This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 21 The Harmonic 0scillut0r 21—1 Linear differential equations In the study of physics, usually the course is divided into a series of subjects,
such as mechanics, electricity, optics, etc., and one studies one subject after the
other. For example, this course has so far dealt mostly with mechanics. But a
strange thing occurs again and again: the equations which appear in different
ﬁelds of physics, and even in other sciences, are often almost exactly the same, so
that many phenomena have analogs in these different ﬁelds. To take the simplest
example, the propagation of sound waves is in many ways analogous to the propaga
tion of light waves. If we study acoustics in great detail we discover that much of
the work is the same as it would be if we were studying optics in great detail. So
the study of a phenomenon in one ﬁeld may permit an extension of our knowledge
in another field. It is best to realize from the ﬁrst that such extensions are possible,
for otherwise one might not understand the reason for spending a great deal of
time and energy on what appears to be only a small part of mechanics. The harmonic oscillator, which we are about to study, has close analogs in
many other ﬁelds; although we start with a mechanical example of a weight on a
spring, or a pendulum with a small swing, or certain other mechanical devices, we
are really studying a certain diﬂerential equation. This equation appears again
and again in physics and in other sciences, and in fact it is a part of 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 ﬂowing back and forth in an electrical circuit; the vibrations of 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
with the food supply and the poisons the bacteria produce; 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 whyxwe study the mechanical oscillator
in such detail. The equations are called linear diﬂerential equations with constant
coeﬁ‘icients. A linear diﬁ‘erential equation with constant coefficients is a differential
equation consisting of a sum of several terms, each term being a derivative of the
dependent variable with respect to the independent variable, and multiplied by
some constant. Thus an dnx/dt" + an_1aln—lx/dt”_1 + + a1 dx/dt + aox = f(t) (21.1) is called a linear differential equation of order n with constant coefﬁcients (each
a, is constant). 21—2 The harmonic oscillator Perhaps the simplest mechanical system whose motion follows a linear differ
ential equation with constant coefﬁcients is a mass on a spring: ﬁrst 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 x, and we shall also suppose that the spring is perfectly
linear, in which case the force pulling back when the spring is stretched is pre
cisely proportional to the amount of stretch. That is, the force is ——kx (with a 21—1 211 Linear differential equations
21—2 The harmonic oscillator 21—3 Harmonic motion and circular
motion 21—4 Initial conditions 21—5 Forced oscillations Fig. 21—1. A mass on a spring: a
simple example of a harmonic oscillator. minus sign to remind us that it pulls back). Thus the mass times the acceleration must equal ——kx:
mtzin/dt2 = —kx. (21.2) For simplicity, suppose it happens (or we change our unit of time measurement)
that the ratio k/m = 1. We shall ﬁrst study the equation d2x/dt2 = —x. (21.3) Later we shall come back to Eq. (21.2) with the k and m 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 ﬁnd
the motion. By numerical integration we found a curve (Fig. 9—4) which showed
that if m 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 it just
keeps going up and down—it oscillates. When we calculated the motion numer~
ically, we found that it went through the equilibrium point at t = 1.570. The
length of the whole cycle is four times this long, or to = 6.28 “see.” 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 differ
entiated twice, is equal to itself with a minus sign. (There are, of 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 x = cos I. If we differentiate
this we ﬁnd dx/dt = —sint and d2x/dt2 = —cost = —x. The function x =
cos I starts, at t = 0, with x = l, and no initial velocity; that was the situation
with which we started when we did our numerical work. Now that we know that
x cos t, we can calculate a precise value for the time at which it should pass
x 0. The answer is t = 1r/2, or 1.57108. We were wrong in the last figure
because of the errors of numerical analysis, but it was very close! Now to go further with the original problem, we restore the time units to
real seconds. What is the solution then? First of all, we might think that we can
get the constants k and m in by multiplying cos t by something. So let us try the
equation x = A cos I; then we ﬁnd dx/dt = —A sin I, and ale/a’t2 = —A
cos t = —x. Thus we discover to our horror that we did not succeed in solving
Eq. (21.2), but we got Eq. (21.3) again! That fact illustrates one of the most
important properties of linear differential equations: if we multiply a solution of
the equation by any constant, it is again a solution. The mathematical reason for
this is clear. If x is a solution, and we multiply both sides of the equation, say by A,
we see that all derivatives are also multiplied by A, and therefore Ax is just as good
a solution of the original equation as x 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 twice
as much, the resulting acceleration is twice as great, the velocity it acquires in a
given time is twice as great, the distance covered in a given time is twice as great;
but it has to cover twice as great a distance in order to get back to the origin
because it is pulled down twice as far. So it takes the same time to get back to the
origin, irrespective of the initial displacement. In other words, with a linear equa
tion, the motion has the same time pattern, 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 satisﬁes the same equation, but not a different equation.
After a little cut and try to get to an equation with a different constant multiplying
x, we ﬁnd that we must alter the scale of time. In other words, Eq. (21.2) has a
solution of the form H II x = cos wot. (21.4) (It is important to realize that in the present case, (.00 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 w is that we are
going to have more omegas before long; let us remember that (do refers to the
natural motion of this oscillator. Now we try Eq. (21.4) and this time we are more
successful, because dx/dt = —o.>0 sin wot and d2X/d12 = —w§ cos wot = —w(2)x. 21—2 So at last we have solved the equation that we really wanted to solve. The equation
de/dt2 = —ng is the same as Eq. (21.2) if co?) = k/m. The next thing we must investigate is the physical signiﬁcance of we. We
know that the cosine function repeats itself when the angle it refers to is 27r. So
x = cos wot will repeat its motion, it will go through a complete cycle, when the
“angle” changes by 27r. The quantity wot is often called the phase of the motion.
In order to change wot by 21r, the time must change by an amount t5, called the
period of one complete oscillation; of course to must be such that woto = 211'.
That is, woto must account for one cycle of the angle, and then everything will
repeat itself—if we increase t by to, we add 21r to the phase. Thus to = 21r/w0 = 27r\/ m/k. (21.5) Thus if we had a heavier mass, it would take longer to oscillate back and forth on
a spring. That is because it has more inertia, and so, while the forces are the same,
it takes longer to get the mass moving. Or, if the spring is stronger, it will move
more quickly, and that is right: the period is less if the spring is stronger. Note that the period of oscillation of a mass on a spring does not depend in
any way on how it has been started, how far down we pull it. The period is deter
mined, but the amplitude of the oscillation is not determined by the equation of
motion (21.2). The amplitude is determined, in fact, by how we let go of it, by
what we call the initial conditions or starting conditions. Actually, we have not quite found the most general possible solution of Eq.
(21.2). There are other solutions. It should be clear why: because all of the cases
covered by x = (1 cos wot 'start with an initial displacement and no initial velocity.
But it is possible, for instance, for the mass to start at x = O, and we may then
give it an impulsive kick, so that it has some speed at t = 0. Such a motion is not
represented by a cosine—it is represented by a sine. To put it another way, if
x = cos wot is a solution, then is it not obvious that if we were to happen to walk
into the room at some time (which we would call “t = 0”) and saw the mass as
it was passing x = 0, it would keep on going just the same? Therefore, x = cos
wot cannot be the most general solution; it must be possible to shift the beginning
of time, so to speak. As an example, we could write the solution this way: x =
a cos w0(t — t1), where t1 is some constant. This also corresponds to shifting the
origin of time to some new instant. Furthermore, we may expand cos (wot + A) = cos wotcosA — sin wot sin A, and write
x = A cos wot + Bsin wot, where A = a cos A and B = —a sin A. Any one of these forms is a possible
way to write the complete, general solution of (21.2): that is, every solution of the
differential equation d2x/dt2 = —w(2,x that exists in the world can be written as (a) x = acos w0(t — t1),
or (b) x = acos (wot + A), (21.6)
or (c) x = A cos wot + Bsin wot. Some of the quantities in (21.6) have names: wo is called the angular frequency;
it is the number of radians by which the phase changes in a second. That is deter
mined by the differential equation. The other constants are not determined by the
equation, but by how the motion is started. Of these constants, a measures the
maximum displacement attained by the mass, and is called the amplitude of oscilla
tion. The constant A is sometimes called the phase of the oscillation, but that is a
confusion, because other people call wot + A the phase, and say the phase changes
with time. We might say that A is a phase shift from some deﬁned zero. Let us put
it differently. Different A’s correspond to motions in different phases. That is
true, but whether we want to call A the phase, or not, is another question. 21—3 A Fig. 212. A particle moving in a
circular path at constant speed. Fig. 21—3. Demonstration of the
equivalence between simple harmonic
motion and uniform circular motion. 21—3 Harmonic motion and circular motion The fact that cosines are involved in the solution of Eq. (21.2) suggests that
there might be some relationship to circles. This is artiﬁcial, of course, because
there is no circle actually involved in the linear motionit just goes up and down.
We may point out that we have, in fact, already solved that differential equation
when we were studying the mechanics of circular motion. If a particle moves in a
circle with a constant speed 12, the radius vector from the center of the circle to the
particle turns through an angle whose size is proportional to the time. If we call
this angle 0 = vt/R (Fig. 21—2) then d6/dt = (.00 = v/R. We know that there is
an acceleration a = v2/R = 003R toward the center. Now we also know that the
position x, at a given moment, is the radius of the circle times cos 0, and that y is
the radius times sin 9: x=Rcos0, y=Rsin0. Now what about the acceleration? What is the x~component of acceleration,
d2x/d22? We have already worked that out geometrically; it is the magnitude of
the acceleration times the cosine of the projection angle, with a minus sign because
it is toward the center. ax = —acos 0 = —w2R cos0 = —w2x. (21.7) In other words, when a particle is moving in a circle, the horizontal component of
its motion has an acceleration which is proportional to the horizontal displacement
from the center. Of course we also have the solution for motion in a circle:
x = R cos wot. Equation (21.7) does not depend upon the radius of the circle,
so for a circle of any radius, one ﬁnds the same equation for a given we. Thus,
for several reasons, we expect that the displacement of a mass on a spring will turn
out to be proportional to cos mot, and will, in fact, be exactly the same motion as
we would see if we looked at the xcomponent of the position of an object rotating
in a circle with angular velocity coo. As a check on this, one can devise an experi
ment to show that the upanddown motion of a mass on a spring is the same as
that of a point going around in a circle. In Fig. 21—3 an are light projected on a
screen casts shadows of a crank pin on a shaft and of a vertically oscillating mass,
side by side. If we let go of the mass at the right time from the right place, and if
the shaft speed is carefully adjusted so that the frequencies match, each should
follow the other exactly. One can also check the numerical solution we obtained
earlier with the cosine function, and see whether that agrees very well. Here we may point out that because uniform motion in a circle is so closely
related mathematically to oscillatory upanddown motion, we can analyze
oscillatory motion in a simpler way if we imagine it to be a projection of something
going in a circle. In other words, although the distance y means nothing in the
oscillator problem, we may still artiﬁcially supplement Eq. (21.2) with another
equation using y, and put the two together. If we do this, we will be able to analyze
our onedimensional oscillator with circular motions, which is a lot easier than
having to solve a differential equation. The trick in doing this is to use complex
numbers, a procedure we shall introduce in the next chapter. 21—4 Initial conditions Now let us consider what determines the constants A and B, or a and A. Of
course these are determined by how we start the motion. If we start the motion
with just a small displacement, that is one type of oscillation; if we start with an
initial displacement and then push up when we let go, we get still a different
motion. The constants A and B, or a and A, or any other way of putting it, are
determined, of course, by the way the motion started, not by any other features of
the situation. These are called the initial conditions. We would like to connect the
initial conditions with the constants. Although this can be done using any one
of the forms (21.6), it turns out to be easiest if we use Eq. (21.6c). Suppose that at
t = 0 we have started with an initial displacement x0 and a certain velocity 00. 214 This is the most general way we can start the motion. (We cannot specify the
acceleration with which it started, true, because that is determined by the spring,
once we specify x0.) Now let us calculate A and B. We start with the equation
for x, x = A cos wot + Bsin wot. Since we shall later need the velocity also, we diﬂerentiate x and obtain
2) = —w0A sin wot + (003 cos wot. These expressions are valid for all 1, but we have special knowledge about x and
v at t = 0. So if we put t = 0 into these equations, on the left we get x0 and v0,
because that is what x and v are at t = 0; also, we know that the cosine of zero is
unity, and the sine of zero is zero. Therefore we get x0=Al+B0=A
and
00 = —w0A0+wOBl = woB. So for this particular case we ﬁnd that
A = X0, B = 00/600. From these values of A and B, we can get a and A if we wish. That is the end of our solution, but there is one physically interesting thing
to check, and that is the conservation of energy. Since there are no frictional
losses, energy ought to be conserved. Let us use the formula x = acos (wot + A);
then
2) = woa sin (wot + A). Now let us ﬁnd out what the kinetic energy T is, and what the potential energy
U is. The potential energy at any moment is %kx 2, where x is the displacement and
k is the constant of the spring. If we substitute for x, using our expression above,
we get U = %kx2 = %ka2 c032 (wot + A). Of course the potential energy is not constant; the potential never becomes negative,
naturally—there is always some energy in the spring, but the amount of energy
ﬂuctuates with x. The kinetic energy, on the other hand, is §mv2, and by sub
stituting for v we get T = %mv2 = §mw§a2 sin2 (wot + A). Now the kinetic energy is zero when x is at the maximum, because then there is no
velocity; on the other hand, it is maximal when x is passing through zero, because
then it is moving fastest. This variation of the kinetic energy is just the opposite
of that of the potential energy. But the total energy ought to be a constant. If
we note that k = mwﬁ, we see that T + U = %mwga2 [cos2 (wot + A) + sin2 (wot + A)] = §mw§a2. The energy is dependent on the square of the amplitude; if we have twice the
amplitude, we get an oscillation which has four times the energy. The average
potential energy is half the maximum and, therefore, half the total, and the average
kinetic energy is likewise half the total energy. 21—5 Forced oscillations Next we shall discuss the forced harmonic oscillator, i.e., one in which there is
an external driving force acting. The equation then is the following: mdzx/dt2 = —kx + F(t). (21.8)
215 We would like to ﬁnd out what happens in these circumstances. The external driv
ing force can have various kinds of functional dependence on the time; the ﬁrst
one that we shall analyze is very simple—we shall suppose that the force is oscillat
ing: F(t) = F0 cos wt. (21.9) Notice, however, that this 0.) is not necessarily wot we have Lo under our control;
the forcing may be done at different frequencies. So we try to solve Eq. (21.8)
with the special force (21.9). What is the solution of (21.8)? One special solution,
(we shall discuss the more general cases later) is x = Coos wt, (21.10) where the constant is to be determined. In other words, we might suppose that if
we kept pushing back and forth, the mass would follow back and forth in step
with the force. We can try it anyway. So we put (21.10) into (21.9), and get —mw2C cos wt = —mwﬁC cos cot + F0 cos wt. (21.11) We have also put in k = mwg, so that we will understand the equation better at the end. Now because the cosine appears everywhere, we can divide it out, and
that shows that (21.10) is, in fact, a solution, provided we pick C just right. The
answer is that C must be C = Fo/m(wﬁ — (02). (21.12) That is, m oscillates at the same frequency as the force, but with an amplitude
which depends on the frequency of the force, and also upon the frequency of the
natural motion of the oscillator. It means, ﬁrst, that if (.0 is very small compared
with (.00, then the displacement and the force are in the same direct...
View
Full Document
 Spring '09
 LeeKinohara
 Physics, Energy, Kinetic Energy

Click to edit the document details