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Unformatted text preview: CHAPTER HARMONIC
MOTION Another question concerns the oscillations of pendulums, and it falls into
two parts. One is whether all oscillations, large, medium, and small, are truly
and precisely made in equal times. The other concerns the ratio of times for
bodies hung from unequal threads; the times of their vibrations, I mean. . . . As
to the prior question, whether the same pendulum makes all its oscillations ~
the largest, the average, and the smallest — in truly and exactly equal times, I
submit myself to that which I once heard from our Academician [Galileo]. He
demonstrated that the moveable which falls along chords subtended by every
arc [of a given circle] necessarily passes over them all in equal times. . . t As to the ratio of times of oscillations of bodies hanging from strings of
different lengths, those times are as the square roots of the string lengths; or
should we say that the lengths are as the doubled ratios, or squares, of the
times. Galileo Galilei, Two New Sciences (1638) 20.1 FINDING A CLOCK THAT WOULDN’T GET SEASICK Navigation has provided one of the most persistent motives for measuring time accurately.
All navigators depend on continuous time information to ﬁnd out where they are and to
chart their course. But until about two centuries ago, no one was able to make a clock
that could keep time accurately at sea. 379 380 HARMONIC MOTION Early travelers noticed that the North Star, unlike other stars, does not change its
position with respect to the earth; it appears to be suspended in the northern sky. The
farther northward they traveled, the higher in the sky the North Star appeared; at the
North Pole it would be directly overhead. By measuring the elevation of the North Star
above the horizon with a sextant, a navigator can determine the distance from the North Pole and the latitude, as Fig. 20.1a illustrates. To Pole Star i parallel of
latitude equatorial plane Longitude west of Gyeenwld‘ (a) (b) Figure 20.1 (a) Latitude determined by position of the North Star.
(b) Longitude requires knowing a reference time. On account of the earth’s rotation, however, charting a course east or west presented
a more complex problem. Only by knowing the time very accurately was a navigator
able to calculate longitude. Because of the earth’s rotation, the sun appears to travel
across the sky from east to west at the rate of one degree in four minutes. (It takes the
sun 24 hours to move once around the earth, so it must move through 360° in 24 hours,
which is 360/24, or 15° in 1 hour, or 1° in 4 minutes.) Now if a navigator determines
the local time from the position of the sun and has a clock which very accurately tells
the time in Greenwich, England (through which, by international agreement, the zero
longitude line runs), he can easily determine his longitude. For every four minutes his
clock, showing Greenwich time, differs from the local time, he is one degree of longitude
away from Greenwich. Even at night, a navigator can determine the longitude by using
star charts to determine the local time and by comparing that time to Greenwich time.
What is essential is an accurate timepiece. The earliest timekeeping devices, built by the ancient Egyptians, consisted of an
alabaster bowl, wide at the top and narrow at the bottom, which had horizontal markings
on the inside to tell the time. As water dripped out through a hole in the bottom of the
bowl, successive lines were exposed. For centuries the basic design of the water clock
remained unchanged; Galileo used a water clock in his fertile experiments with balls
rolling down inclined planes. 20.2 SIMPLE HARMONIC MOTION 381 Sometime between the eighth and eleventh centuries, Chinese artisans constructed a
water clock that had the characteristics of a mechanical clock. Falling water powered a
wheel that contained small cups around its rim. As a cup filled with water, it became
heavy enough to trip a lever which allowed the next cup to move into place and advance
the wheel by a step. In thirteenthcentury Europe, variations of the Chinese water clock
became popular. Aside from the fact that these clocks did not keep very good time, they
also tended to freeze in the European winters. Sand clocks (hourglasses) introduced in the fourteenth century avoided the problem
of freezing, but because of the weight of sand, they were limited to measuring short
intervals of time. One of the chief uses of the hourglass was to determine a ship’s position
by “dead reckoning.” Sailors would throw a log overboard with a long rope attached to
it, and then count knots, which were tied in the rope at equal intervals, as the rope played
out for a specific amount of time. In this way sailors could crudely estimate the speed,
or knots, at which the ship was moving. By knowing their speed and how long they had
traveled in a certain direction, they could track their position. The first truly mechanical clocks were built in the fourteenth century and consisted
of pulleys and weights with escapements, similar to presentday cuckoo clocks. The
accuracy of these early mechanical clocks depended on the friction between parts, the
driving weights, and the skill of the craftsman constructing it. No two clocks would show
the same time let alone keep accurate time. What was needed was some sort of periodic,
repeating device whose frequency was essentially a property of the device itself. 20.2 SIMPLE HARMONIC MOTION The event that led to accurate timepieces was the analysis of periodic or harmonic motion
— motion that repeats itself in equal intervals of time. When an object moves back and
forth over the same path in harmonic motion, we say that it is oscillating. We now
explore this type of motion. , In Chapter 14, we discussed the stability of an object subject to a force F: —kx, which acts on the object whenever it moves away from the equilibrium point x = 0.
This force, for example, describes the behavior of a spring. Associated with this force
is the potential energy U = ya? We used this potential energy as a model to study stability and discovered that it typified
many systems that oscillate: a marble rolling in the bottom of a bowl, a mass oscillating
at the end of a spring, a swinging pendulum, a vibrating guitar string, and even a complex
political or ecological system ﬂuctuating. All these systems have the property that if they
are disturbed from equilibrium, the restoring force that acts on them tends to move them
back into equilibrium. These systems have another property in common too: when disturbed from equilib
rium, the system tends to return to equilibrium by the action of the restoring force, but
they overshoot that point because of inertia. After the overshoot the restoring force again
acts to return the system to equilibrium. The result is that the system winds up oscillating 382 HARMONIC MOTION back and forth, like a marble in a bowl, a mass on a spring, and a guitar string. Figure
20.2 shows “snapshots” of a mass oscillating on the end of a spring. The horizontal displacement of the mass, plotted as a function of time, traces out a path resembling a
sine or cosine curve, as suggested by Fig. 20.2. Figure 20.2 Snapshots of a mass oscillating on the end of a spring. To determine this curve mathematically we apply Newton’s second law, F = ma.
Since the acceleration is the second derivative of the displacement, a = de/dtz, and we are considering the spring force F = —kx, we have
dzx
— = —kx.
m dt2
Dividing by the mass and explicitly denoting x as a function of time, we obtain
d2 _ k
lit5x“) — — We). (201) We already believe that the solution of this differential equation, x(t), whose second derivative is proportional to the negative of the function itself, will be a function that
oscillates back and forth with time. In Chapter 3, we encountered two functions which have this property: the sine and
cosine functions, To recapitulate, d
E sin 9 = cos 6, (3.6) and 20.2 SIMPLE HARMONIC MOTION 383 d % cos 6 = ——sin 6. (3.7)
From these derivatives we can verify that d2 E sin 6 = sin 6, d2 @ cos 6 = —cos 6. The sine and cosine functions satisfy the type of differential equation we have, but
how are they and 6 related to x(t)? As time goes on, the oscillating mass traces out a
sine curve on moving paper, as Fig. 20.2 illustrates. This suggests that x(t) = A sin em, (20.2) where A is some constant and 0(t) some angle that depends on time t. The cosine function
is also a solution, but we’ll discuss it later. In Example 1 below we use the differential
equation (20.1) to show that we can arrange matters so that 00‘) increases uniformly with
time, 6(t) = (Dot, where (no is positive constant (called the angular frequency) given by (no = ~ , (20.3)
m
and that A (called the amplitude) is the largest positive value x(t) can have. Thus a solution
of the differential equation (20.1) is x(t) = A sin wot, where (no = \/k/—m, and simple harmonic motion is a result of a linear restoring force
(F = —kx). In our discussion of uniform circular motion in Chapter 9, we had precisely the same
connection between the angle which locates an object in such motion and time. This is
no accident. As Figure 20.3 illustrates, the shadow of an object undergoing uniform Figure 20.3 The shadow of a peg on an object executing uniform
circular motion exhibits simple harmonic motion. 384 HARMONIC MOTION circular motion executes exactly the same motion as the oscillating mass and spring
system. The oscillating mass is one component of uniform circular motion. In Chapter 9, we called (00 the angular speed. When we use it to describe harmonic
motion, we call it the angular frequency. The angular frequency is the angle (in radians)
an object in circular motion moves through per unit time. This angular frequency is
intimately related to the number of oscillations a corresponding object in harmonic motion
makes per second for the reason we saw earlier. Here’s the connection: An object in
circular motion moves through 211' radians before it returns to where it started; in other
words, it moves through 211' radians in one cycle. The frequency f is the number of cycles
an object in harmonic motion completes per second. Since 211' radians corresponds to
one cycle, f = (no/(2w). (20.4) In SI units, frequency is measured in hertz, abbreviated HZ, where 1 Hz : 1 cycle/
second. The time required to complete one oscillation, known as the period T, is simply the
reciprocal of the frequency: T = l/f. (20.5) Figure 20.4 illustrates the period of a sinusoidal function. Figure 20.4 Period and amplitude of a sinusoidal function. # Example 1
Show that if a function x(t) = A sin 6(t) (20.2)
satisfies the differential equation d2 _ k 3x0) — —;x(t), (20.1)
then with a suitable choice of initial conditions we have 6(t) = (not, where 000 = Vk/m and A is the largest value of x(t). 20.2 SIMPLE HARMONIC MOTION 385 Differentiate (20.2) with the help of the chain rule to get dx dx d9 (19
Z — E E — A COS 90) dt . (20.6) Differentiate again using both the product rule and the chain rule to obtain
43;:
dt2 To satisfy (20.1) we want the right—hand side of (20.7) to reduce to k
— £x(t) = — A sin 0(t).
m m 2
d9 d29
= —A sin 6(t) (E) + A cos 90);; . (20.7) Therefore it suffices to choose 0(t) so that d6 2 k d20
A<dt> — mA and Adtz ~0.
The ﬁrst of these implies (de/dt 2 = k/m, so de/dt = two, where 000 = Vk/m. If we
want 0 to increase with time we choose the plus sign and obtain dO/dt = (90, which,
incidentally, implies dZG/alt2 = 0. Therefore 6(1) = (not + C, where C is a constant. If
the spring is in the relaxed position (as shown in Fig. 20.2) at time t = 0, then 6 = 0
when t = 0 and we must choose C = 0. Thus x(t) = A sin (not. Since the largest value of sin 0 is 1, the factor A represents the largest value of the
displacement x(t). N ow to interpret the results physically. The angular frequency depends on the physical
characteristics of the system, namely, the spring constant k and the mass m. The stiffer
the spring, the larger the value of the spring constant, and by Eq. (20.3), the larger the
number of oscillations in one second. In other words, stiffer springs make the system
oscillate more rapidly. This makes sense because a stiffer spring exerts a greater force
and tends to accelerate the mass more. Equation (20.3) also tells us that the greater the
mass, the slower the oscillations. We expect that larger values of m would lead to slower
oscillations because of inertia. Because the frequency depends only on the physical
characteristics of a particular mass and spring system, we refer to (no as the natural
angular frequency; it is the angular frequency at which the system will naturally oscillate.
The frequency of the oscillations is independent of the amplitude A. Once we have x(t) for a particular system, we can find everything there is to know
about the motion of that system. For example, to ﬁnd the velocity of the mass, we need
only take dx/dt; we already did this in Eq. (20.6), which now becomes dx — = A S t. 20.8
dt (.00 C0 (.00 ( ) Similarly, the acceleration at any instant is found by differentiating the velocity, 386 HARMONIC MOTION d2
d—tf = —w§A sin (not = —m§x. (20.9) The type of oscillatory motion we are studying is called simple harmonic motion.
And an object which executes this type of motion is called a simple harmonic oscillator,
abbreviated as SHO. The term simple refers to the absence of external forces such as
friction or viscosity. The word harmonic refers to music: musical instruments generally
vibrate harmonically. As we’ve already seen, the world is full of simple harmonic os
cillators. Once we understand completely the SH0, we can put it in our pocket, and
explore the world of problems it solves. Example 2
A particle of mass 0.25 kg undergoes simple harmonic motion with an amplitude of 0.15
m and a frequency of 100 Hz. What is (21) its angular frequency? (b) the spring constant? (0) its maximum velocity? (d) its maximum acceleration? (a) Using coo = 21rf, we have 000 = 630 rad/s.
(b) Knowing that the natural angular frequency (:03 = k/m, we have k = mung = (0.25 kg)(630 rad/s)2 = 9.9 X 104 N/m. (To end up with units of N/m, we drop the radians because radians are dimensionless.)
(c) From Eq. (20.8), v(t) = Awo cos (not, ' (20.8) we see that because the maximum value of the cosine is l, the maximum value of the
velocity is Umax = Awo = (630 rad/s)(0.15 m) = 94 m/s. ((1) For the same reason as in (c), we see that Eq. (20.9) implies that the maximum
acceleration is amax = Awg = (630 rad/s)(0.15 m) = 6.0 X 104 m/s2. Questions
1. Give some additional examples of harmonic motion. 2. For a particle undergoing simple harmonic motion, where (at either the equilibrium
point or the endpoints) does the maximum value occur in the
(a) force acting on it, (b) speed, (c) acceleration? 3. Any real spring, of course, has mass. If the mass of a spring were taken into ac 20.3 ENERGY CONSERVATION AND SIMPLE HARMONIC MOTION 387 count, how would this change the frequency of a massandspring system? Give a
qualitative argument. 4. What changes could you make in a simple harmonic oscillator to double its
(a) maximum speed, (b) maximum acceleration? 5. A small (0.10kg) mass is attached to the bottom of an unstretched vertical spring
and set into motion. If the maximum speed of the mass is 0.20 m/s and its maxi
mum acceleration is 0.5 m/sz, find (a) the spring constant,
(b) the amplitude of the oscillations,
(c) the frequency of the oscillations. 6. A 2.5kg block hangs from a spring. If a 0.5kg mass is attached to the block, the
spring stretches an additional 0.05 m. Determine the frequency of oscillations if
only the 2.5kg block is attached to the spring. 7. A mass is attached to a spring of spring constant k. If the spring is cut in half and
the same mass suspended from one of the halves, how are the frequencies of oscil
lation related, after and before the spring is cut? 8. Two springs of spring constants k1 and k2 are attached to a block of mass m as
shown. What will be the frequency of oscillations? 20.3 ENERGY CONSERVATION AND SIMPLE HARMONIC MOTION In Chapter 14, we argued that an object near a potentialenergy minimum would oscillate
because potential energy would turn into kinetic energy, which in order to be conserved
would turn back into potential energy. That was how we explained why the object doesn’t
stop at the equilibrium point where there’s no force acting on it. Conservation of energy
led us to expect oscillations in the first place. Now we’ve obtained the same result without
mentioning energy at all. To connect the two approaches, let’s examine the energy of a
simple harmonic oscillator.
We already know that the potential energy is given by U(x) = gkxz. (14.1) From the displacement x(t) given by Eq. (20.2) we can express the potential energy as
a function of time: U(t) = ékAZ sin2 (not. (20.10) The kinetic energy of the system is émvz, where the velocity v is related to x(t) through
v = dx/dt. We already found this derivative: 388 HARMONIC MOTION dx
E = Awo cos (not, (20.8) so the kinetic energy as a function of time is
2
1 2 1 d"
K(t) = imv = am it , K(t) = mung/12 cos2 wot. (20,11) 1
5
Therefore the total energy of the system is E=K+U, E = émooéA2 cos2 (not + ékA2 sin2 (not. At first glance it appears as if the total energy is not constant because it seems to
depend on t. To show that the total energy is indeed constant, we recall that (03 = k/m,
and substitute this into the expression for E: E = %m(k/m)A2 cos2 (not + ékAZ sin2 (not, E = ékAZ(cos2 (not + sin2 (not). The term in parentheses, according to trigonometry, is always equal to one, so we have
the result = gkAl, (20.12) which is a constant. Figure 20.5 shows graphs of U(t), K(t), and the total energy E. Figure 20.5 Graphs of potential, kinetic, and total energy for a
simple harmonic oscillator. Example 3
A spring of spring constant 20 N/m hangs unstretched. A 0.20—kg mass is attached to
the free end and released. Find (a) how far below the initial position the mass descends, (b) the energy of the oscillator. 20.3 ENERGY CONSERVATION AND SIMPLE HARMONIC MOTION 389 When the mass is released, its gravitational potential energy is converted into
potential energy stored in the spring, and it momentarily stops descending when the two
energies are equal. Let d be the distance the mass falls. Then its initial potential energy
is mgd and its ﬁnal energy is ékdz. Equating these two energies ékd2 = mgd,
we can solve for d:
d = 2mg/k = 2(0.20 kg)(9.8 m/s2)/(20 N/m) = 0.20 m. This result makes sense because the greater the mass the more inertia the object has, the
greater the force needed to stop it, and the farther it stretches the spring; the stiffer the
spring, the less it will stretch. The amplitude of the oscillations will be half the distance it falls because it oscillates
equally up and down about that point. Therefore, substituting A = 0.10 m into Eq.
(20.12), we find = gkA2 = 50.20 N/m)(0.10 m)2 = 0.001 J. Equation (20.12) represents the initial amount of energy put into the oscillator by an
outside agent performing work, as when you stretch a spring, or pluck a guitar string.
As a result of friction that energy is gradually transformed to heat, so the energy of a
real harmonic oscillator decreases with time and the amplitude of successive oscillations
diminishes. But the frequency of the oscillations, being independent of the amplitude,
remains constant; only the amplitude of successive oscillations decreases. Robert Hooke, after whom the spring force law is named, understood the essential
feature of spring oscillations — that even in the presence of friction, the frequency remains
constant. In the 16505 he experimented with the idea of using a metal spring to regulate
the frequency of a clock. The ﬁrst springcontrolled clock was built, however, by Christian
Huygens, a Dutch physicist. His idea was to use a spiral spring — the type still used
today in mechanical watches. Questions 9. When a simple harmonic oscillator is at onethird of its maximum displacement,
what fraction of its total energy is in kinetic energy? 10. A simple harmonic oscillator of mass 0.40 kg has a frequency of 15 Hz and a
total energy of 30 J. What is the amplitude of its oscillations? 11. Express the total energy of a simple harmonic oscillator, Eq. (20.12), in terms of
its maximum velocity. 12. From conservation of energy, show that the instantaneous velocity of a simple har
monic oscillator is given by U(t) 2 two V142 — x2, where x = x(t) is the instantaneous displacement. 390 HARMONIC MOTION 20.4 INITIAL CONDITIONS When we searched for a solution to the simple harmonic oscillator, we found one of the
form x(t) = A sin (not. But is it the only solution? It is easy to verify that another function, x(t) = B cos (not,
satisfies the same differential equation of simple harmonic motion, d2 k
bit—2x“) = ‘gxm, (201) where B is any constant. And we can form another solution to the equation by taking the
sum of the two solutions: x(t) = A sin (not + B cos (not. (20.13) To make matters worse, there are other possibilities, like x(t) = C sin (wot + (1)), where
(I) is a constant angle. However, it can be shown that this and any other possibility can
be converted into a sum of sine and cosine functions by a suitable choice of A and B.
For this reason we say that Eq. (20.13) represents the general solution. Example 4
Verify that x(t) = A sin (not + B cos wot, where (1)3 = k/m, satisfies Eq. (20.1).
Calculating the first derivative, we have dx . E = Awo cos wot — Bwo sm wot,
and taking the derivative again, we get dzx , E = —Aw3 sm (not — Bwé cos (not.
Writing (:03 = k/m we have dzx k . —2 = ——(A s1n (not + B cos (not). dt m Recognizing that the term inside the brackets is x(t), we see that this solution also satisfies
Eq. (20.1). Now that we have many solutions to the equation of simple harmonic motion, how
do we choose one which describes a particular case? One way to find out is to ask where
the oscillator is, and how fast it is moving at some time; a convenient choice of time is
t = 0 (when we start our clock). This will identify a particular solution. To illustrate
how this works, let’s begin with the general solution x(t) = A sin (not + B cos wot. (20.13) 20.4 INITIAL CONDITIONS 391 The two constants A and B are yet undetermined. (Remember that the natural angular
frequency (.00 is determined by the massandspring constant of the oscillator.) But by
specifying the displacement and velocity of the oscillator at time t = 0, we will determine the constants A and B.
Now suppose that, as Fig. 20.6 illustrates, at t = O the oscillator has a displacement x(0) and zero velocity, 12(0) = 0. According to Eq. (20.13), we have
x(0) = A sin 0 + B cos 0. equilibrium
position I x(0)
Figure 206 Initial conditions for a simple harmonic oscillator. Since sin 0 = 0 and cos 0 = 1, this condition requires thatB # x(0), so we’ve determined one of the constants, B.
Before we impose the second condition, that involving the initial velocity, we need to calculate the velocity at any instant: dx
v(t) = E = Awo cos wot — Bwo sin (not. (20.14) Setting the velocity equal to 0 at t = 0, we get
11(0) = 0 = Awo cos 0 — Bwo sin 0,
which implies
A000 = 0, hence A = 0. So we’ve determined A as well. Therefore, the position at any time for
an oscillator that starts out in this particular way is x(t) = x(0) cos (not. By specifying the initial conditions — the displacement and velocity of the oscillator
at time t = 0 — we can determine the constants that specify the simple harmonic motion
for a particular oscillator. We’ve seen another example of this process in Chapter 6, when
we found the general description of a projectile and imposed initial conditions on position
and velocity to determine a specific trajectory. In some respects the example we just solved is typical of what happens in general.
Somewhere in the process of solving the differential equation, two integrations are required
(one to go from acceleration to velocity, the second to go from velocity to position).
Accompanying the two integrations are two constants, which here appear in the solution
as A and B. By imposing initial conditions, we determine those constants and specify
the particular solution. 392 HARMONIC MOTION There is no need to memorize the specific solution for every simple harmonic os
cillator. All you need to know is x(t), Eq. (20.13), and how to apply the specific initial
conditions for a particular problem. Example 5
A 0.5kg mass is attached to a spring with spring constant k = 120 N/m and set into
motion. A clock is started when the oscillator has a displacement x(0) = 0.3 m and a
velocity 12(0) = 6.0 m/s. What is the displacement of the oscillator at time t? Starting with the general solution, Eq. (20.13), x(t) = A sin wot + B cos (not,
we set x(0) = 0.3 m, which implies
0.3m = Asin0 + BcosO, or that B = 0.3 m.
Taking the derivative of x(t) and setting it equal to 11(0), we have 11(0) = 6.0 m/s = (DOA cos 0 — (noB sin 0. Solving for A and using too = Vk/m = 15 rad/s, we ﬁnd A = 0.4 m. Now that we
know the two constants A and B, we can describe the displacement of the oscillator at
any instant as x(t) = (0.4 m)sin(15t) + (0.3 m)cos(15t), where t is in seconds. By understanding general principles and working out specific examples, we gain not
only additional insights into the way things work, but often obtain technological im
provements as well. In 1713 the British government offered a prize of £20,000 to anyone
who could build a clock accurate enough to enable a seafarer to determine longitude to
within onehalf of a degree (about 35 miles). Among the many dexterous craftsmen who
sought to win the ample award was an English clockmaker, John Harrison. For 40 years
he struggled to construct a springdriven clock which could cope with rolling seas,
temperatureinduced expansion and contraction, and the corrosive salt spray. Finally in
1761 he sent his son on a voyage to Jamaica to test his clock, but only after the government
forced him to build an identical model lest the original be lost at sea. His masterpiece
was a technological triumph — it allowed the navigator to determine longitude to within
onethird of a degree. Questions 13. At time t = 0 an oscillator having a natural frequency of 35 Hz has a displace
ment x(0) = 0 and a velocity 11(0) = 20 m/s. (3) What is the displacement of the oscillator at any time?
(b) Find the maximum velocity of the oscillator. 20.5 THE SIMPLE PENDULUM 393 14. Suppose that initially an oscillator has a displacement of 0.25 m, a velocity of
— 10 m/s, and a frequency of 10 Hz. (a) What is its amplitude?
(b) Find the displacement at any instant. 15. Show that x(t) as given in Eq. (20.13) leads to energy conservation. 20.5 THE SIMPLE PENDULUM A very special and important aspect of simple harmonic motion is that the angular
frequency too does not depend on the amplitude (A or B) of the motion. This independence
means that the time for each complete cycle, the period T, also does not depend on the
amplitude; if the oscillator makes large oscillations, it moves rapidly, and if it makes
smaller oscillations it moves more slowly. Even in the real world, where friction makes
oscillations die down, the oscillator always takes the same amount of time for each cycle.
This is the characteristic that allows a simple harmonic oscillator to be used as a timing
device. The discovery of this fact led immediately to the invention of the first accurate
Clocks. Even today, wristwatches which are accurate to within a few seconds per month
use as timekeeping devices a kind of harmonic oscillator — a quartz crystal. But earlier clocks used a different oscillator — the pendulum. Galileo made the crucial
discovery that a pendulum takes the same time per swing, even as its motion dies down,
and he thereby laid the groundwork for improved timekeeping. In the Two New Sciences
he eloquently summarized his observations. Folklore places Galileo’s discovery in the Duomo, or Cathedral, in Pisa. The famous
Leaning Tower of Pisa is actually the bell tower of the magnificent, highceilinged
cathedral. Hanging from the ceiling on a long cable is a lamp, which one day Galileo
supposedly noticed swinging back and forth, probably just after it had been lit. Timing
the swings by comparing them to his own pulse, Galileo realized that they always took the same time even as the swings became smaller and smaller.
This famous lamp, called Galileo’s lamp, still hangs in the Cathedral at Pisa. How
ever, there’s one thing wrong with the tale: the church’s records show that the lamp was installed in the 16505, ten years after Galileo’s death. Setting fables aside, let’s analyze the motion of a simple pendulum to find out
precisely what factors determine the period. The idea is to use Newton’s second law to
find a differential equation that describes the motion of a pendulum and cast it into the form of Eq. (20.1),
E
dt2 In other words, we need to find an equation which says that the second derivative of the oscillating quantity is proportional to the negative of the quantity. Then, without any further trouble, we’ll know that the quantity multiplying —x (the position occupied by 008) is the square of the angular frequency with which a pendulum swings. In this case
it will turn out that the oscillating quantity is not the displacement but rather the angle of the pendulum cord with respect to the vertical. = —w3x. (20.1 ') 394 HARMONIC MOTION Let’s consider the simple pendulum, like the one shown in Fig. 20.7a. We call it
simple because we are idealizing the pendulum as a point of mass m at the end of a
massless string of length L. An analysis based on Newton’s second law, however, is not
limited to this case; it’s only easier. Suppose we start the pendulum in motion by pulling
it aside and releasing it. When the pendulum moves through an angle 6 from the vertical,
it goes through a distance 6L along a circular arc from its equilibrium position. The force
responsible for tending to restore the pendulum to its equilibrium position (hanging straight
down) is its weight mg. We can resolve this force into components parallel and perpen
dicular to the string, as illustrated in Fig. 20.7b. The perpendicular component, the one
that is always tangent to the circular arc, causes the pendulum to accelerate back to its
equilibrium position; from Fig. 20.7b we see that this component is —mg sin 6. The
acceleration along this path is the second derivative of the displacement along the circular
arc: m /
g /mgsin0 (a) (b)
Figure 20.7 (a) The simple pendulum. (b) Force diagram for pendulum.
d2 d26
— L6 = L— .
dt2( ) dt2
Therefore, along the arc Newton’s second law implies
c126
mLW = —mg sin 0.
Canceling the mass and dividing through by L, we get
dze s‘ 6
—— = — — 1n .
(it2 This last result is not the equation for a simple harmonic oscillator; the second
derivative of the displacement (here 6) is not proportional to — 6 but to —sin 6. Fur
thermore, it can be shown that no elementary function will satisfy this differential equation. The intrepid physicist, however, is not daunted by such minor obstacles. If physicists
only solved problems that they knew how to solve exactly, they would accomplish very
little. The essence of practical physics is to ignore what is unimportant and to approximate.
As Table 20.1 indicates, when expressed in radians, 6 is approximately equal to sin 6;
the smaller the angle, the closer the agreement. Even for an angle of 11/4 rad (45°) the 20.5 THE SIMPLE PENDULUM 395 difference between 0 and sin 0 is only about 10%. Therefore, as long as we only consider
small oscillations, we can safely replace sin 0 by 0 in our equation: dZO g
— = — —e. .
dt2 L (2015) Now this equation is exactly like the simple harmonic oscillator equation; the variable
now is 0, but that doesn’t matter. We know that the solution is 8(t) = 60 cos wot, (20.16) where 60 is the amplitude determined from the initial conditions. This solution is rea
sonably good if the amplitude of the swings is small. Table 20.1 Comparison of 6 (in rad) with
sin 6 for Small Angles 8 (deg) 0 (rad) sin 8
0 0 0
0.5730 0.0100 0.0100
5.730 0.1000 0.0998
11.459 0.2000 0.1987
17.189 0.3000 0.2955
45.000 17/4 = 0.7854 0.7071 By comparing Eq. (20.15) with the simple harmonic oscillator equation, we see that
the angular frequency of the oscillations is we = V g/L, (20.17)
which means that the period is
T = 217/000 2 271' VL/g. Consequently, on any given planet, the frequency of a simple pendulum depends only
. on its length. Unlike a mass on a spring, for which the frequency Vk/m does depend on
the mass, the frequency of a pendulum is independent of its mass. The reason the natural frequency is independent of the mass is exactly the same
reason that the acceleration of a falling body on the surface of the earth doesn't depend
on its mass: through Newton‘s second law, F = ma, and the universal law of gravity,
F = GmMe/Rez, the mass m cancels. The ingenious Isaac Newton used pendulums of
different masses to test this cancelation with a precision of one part in a thousand. Because 396 HARMONIC MOTION pendulums of identical length but different mass have equal frequencies, this proves
exactly the same law as dropping a penny and a feather in a vacuum. Newton realized
that the pendulum experiment works without being in a vacuum and is easier to observe. Example 6
A pendulum has an amplitude of 20° and a length of 2.0 m. Find (a) its natural frequency, (b) its maximum velocity.
Using Eq. (20.17), we ﬁnd we = Vg/L = V(9.8 m/sZ)/(2.0 m) = 2.2 rad/s. Differentiating Eq. (20.16), we obtain d_0 = —0(n sinwt.
dt 00 o The velocity of the pendulum along the circular arc is given by v = L de/dt, so the
maximum speed is
vmax = eowoL = (20°)(1'r rad/180°)(2.2 rad/s)(2.0 m), which turns out to be 1.5 m/s. Note that one must express 00 and too in radians, not
degrees, to get the right answer. W Questions
16. What must be the length of a pendulum to have a period of 1.0 s? 17. A pendulum has a period of 4.0 s on the surface of the earth. What will be its
period on the surface of the moon? 18. A certain pendulum has a period of 3.0 s. How will the period change if the
length is increased by 60%? If the length is decreased by 60%? 19. Qualitatively argue, taking into account the mass of the rod on which a pendulum
swings, how the frequency would change from that of a simple pendulum. 20. A simple pendulum of length 1.50 m makes 80 oscillations in 200 5. What is the
local acceleration due to gravity? 20.6 A FINAL WORD We started out to study harmonic oscillation, a motion executed by various things, like
pendulums and guitar strings. Understanding such periodic motion was crucial to the
development of accurate timepieces. In our analysis we had to ignore air resistance and
friction — idealizations we have often made. Yet for the pendulum, we approximated
even further when we found out that it is not quite a harmonic oscillator, because its
motion is along a circular arc. Is physics imprecise? 20.6 A FINAL WORD 397 Many people believe that physicists seek the most fundamental and precise equations
that govern the behavior of the universe. But in fact, physicists don’t have completely
universal equations at their disposal. Newton’s laws aren’t such principles; his laws don’t
accurately describe objects as small as atoms or as large as galaxies. And although we
understand atoms (quantum theory) and galaxies (general theory of relativity), we don’t
have one fundamental set of laws that explains both at the same time. Many physicists,
though, search for such a law and believe that it soon will be within their grasp. Suppose, however, that we already knew the fundamental laws that govern the
universe. What would we do then? The obvious approach would be to write down those
equations and find all the solutions. This would be excruciatingly difficult, since the laws
presumably would be expressed as differential equations. But in principle it would seem
it could be done; a differential equation can be solved numerically by a sufficiently
powerful computer even if it is impossible to express the solution analytically by formulas.
So, if we found all the solutions, we would unlock all the secrets of the universe. Or
would we? Solving the equations of the universe numerically is something we would not want
to do even if we could. The reason is very simple. The computer printout would be as
complicated as the universe itself — and we already have the universe! What we want
from physics is not the precise numerical results that describe exactly how everything
behaves. Instead, what we seek is something much more subtle. We want understanding,
insight, and at best a kind of trained and dependable intuition about why things work the
way they do. In studying the differential equation of the simple harmonic oscillator we gained an
understanding of how some things work, even though we don’t know of any physical
system that precisely satisfies this equation. Yet, as we look at the world around us,
mentally armed with this equation and its solutions, we begin to see everywhere examples
of things that we know have this equation buried somewhere deep in their behavior. Our
understanding of how things work has been inexpressibly enriched once we grasp the
idea of extracting from complicated phenomena simple, underlying elements. Harmonic
motion is often one of those elements. The road to insight often does not go through
meticulous, complete, and precise description. It usually starts out in quite a different
direction, passing first through crude but clever estimations and approximations. ...
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 Fall '07
 Goodstein

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