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In order to change œgÝ by 27, the time must change by an amount íạ, called
the per7od of one complete oscillation; of course #o must be such that ¿go = 27.
That is, go must account for one cycle of the angle, and then everything will
repeat itself—If we Increase ý by #o, we add 2z to the phase. Thus
tọ = 2/œo = 2mm. (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. Ôr, ïf 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 hou ¡it has been started, how far down we pull ít. The period
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1s determined, but the amplitude of the oscillation is no determined by the
cquation of motion (21.2). The amplitude 2s determined, in fact, by how we let
go of it, by what we call the zn#tial condiHions or starting conditions.
Actually, we have not quite found the most general possible solution of
Edq. (21.2). There are other solutions. It should be clear why: because all of the
cases covered by # = øcos /œg start with an initial displacement and no initial
velocity. But it is possible, for instance, for the mass to start at z = 0, and we
may then give it an impulsive kick, so that it has some speed at ¿ = 0. Such
a motion is not represented by a cosine——it is represented by a sine. 'o put 1§
another way, iÝ — cosœg# 1s a solution, then is it no obvious that if we were
to happen to walk into the room at some từne (which we would call “¿ = 0”)
and saw the mass as it was passing z = 0, ¡it would keep on goïng just the same?
Therefore, ø = cosœo cannot be the most general solution; it must be possible
to shift the beginning of tỉme, so to speak. As an example, we could write the
solution this way: # = øœcosœg(# — tị), where íq is some constant. "This also
corresponds to shifting the origin of time to some new instant. Eurthermore, we
may expand
cos (œo# + A) = cosugf# eos Á — sin „g£ sỉn A,
and write
œ= Acosœg£ + Bsin œg#,
where 4 = øcos A and = —asin 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 đ?z/df? = —„ÿz that exists in the world can be written as
(a) % = acOS0g(È — #1),
(b) % = acos (0£ + A), (21.6)
(c) %= Acosoo£ + Bsìn uot.
Some of the quantities in (21.6) have names: œọ is called the angular [requencU;
it is the number of radians by which the phase changes in a second. “That 1s
determined by the diferential equation. The other constants are not determined
by the equation, but by how the motion is started. Of these constants, œ measures
the maximum displacement attained by the mass, and is called the ampiitude
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of oscillation. "The constant A is sometimes called the phase of the oscillation,
but that is a confusion, because other people call «¿o£ + A the phase, and say
the phase changes with time. We might say that A is a phase shúf† from some
defned zero. Let us put it diferently. Diferent A?s correspond to motions in
diferent phases. 'That ¡is true, but whether we want to call A £he phase, or not,
1s another question.
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 acbually involved in the linear motion—i% just goes up and down.
W©e may point out that we have, in fact, already solved that diferential equation
when we were studying the mechanics of circular motion. lf a particle moves In
a circle with a constant speed 0, the radius vector from the center of the cirele
to the particle turns through an angle whose size is proportional to the time. lÝ
we call this angle Ø = œt/R (Fig. 21-2) then đØ/đf = œạ = 0/R.. We know that
there is an acceleration a = 02/ = uŸR toward the center. Now we also know
that the position z, at a given moment, is the radius of the cirele times cos ổ,
and that is the radius times sin 0:
z= Rcos0, ụ= Rsin0.
Now what about the acceleration? What is the z-component of acceleration,
d2z/dt?? 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.
đ„ = —acoS = —wg]#cos 0 = —u0%. (21.7)
Fig. 21-2. A particle moving ¡In a circular path at constant speed.
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In other words, when a particle is moving ín a cirele, the horizontal component of
10s 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:
% = Rcosuot. Equation (21.7) does not depend upon the radius oŸ the circle, so
for a circle of any radius, one fnds the same equation for a given œọ. Thus, for
several reasons, we expect that the displacement of a mass on a spring will turn
out to be proportional to cosœg#, and will, in fact, be exactly the same motion
as we would see if we looked at the z-component of the position of an object
rotating in a circle with angular velocity œo. As a check on this, one can devise
an experiment to show that the up-and-down motion of a mass on a spring is the
same as that ofa poïnt goïing around in a cirele. In Eig. 21-3 an arc 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 ïf 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.
Light 1
Projector
Screen
Fig. 21-3. Demonstration of the equivalence between simple harmonIc
motion and uniform circular motion.
Here we may point out that because uniform motion in a cirele is so closeÌy
related mathematically to oscillatory up-and-down motion, we can analyze oscil-
latory motion in a simpler way if we imagine it to be a projection oŸ something
goïing in a circle. In other words, although the distance means nothing in the
oscillator problem, we may still artificially supplement Eq. (21.2) with another
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equation using , and put the two together. If we do this, we will be able to
analyze our one-dimensional oscillator with circular motions, which is a lot easier
than having to solve a diferential equation. The trick in doing this is to use