency of 0.800 Hz. Determine the frequency in the following cases. (a) The mass is doubled. (b) The amplitude is tripled. 29. Identify which of the following examples is SHM and which is not. Explain. (a) a bouncing ball (b) a hockey player moving a puck back and forth with his stick (c) a plucked guitar string Consolidate Your Understanding Create your own summary of oscillatory motion, simple harmonic motion, restoring force, and mechanical resonance by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed above and the Learning Outcomes on page 342. 1. Prepare a quick lesson that you could use to explain Hooke’s law to a peer using the following terms: restoring force, displacement, linear relationship. 2. Construct a two-column table with the title “Mass-spring System.” The first column has the heading, “Factors Affecting Period” and the second column has the heading, “Factors Not Affecting Period.” Categorize the following factors into the two columns: mass, spring constant, amplitude, restoring force, velocity. Think About It Review your answers to the Think About It questions on page 343. How would you answer each question now? e TEST To check your understanding of oscillatory motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 391 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 392 C H A P T E R 8 Key Concepts In this chapter you will learn about: mechanical waves — longitudinal and trans
verse universal wave equation reflection interference acoustical resonance Doppler effect Learning Outcomes When you have finished this chapter, you will be able to: Knowledge describe how transverse and longitudinal waves move through a medium explain how the speed, wavelength, frequency, and amplitude of a wave are related describe how interference patterns can be used to determine the properties of the waves explain the Doppler effect describe the difference between transverse and longitudinal waves describe how waves are reflected explain the relationship between rays and waves apply the universal wave equation to explain how frequency, wavelength, and wave velocity are related explain the effects of constructive and destructive interference Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems 392 Unit IV Mechanical waves transmit energy in a variety of ways. Figure 8.1 What do bats and dolphins have in common? The phrase “blind as a bat” states a common fallacy. Bats have some vision using light, but when placed in pitch-black rooms crisscrossed with fine wires, they can easily fly around and unerringly locate tiny flying insects for food. Dolphins have shown that they can quickly locate and retrieve objects even when they are blindfolded. We usually assume that vision requires light but both bats and dolphins have evolved the ability to “see” using sound waves. Research in science and technology has developed “eyes” that enable humans also to see using sound waves, that is, navigate with senses other than sight. Medicine uses ultrasound (frequencies above the audible range) to look at objects such as a fetus or a tumour inside the body. Submarines can circumnavigate the globe without surfacing by using sound waves to explore their underwater environment. In Chapter 6, you studied how mass transfers energy when it moves through space. Waves, on the other hand, are able to transmit vast quantities of energy between two places without moving any mass from one location to another. Radio waves carry information, sound waves carry conversations, and light waves provide the stimulus for the cells that enable vision. This chapter introduces you to the nature and properties of waves. By experimenting with various forms of wave motion, you will learn about this common, but often misunderstood method of energy transmission. 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 393 8-1 QuickLab 8-1 QuickLab Fundamental Properties of Wave Motion Problem To determine properties of waves in a ripple tank. Materials ripple tank and apparatus for
its operation dowel ( 1.5 cm in diameter) 2 stopwatches ruler, two paper clips light and stand to project waves onto screen screen (a large sheet of newsprint works well) Procedure 1 Set up the ripple tank as shown in Figure 8.2. The water should be about 1 cm deep. Make sure that energy-absorbing buffers are placed around the edge of the tank to prevent unwanted reflections. Check your assembly with your instructor. light source Figure 8.2 paper screen 2 (a) Place a tiny spot of paper in the middle of the ripple tank. (b) Dip the end of your finger once into the water about the middle of the ripple tank to create a single, circular, wave front. Observe the speck of paper as the wave front passes it. 3 (a) On the screen, place the two paper clips at a measured distance apart, 30–40 cm. (b) Position your finger so that its shadow is over one of the paper clips and generate another single wave front. (c) Using a stopwatch, measure the time for the wave to travel from one paper clip to the other. Record the distance and time. Calculate the speed of the wave. Do a few trials for accuracy. 4 (a) Place the dowel horizontally in the water near one edge of the tank. Tap the dowel gently and observe the wave front. Sketch and describe the motion. (b) Position the paper clips in the wave’s path and measure the speed of the straight wave front. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves. Questions 1. When a wave front passes the speck of paper, what motion does the paper make? Does it move in the same or the opposite direction to the motion of the wave front? What does that tell you about the motion of the water as the wave moves through it? 2. On your sketches, draw several vector arrows along the fronts to indicate the direction in which they are moving. What is the angle between the line of the wave front and its motion? In Procedure 4(a), what is the angle between the edge of the dowel and the direction of the motion of the wave front? Sketch what you observe. Describe the motion. 3. Which wave front moves faster, the circular wave front or the straight wave front
? Think About It 1. What differences and similarities are there between the ways energy is transmitted by waves and by matter? e SIM Find out more about waves in ripple tanks. Go to www.pearsoned.ca/school/ 2. What assumptions must be made to use water waves as a model for physicssource. sound waves? Discuss your answers in a small group and record them. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 8 Mechanical waves transmit energy in a variety of ways. 393 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 394 8.1 The Properties of Waves info BIT In December of 2004, an earthquake near the island of Sumatra set off a tsunami that is estimated to have had more than 2 petajoules (1015 J) of energy. This tsunami, the most powerful in recorded history, took over 225 000 lives and did untold billions of dollars in damage to the economies and the environments of the countries that border on the Indian Ocean. info BIT On the day of the tsunami of 2004 that devastated Phuket, Thailand, people travelling in a ferry in deep water offshore from Phuket felt only a greater than usual swell as the wave passed them by. Figure 8.3 Surfers use a wave’s energy to speed their boards across the water. medium: material, for example, air or water through which waves travel; the medium does not travel with the wave wave: disturbance that moves outward from its point of origin, transferring energy through a medium by means of vibrations equilibrium position: rest position or position of a medium from which the amplitude of a wave can be measured crest: region where the medium rises above the equilibrium position trough: region where the medium is lower than the equilibrium position When a surfer catches a wave, many people assume that the forward motion of the surfer is the result of the forward motion of the water in the wave. However, experimental evidence indicates that in a deep-water wave the water does not, in general, move in the direction of the wave motion. In fact, the surfer glides down the surface of the wave just as a skier glides down the surface of a ski hill. Like the skier, the surfer can traverse across the face of the hill as well as slide down the hill. But, unlike the ski hill, the water in the wave front is constantly rising. So
, even though the surfer is sliding down the front of the wave he never seems to get much closer to the bottom of the wave. It is a common misconception that the water in a wave moves in the direction in which the waves are travelling. This may be because waves arriving at the shoreline move water to and fro across the sand. As you will see, this movement is a feature of the interaction of the wave with the sloping shoreline rather than the actual motion of the wave itself. In deep water, there is only very limited lateral motion of water when a wave moves past a particular point. 394 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 395 Waves and Wave Trains When a stone is thrown into a still pond or lake, a ripple moves outward in ever-enlarging concentric circles (Figure 8.4). The water is the transporting medium of the wave and the undisturbed surface of the water is known as the wave’s equilibrium position. Regions where the water rises above the equilibrium position are called crests and regions where the water is lower than its equilibrium position are called troughs. In the crest or trough, the magnitude of greatest displacement from the equilibrium is defined as the waves’ amplitude (A). A complete cycle of a crest followed by a trough is called a wavelength; its symbol is the Greek letter lambda, (Figure 8.5). λ A crest A equilibrium position λ trough Figure 8.5 Properties of a wave wavelength wavelength A wave front moving out from the point of origin toward a barrier is called an incident wave. A wave front moving away from the barrier is called a reflected wave, while a series of waves linked together is a wave train. The concept of a wave train implies a regular repetition of the motion of the medium through which the wave travels. As a result, many parts of the medium are moving in a motion that is identical to the motion of other points on the wave train. At these points, the waves are said to be in phase (Figure 8.6). λ B λ D E A C λ A and B are in phase C, D, and E are in phase Figure 8.6 In-phase points along a wave train have identical status relative to the medium and are separated by one wavelength, λ. Instead of creating individual pulses by hand in a ripple tank, you may use a wave generator to
create a continuous series of crests and troughs forming a wave train. Wave generators can act as a point source similar to the tip of a finger, or as a straight line source, similar to a dowel. In 8-1 QuickLab you measured the speed of a single pulse by observing its motion. However, because it is impossible to keep track of a single wave in a wave train, to measure the speed of a wave train requires a greater understanding of the properties of waves. crests crests troughs troughs Figure 8.4 Many of the terms used to describe wave motions come from the observation of waves on the surface of water. e WEB To learn more about experiments using ripple tanks, follow the links at www.pearsoned.ca/school/ physicssource. amplitude: the distance from the equilibrium position to the top of a crest or the bottom of a trough wavelength: the distance between two points on a wave that have identical status. It is usually measured from crest to crest or from trough to trough. wave front: an imaginary line that joins all points reached by the wave at the same instant incident wave: a wave front moving from the point of origin toward a barrier reflected wave: a wave front moving away from a barrier wave train: a series of waves forming a continuous series of crests and troughs point source: a single point of disturbance that generates a circular wave Chapter 8 Mechanical waves transmit energy in a variety of ways. 395 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 396 8-2 Inquiry Lab 8-2 Inquiry Lab Wave Trains in a Ripple Tank, Part 1: Reflecting Waves In this ripple tank experiment, the properties of a twodimensional wave train are analyzed. Question How do the incident and reflected wave trains interact when wave trains reflect from a straight barrier? Materials and Equipment ripple tank, including the apparatus for its operation straight barrier wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) Variables In this experiment you are to observe the directions of motion of the incident waves and reflected waves and how these directions are related to each other. Other variables to be observed are the interactions that occur when the incident and reflected wave trains move in different directions through the same point in the ripple tank. As you observe the wave motions you should identify which are the controlled variables, manipulated variables, and responding variables. General Procedure 1 (a
) Set up the ripple tank as shown in Figure 8.7. (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork light source generator motor generator support point sources To power supply paper screen Figure 8.7 Procedure 1 (a) Place the point-source wave generator at one edge of the ripple tank and the straight barrier at the other edge. The shadow of both the barrier and the source should be visible on the screen. (b) Use the point-source wave generator to create a continuous wave train in the ripple tank. Observe what happens to the incident wave train when it meets the reflected wave train. (c) Make a sketch of your observations. Wave trains are a bit tricky to observe at first. Discuss what you see with your team members. When you have reached a consensus, write a brief description of your observations. On your sketch, place vector arrows along an incident and a reflected wave front to indicate the direction and speed of their motions. 2 (a) Set up the straight-line wave generator at one edge of the ripple tank. Place the barrier at the other edge parallel to the generator. The shadows of both the generator and the barrier should be visible on the screen. (b) Start the generator to create a continuous wave train. Observe what happens when the reflected wave train moves back through the incident wave train. Draw diagrams and write a description of the observations. Again, draw vector arrows along incident and reflected wave fronts to indicate their relative velocities. 396 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 397 3 Move the barrier so that it is at an angle of about 30° 2. (a) When the barrier is parallel to the straight wave to the generator and repeat step 2 (b). 4 Set the barrier so that the angle between it and the generator is about 60º and repeat
step 2 (b). Analysis 1. (a) When the incident wave train created by the point-source generator is passing through the reflected wave train, what happens to the waves in the region where they overlap? (b) Can you see the direction of the motion for both the incident and reflected wave trains? generator, what pattern do you observe when the reflected waves are moving back through the incident waves? (b) In which direction does the pattern seem to be moving? Can you see the direction of the motion for both the incident and reflected wave trains? 3. Answer question 2 for the set-up when the barrier is at an angle to the straight wave generator. 4. In all cases above, how does the spacing of the waves in the reflected wave train compare to the spacing of the waves in the incident wave train? Waves and Rays When waves in a ripple tank are viewed from above (Figure 8.8) the wave fronts appear as a set of bright and dark bands (crests and troughs). When we draw wave trains as seen from above, we use a line to represent a wave front along the top of a crest. The point halfway between two lines is the bottom of a trough. A series of concentric circles represents the wave train generated by a point source. ray: a line that indicates only the direction of motion of the wave front at any point where the ray and the wave intersect rays wavelength wave source crests troughs velocity vectors Figure 8.8 View of a ripple tank from above Figure 8.9 A point source generates waves that move outward as concentric circles with the source at their centre Waves are in constant motion. At all points on a wave front, the wave is moving at right angles to the line of the crest. There are two ways to indicate this (Figure 8.9). You could draw a series of vector arrows at right angles to the wave front with their length indicating the speed of the wave. Or, you could draw rays, lines indicating only the direction of motion of the wave front at any point where the ray and the wave front intersect. The rays in Figure 8.9 are called diverging rays since they spread out as they move away from the origin. When rays diverge, it indicates that the energy given to the wave at its source is being spread over a larger and larger area. This is why, as sound moves away from a point source such as a bell, the volume decreases with the square of the distance. e WEB To learn more about
the mathematical relationship between the volume of sound and the distance from the source, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 397 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 398 light rays trough crest water ripple tank bottom paper screen bright areas under crests dark areas under troughs Figure 8.10 Sketch of a wave showing light refracting through a crest and a trough When waves in a ripple tank are projected onto a screen below, the wave fronts appear as a set of bright and dark bands. It may seem logical that the light and dark bands seen on the screen below the ripple tank result from the differences in water depth between the crests and the troughs. But that difference is only about a millimetre and cannot account for the high contrast in light seen on the screen. In fact, a crest acts like a converging lens to concentrate the light, creating a bright bar. A trough acts like a diverging lens to spread the light out, making the area under the trough darker (Figure 8.10). You will learn more about light refraction in Unit VII of this course. Reflection of a Wave Front When a wave front is incident on a straight barrier, it reflects. The direction the wave travels after reflection depends on the angle between the incident wave front and the barrier. A circular wave front, as generated by a point source, S, produces the simplest reflection pattern to explain. In this case, the reflected wave follows a path as if it had been generated by an imaginary point source S, at a position behind the barrier identical to that of the actual point source in front of the barrier (Figure 8.11). Now consider an incident wave front created by a straight wave generator (Figure 8.12). The straight wave front also reflects as if the reflected wave had been generated by an imaginary generator located behind the barrier. The position of the imaginary generator behind the barrier is equivalent to the position of the real generator in front of the barrier. The incident wave front and the reflected wave front are travelling in different directions, but the angle between the incident wave front and the barrier must be identical to the angle between the reflected wave front and the barrier. e SIM Find out more about the ways waves reflect. Go to www.pearsoned.ca/school/ physicssource. S imaginary source rays from the imaginary source reflecting surface of barrier
S real source imaginary incident wave generated by S reflected portion of incident wave incident wave generated by S imaginary straight wave generator reflected wave front real straight wave generator incident wave from imaginary generator reflecting surface of barrier incident wave front Figure 8.11 When circular waves reflect from a straight barrier, the reflected waves seem to be moving away from an imaginary source. Figure 8.12 When straight waves reflect from a straight barrier, the angle between the reflected wave front and the barrier must be equal to the angle between the incident wave front and the barrier. 398 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 399 M I N D S O N Waves Can Have Curls Too When waves travel in deep water, their shape is similar to the waves in a ripple tank. But as waves near the shoreline they change shape and develop what is known as a curl in which the top of the wave falls in front of the wave. Recall Figure 8.3 on page 394. Explain the causes of a wave’s curl in terms of its motion. 8-3 Inquiry Lab 8-3 Inquiry Lab Wave Trains in a Ripple Tank, Part 2: Wave Speed and Wavelength Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this ripple tank experiment, the properties of a twodimensional wave train are further analyzed. Question What effect does a change in speed have on wave trains? Materials and Equipment ripple tank, including the apparatus for its operation wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) two small blocks of wood about 8 mm thick Variables In this lab you will be observing how water depth affects the properties of waves. The variables that might be affected by changes in the depth are speed, frequency, wavelength, and direction. As you make your observations, consider which of the variables are controlled variables, manipulated variables, and responding variables. General Procedure 1 (a) Set up the ripple tank as shown in Figure 8.7 (page 396). (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is
a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Procedure 1 Place small pads (about 8 mm thick) under the legs along one edge of the ripple tank so that the water gets shallower toward that edge. The water should be less than 1 mm deep at the shallow edge. 2 (a) Near the deep edge, create a wave train using the point-source generator. (b) Observe what happens to the wave fronts as the wave train moves toward the shallow edge. Discuss your observations with your team members. Sketch and briefly describe your observations. (c) Place vector arrows along several of the wave fronts to indicate the direction and speed of the wave fronts as they move into shallow water. 3 (a) Set up the straight-line wave generator at the deep edge of the tank. (b) Turn on the generator and observe the wave train as it moves into the shallow water. (c) Sketch your observations and describe the motion of the wave train as it moves into shallow water. Use vector arrows along the wave fronts to assist you in your descriptions. 4 (a) Now place the pads under the legs of the ripple tank on an edge that is at a right angle to the position of the straight-line wave generator. (b) Use the straight-line wave generator to create a wave train. (c) Observe the wave train as it travels across the tank. Discuss your observations with your team members. Sketch and write a brief description of what you saw to accompany your sketch. Use vector arrows drawn along several of the wave fronts to indicate their relative velocity as they move into the shallow water. Chapter 8 Mechanical waves transmit energy in a variety of ways. 399 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 400 Analysis 2. What do you think causes the observed changes? 1. For each of the trials, when the waves moved from deep to shallow edge (or vice versa), comment on the kinds of changes you observed. 3. When the straight wave fronts moved across the tank at right angles to the change in the depth of the water, was the shape of the wave front affected? (a) Were the wavelengths of the incident waves affected as they moved into shallow water? 4. What properties of waves are affected as the waves move from deep to shallow water? (b) Was the shape of the
wave fronts affected as they entered shallow water? (c) If so, how did the shape of the wave fronts change as they changed speed? 5. When a water wave moves toward a beach, how would the change in the depth of the water affect the motion of the wave? 8.1 Check and Reflect 8.1 Check and Reflect Knowledge 1. If a wave pattern is created by a point source, what is the nature of the ray diagram that would represent the wave fronts? 2. When a wave front reflects from a barrier, what is the relationship between the direction of the motions of the incident and reflected wave fronts? Applications 3. The sketch shows a ray diagram that represents the motion of a set of wave fronts. If you were observing these wave fronts in a ripple tank, describe what you would see. rays 4. Draw a diagram of a set of straight wave fronts that are incident on a straight barrier such that the angle between the wave fronts and the barrier is 40˚. Draw the reflected wave fronts resulting from this interaction. How do the properties (speed, wavelength, and amplitude) of the reflected wave compare with the properties of the incident wave? Use a wavelength of about 1 cm in your diagram. Extensions 5. Reflection of light is the essence of how we use mirrors to see images. What does the reflection of waves in a ripple tank tell you about the formation of images? Hint: Think of where the reflected waves in the ripple tank seem to originate. 6. When a sound travels in water, the speed of the sound depends on the temperature of the water. If the sonar ping emitted by a submarine has a wavelength of 2.50 m, what happens to that wavelength when it enters a region where sound travels faster? e TEST To check your understanding of the properties of waves, follow the eTest links at www.pearsoned.ca/ school/physicssource. 400 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 401 8.2 Transverse and Longitudinal Waves Did you ever have a Slinky™ toy when you were a child? When a Slinky™ is stretched out along the floor and oscillated from side to side across its axis (centre line), or forward and back along its axis, mechanical waves are transmitted along its length. The sideways oscillations set up a transverse wave while those along the axis set up a longitudinal wave as shown in Figure 8
.13. In this section we will consider the characteristics of such waves. Transverse Pulses info BIT The ever-popular Slinky™ was invented in 1945 by Richard James, a naval engineer working on tension springs. The name comes from the Swedish for “sleek” or “sinuous.” Each Slinky™ is made from 80 feet (24.384 m) of wire. A pulse moving through a spring is a good introduction to the way a wave moves through a medium. An ideal spring is one that allows a pulse to travel through it without loss of energy. By definition, a pulse is just the crest or the trough of a wave; its length is one-half a wavelength. The spring provides a medium in which the motion of a pulse can be observed from the side. Initially, the spring is in its equilibrium position. When you flip the spring sharply to the side and back, the motion of your hand sets up a series of sequential motions in the coils of the spring. Each coil imitates, in turn, the motion of the hand. This results in a transverse pulse (Figure 8.14) that moves along the spring. As a pulse moves along a spring, the coils of the spring move at right angles to the direction of the pulse’s motion. Compare v hand and v pulse in Figure 8.14. At the front of the pulse, the coils are moving away from the spring’s equilibrium position toward the point of maximum displacement from the equilibrium. In the trailing edge of the pulse, the coils are moving back toward the equilibrium position. Hand starts at equilibrium position of spring. Front of pulse starts to move along spring. Hand is at maximum amplitude. Hand continues to move up. vhand 0 vpulse vhand 0 v A Hand continues to move down. Figure 8.13 (a) A transverse pulse (b) A longitudinal pulse Arrows indicate the direction of the medium. The pulses are moving through the springs toward the bottom of the page. pulse: a disturbance of short duration in a medium; usually seen as the crest or trough of a wave e WEB To learn more about the forces operating in an oscillating spring, follow the links at www.pearsoned.ca/school/ physicssource. amplitude A As the hand moves toward the equilibrium position, the amplitude of the pulse moves along the spring. Pulse is complete when the hand is at equilibrium position. Figure 8.14 When you move your hand you
set up a sequence in which the coils of the spring imitate the motion of your hand. This creates a moving pulse. l vhand 0 Chapter 8 Mechanical waves transmit energy in a variety of ways. 401 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 402 Energy Changes During the Movement of a Pulse Along the pulse, energy is stored in the form of both elastic potential energy and kinetic energy. As a section of the spring moves from the equilibrium position to the top of the pulse, that section has both kinetic energy (it is moving sideways relative to the direction of the pulse) and elastic potential energy (it is stretched sideways). At the point on the pulse where the displacement is greatest the coils are, for an instant, motionless. Then, the tension in the spring returns the coils to their equilibrium position. No blurring. This indicates: Ek Ep 0 maximum 0 0 Ek Ep Figure 8.15 A transverse pulse is generated when a spring is given a sharp flip to the side. Arrows indicate the direction of motion of the coils. Can you determine which way the pulse is moving? In Figure 8.15 the blurring on the front and back segments of the pulse indicates the transverse motion and the presence of kinetic energy as well as elastic potential energy. At the top, there is no blurring as the coils are temporarily motionless. At that instant that segment of the spring has only elastic potential energy. As it returns to its equilibrium position, the segment has, again, both kinetic and potential energies. The energy in a pulse moves along the spring by the sequential transverse motions of the coils. Recall from section 6.3 that a pendulum, along the arc of its path, has both kinetic and potential energy, but at the point where the pendulum’s displacement is greatest, all the energy is in the form of potential energy. Thus, the energy of an oscillating pendulum is equivalent to its potential energy at the point where its displacement is greatest. Similarly, the amplitude of the wave in an experimental spring can be used to determine the quantity of energy that is stored in the pulse. Concept Check You generate a pulse in a Slinky™ stretched out on the floor. If you wish to, you could give the next pulse more energy. How would you do that? info BIT When considering sound, amplitude determines loudness. 402 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08
2:23 PM Page 403 8-4 Inquiry Lab 8-4 Inquiry Lab Pulses in a Spring, Part 1: Pulses in an Elastic Medium Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study how a pulse moves through a medium. Question What are the mechanics by which pulses move through a medium? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). 5 On your sketch of the pulse, label the following parts: — The amplitude (A) is the perpendicular distance from the equilibrium position of the spring to the top of the pulse. — The pulse length (l) is the distance over which the spring is distorted from its equilibrium position. NOTE: When you stand at the side of the spring, the pulse seems to move past you very quickly, almost as a blur. Watching from the end of the spring may make it easier to observe the details of the motion. Materials and Equipment 6 Make a longitudinal pulse by moving your hand light spring metre-stick or measuring tape stopwatch masking tape CAUTION: A stretched spring stores considerable amounts of elastic potential energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. If you allow the spring to gently unwind as you are walking you will prevent the spring from tying itself into a knot. Procedure 1 Have one team member hold the end of the spring while another stretches it until it is moderately stretched (about 5–6 m). 2 Place strips of masking tape on the floor at either end of the spring to mark this length. Near the middle of the spring, attach a strip of tape about 5 cm long to one of the coils. 3 Have one of the people holding the spring generate a transverse pulse. Generate the pulse by moving your hand sharply to one side (about 60–75 cm) and back to its original position. This is a transverse pulse since its amplitude is perpendicular to the direction of its motion. 4 Sketch the pulse. Indicate the motions of the pulse and the coils using vector arrows. Observe the motion of the tape at the middle of the spring to assist in these observations. Generate more pulses until you understand the nature of the motion of the pulse in the spring. sharply toward the person holding the spring at the other end,
and then back to its original position. This pulse is called a longitudinal pulse, because its amplitude is along the direction of its motion. Repeat the pulse a few times to determine the nature of the motion of the spring as the pulse moves through it. Sketch and describe the motion of the coils as the pulse moves along the spring. Analysis 1. What determines the amplitude of the transverse pulse? The longitudinal pulse? 2. Does the pulse change shape as it moves along the spring? If so, what causes the change in shape of the pulse? Would you expect the pulse shape to change if this were an isolated system? 3. How is the reflected pulse different from the incident pulse? If this were an isolated system, how would the reflected pulse differ from the incident pulse? 4. Describe the motion of the strip of tape at the middle of the spring as the pulse passes it. Does the tape move in the direction of the pulse? 5. How does the motion of the medium relate to the motion of the pulse? 6. How does the pulse transfer energy from one end of the spring to the other? 7. The motion of a pulse in the spring requires you to make assumptions about the motion of an ideal pulse. What assumptions must you make to create a model of how a transverse pulse moves through an elastic medium? Chapter 8 Mechanical waves transmit energy in a variety of ways. 403 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 404 v Reflection of Pulses from a Fixed Point incident pulse reflected pulse Figure 8.16 Reflection from the fixed end of a spring causes the pulse to be inverted. e SIM Find out about similarities and differences between transverse and longitudinal waves. Go to www.pearsoned.ca/ school/physicssource. When a pulse (or wave) is generated in a spring it soon arrives at the other end of the spring. If that end is held in place, the total pulse reflects from the end and travels back toward the source. The reflected pulse is always inverted relative to the incident pulse (Figure 8.16). In an ideal medium, the other properties of the pulse (amplitude, length, and speed) are unaffected by reflection. These properties of the reflected pulse are identical to those of the incident pulse. When a wave train is generated in the spring, the crests of the incident wave are reflected as troughs while the troughs of the incident wave are reflected as crests. v Long
itudinal Waves If, instead of moving your hand across the line of the spring, you give the spring a sharp push along its length, you will observe that a pulse moves along the spring. This pulse is evidence of a longitudinal wave. The pulse is seen as a region where the coils are more tightly compressed followed by a region where the coils are more widely spaced. These two regions are called, respectively, a compression and a rarefaction and correspond to the crest and trough in a transverse wave. In the case of a longitudinal wave, the coils of the spring oscillate back and forth parallel to the direction of the motion of the wave through the medium (Figure 8.17). But, as with transverse waves, once the wave has passed through, the medium returns to its original position. Once again, energy is transmitted through the medium without the transmission of matter. Figure 8.17 Longitudinal waves are formed when the source oscillates parallel to the direction of the wave motion. spring in equilibrium position Project LINK What aspect of your seismograph will relate to the ideas of compression and rarefaction in a longitudinal wave? longitudinal wave in the spring vwave vcoils vcoils vcoils vcoils rarefaction rarefaction compression compression wavelength 404 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 405 8-5 Inquiry Lab 8-5 Inquiry Lab Pulses in a Spring, Part 2: Speed, Amplitude, and Length Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study the speed, amplitude, and length of pulses. You will set up the experiment similarly to 8-4 Inquiry Lab on page 403. Question What is the relationship between the amplitude, length, and speed of a pulse? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). Materials and Equipment light spring stopwatch metre-stick or measuring tape masking tape Procedure 1 (a) Measure the speed of a transverse pulse as it moves along the spring. Have the person creating the pulse “count down” so that team members with stopwatches can time the pulse as it moves toward the other end. Measure the time from the instant the front edge of the pulse leaves the hand of the person generating it until the front edge arrives at the hand at the other
end. Do this a few times to establish a consistent value. Record your results. Use the time and the distance between the hands to calculate the speed of the pulse. (b) Generate pulses by moving your hand to the side and back at different speeds (more quickly or more slowly). Measure the speed of each of these pulses. 2 Have the person holding one end of the spring move so that the spring is stretched about 1 m farther. (Do not overstretch the spring.) Generate a pulse and measure the speed of the pulse in the spring at this higher tension. Carefully walk the spring back to the length used initially. 3 Make a transverse pulse by moving your hand a different distance sideways. Try to keep the time used to make the pulse the same as before. Repeat this a few times to observe changes in the pulse. Record your observations. 4 Now make several transverse pulses by moving your hand to a given amplitude but change the speed at which you move. Repeat a few times and record your observations. Analysis 1. Does the speed at which you moved your hand to generate a pulse affect the speed of the pulse? 2. When the spring was stretched to a greater length, what happened to the speed of the pulse? 3. What controls the amplitude (A) of the pulse? Can you create pulses with equal lengths but different amplitudes? 4. What controls the length (l ) of the pulse? Can pulses of equal amplitudes have different lengths? 5. What is the relationship between the length of the pulse and the speed (v) of the pulse in the medium? 6. Does the length of a pulse affect its larger amplitude or vice versa? Explain why or why not. 7. Does the energy in a pulse seem to depend on its amplitude or its length? Give reasons for your decision. Consider what changes occur as the pulse moves through the spring. 8. What determines the speed, the length, and the amplitude of the pulse? 9. What aspect of wave motion in water can you simulate by changing the tension in a spring? 10. What do your findings for the relationship of the amplitudes and lengths of pulses in springs tell you about the relationship between the amplitudes and wavelengths of waves in water? 11. Sound is often referred to as a wave. What aspect of a sound would relate to (a) the amplitude and (b) the wavelength of its waves? Chapter 8 Mechanical waves transmit energy in a variety of ways. 405 08-PearsonPhys20
-Chap08 7/24/08 2:23 PM Page 406 Pulse Length and Speed Figure 8.18 The length (l) of the pulse depends on the speed (v) of the pulse and the time (t) taken to complete the pulse. e WEB To learn more about the way the structures of the human ear transfer sound waves, follow the links at www.pearsoned.ca/school/ physicssource. info BIT When a wave moves across the surface of water, the water moves between crests and troughs by localized circular motions. This local circular motion moves water back and forth between a trough and the adjacent crest. direction of motion of surface vwave vmedium The speed of the pulse depends on the medium. If you stretch the spring so that the tension increases, then the speed of the pulse increases. Relaxing the tension causes the speed to decrease. The speed of the pulse in the spring also determines the length of the pulse. vhand 0 Hand starts at equilibrium position of spring. vhand 0 v A Front of pulse starts to move along spring. vpulse Hand is at maximum amplitude. amplitude A Pulse is complete when the hand is at equilibrium position. l vhand 0 The instant you start to move your hand to generate a pulse, the disturbance begins to move along the spring at a constant speed, v. Assume that the time it takes to move your hand to create the complete pulse is t. By the time your hand returns the spring to its equilibrium position, the front of the pulse will have travelled a distance d, which can be defined as the length l of the pulse (Figure 8.18). Remember d v and, therefore t d vt l vt Waves and the Medium A solid such as a spring is an elastic medium and can store elastic potential energy by stretching longitudinally or transversely. Typically, the way that fluids (liquids and gases) store elastic potential energy is by being compressed. Therefore, waves within fluids are typically longitudinal waves, known as pressure waves. This is the principle used in engines and aerosol sprays. As compressions and rarefactions move through a fluid, the motion of the molecules in the fluid is very similar to the motion of the coils when a longitudinal wave moves through a spring. For water to transmit energy as a transverse wave, the waves must be displaced vertically, but in liquids the vertical displacement cannot be a form of elastic potential energy. Thus, transverse waves can be transmitted only at the
surface of water, or other liquids, where the waves are the result of gravitational potential energy rather than elastic potential energy. motion of water within wave Figure 8.19 406 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 407 M I N D S O N Wave Motion in Fluids Water can transmit both transverse (surface) waves and longitudinal (internal) waves such as sound. We know that sound waves in gases are longitudinal waves. Is it possible to create a transverse wave in a gas? Why or why not? Consider how transverse waves are created in liquids. Example 8.1 To create a pulse in a fixed ideal spring, you move your hand sideways a distance of 45 cm from the equilibrium position. It takes 0.80 s from the time you begin to move your hand until it returns the spring to its equilibrium position. If the pulse moves at a speed of 2.5 m/s, calculate the length of the pulse and describe the incident pulse and reflected pulse that pass through the midpoint of the spring. Practice Problems 1. A pulse is generated in a spring where it travels at 5.30 m/s. (a) If the time to generate the pulse is 0.640 s, what will be its length? (b) How does the speed of the pulse affect its amplitude? 2. A pulse moves along a spring at a speed of 3.60 m/s. If the length of the pulse is 2.50 m, how long did it take to generate the pulse? 3. A pulse that is 1.80 m long with an amplitude of 0.50 m is generated in 0.50 s. If the spring, in which this pulse is travelling, is 5.0 m long, how long does it take the pulse to return to its point of origin? 4. A spring is stretched to a length of 6.0 m. A pulse 1.50 m long travels down the spring and back to its point of origin in 3.6 s. How long did it take to generate the pulse? Answers 1. (a) 3.39 m (b) It does not; they are independent. 2. 0.694 s 3. 2.8 s 4. 0.45 s Given A 45 cm 0.45 m t 0.80 s v 2.5 m/s Required (a) length of the pulse (b) description of incident pulse passing
the midpoint of the spring (c) description of reflected pulse passing the midpoint of the spring Analysis and Solution (a) The length of the pulse can be found using l vt. l vt (0.80 s) 2.5 m s 2.0 m (b) The spring is defined as an ideal spring, so the amplitude of the pulse is constant. The amplitude at all points on the spring will be the same as at the source. Therefore, A 0.45 m At the midpoint of the spring, the amplitude of the incident pulse is 0.45 m, its length is 2.0 m, and its speed is 2.5 m/s. (c) Reflection inverts the pulse but does not change any of its properties. The reflected pulse is identical to the incident pulse except that it is inverted relative to the incident pulse. Paraphrase and Verify (a) The length of the pulse is equal to 2.0 m. (b) In an ideal spring the amplitude of the pulse is constant. (c) When pulses are reflected from a fixed end of a spring they are inverted. Chapter 8 Mechanical waves transmit energy in a variety of ways. 407 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 408 direction of pulse motion Waves Are a Form of Simple Harmonic Motion path of hand motion equilibrium position of spring Figure 8.20 Simple harmonic motion generates a wave train in the form of a sine curve. spring If you move your hand from side to side in simple harmonic motion, as indicated in Figure 8.20, transverse waves are generated in the spring. When a transverse wave moves through a medium, the motion of the medium may seem, at first, quite complex. In a transverse wave, each segment of the medium simply oscillates in simple harmonic motion about its equilibrium position in the direction perpendicular to the direction of the wave motion. This simple harmonic motion is transferred sequentially from one segment of the medium to the next to produce the motion of a continuous wave. Universal Wave Equation Pulses provide a useful tool to introduce the nature of waves. However, in nature, sound and light are wave phenomena rather than pulses. In this section, we will begin to shift the emphasis to the properties of waves. Whereas the letter l is used to indicate the length of a pulse, the Greek letter lambda, λ, is used to indicate wavelength. The terms crest and trough come from the description
of water waves but are used throughout wave studies. For a water wave, the crest occurs where the medium is displaced above the equilibrium position, while a trough is the region displaced below the equilibrium position. However, for media such as springs, the terms crest and trough merely refer to two regions in the medium that are displaced to opposite sides of the equilibrium position (Figure 8.20). Other variables used in wave studies (frequency–f, period–T, amplitude–A) come from and have the same meanings as in your study of simple harmonic motion in section 7.2. The period (T) is the time taken to generate one complete wavelength. Since two pulses join to create one wave, the period for a wave is twice the time required to generate a pulse. Therefore, the wavelength of a wave is twice the length of a pulse. With this in mind, the relationship between wavelength, speed, and period is the same for waves as it is for pulses. That is, λ vT rather than l vt. For periodic motion, T 1. f The equation for wavelength now can be written as λ v f or v f λ. The latter form is known as the universal wave equation. 408 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 409 Constant Frequency, Speed, and Wavelength In 8-3 Inquiry Lab, you investigated what happened to a wave train as it moved from deep to shallow water. Changes occurred because the speed in shallow water was slower than it was in deep water. Since the frequency of the waves as they moved from deep to shallow water was unchanged, the reduction in speed was, as predicted by the universal wave equation, accompanied by a reduction in wavelength (Figure 8.21). For a constant frequency, the ratio of the velocities is the same as the ratio of the wavelengths. λ 2 v2 boundary between deep and shallow water straight wave generator λ 1 v1 shallow water (slow speed) wave fronts deep water (faster speed) ripple tank Figure 8.21 When the frequency is constant, a change in speed results in a change in wavelength. When waves change speed, they often change direction as well. You will study this phenomenon further in Unit VII. Example 8.2 To generate waves in a stretched spring, you oscillate your hand back and forth at a frequency of 2.00 Hz. If the speed of the waves in the spring is
5.40 m/s, what is the wavelength? Given v 5.40 m/s f 2.00 Hz Required wavelength Analysis and Solution The variables (v, f, λ) are related by the universal wave equation. v f λ v λ f m 5.40 s 2.00 Hz 5.40 m s 2.00 1 s 2.70 m Paraphrase and Verify The wavelength is 2.70 m. Practice Problems 1. Orchestras use the note with a frequency of 440 Hz (“A” above middle “C”) for tuning their instruments. If the speed of sound in an auditorium is 350 m/s, what is the length of the sound wave generated by this frequency? 2. A submarine sonar system sends a burst of sound with a frequency of 325 Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 8.50 s. If the wavelength of the sound is 4.71 m, how far away is the rock face? 3. A fisherman anchors his dinghy in a lake 250 m from shore. The dinghy rises and falls 8.0 times per minute. He finds that it takes a wave 3.00 min to reach the shore. How far apart are the wave crests? Answers 1. 0.795 m 2. 6.51 km 3. 10 m Chapter 8 Mechanical waves transmit energy in a variety of ways. 409 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 410 M I N D S O N Wavelength, Frequency, and Speed • Walk side by side with a partner at • With both students keeping their the same speed. One student should take long steps while the other takes very short steps. 1. If the two students maintain their pace, what is the relationship between the frequency and the length of their steps? steps the same length as in the first trial, walk so that your steps are in phase (take steps at the same time). 2. When the two students walk in phase, what is the effect of taking shorter steps? What is the relationship between speed and step length? 8.2 Check and Reflect 8.2 Check and Reflect Knowledge 1. Explain the relationship between the motion of a transverse wave and the motion of the medium through which it moves. 2. Explain how the medium moves when a longitudinal wave passes through it. 3. What is the difference between a transverse and a longitudinal wave
? 4. What determines the amount of energy stored in a wave? Applications 5. Sound waves travel through seawater at about 1500 m/s. What frequency would generate a wavelength of 1.25 m in seawater? 6. Temperature changes in seawater affect the speed at which sound moves through it. A wave with a length of 2.00 m, travelling at a speed of 1500 m/s, reaches a section of warm water where the speed is 1550 m/s. What would you expect the wavelength in the warmer water to be? 7. A speaker system generates sound waves at a frequency of 2400 Hz. If the wave speed in air is 325 m/s, what is the wavelength? 8. When you generate a wave in a spring, what is the relationship between the frequency, wavelength, and amplitude? 410 Unit IV Oscillatory Motion and Mechanical Waves 9. Two tuning forks are generating sound waves with a frequency of 384 Hz. The waves from one tuning fork are generated in air where the speed of sound is 350 m/s. The other tuning fork is generating sound under water where the speed of sound is 1500 m/s. Calculate the wavelength for the sound (a) in air, and (b) in water. (c) Would you hear the same musical note under water as you did in air? Extensions 10. A radio speaker generates sounds that your eardrum can detect. What does the operation of the speaker and your eardrum suggest about the nature of sound waves? How does the nature of the medium (air) through which sound travels support your assumptions? radio speaker direction of oscillation e TEST eardrum direction of oscillation To check your understanding of transverse and longitudinal waves, follow the eTest links at www.pearsoned.ca/school/physicssource. 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 411 8.3 Superposition and Interference Superposition of Pulses and Interference When waves travel through space it is inevitable that they will cross paths with other waves. In nature, this occurs all the time. Imagine two people who sit facing each other and are speaking at the same time. As each person’s sound waves travel toward the other person they must meet and pass simultaneously through the same point in space (Figure 8.22). Still, both people are able to hear quite plainly what the other person is saying. The waves obviously were able to pass through each
other so that they reached the other person’s ears unchanged. How waves interact when they cross paths is well understood. When you observe two waves crossing in the ripple tank, things happen so quickly that it is difficult to see what is happening. Still, it is plain that the waves do pass through each other. By sending two pulses toward each other in a spring, it is easier to analyze the events. It is helpful to imagine that the spring in which the pulses are travelling is an ideal, isolated system. The pulses then travel without loss of energy. First, consider two upright pulses moving through each other. When two pulses pass through the same place in the spring at the same time, they are said to interfere with each other. In the section of the spring where interference occurs, the spring takes on a shape that is different from the shape of either of the pulses individually (Figure 8.23). vA pulse A vB pulse B actual position of spring region of pulse overlap pulse A pulse B pulse A x x y y pulse B z z pulse A pulse B Region of overlap Original position of pulse A Original position of pulse B Resultant sound waves from girl sound waves from boy Figure 8.22 When two people talk simultaneously, each person’s sound waves reach the other person’s ears in their original form. interference: the effect of two pulses (or two waves) crossing within a medium; the medium takes on a shape that is different from the shape of either pulse alone Figure 8.23 When two upright pulses move through each other, the displacement of the resultant pulse is the sum of the displacements of pulse A and pulse B. If at any point in the region of overlap, the displacement of one pulse, shown here as x, y, and z, is added to the displacement of the other, the displacement of the resultant pulse is increased. This is called constructive interference. Chapter 8 Mechanical waves transmit energy in a variety of ways. 411 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 412 principle of superposition: the displacement of the combined pulse at each point of interference is the sum of the displacements of the individual pulses constructive interference: the overlap of pulses to create a pulse of greater amplitude destructive interference: the overlap of pulses to create a pulse of lesser amplitude e WEB Find out more about superposition of pulses. Follow the links at www.pearsoned.ca/ school/physicssource. The new shape that the
spring takes on is predicted by the principle of superposition. This principle, based on the conservation of energy, makes it quite easy to predict the shape of the spring at any instant during which the pulses overlap. The displacement of the combined pulse at each point of interference is the algebraic sum of the displacements of the individual pulses. In Figure 8.23 the two pulses have different sizes and shapes and are moving in opposite directions. The displacement of a pulse is positive for crests and negative for troughs. Since in Figure 8.23 both displacements are positive, at any point where the two pulses overlap, the displacement of the resultant pulse is greater than the displacements of the individual pulses. When pulses overlap to create a pulse of greater amplitude, the result is constructive interference (Figure 8.24). Now consider the case when an inverted pulse meets an upright pulse. The displacement of the inverted pulse is a negative value. When the displacements of these pulses are added together, the displacement of the resultant pulse is smaller than the displacement of either pulse. When pulses that are inverted with respect to each other overlap to create a pulse of lesser amplitude, the result is destructive interference (Figure 8.25). resultant pulse pulse A pulse B Figure 8.24 Constructive interference pulse A Figure 8.25 Destructive interference resultant pulse pulse B 412 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 413 Figure 8.26 shows a special case of destructive interference. Two pulses that have the same shape and size are shown passing through each other. Because the pulses are identical in shape and size, their displacements at any position equidistant from the front of each pulse are equal in magnitude but opposite in sign. At the point where the two pulses meet, the sum of their displacements will always be zero. At the instant when these two pulses exactly overlap, the displacement at all points is zero and the pulses disappear. The resultant is a flat line. Immediately following this instant, the pulses reappear as they move on their way. The Inversion of Reflected Pulses in a Fixed Spring The principle of superposition explains why pulses are inverted when they reflect from the fixed end of a spring (Figure 8.27). Because the end of the spring is fixed in place, at that point the sum of the displacements of the incident pulse and the reflected pulse must always be zero. Thus, at the point of reflection, the displacement
of the reflected pulse must be the negative of the incident pulse. Hence, the reflected pulse must be inverted relative to the incident pulse. info BIT Since, at the point of reflection in the spring the system is basically an isolated system, all the energy in the incident pulse must be carried away by the reflected pulse. actual position of spring original incident pulse reflected segment of incident pulse Pulse arrives at fixed end of the spring. vP vP actual position of spring vA pulse A point where pulses meet pulse B vB region of overlap pulse B pulse B vB pulse A x x y y z z pulse A vA pulse A pulse B pulse A vA pulse B vB Region of overlap Original position of pulse A Original position of pulse B Resultant Figure 8.26 When pulses that are inverted with respect to each other overlap, the displacement of one pulse is reduced by the displacement of the other pulse. At any point in the region of overlap, the displacement of Pulse B, shown here as x, y, and z, reduces the displacement of Pulse A to produce the resultant. This is called destructive interference. vP vP Reflected pulse leaves fixed end of the spring. Figure 8.27 If the end of the spring is fixed, the reflected pulse must be inverted relative to the incident pulse. Chapter 8 Mechanical waves transmit energy in a variety of ways. 413 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 414 8-6 Inquiry Lab 8-6 Inquiry Lab Interference of Waves Questions 1 What happens when two pulses pass through the same point in a medium? 2 How can two waves, moving in opposite directions, exist simultaneously in the same space? 3 What causes a standing wave? Materials light spring heavy spring masking tape stopwatch tape measure or metre-stick CAUTION: A stretched spring stores considerable amounts of elastic energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. Allow the spring to gently unwind as you are walking to prevent the spring from tying itself into a knot. Variables Part 1: In this part you are concerned with the amplitudes and lengths of the pulses. Observe these variables before, during, and after the period in which they interfere with each other. Part 2: In this part you will explore the relationship between the frequency and the standing wave pattern generated in a spring. From the structure of the standing wave
pattern and the length of the spring, the wavelength and the speed of the standing wave are easily calculated. For both parts, identify which are the controlled variables, manipulated variables, and responding variables. Part 1: Superposition and Interference of Pulses 1 (a) Place two parallel strips of tape on the floor about 5 m apart. Measure and record the distance between them. Use these tapes to maintain a constant length for the spring while it is stretched. Attach a third strip of tape about 5 cm long to one of the coils near the middle of the spring as a marker. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork (b) Have the team member holding one end of the spring generate a transverse pulse. When this pulse reaches the fixed end of the spring, have the same team member generate a second similar pulse. Try to generate the second incident pulse so that it meets the reflection of the first pulse at the strip of tape near the middle of the spring. Focus on the nature of the spring’s motion while the pulses interact. This complex interaction occurs quite quickly and may need to be repeated a few times until you are confident that you can see what is happening. Discuss the observations with your team members. (c) Record your observations in sketches and writing. 2 (a) Again, have one team member generate two pulses. This time, however, generate the second pulse so that it is on the opposite side of the spring (i.e., inverted) to the first pulse. The second pulse will now be on the same side of the spring as the reflected pulse. Again, time the pulses so that they meet near the centre of the spring. (b) Observe how these pulses interact when they meet at the centre of the spring. Discuss what you think is happening with the other members of your team. Analysis 1. When pulses on opposite sides of the spring meet, does the amplitude increase or decrease in the region of overlap? 2. When pulses on the same side of the spring meet, does the amplitude increase or decrease in the overlap region? Part 2: Standing Waves Procedure 1 (a) Have a team member at one end of the spring create a double wave (a series of four pulses) by oscillating his or her hand back and forth twice across the spring’s equilibrium position. (b) Observe what happens as this wave travels back and forth along the spring. Pay particular attention to what happens when the reflected
portion of the wave is passing through the incident wave. Discuss your observations with your lab team to come to a consensus on what is occurring. 414 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 415 (c) Record your observations. Keep in mind what Analysis you observed when the pulses crossed in Part 1 of the lab. 2 (a) Now create a steady wave train by moving your hand back and forth. Try to find the frequency such that the spring oscillates in two segments about its midpoint. If, at first, there are more than two segments, then reduce the frequency slightly. If, at first, the spring is oscillating as only one segment, then increase the frequency until the second segment appears. Once the spring begins to oscillate as two segments, maintain that frequency. (b) Measure and record the frequency of oscillation by timing ten oscillations. Since you know the length of the spring (the distance between the tapes you placed on the floor), record the length of a wave for this frequency. Each half of the spring is a pulse so that, in this mode, the wavelength is equal to the length of the spring. (c) Record the data obtained in step 2(b) in a table. Use column headings: trial number, number of segments, frequency, wavelength, and speed. 3 (a) Begin with the frequency at which the spring oscillates in two halves and gradually increase the frequency. (b) Describe what happens when you try to maintain a slight increase in the frequency. Keep increasing the frequency until a new oscillation pattern is established. Measure the frequency for this pattern. Record your results in your table of data. (c) Starting from this frequency, gradually increase the frequency until a new pattern of oscillation is found. Once the new wave pattern is established, measure its frequency and record your measurements in your data table. 1. When you created a sustained wave so that the spring oscillated as a stable pattern, in which direction did the waves move? Why do you think that is the case? Does this tell you why this pattern is known as a standing wave? 2. (a) Two segments of a standing wave are equal to one wavelength. For each trial recorded in the table of data, calculate the wavelength of the standing wave. (b) For each trial, use the universal wave equation to calculate the speed of the waves in the spring. 3.
To what does the speed of a standing wave refer? 4. Express the frequencies, for the different trials recorded in your data table, as ratios using simple whole numbers. Compare these ratios to the number of segments in which the spring oscillates for each trial. NOTE: The parts of a standing wave that remain motionless are called nodes or nodal points. The midpoints of the parts that oscillate back and forth are called antinodes. Each segment that contains an antinode is simply a pulse or one-half a wavelength. In a standing wave two adjacent segments are required to complete one wavelength. 5. Once a standing wave is established in the spring, what do you notice about the amplitude of the oscillations you use to sustain the wave compared with the amplitude of the antinodes? What explanation might exist for the difference in these two amplitudes? 6. Beginning at the fixed end of the spring, describe the locations of the nodes (points that remain motionless) and antinodes (midpoints of the parts that oscillate back and forth) along the spring in terms of wavelength. 7. How does the principle of superposition explain what must be happening at the antinodes of a standing wave? 4 If time permits, change to the heavier spring and repeat steps 2 and 3. 8. What relationship exists between the wavelength of a standing wave and the frequency creating the wavelength? CAUTION: Be very careful not to accidentally release the heavy spring while it is stretched. It will contain a large quantity of elastic potential energy and may seriously injure someone. To relax the tension in the spring, walk one end of the spring slowly toward the other end. 9. Go back to your observations in Part 1 of 8-2 Inquiry Lab. When a train of straight waves parallel to the barrier was reflected back through the incident wave train, did you observe a standing wave? 10. If you could generate a standing wave for sound, what do you think would be the nature of the sound at the location of an antinode? a node? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 415 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 416 M I N D S O N Total Destruction? At the instant when two pulses “completely destroy” each other, the spring is in its equilibrium position
. How is it possible for the two pulses to reappear as if from nothing? Where does the energy in the pulses go when the sum of the amplitudes is zero? Hint: It might help to think of the spring in terms of a system. Standing Waves and Resonance When two wave trains with identical wavelengths and amplitudes move through each other (Figure 8.28), the resulting interference pattern can be explained by using the principle of superposition. When crests from the two waves or troughs from the two waves occupy the same point in the medium, the waves are in phase. Waves that are in phase produce constructive interference. When a crest from one wave occupies the same point in the medium as a trough from a second wave, we say that these waves are out of phase. Out-of-phase waves produce destructive interference. As the two wave trains pass through each other in opposite directions, they continually shift in and out of phase to produce a wave that seems to oscillate between fixed nodes, rather than move through the medium. wave A wave B wave A B vA vB A λ1 a) (b) (c) (d) (e) (f) actual position of medium at area where waves overlap constructive interference points at which only destructive interference occurs e MATH To graphically analyze the superposition of waves that are in or out of phase, visit www.pearsoned.ca/school/ physicssource. e WEB Find out more about the superposition of waves. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.28 The diagrams show how waves travelling in opposite directions interfere as they move through each other. 416 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 417 • (a) Point A is the initial point of contact between the two wave trains shown in blue and purple. The crest from the purple wave train and the trough from the blue wave train arrive at point A at the same instant. • (b) The two identical waves have moved a distance of 1 λ in opposite 4 directions. This overlap results in destructive interference and the spring is flat in the region of overlap. The position of the spring where the two waves overlap, the resultant, is shown in red. • (c) Each wave has moved a further 1 λ along the spring. Now the 4 waves are exactly in phase and constructive interference
occurs. The regions to the left and right of point A show a crest and a trough, respectively, with displacement of the resultant being twice that of the blue or purple waves. • Every time the wave trains move a further 1 λ along the spring, the 4 interference changes from constructive to destructive and vice versa. At point A, only destructive interference occurs. The magnitudes of the displacements of the waves arriving at point A are always equal but opposite in sign. As the waves continue to move in opposite directions, the nature of the interference continually changes. However, at point A and every 1 λ from point A, there are points at which only destructive 2 interference occurs. These are called nodal points or nodes. Between the nodes, the spring goes into a flip-flop motion as the interference in these areas switches from constructive (crest crossing crest) to destructive (crest crossing trough) and back to constructive interference (trough crossing trough). The midpoints of these regions on the spring are called antinodes. The first antinode occurs at a distance λ on either side of A, and then at every 1 of 1 λ after that point. Because 2 4 the wave seems to oscillate around stationary nodes along the spring, it is known as a standing wave. Standing waves are also seen in nature; an example is shown in Figure 8.29. Standing Waves in a Fixed Spring When you generate a wave train in a spring that is fixed at one end, the reflected wave train must pass back through the incident wave train. These two wave trains have identical wavelengths and nearly identical amplitudes. For incident and reflected wave trains, the initial point of contact is by definition the fixed point at which reflection occurs. This means that the endpoint of the spring is always a nodal point and, as shown in Figure 8.30, nodes occur every 1 λ from that point with antinodes 2 between them. λ1 2 λ1 2 λ1 2 λ1 2 λ1 2 λ1 2 node: a point on a spring or other medium at which only destructive interference occurs; a point that never vibrates between maximum positive amplitude and maximum negative amplitude; in a standing wave nodes occur at intervals of 1 λ 2 antinode: a point in an interference pattern that oscillates with maximum amplitude; in a standing wave antinodes occur at intervals of 1 λ 2 standing wave: a condition in a spring or other medium in which a wave seems
to oscillate around stationary points called nodes. The wavelength of a standing wave is the distance between alternate nodes or alternate antinodes. Figure 8.29 Standing waves occur in nature. This photograph shows a standing wave in a stream crossing a sandy beach in Scotland. e WEB Find out about the details of a standing wave in a spring. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.30 In a spring with a fixed end, a standing wave must contain a whole number of antinodes. Nodes occur every half-wavelength from the ends. Antinodes oscillate between shown positions. Nodes remain motionless. antinode Chapter 8 Mechanical waves transmit energy in a variety of ways. 417 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 418 e WEB To learn more about the Tacoma Narrows Bridge collapse, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.31 Resonance, caused by wind, set up a standing wave that destroyed the Tacoma Narrows Bridge. resonance: an increase in the amplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave e WEB To learn more about the giant shock absorbers added to the Millennium Bridge, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.32 The tone produced when you blow across the top of an open bottle depends on the length of the air column. Resonant Frequencies When a standing wave is present in a spring, the wave reflects from both ends of the spring. There must be a nodal point at both ends with an integral number of antinodes in between. The spring “prefers” to oscillate at those frequencies that will produce a standing wave pattern, called the resonant frequencies for the spring. When the generator is oscillating at a resonant frequency, the energy is added to the spring in phase with existing oscillations. This reinforces and enhances the standing wave pattern. The added energy works to construct waves with ever-larger amplitudes. If the generator is not oscillating at a resonant frequency of the medium, the oscillations tend to destroy the standing wave motion. Amplitude and Resonance Perhaps the most impressive display of a standing wave occurred when resonance set up a standing wave in the bridge across the Tacoma Narrows in the state
of Washington (Figure 8.31). Opened in November 1940, the bridge was in operation only a few months before resonance ripped it apart. More recently, in June 2000, the newly opened Millennium Bridge in London had to be closed for modifications when the footsteps of pedestrians set up resonance patterns. Anyone who has ever “pumped up” a swing has used the principle of resonance. To increase the amplitude of its motion, the swing must be given a series of nudges in phase with its natural frequency of motion. Each time the swing begins to move forward, you give it a little push. Since these little pushes are produced in resonance with the swing’s natural motion, they are added to its energy and the amplitude increases. If you pushed out of phase with its natural motion, the swing would soon come to rest. Concept Check Why does it take so little energy to sustain a standing wave in a spring? Concept Check Resonating Air Columns All wind instruments use the principle of resonance to produce music. The simplest example of resonance in music is the note produced when you blow over the top of a bottle (Figure 8.32). Blowing across the top of the bottle oscillates the air in the bottle and generates a standing wave. This standing wave is like the waves travelling in a spring, but unlike a spring that is fixed at both ends, the air column is fixed only at the end where reflection occurs and is free to oscillate at the open end. The resonant frequency of the note produced depends on the length of the air column because, to resonate, the standing wave must have a node at the closed end of the bottle and an antinode at the open end (Figure 8.33). 418 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 419 Closed-Pipe or Closed-Tube Resonance NR R resonance heard NR no resonance heard When a wave source is held at the open end of a pipe, it sends down a wave that reflects from the closed end of the pipe and establishes a standing wave pattern. The sound one hears depends on the length of the air column in the pipe relative to the length of the standing wave. If an antinode occurs at the open end of the pipe (Figure 8.33 (a) and (c)), a point of resonance (resulting from constructive interference) occurs at the open end of the pipe and the sound appears to be amplified. This phenomenon
is known as closed-pipe or closed-tube resonance. However, if the open end of the pipe coincides with the position of a node (destructive interference), then almost no sound can be heard because the source (tuning fork) and the standing wave are out of phase (Figure 8.33 (b) and (d)). (b) (a) R NR R (c) (d) Figure 8.33 Resonance series. A tuning fork sets up a standing wave in the air column. The volume of the sound one hears will vary depending on whether there is an antinode, (a) and (c), or a node, (b) and (d), at the end of the pipe. Nodes and Antinodes in Closed-Pipe Resonance In the air column, nodes are located every half-wavelength from the end at which the wave is reflected, just as they are in a standing wave in a spring. If the pipe length is equal to any multiple of 1 λ, there will 2 be a node at the upper end of the pipe, and destructive interference λ, 4 λ, 3 λ, 2 will occur. Thus, when the air column is 1 λ, … in length, 2 2 2 2 little or no sound will be heard. Antinodes in the air column are located one quarter-wavelength from the end of the pipe where reflection occurs, and then every halfwavelength from that point. Thus, resonance is heard when the pipe is λ, 5 λ, 3 1 λ, … long. When resonance is heard for an air column closed 4 4 4 at one end, we know that the open end of the column coincides with the location of one of the antinodes. This information can be used to measure the wavelength of sound in gases. If the frequency of the sound is known, then the wavelength can be used to calculate the speed of sound in the gas. Concept Check Is the volume of a sound related to speed, wavelength, amplitude, or frequency of the wave? What evidence is there to support your answer? Example 8.3 A tuning fork with a frequency of 384 Hz is held above an air column. As the column is lengthened, a closed-pipe resonant point is found when the length of the air column is 67.5 cm. What are possible wavelengths for this data? If the speed of sound is known to be slightly greater than 300 m/s, what
is (a) the actual wavelength, and (b) the actual speed of sound? Chapter 8 Mechanical waves transmit energy in a variety of ways. 419 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 420 Practice Problems 1. A tuning fork of frequency 512 Hz is used to generate a standing wave pattern in a closed pipe, 0.850 m long. A strong resonant note is heard indicating that an antinode is located at the open end of the pipe. (a) What are the possible wavelengths for this note? (b) Which wavelength will give the most reasonable value for the calculation of the speed of sound in air? 2. A tuning fork with a frequency of 256 Hz is held above a closed air column while the column is gradually increased in length. At what lengths for this air column would the first 4 resonant points be found, if the speed of sound is 330 m/s? 3. A standing wave is generated in a spring that is stretched to a length of 6.00 m. The standing wave pattern consists of three antinodes. If the frequency used to generate this wave is 2.50 Hz, what is the speed of the wave in the spring? 4. When a spring is stretched to a length of 8.00 m, the speed of waves in the spring is 5.00 m/s. The simplest standing wave pattern for this spring is that of a single antinode between two nodes at opposite ends of the spring. (a) What is the frequency that produces this standing wave? (b) What is the next higher frequency for which a standing wave exists in this spring? Answers 1. (a) 3.40 m @ 1.74 103 m/s; 1.13 m @ 580 m/s; 0.680 m @ 348 m/s; 0.486 m @ 249 m/s (b) 0.680 m 2. 0.322 m, 0.967 m, 1.61 m, 2.26 m 3. 10.0 m/s 4. 0.313 Hz, 0.625 Hz Given f 384 Hz l 67.5 cm 0.675 m Required wavelength and speed of sound Analysis and Solution λ, 5 λ, 3 The resonant point might represent 1 λ,…, etc., for 4 4 4 this tuning fork. Assume that 67.5 cm is the first resonant point; that means 67.5 cm is 1
λ. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 1 λ. Therefore, 4 λ 4l v fλ 4(0.675 m) 2.70 m (384 Hz)(2.70 m) 1037 m/s 1.04 103 m/s This value is larger than the speed of sound in air. If the speed of sound is not of the proper order of magnitude, then assume that the resonant point is the second point of resonance and that 67.5 cm is 3 λ. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 3 λ. Therefore, 4 l λ 4 3 75 m) (384 Hz)(0.900 m) 4(0.6 3 v fλ 0.900 m 345.6 m/s 346 m/s This is a reasonable speed for sound in air. Complete the analysis by assuming that l 5 λ. Therefore, 4 l λ 4 5 75 m) (384 Hz)(0.540 m) 4(0.6 5 v fλ 0.540 m 207.4 m/s 207 m/s This value is less than the speed of sound in air. Paraphrase and Verify The calculations for the speed of sound indicate that the data must have been for the second point of resonance. This assumption gives the speed for sound of 346 m/s. The assumption that the pipe length is for the first resonant point results in a speed about three times that of sound. The assumption that the pipe length is for the third resonant point produces a speed less than 300 m/s. 420 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 421 8-7 Inquiry Lab 8-7 Inquiry Lab Measuring the Speed of Sound Using Closed-pipe Resonance Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork When a sound wave travels down a closed pipe, the incident wave reflects off the end of the pipe and back toward the source. The interaction of the incident and reflected waves sets up an interference pattern inside the pipe, known as a standing wave. This standing wave can be used to determine the wavelength of the sound. Problem What is the speed of sound in air? Variables The universal wave equation relates the speed (v) of a wave to its frequency (f )
and wavelength (). The wavelength is determined from the length of the pipe (l ) and the number of the resonant point as counted from the reflecting surface. Materials and Equipment tuning forks and tuning fork hammer or an audio frequency generator glass or plastic pipe tall cylinder Procedure 1 Assemble the apparatus as shown in Figure 8.34. metrestick tuning fork open-ended pipe tall cylinder water Figure 8.34 2 Place the pipe in the water-filled cylinder so that the column of air in the pipe is quite short. 3 Strike the tuning fork with the hammer. 4 Hold the tuning fork as shown over the end of the pipe and lift the pipe slowly so that the length of the column of air in the pipe increases. 5 As you approach a point where the volume of the sound increases, move the pipe slowly up and down to find the point where the resonance is greatest. Strike the tuning fork as often as necessary to maintain the sound source. 6 Determine the length of the column of air that gives the greatest resonance and record it in a table like Table 8.1. Measure the length from the surface of the water to the location of the tuning fork. 7 Beginning with the column of air at the previously recorded length, gradually increase the length until you have determined the length of the air column that gives the next point of resonance. Record this length. 8 Repeat step 7 to find the length of the column for the third resonant point. Table 8.1 Column Length and Resonance Frequency f (Hz) Length of column at first resonant point l1 (m) Length of column at second resonant point l2 (m) Length of column at third resonant point l3 (m) Analysis 1. In terms of wavelength, how far is each of the first three resonant points from the reflecting surface of the water at the bottom of the air column? 2. Calculate the wavelength of the sound from the tuning fork for each resonant point. Record your answers in a table similar to Table 8.2. Calculate the speed of sound for each of the wavelengths. 3. When you calculate the wavelength for different resonant points, do the answers agree? If not, what might cause the differences? 4. Why should you start with a short column of air and increase its length if you are to be sure that you have correctly determined the wavelength? Chapter 8 Mechanical waves transmit energy in a variety of ways. 421 08-PearsonPhys20-Chap08 7/24/08 2:23 PM
Page 422 ▼ Table 8.2 Resonant Points, Wavelength, and Speed of Sound First Resonant Point Second Resonant Point Third Resonant Point Frequency f (Hz) Wavelength 4l (m) Speed v (m/s) Wavelength 4l/3 (m) Speed v (m/s) Wavelength 4l/5 (m) Speed v (m/s) 5. Why should you measure the length of the column from the reflecting surface to the tuning fork rather than to the top end of the pipe? 6. What is the speed of sound at room temperature? 7. Investigate the effect that air temperature has on the speed of sound. Use hot water or ice water to modify the temperature of the air in the column. Compare the measured speed of sound for at least three temperatures. Suspend a thermometer down the pipe to determine the temperature of the air in the pipe. Plot a graph of the measured speed of sound versus the temperature. Does the graph suggest a linear relationship? Can you use the graph to predict the speed of sound at other temperatures? 8. An alternative technique to determine the speed of sound is to measure the time for an echo to return to you. Stand a measured distance from a wall or other surface that reflects sound. Create an echo by striking together two hard objects, such as metal bars, or, perhaps beating a drum. Listen for the echo. Once you have established an approximate time for the echo to return, strike the bars in a rhythm so that they are in phase with the echo. Have a team member count the number of beats in 1 min. The period of this frequency is the time required for the sound to travel to the wall and back. Use that data to calculate the speed of the sound. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. fundamental frequency: the lowest frequency produced by a particular instrument; corresponds to the standing wave having a single antinode, with a node at each end of the string Music and Resonance Complex modes of vibration give instruments their distinctive sounds and add depth to the musical tones they create. A string of a musical instrument is simply a tightly stretched spring for which the simplest standing wave possible is a single antinode with a node at either end. For this pattern, the length of the string equals one-half a wavelength and the frequency produced is called the fundamental frequency (Figure 8.35(a)). fundamental frequency equilibrium position
info BIT Assume that the fundamental frequency is f. In physics and in music, the frequency 2f is called the first overtone; 3f is the second overtone, and so on. These frequencies are said to form a harmonic series. Thus, physicists may also refer to the fundamental frequency (f ) as the first harmonic, the frequency 2f as the second harmonic, the frequency 3f as the third harmonic, and so on. Figure 8.35 (a) standing wave with an antinode at the centre of the string. The fundamental frequency of a vibrating string oscillates as a This is the lowest frequency produced by a particular instrument. But other standing wave patterns can exist in the string at the same time as it oscillates at its fundamental frequency. By plucking or bowing a string nearer its end than its middle, the string is encouraged to vibrate with multiple frequencies. The frequencies above the fundamental 422 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 423 frequency that may exist simultaneously with the fundamental frequency are called overtones. Figures 8.35 (b) and (c) show the shape of a string vibrating in its first and second overtones, respectively. Figure 8.36 shows a violinist bowing and fingering the strings of her violin to produce notes. 1st overtone without the fundamental frequency equilibrium position fundamental frequency 1st overtone with the fundamental frequency equilibrium position Figure 8.36 The violinist’s fingering technique changes the length of the string and thus changes the fundamental frequency of vibration. overtone: any frequency of vibration of a string that may exist simultaneously with the fundamental frequency The first overtone has the form of a standing wave with two Figure 8.35 (b) antinodes. A node exists at the midpoint of the string. The lower portion of the diagram shows a string vibrating with both the fundamental frequency and the first overtone simultaneously. 2nd overtone without the fundamental frequency equilibrium position fundamental frequency 2nd overtone with the fundamental frequency equilibrium position Figure 8.35(c) The lower portion of the diagram shows a string vibrating with both the fundamental frequency and second overtone. The vibration that produces the second overtone has three antinodes. The actual form of a vibrating string can be very complex as many overtones can exist simultaneously with the fundamental frequency. The actual wave form for a vibrating string is the result

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