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part of mechanics.
The harmonic oscillator, which we are about to sbudy, has close analogs in
many other fields; although we star with a mechanical example oŸ a weight on a
spring, or a pendulum with a small swing, or certain other mechanical devices, we
are really studying a certain điƒƒeremtial cquation. This equation appears again
and again in physics and in other sciences, and in fact it is a part oŸ so many
phenomena that its close study is well worth our while. Some of the phenomena
involving this equation are the oscillations of a mass on a spring; the oscillations
of charge Ñowing back and forth in an electrical circuit; the vibrations oŸ a tuning
fork which is generating sound waves; the analogous vibrations of the electrons
in an atom, which generate light waves; the equations for the operation of a
servosystem, such as a thermostat trying to adjust a temperature; complicated
interactions in chemical reactions; the growth of a colony of bacteria in interaction
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with the food supply and the poisons the bacteria produece; foxes eating rabbits
eating grass, and so on; all these phenomena follow equations which are very
similar to one another, and this is the reason why we study the mechanical
oscillator in such detail. 'Phe equations are called znear djfƒerential cquations
tuïth constant coefficien#s. A linear diferential equation with constant coefficients
1s a diferential equation consisting of a sum of several terms, each term beïng a
derivative of the dependent variable with respect to the independent variable,
and multiplied by some constant. Thus
dụ đa (đP + aụ— d0 1a/dÉfCT + ccc + ai de dt + ag# = ƒ) — ð)
1s called a linear diferential equation of order ø with constant coefficients (each
đ¿ 1s constant).
21-2 The harmonic oscillator
Perhaps the simplest mechanical system whose motion follows a linear difer-
ential equation with constant coeflicients is a mass on a spring: frst the spring
stretches to balance the gravity; once it is balanced, we then discuss the vertical
displacement of the mass from its equilibrium position (Fig. 21-1). We shall
call this upward displacement z, and we shall also suppose that the spring 1s
perfectly linear, in which case the force pulling back when the spring is stretched
1s precisely proportional to the amount of stretch. "hat is, the force is —kø
(with a minus sign to remind us that it pulls back). Thus the mass times the
acceleration must equal —kz:
md°+/dt2 = —ka. (21.2)
L9
Fig. 21-1. A mass on a spring: a simple example of a harmonic
oscillator.
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Eor simplicity, suppose it happens (or we change our unit of time measurement)
that the ratio k/n = 1. We shall fñrst study the equation
d°+/di? = —z. (21.3)
Later we shall come back to Bq. (21.2) with the & and rn explicitly present.
We have already analyzed Eq. (21.3) in detail numerically; when we first
introduced the subject of mechanics we solved this equation (see Eq. 9.12) to
fnd the motion. By numerical integration we found a curve (Eig. 9-4) which
showed that 1Í rm was initially displaced, but at rest, it would come down and go
through zero; we did not then follow it any farther, but of course we know that
1t just keepbs going up and down——It osc/lÏates. When we calculated the motion
numerically, we found that it went through the equilibrium poïnt at ý = 1.570.
The length of the whole cycle is four times this long, or #o = 6.28 “sec.” This
was found numerically, before we knew much calculus. We assume that in the
meantime the Mathematics Department has brought forth a function which,
when differentiated twice, is equal to itself with a minus sign. (There are, oŸ
course, ways of getting at this function in a direct fashion, but they are more
complicated than already knowing what the answer is.) The function is # = cosứ.
Tf we differentiate this we fnd đz/đt = — sin£ and d”z/đt? = — cost = —z. The
function # = cos£ starts, at ứ —= 0, with z = 1, and no initial velocity; that was
the situation with which we started when we did our numerical work. Now that
we know that # = cosứ, we can calculate a prec¿se value for the time at which it
should pass z = 0. The answer is ý = Z/2, or 1.57080. We were wrong in the lasb
figure because of the errors of numerical analysis, but it was very closel
Now to go further with the original problem, we restore the time units to real
seconds. What is the solution then? First ofall, we might think that we can get the
constants & and ?m in by multiplying cos ý by something. So let us try the equation
œ = Acosf; then we fnd dz/dt = — Asinf, and đ?z/d‡2 = —Acost = —z. Thus
we discover to our horror that we diỉd not succeed in solving Eq. (21.2), but we
gọt Eq. (21.3) again! That fact illustrates one of the most important properties
of linear diferential equations: ?ƒ e rmuliipl a solulion oƒ the equalion DỤ ang
constant, ft ís again œ solulion. The mathematical reason for this is clear. IÍ ø is
a solution, and we multiply both sides of the equation, say by 4, we see that all
derivatives are also multiplied by 4, and therefore 4z is just as good a solution
of the original equation as ø was. The physics of it is the following. If we have a
weight on a spring, and pull it down twice as far, the force is Ewice as much, the
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resulting acceleration is twice as great, the velocity it acquires in a given tỉme is
twice as preat, the distance covered in a given time is twice as great; but it has
to cover ÿwice as great a distanee in order to get back to the origin because 1 1s
pulled down twice as far. So 1% takes the sdrne tữne to get back to the origin,
irrespective of the initial displacement. In other words, with a linear equation,
the motion has the same f#ữne paitern, no matter how “strong” it is.
That was the wrong thing to do—it only taught us that we can multiply
the solution by anything, and it satisfes the same equation, but not a diferent
cquation. After a little cut and try to get to an equation with a diferent constant
multiplying z, we fnd that we must alter the scale of fzme. In other words,
Eq. (21.2) has a solution oŸ the form
% = COSUgÝ. (21.4)
(It is important to realize that in the present case, œo is not an angular velocity
of a spinning body, but we run out of letters if we are not allowed to use the same
letter for more than one thing.) The reason we put a subscript “0” on œ is that we
are going to have more omegas before long; let us remermber that œọ refers to the
natural motion of this oscillator. Now we try Eq. (21.4) and this time we are more
successful, because đ#/đf = —øg sin œo£ and d2z/đt? = —u cosuoŸ = —u§z. 8o
at last we have solved the equation that we really wanted to solve. 'The equation
d3z/dt? = —u§z is the same as Eq. (21.2) IŸ ø8 = k/m.
The next thing we must investigate is the physical signifcance oŸ œạ. We
know that the cosine function repeats itself when the angle it refers to is 2. So
% = cosug# will repeat its motion, ¡% will go through a complete cycle, when the
“angle” changes by 2z. The quantity œg is often called the phøse of the motion.