mazur ch15

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CONCEPTS

15.1 Periodic motion and energy15.2 Simple harmonic motion

15.3 Fourier’s theorem15.4 Restoring forces in simple

harmonic motion

QUANTITATIVE TOOLS

15.5 Energy of a simpleharmonic oscillator

15.6 Simple harmonic motionand springs

15.7 Restoring torques15.8 Damped oscillations

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C H A P T E R 1 5

Periodic motion

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2 CHAPTER 15 Periodic motion Conceptspost

Figure 15.1 A cart fastened to a spring that is anchored to apost executes a periodic back-and-forth motion when it is re-

leased after being compressed.

A ny motion that repeats itself at regular time in-tervals is called periodic motion. The motion of,say, the Moon revolving around Earth or Earth

revolving around the Sun is one type of familiar peri-odic motion, as is the rotation of the hands of a clock.In this chapter, we are interested in a particular type of

periodic motion, namely back-and-forth periodic mo-tion. Such motion, called either vibration or oscillation

(the two words mean the same thing), is common at allscales in the universe. A rocking chair, a swing, the pen-dulum of a grandfather clock, the strings on a guitar,the wings of a mosquito all oscillate. At the atomiclevel, atoms oscillate inside solids. At the cosmic level,the entire universe may oscillate in an ever-repeatingcycle of expansion and contraction. The physical na-ture of these periodic motions are very different fromone another, and yet they are all closely related andtheir mathematical descriptions are always of the same

form, differing only in the quantities involved.In this chapter, we first develop a model for the simplest type of oscillation, called simple harmonic mo-

tion, and relate it to a periodic motion we are alreadyfamiliar with: rotational motion. Then we show thatany oscillation can be described in terms of simple har-monic motions. After discussing a number of examples,we discuss the effects of energy dissipation during os-cillations.

15.1 Periodic motion and energy

Figure 15.1 shows the periodic motion of a spring-cartsystem. The system, initially at rest, is compressed andthen released from rest. After being released, the cartoscillates back-and-forth along the low-friction track,alternatingly compressing and stretching the spring.

15.1 (a) List the forces exerted on the spring-cart system of Figure 15.1 right after it is releasedand draw a free-body diagram for each object inthe system. (b) Which of these forces do work onthe system as it oscillates? (c) As the hand push-

es on the cart and compresses the spring, is thework done by the cart on the spring positive, neg-ative, or zero? (d) How does the work done bythe cart on the spring compare with that done bythe spring on the cart?

To understand why the motion of the cart in Figure15.1 is periodic and why the cart doesn’t simply returnto its initial position and remain there, let us look at the

mechanical energy in the cart-spring system. As thespring is compressed by the hand, elastic potential ener-gy is stored in the system. Just before the cart is releasedthe system is at rest, so the mechanical energy of the sys-tem consists entirely of potential energy (Figure 15.2a).After the cart is released, the spring-cart system is closed— none of the forces exerted on it does any work on it— and so the energy of the system must remain con-stant.

In order to analyze the system’s energy changes in

more detail, we choose an x axis that points in the ini-tial direction of motion of the cart and that has its ori-gin at the center of the cart’s equilibrium position. Asthe expanding spring speeds up the cart in the positive

x direction, the kinetic energy increases and the poten-tial energy decreases (Figure 15.2b). When the cartreaches the equilibrium position at x = 0 the potentialenergy is zero (Figure 15.2c). The cart, however, doesnot stop at this position because its kinetic energy is

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15.1 Periodic motion and energy

0 x

(a) 0.00 s

U sK

U sK

(b) 0.09 s

(h) 0.63 s

U sK

U sK

(i) 0.72 s

U sK

(c) 0.18 s

U sK

(d) 0.27 s

U sK

(e) 0.36 s

U sK

( f ) 0.45 s

U sK

( g) 0.54 s

Figure 15.2 Periodic motion is the result of a continuous exchange between kinetic and potential energy.

not zero. The cart therefore overshoots the equilibposition and begins stretching the spring. As the spstretches, it exerts on the cart a restoring forceslows the cart down and converts kinetic energy to elastic potential energy (Figure 15.2d). This coues until the kinetic energy falls to zero, at which p

the cart reverses its direction of travel (Figure 15The potential energy is converted back to kinetic egy (Figure 15.2 f ), and the cart again returns to the librium position, this time travelling leftward (Fi15.2 g). The potential energy is zero again, but witcart’s kinetic energy being nonzero, it overshootequilibrium position once again. Provided no enis dissipated, the system returns to its initial state (pare Figures 15.2a and 15.2i). A new cycle beginsthe cart repeats its back-and-forth motion.

15.2 (a) In Figure 15.2e, the cart’s displacemfrom the equilibrium position is maximumthe cart’s acceleration at that instant positnegative, or zero? (b) At which instant(s) in Fure 15.2 is the magnitude of the cart’s acceletion largest? At which instant(s) is it smalle(Use the blue position curve to answer thquestions.)

The time interval it takes to complete a full cof the motion is the period T (see Section 11.1)the motion shown in Figure 15.2 the period is equ

the time elapsed between (a

) and (i):

T = 0.72 sinverse of the period, called the frequency of the

tion, , gives the number of cycles compl

per second. In Figure 15.2, f = 1/(0.72 s) = 1.4 s–1

so the cart completes 1.4 cycles each second.The magnitude of the cart’s maximum displace

from the equilibrium position is called the ampli A of the periodic motion. The amplitude is relatethe mechanical energy of the system. In Figure 15.example, the larger the initial compression of the spthe larger the amplitude and the larger the initial potial energy.

In practice, periodic motion in mechanical sysdoes not continue indefinitely because some eneralways dissipated (friction in the cart’s bearings, air tance, heating of the spring). This dying out of permotion due to dissipation of energy is called dam

Initially, we shall neglect damping and with this simcation, we find that

Periodic motion is characterized by a continuo

conversion between potential and kinetic energy

a closed system.

f 1T

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4 CHAPTER 15 Periodic motion Concepts

All systems that exhibit this behavior have one com-mon feature: a restoring interaction that tends to returnthe system to equilibrium. Some examples are illustrat-ed in Figure 15.3. Each of these systems executes period-ic motion when disturbed from its equilibrium position,and in each case the amplitude of the motion is a mea-

sure of the mechanical energy of the system.

15.3 For each system in Figure 15.3, identify (a)the restoring force and (b) the type of potentialenergy associated with the motion.

15.2 Simple harmonic motion

Measuring the period of any oscillating system revealsa surprising fact: when the amplitude of the motion is

not too large, the period is independent of the ampli-tude. For example, an object suspended from a springhas a fixed period of oscillation regardless of its ampli-tude (Figure 15.4). The period of a pendulum, too, is in-dependent of the amplitude of its swing, which is whypendulums are used to regulate the movement of me-chanical clocks. Likewise, when you pluck a guitar string,the pitch of the sound, which is determined by thestring’s period of oscillation, is the same regardless of how far back you pull the string.

A system exhibiting equal periods for all amplitudesis called isochronous. Provided their amplitudes are

not too large, all the systems shown in Figure 15.3 are

isochronous. Individual systems have different periodsthat are determined by the properties of the system,but any given system has a period that is essentially in-dependent of amplitude.

The graph of displacement versus time for anisochronous oscillating system is shown in Figure 15.5.

The x(t ) curves for an isochronous oscillation are allsine functions. For this reason they are referred to as si-

nusoidal curves. Any periodic motion that yields a si-nusoidal x(t ) curve is called simple harmonic motionand a system executing such motion is called a simple

harmonic oscillator . For small amplitudes all the sys-tems shown Figure 15.3 yield this type of x(t ) curve,and so all are examples of simple harmonic motion.The motion is called harmonic because in musical in-struments harmonic motion is responsible for thesound emitted by the instruments.

The fact that simple harmonic motion is isochro-

nous allows us to draw some important conclusions.Consider, for example, the two isochronous x(t)

curves in Figure 15.5, one with twice the amplitude of the other. These could be the curves of two identicalpendulums or any other of the systems shown in Fig-ure 15.3, with one oscillating object moving twice asfar out as the other. In one period, object 1 alwayscovers twice the distance that object 2 covers. Becausethe periods are the same, the velocity of object 1 mustalways be twice as large as that of object 2 at the sameinstant in the periodic motion. Neither motion hasconstant velocity or constant acceleration — the

velocities and accelerations change constantly. How-

(a)

(c) (d )

(b)

Figure 15.3 Examples of oscillating systems: (a) pendulum, (b) ruler on desk, (c) ball on curved surface, (d) taut cello string.

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15.2 Simple harmonic motion

object 1

turning point

turning point

object 20

A12

A1

2

! A

+ A

x

14T

12T

34T

T

Figure 15.5 Positions-versus-time graphs for two isochroobjects.

7

4T

3

2T

5

4T

3

4T

1

2T

1

4T T 0

2 A1

2 A2

Figure 15. 4 The oscillation of an object suspended from a spring is isochronous: the period T does not depend on the atude of the oscillation.

ever, in order for the velocity of object 1 always to betwice that of object 2, the acceleration of object 1must, at each instant, be twice as large as the acceler-ation of object 1.

As you saw in Checkpoint 15.2, the acceleration islargest when the displacement from the equilibriumposition is largest (that is to say, at the turning points

which are the positions where the object turns around).This tells us that the acceleration of object 1 at its turn-

ing point x1 = A is twice as large as that of object 2 atits turning point . Given that the two objectsare identical, we conclude that the acceleration of anisochronous object is proportional to the displacementfrom the equilibrium position. Because the accelera-tion of an object is proportional to the vector sum of the forces exerted on it, this tells us that the restoringforce exerted on the oscillating object is also propor-tional to the displacement from the equilibrium posi-tion. We already know that springs exert linearrestoring forces (Hooke’s law, Section 8.9). The obser-vation that all systems in Figure 15.3 are isochronous

tells us, then, that the restoring forces in all these sys-tems must also be linearly proportional to displace-ment from the equilibrium position.

A object executing simple harmonic motion is sub-

ject to a linear restoring force that tends to return

the object to its equilibrium position and is linear-

ly proportional to the object’s displacement from

x2 12 A

its equilibrium position.

15.4 Suppose the spring in Figure 15.1 is copressed twice as much. (a) By how much doesmechanical energy of the spring-cart systemcrease? (b) What is the relationship between amplitude of the oscillation and the mechanenergy in the oscillating system?

Simple harmonic motion is closely related to riodic motion we have already studied: circulartion at constant speed. To see how the tworelated, imagine projecting the shadow of a ssphere in circular motion at constant speed onscreen perpendicular to the plane of the motion

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(a)

(b)

Figure 15.7 Experimental demonstration of the correspondence between circular motion at constant speed and simple har-monic motion for (a) an object suspended from a spring and (b) a pendulum.

A

shadow

screen

light

A sin A sin t

= t

ω

Figure 15.6 A small black sphere moves at constant speed ina circle in the vertical plane. A light source to the left of thesphere casts a shadow of the sphere onto the screen. As thesphere moves in its circle, the shadow moves up and down insimple harmonic motion.

ures 15.6 and 15.7). While the sphere sweeps out acircular trajectory, its shadow oscillates up and downon the screen. The position of the shadow at any in-stant is given by the vertical component of thesphere’s position vector at that instant. If the spherebegins on the horizontal axis and moves with a con-stant rotational speed , the angle between thesphere’s position vector and the horizontal axis growssteadily with time: . The position of thesphere’s shadow on the screen is then described by

, where A is the radius of the sphere’s trajectory. Because it is described by a sinusoidal func- sin( t )

t

* After the French mathematician Jean-Baptiste-Joseph Fourier(1768–1830)

tion, the projection of circular motion at constantspeed is simple harmonic motion.

The correspondence between circular motion atconstant speed and simple harmonic motion can bedemonstrated experimentally by projecting the shad-ows of an object in simple harmonic motion and an ob-

ject in circular motion at constant speed on a screen, asin Figure 15.7. If the rotational speed is adjusted so thatits period equals that of the oscillation, the shadows of the two objects move synchronously.

15.5 What are (a) the direction of the velocityof the shadow on the screen in Figure 15.6 and(b) the direction of the shadow’s acceleration?

15.3 Fourier’s theorem

Simple harmonic motion is important because, as Fig-ure 15.3 suggests, it is very common. Another reason itis important is that according to a rule known as Fouri-

er’s theorem,* any periodic function can be written as asum of sinusoidal functions of frequency ,where is an integer and T is the period of the func-tion. This means that any periodic motion can be treat-ed as a superposition of simple harmonic motions. So if

n!1 f n nT

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15.3 Fourier’s theorem

A1

A2

A3

A4

O r i g i n a l

H a r m o n i c s

S u m o

f H a r m o n i c s

A1h1(t )

(a)

(b) (c)

A1h1(t ) + A2h2(t )

A1h1(t ) + A2h2(t ) + A3h3(t )

A1h1(t ) + A2h2(t ) + A3h3(t ) +

t

T

x

f 1 = 1/T

f 2 = 2 f 1

f 3 = 3 f 1

f 4 = 4 f 1

A1sin(2 f 1t ) = A1h1(t )π

A2 sin(2 f 2t ) = A2h2(t )π

A3 sin(2 f 3t ) = A3h3(t )π

hn(t ) = sin(2 f n t ) f n = n/T

Harmonics:

π t

t

A4 sin(2 f 4t ) = A4h4(t )π

x x

Figure 15.8 Fourier analysis of a periodic function into sinusoidal functions. The period of each function is indicated bshading. (a) Original periodic function. (b) Set of harmonics used to describe the function in (a). (c) Successive sum omonics showing an increasingly improving match to the original periodic function.

* In general, an infinite sum of sines and cosines is requibreak down arbitrary periodic functions. Only periodic funfor which (that is to say, they are odd inas the one shown in Figure 15.8) can be analyzed with just

(which are also odd in t ).

f ( t ) f (t )

we understand simple harmonic motion, we can dealwith any other type of periodic motion by breaking itdown into its simple harmonic components. Because anymotion that is nonrepetitive can be thought of as peri-odic motion having an infinitely long period, Fourier’stheorem has broad implications: the time-dependent

motion of any system can be expressed in terms of sinu-soidal functions.Consider, for example, the periodic motion represent-

ed by the nonsinusoidal curve in Figure 15.8a. Althoughvery different from simple harmonic motion, this mo-tion can be expressed as the sum of a few simple har-monic components, shown in the sinusoidal curves of Figure 15.8b. The first component has the same fre-quency as the original periodic function; thisfrequency, the lowest in the sum, is called the funda-

mental frequency or first harmonic. All other compo-nents, called higher harmonics, are at integer multiples

of the fundamental frequency: .*

Whenthese higher harmonics are added to the first harmon- f n nf 1 nT

f 1 1T

ic, the shape of the resulting periodic function is nnusoidal, but the period remains the same (Fi15.8c). By adjusting the amplitude of each harmonis possible to make the sum of harmonics fit the onal periodic function.

The breaking down of a function into harm

components is called Fourier analysis, and the resusum of sinusoidal functions is called a Fourier s

Figure 15.9 shows one way of graphically representhe Fourier series: the amplitudes An of all the harmics are plotted against the frequencies of the harmics. As you saw in Checkpoint 15.4, the energysimple harmonic oscillator is proportional to the sqof the amplitude. For this reason, it is customary to

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Figure 15.10 Spectrum

analyzer display on astereo component.

T T min

x

f

f 1 = 1 /T

f max = 1 /T min

(a)

(b)

( An)2

2T

4T

6T

t

Figure 15.11 Relationship between a periodic function andits spectrum.

t

Fourieranalysi

Fouriersynthesis

An

(a)

(b)

( An)2

(c)

f

f

2T

4T

6T

x

2T

4T

6T

Figure 15.9 (a) Periodic function and (b) amplitudes of theharmonic functions in the Fourier series of the functionshown in (a). (c) Spectrum of the periodic function, obtainedby plotting the squares of the amplitudes.


An2 as a function of frequency, as shown in Figure 15.9c;

such a plot is called the spectrum of the periodic function.

The spectrum shows the contribution of each harmonic to the original periodic function. A similartype of visual display of frequency content is found onsome stereophonic components; displays such as theone shown in Figure 15.10 provide a visual representa-tion of the frequency content of the sound beingplayed.

Figure 15.11 shows some important relationshipsbetween a periodic function and its spectrum. The pe-riod T of the function determines the frequency of the first harmonic. In other words, the duration of the slowest component of the function determinesthe lowest frequency in the spectrum. Conversely,the duration T min of the fastest component in thefunction determines the highest frequency in the spectrum.

Fourier analysis has many applications, ranging fromelectronic signal processing to chemical analysis andvoice recognition. The ear functions as a Fourier ana-lyzer because the inner ear breaks down the oscilla-tions produced by sound into the simple harmoniccomponents, each producing a nerve impulse propor-tional to the amplitude of that component.

The inverse of Fourier analysis — the creation of periodic functions by adding sinusoidal functions to-gether — is called Fourier synthesis. A musician’s elec-tronic synthesizer, for example, produces a sum of harmonics that add up to produce different periodicfunctions. When the amplitudes of the different har-monics are adjusted so that they match the frequencycontent of real instruments, the synthesizer simulatesthe sounds of many musical instruments.

15.6 (a) What does the spectrum of a single sinu-soidal function of period T look like? (b) As T isincreased, what change occurs in the spectrum?

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0 x

stable unstable neutral

x1

x2

x3

ΣF →

ΣF →

ΣF →

ΣF x = 0

ΣF x

ΣF x = 0

Figure 15.12 Whenever the vector sum of the forces exert-ed on an object is zero, the object is in translational equilib-rium. Three types of equilibrium are represented in thisgraph: stable, unstable, and neutral.

15.4 Restoring forces in simpleharmonic motion

Periodic motion takes place about a position of trans-lational or rotational equilibrium (Section 12.5) and

requires a restoring force that tends to return the object to the equilibrium position.Consider, for example,an object moving along an x axis and subject to the vec-tor sum of forces whose x component is shown graph-ically in Figure 15.12. The object is in translationalequilibrium at each position where the vector sum iszero: at x1, x2, and all values of x > x3. Can the object os-cillate about any of these positions? From the graphwe see that is positive to the left of x1, and so thevector sum points toward x1. To the right of x1

is negative and so the vector sum of forces againpoints toward x1. So the forces exerted on the object

tend to return the object to x1 whenever it moves awayfrom x1. Around x1 the vector sum of forces “restore”the translational equilibrium and so the object can os-cillate around x1.

Because the vector sum of the forces always push-es the object back to x1, this equilibrium position is saidto be stable. You can visualize this position by imagin-ing a ball at the bottom of a bowl: when the ball is displaced in any direction, a restoring force accelerates

F x

F

F x

15.4 Linear restoring forces

it back toward the bottom.On either side of equilibrium position x2, the d

tion of the vector sum of the forces is such that th ject accelerates away from the equilibrium posiSuch an equilibrium position is called unstable becthe forces tend to amplify any disturbance from

librium. Imagine balancing a ball on top of a hillslightest displacement of the ball in any directienough to make it accelerate away from the equium position.

Positions for which x > x3 represent a neutral

librium: moving the ball in any direction has no eon its subsequent acceleration.

What all this tells us is that objects tend to oscabout stable equilibrium positions because the foabout such positions tend to restore the equilibrA consequence of the restoring forces about sequilibrium positions is that,

In the absence of friction a small displacementa system from a position of stable equilibriu

causes the system to oscillate.

15.7 (a) Are there any equilibrium positionthe force-versus-distance graph in Figure 15.13so, is the equilibrium stable, unstable, or neutr(b) Compare the magnitude of the restoring fofor equal displacements on either side of x

Figure 15.13a.

Figure 15.13 shows why so many systems mosimple harmonic motion. As we saw in section 15.2ple harmonic motion requires linear restoring forcthat is, forces that are linearly proportional to theplacement from the equilibrium position. In Check15.7b you saw that the restoring force about the sequilibrium position xo in Figure 15.13 is not liHowever, as illustrated by the successive enlargemin Figure 15.13, the curve, which gives the relationbetween force and position, becomes indistinguishfrom a straight line for small displacements fromequilibrium position xo (say, within the shaded regIn other words,

For sufficiently small displacements away from t

equilibrium position, restoring forces are alwa

linearly proportional to the displacement. Cons

quently, for small displacements objects execu

simple harmonic motion about any stable equili

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part 2part 1

stretched

compressed

inside of curve

outsideof curve

(a)

(b) (c)

part 2 par t 1

→

F 21c

→

F 21c

Figure 15.15 (a) A pole is held fixed at one end and bent atthe other end. (b) The material of which the pole is made iscompressed in some places and stretched in others. (c) The result is a restoring torque that tends to return the pole toits original shape.

part 2 part 3part 1

part 2 part 3part 1

(a)

(b)

(c)part 2

→

F 12 c

→

F 32 c

Figure 15.14 (a) A taut string pulled upward from its

equilibrium position. (b,c) Any small part of the displacedstring experiences forces exerted by its left and right neigh-bors that do not add to zero, causing a nonzero restoringforce.

rium position.

The restoring forces in the examples in Figure 15.3fall into two categories: elastic (b, d) and gravitational(a, c). We saw in Chapter 8 that, for small displacements

from the equilibrium position, the elastic restoringforce of a spring is linearly proportional to displace-ment. The same is true for other elastic restoring forces.Figure 15.14, for example, shows a string displaced fromits equilibrium position. Each part of the stretchedstring is subject to elastic (contact) forces exerted bythe parts immediately adjacent to it. Part 2 in the mid-dle of the string in Figure 15.14b, for instance, is subjectto a force exerted by part 1 and a force exerted by part3. In the equilibrium position, when the string isstraight, the elastic forces exerted by parts 1 and 3 onpart 2 add to zero. As the string is displaced, however,

the elastic forces exerted by the two neighboring partsno longer line up (Figure 15.14c) and their vector sumprovides a restoring force that tends to return thestring to its equilibrium position. For small displace-ments, as we just saw, this restoring force is linearlyproportional to the displacement from the equilibriumposition.

The same line of reasoning applies to the restoringforces that arise when materials are twisted or bent. Fig-

ure 15.15 shows that bending a piece of material com-presses the surface that lies on the inside of the curve andstretches the surface that lies on the outside of the curve.The resulting internal elastic forces cause torques on eachpart of the material, torques that tend to return the ma-terial to its equilibrium shape. For small displacements,the internal elastic forces are linearly proportional to the

displacement, and consequently the restoring torque, too,

x10

0 x

x

xo

x

(a)

(c)

(b)

x10

ΣF x

Figure 15.13 For sufficiently small displacements near a sta-ble equilibrium position, the restoring force is linear in thedisplacement.

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15.4 Linear restoring forces

s

l

x

(a) (b)

F Eb⊥ G

→

→

→

F sb

c

F Eb//

G →

F Eb

G

Figure 15.16 (a) For small angles the horizontal displace-

ment x of a pendulum is almost the same as the length of thearc s along which it is displaced. (b) Free-body diagram forthe pendulum bob.

is linearly proportional to the (rotational) displacement.Let us next examine a pendulum, where the restor-

ing force is due to gravity. Figure 15.16 shows a pendulum bob of mass m suspended from a string of length l . If the bob is much more massive and muchsmaller than the string supporting the bob, we can ig-nore the mass of the string and treat the bob as a par-ticle. Such a pendulum is called a simple pendulum. Thebob is pulled back over an arc of length . As the free-body diagram in Figure 15.16b shows, the bob is subjectto two forces: a contact force exerted by the string anda gravitational force. The restoring force is provided bythe component of the gravitational force perpendic-ular to the string, . From Figure 15.16a, we seethat the magnitude of is equal to andthat . Provided the rotational displace-ment is small, the bob’s horizontal displacement x isapproximately equal to the arc length s, x s, and so

Because the rotational position is definedby , we see that, for small angles,For small angles, therefore, the magnitude of therestoring force exerted on the bob is

= mg( s/l ) (mg/l ) x. The restoring force istherefore linearly proportional to the horizontal dis-placement x of the bob from its equilibrium position.

! Example 15.1: Displaced string

Show that for small displacements the restoring force

mg sin mg

sin . sl

sin sl .

F G

Eb mg sin sin xl

F G

Eb

s

exerted on part 2 in the middle of the displaced sin Figure 15.14 is linearly proportional to displacemof that part from its equilibrium position.

Getting started: Figure 15.14c shows the forceerted by parts 1 and 3 on part 2 when the string isplaced from its equilibrium position. I’ll assume tforces are much larger than the force of gravity onpiece, so that I can ignore gravity in this problem.force that pulls the string away from the equilibposition is not shown, meaning it has been releafter being pulled away from the equilibrium pos

I begin by making a free-body diagram and ching a set of axes (Figure 15.17). The x componenthe two forces cancel; the sum of the y componethe restoring force. The magnitude of the y componis determined by the angle between the elastic fo

and and the x axis.I also make a sketch of the displaced string, sho

the displacement of part 2 from its equilibriumsition. I denote the length of the string in its equium position by l and that of the displaced string b

Devise plan: The forces and are equmagnitude and their y components are determinesin which is equal to . If the displacemesmall, I can assume that the length of the string dn’t change very much from its equilibrium value so

. I can then also consider and to bestant. Using this information, I can express the reing force in terms of the displacement .

Execute plan: From my sketch I see that the reing force is . Because and

l l F c

12 F c

32

y(12 l )

2 F c

12 F c

32

y

F

c12 F

c32

1

y

3

F c

32F c

12F c12 y F c32 y

st r i ng

displacement

equilibrium po

part 2

y

12 l

y

l'12

F F 12xc FF

F F 12yc F F 32y

c →

→ →

→→

F 12c

Figure 15.17

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equal in magnitude, I can write the sum of the y com-ponents as . I also know that

. Combining these tworelationships, I obtain for the magnitude of the restor-ing force: . For smalldisplacements, the term in parentheses is constant and

so the restoring force is, indeed, proportional to thedisplacement .

Evaluate result: The answer confirms a more gen-eral principle: close to equilibrium any restoring forceis proportional to the displacement. I made two as-sumptions to derive my answer. The first is that gravi-ty can be ignored. Indeed taut strings tend to bestraight, indicating that gravity (which would make thestring sag) doesn’t play an appreciable role. The otherassumption I made was that the length of the string

doesn’t change much. This assumption is also justifiedbecause the displacement of a string tends to beseveral orders of magnitude smaller than the stringlength l .

15.8 Using a calculator, determine the percenterror in the approximation for polarangles of 1°, 5°, 10°, and 20°.

In Section 15.1, I showed how oscillations arise froma continuous conversion between kinetic energy andpotential energy. Another way to look at oscillationsis to say that

Oscillations arise from an interplay between

inertia and a restoring force.

When an object that is subject to a restoring force isdisplaced from its equilibrium position, the restoringforce accelerates it back toward the equilibrium position. Once it gets there, however, it has a nonzerovelocity, and its inertia causes it to overshoot the equilibrium position.

This picture of an interplay between inertia andrestoring force allows us to make some qualitative pre-

sin

y

4

y

F restoring 2F c12 y (4F c12l ) y

y(12l ) y(1

2l )

2F c12 y 2F c12 sin

sin

dictions about the period of an oscillator. The largerthe mass of the oscillating object, the harder it is to accelerate, and so the longer the period.* Increasingthe restoring force has the opposite effect: the largerthe restoring force, the larger the object’s accelerationand the shorter the period — a stiff spring causes a

much more rapid oscillation than a soft, sloppy one.The period of an oscillating object increases when

its mass is increased and decreases when the mag-

nitude of the restoring force is increased.

This relationship does not hold for pendulums, however, because for a pendulum the restoring forceand mass are proportional to each other. Increasingthe mass of a pendulum bob makes the pendulumharder to accelerate but increases the magnitude of thegravitational force exerted on the pendulum bob —restoring force — in proportion (see Figure 15.16b).

The two effects cancel, and soThe period of a pendulum is independent of the

mass of the pendulum.

15.9 (a) What effect, if any, does increasing thelength l of a simple pendulum have on its peri-od? (b) A pendulum and an object suspendedfrom a spring are taken to the Moon, where theacceleration due to gravity is smaller than thaton Earth. How does the period of the pendulum

on the Moon compare with that on Earth? (c)How does the period of the suspended object onthe Moon compare with that on Earth?


* You can demonstrate this as follows. Position a plastic or thinwooden ruler on a table so about half its length overhangs theedge of the table. Pull the free end down and release, noting therapid oscillations. Next attach an eraser or other small object ontop of the overhanging part of the ruler and repeat the experi-ment. Because of the increased mass, the period is now much

longer.

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[(H1F)]

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S E L F - Q U I Z

14

S E L F - Q U I Z

1. Is an oscillating object in translational equilibrium?2. A pendulum bob swings through a circular arc defined by positions A and D

in Figure 15.18. (a) At A, B, and C, what are the possible directions of the ve-locity of the bob and of the restoring force exerted on it? (b) What is the di-rection of the bob’s acceleration at A and C ?

3. (a) In the cart-spring system shown in Figure 15.19, is the cart’s speed higherat A or at B? (b) At which of these two positions is the restoring force actingon the cart larger? (c) At which of these two positions is the cart's acceleration larger?

4. A T-shaped wooden structure is balanced on a pivot in the two configurations shown in Figure 15.20. Wconfiguration is in stable equilibrium? Which configuration is likely to oscillate?

ANSWERS

1. No. When an object is in translational equilibrium, the sum of the forces exerted on it is zero. An oscilobject is subject to a variable force that accelerates the object back and forth.

2. (a) The velocity is always tangent to the bob’s trajectory; the restoring force always points along the arc tothe equilibrium position C . At A, the bob reaches its maximum displacement, so its velocity must be zer

restoring force points along the arc towards C . At B, the velocity can be in either direction along the arc abob swings back and forth, but the restoring force always points along the arc towards C . At C , the vecan be to the right or left, but the restoring force is zero because this is the equilibrium position. (b) At Aacceleration is tangent along the arc toward C . At C , the bob does have a centripetal radial acceleration ded vertically upward,but no tangential acceleration.

3. (a) Once released from the stretched position shown, the cart speeds up as it moves left toward the equum position, meaning its speed is higher at A than at B. (b) The restoring force is proportional to the dispment from the equilibrium position and so is larger at B than at A. (c) The acceleration is proportional trestoring force, and so it, too, is larger at B.

4. If the structure in configuration A is tipped slightly to the left as shown in Figure 15.21, the gravitationalexerted on the part to the left of the vertical line through the pivot is larger than that exerted on the pathe right of the pivot and so the torques caused by these forces make it tip farther in the same direction,

from the equilibrium position. Conversely, if it is tipped to the right, it tips farther to the right. This confition is unstable, and so the structure does not oscillate. If the structure in configuration B is tipped slightther way, the torques caused by gravitational forces exerted on the parts on either side of the pivot causmove in the opposite direction, back toward the equilibrium position. Configuration B is stable, and so the ture oscillates.

A

BC

D

A B

A B

xmaxeq.

A B

Figure 15.1

Figure 15.19

Figure 15.20

Figure 15.21

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15.5 Energy of a simple harmonic oscillator

position

at t = 0

vertical

component

at t = 0

1 cycle

x x

t t 1

position

at instant t 1

vertical

component

at instant t 1

2

A

ω

Figure 15.22 Refecircle and x(t ) curvesimple harmonic ostor.

15.5 Energy of a simple harmonic oscillator

In Section 15.2, you saw that there is a correspondence between simple harmon-ic motion and circular motion at constant speed. We begin our quantitative de-scription of simple harmonic motion by exploiting this correspondence.

Figure 15.22 shows a sinusoidally varying time-dependent function describingsome simple harmonic motion represented by a rotating arrow of length A whosetip traces out a circle. The circle is called a reference circle, and the arrow is calleda phasor. The phasor rotates in a counterclockwise direction with a constant rota-tional speed . The vertical component of the phasor varies sinusoidally with time(see Figures 15.6 and 15.7).

Let us first establish a relationship between the rotational speed of the phasorand the frequency f of the corresponding simple harmonic motion. If the phasorcompletes one revolution in a period T , its rotational speed is

(15.1)

and the frequency of the corresponding simple harmonic motion is, as noted inSection 15.1,

. (15.2)

The derived SI unit of frequency is the hertz (Hz) after the 19th-century Germanphysicist Heinrich Hertz, who produced the first radio waves:

. (15.3)

Substituting 1/T from Eq. 15.2 into Eq. 15.1, we find

. (15.4)

So the rotational speed of the phasor is larger than the frequency f of the sim-ple harmonic motion by a factor 2. The reason for this factor is as follows: mea-sures the change in rotational position per unit time, and in one revolution the

2 f

1 Hz 1 s1

f 1

T

t

2

T ,

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16 CHAPTER 15 Periodic motion Quantitative Tools

rotational position changes by 2 . Frequency, on the other hand, measures thenumber of cycles per unit time, and one revolution on the reference circle corre-sponds to one oscillation cycle (Figure 15.22).

In the context of oscillations, is often called the angular frequency becauseit has the same unit (s–1 or inverse second) as the frequency f . Keep in mind, how-ever, that the frequency f and the angular frequency are not equal (Eq. 15.4).To facilitate remembering the distinction between the two, we use the SI unit hertzfor f and the SI unit s–1 for .

The rotational position of the tip of the phasor relative to the positive horizon-tal axis is called the phase of the motion (Figure 15.23). The phase is representedby the symbol . For constant angular frequency, the phase increases linearlywith time:

(constant angular frequency), (15.5)

where the constant is the initial phase at t = 0. The phase tells us at whatpoint in the cycle the motion is; the initial phase is the phase at t = 0. Becauserotational position is a unitless number, the phase is unitless too. You determinethe phase by expressing the angle between the phasor and the positive horizontalaxis in radians and dividing by 1 rad.

The vertical component of A in Figure 15.23 can now be written in the form:

(simple harmonic motion), (15.6)

where the amplitude A of the oscillation is equal to the length of the phasor. Equa-

tion 15.6 gives a general expression for the position of a simple harmonic oscilla-tor as a function of time when the equilibrium position is at x = 0 — a pendulumbob, an object suspended from a spring, a piece of taut string. Because the motionis periodic, x(t ) has the same value each time the phase increases by . For acrest (where x(t ) = A and the sine function in Eq. 15.6 has the value 1), the phasecould be , or plus an integer multiple of ; for a trough (where x(t ) = – A andthe sine function in Eq. 15.6 has the value –1), the phase could be , or plusan integer multiple of .2

3 2

3 2

2

2 2

2

i

(t )

(t )

x(t ) A sin (t ) A sin( t i)

i

(t ) t i

A sin (t 1) A sin

i

(phase

at t = 0)

amplitude

x x

+ A

! A

t t 1

(phase at instant t 1)

i

2

(t 1) = t 1 + i

A

"

Figure 15.23 The phase of a simple harmonic oscillator at any instant is given by the angle between the phasor and the pos-itive horizontal axis measured in the counterclockwise direction.

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15.6 Simple harmonic motion and springs

! Exercise 15.2: Oscillation phases

Consider the spring-cart system of Figure 15.1 again. The cart is pulled away fromits equilibrium position in the positive x direction and then released at t = 0. (a)What is the initial phase of the cart’s oscillation? What is the phase of the cart’s os-cillation (b) half a period later and (c) one period later? What are the cart’s initialphases in the following two situations?(d) The cart is at its equilibrium position

when, at t = 0, it is given a sharp blow in the negative x direction so it starts oscil-lating. (e) The cart is first pulled back a distance d in the positive x direction, andthen, at t = 0 it is given a sharp blow, also in the positive x direction, so it carriesout an oscillation of amplitude A = 2d.

Solution: (a) The cart starts at maximum amplitude in the positive x direction, sothe phasor is pointing straight up and the initial phase is = (90°)(2 rad/360°)/(1rad) = /2."

(b) In one period, the phasor completes a full cycle, so half a period later the pha-sor has traversed half a cycle (180°). The angle between the phasor and the hori-zontal axis is thus 90° + 180° = 270°, and thus = (270°)(2 rad/360°)/(1 rad)

= 3 /2."

(c) In one period the phasor has completed a full cycle, so the phase is equal to theinitial phase + = + = ."

(d) If the cart is in the equilibrium position, the phasor must be parallel to the hor-izontal axis and so the phasor is at 0° or 180°. When the phasor is at 0° the cartmoves in the positive x direction; when it is at 180° the cart moves in the negative

x direction, so the initial phase is = (180°)(2 rad/360°)/(1 rad) = ."

(e) The displacement of the cart is given by Eq. 15.6. At t = 0: x(0) = A sin . I amgiven that x(0) = +d, and A = 2d, so +d = 2d sin , or and so = /6

or 5 /6. Because the cart moves in the positive x direction at t = 0, the initial phasemust be = /6."

By differentiating Eq. 15.6 with respect to time, we can obtain general expres-sions for the x components of the velocity and acceleration of a harmonic oscilla-tor:


(simple harmonic motion). (15.8)

Equations 15.6 through 15.8 are the basic kinematic equations for simple harmon-ic motion. We derived them without considering any particular system — every-thing followed from the fact that simple harmonic motion corresponds to theprojection onto a vertical screen of some object moving in circular motion atconstant speed.

a x

d2 x

dt 2 2 A sin( t

i)

v x dx

dt A cos( t i)

i

i

i

i

i

i

5 2

2 2 2

sin i 12 i

(t )

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15.10 (a) What are the algebraic signs of x, v x, and a x when the phaseis between 0 and ? (b) Repeat for .

Substituting Eq. 15.6 into Eq. 15.8, we find that the x component of the accel-

eration of an object in simple harmonic motion is proportional to the displace-ment and points in the direction opposite the displacement (that is, toward theequilibrium position):


Substituting the definition of the acceleration into Eq. 15.9 we obtain a second-order differential equation in the variable x


This equation is called the simple harmonic oscillator equation, because any sys-tem that satisfies this equation undergoes simple harmonic motion. As we shallsee in the next two sections, the symbols may take on different meanings for dif-ferent types of simple harmonic motion, but the mathematical form is always thatgiven in Eq. 15.10.

Multiplying both sides of Eq. 15.9 by the mass of the oscillating object, we have


The left side of Eq. 15.11 is equal to the x component of the vector sum of theforces acting on the oscillating object, ma x = , and so


Because m and are positive constants, Eq. 15.12 tells us that an object in simpleharmonic motion is subject to a force that always points in a direction opposite theobject’s displacement x from the equilibrium position and is linearly proportion-al to x — in other words a linear restoring force as we already concluded in Sec-tion 15.4

Let us next examine the mechanical energy of a simple harmonic oscillator,using the kinematic equations 15.6–8. The work done by the forces exerted on theoscillating object when it is moving from the equilibrium position to position x isfound by integrating the x component of the position-dependent force exerted on

the oscillator (Eq. 9.22).

, (15.13)

where we have substituted the restoring force given in Eq. 15.12. This work caus-es the kinetic energy of the oscillating object to change, and so Eq. 15.13 gives usthe change in kinetic energy . Working out the integration thus yields

ma x m 2 x

2 (t ) t i (t )

3 2

d2 x

dt 2 2 x

F x m 2 x

a x 2 x

F x

W x

xo

sF x( x)dx x

xo

m 2 x dx

K

For the rest of section

15.5, the only really

mportant result is the facthat the energy is

proportional to the square

of the amplitude. The

rest you can gloss over.

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(15.14)

Because the interaction causing the oscillation is reversible, this change in kinet-ic energy must be provided by some form of potential energy. (In the examples wediscussed, the energy was provided by gravitational potential energy or elastic po-

tential energy of a spring). If we include the object causing the oscillation in oursystem, the system must be closed so . Therefore the changein potential energy of the (closed) system is , or, substituting Eq.15.14:

. (15.15).

If we let the potential energy at the equilibrium position be zero, ,we see that the potential energy of the oscillating system is , andso the mechanical energy is

, (15.16)

or, substituting for x from Eq. 15.6 and for v x from Eq. 15.7,


where we have used the fact that for all . Equation 15.17shows that the mechanical energy of a simple harmonic oscillator is constant andproportional to the square of the amplitude, as we concluded earlier.

15.11 (a) Use Eq. 15.15 to determine the maximum potential energy of a

simple harmonic oscillator. (b) Does your answer to (a) agree with Eq. 15.17?


Whenever an object undergoes reversible deformation as a result of stretching,compressing, bending, or twisting, it exerts an elastic restoring force. Consider, forexample, the spring-cart system in Figure 15.24.The spring exerts on the oscillat-ing cart a force (Eq. 8.20), where k is the spring constant and

x – xo is the displacement of the cart from its equilibrium position at xo, where thespring is relaxed.

We can use this information about the restoring force to obtain an equation forthe cart’s motion in terms of the amplitude, angular frequency, and phase of the mo-tion. If we let the origin be at the equilibrium position, xo = 0, and so

(15.18)

The equation of motion for the cart in the horizontal direction is then

(15.19)

F c scx kx .

F c scx k( x xo)

U K

E K U 0

K m 2 x

xo

xdx m 2[12 x2]

x xo

12 m 2 xo

2 12 m 2 x2

U ( x) 12m 2 x2

U ( xo) 0

U U ( x) U ( xo) 12 m 2 x2

12 m 2 xo

2

E K U 12 mv2

12 m 2 x2

E 12 m 2 A2 cos2( t i)

12 m 2 A2 sin2( t i)

12 m 2 A2

cos2 sin2 1

F cx ma x kx ,

→

F sc

c

x xo

eq

Figure 15.24 When a fastened to a spring is paway from its equilibriumtion xo, the spring exerestoring force on the car

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where m is the cart’s mass, or

(15.20)

or . (15.21)

The solution to this simple harmonic oscillator equation is a sinusoidal function x(t ), and comparison with Eq.15.10 tells us that k/m 2. Therefore the angularfrequency of the oscillation is

. (15.22)

The plus sign in front of the square root means that we must take the positivesquare root because is always positive.

The motion of the cart is given by Eq. 15.6,

. (15.23)

Figure 15.25 shows a number of different sinusoidal functions of the form givenby Eq. 15.23 that satisfy Eq. 15.21. All of the functions x(t ) shown in Figure 15.25complete one cycle in a period T and so they have the same angular frequency ,whose value is fixed by properties of the system — the spring constant k and thecart’s mass m (Eq. 15.22). However, the peak heights tell you the curves do notall have the same amplitude A, and the fact that each curve reaches its maximumat different instants tells you that they have different initial phase . Without in-formation in addition to the values of k and m, we cannot determine the ampli-tude A and initial phase .

In order to determine A and , we have to know the position and x compo-nent of the velocity at some instant. The following examples illustrate how to de-termine these quantities for some specific situations.

! Example 15.3: Cart pulled away and released from rest

A cart of mass m = 0.50 kg fastened to a spring of spring constant k = 14 N/m ispulled 30 mm away from its equilibrium position and then released with zero ini-tial velocity. What are the cart’s position and x component of velocity 2.0 s afterbeing released?

Getting started: I begin by making a sketch of the initial condition and repre-

senting the oscillation by a reference circle and a graph of position versus time(Figure 15.26). I choose the positive x axis to be in the same direction as the cart’sinitial displacement with its origin at the equilibrium position. Because the cart isreleased with zero initial velocity, its initial kinetic energy at x = +30 mm is

and so each time it returns to that position it again has K = 0. This tellsme that the cart’s initial displacement is also its maximum displacement, and so

A = 30 mm. Because the cart begins at its maximum positive displacement, Iknow that the oscillation begins with the phasor straight up and so the initial

K 0,

1

i

i

i

ma x m d2 x

dt 2 kx,

d2 x

dt 2

k

m

x

km

x(t ) A sin km t i

x (mm)0 30

x x

t

→ →

v = 0eq

Figure 15.26

T

(t )

t

all three curvesreturn to their

initial value at t = T

Figure 15.25 Four of the infi-nite number of solutions thatsatisfy Eq. 15.21.

important!

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phase is .

Devise plan: To determine the position of the cart at t = 2.0 s, I can use Eq.15.6. This equation contains the angular frequency, which I can determine fromEq. 15.22, and the amplitude and initial phase, which I already determined. To cal-culate the x component of the velocity of the cart, I can use Eq. 15.7.

Execute plan: From Eq. 15.22, I get for the angular frequency of the oscillation:

= 5.3 s–1.

Substituting this angular frequency, t = 2.0 s, A = 30 mm, and intoEq. 15.6, I obtain for the cart’s position at , t = 2.0 s

."

Substituting the same values into Eq. 15.7, I obtain for the x component of the

cart’s velocity

."

Evaluate result: The magnitude of the value I obtained for x is smaller thanthe amplitude, as it should be. The cart moves a distance of 4 A = 0.12 m in onecycle, which takes 2 = 1.2 s, so the cart’s average speed is (0.12 m)/(1.2 s) = 0.10m/s, indicating that the magnitude of the velocity I obtain is reasonable. The phaseat t = 2.0 s is = (5.3 s–1)(2.0 s) + /2 = 12, which corresponds to 12/(2 ) = 1.9cycle and so the phasor is in the fourth quadrant confirming that x should be neg-

ative. Because the cart is moving in the positive x direction when the phasor is inthe fourth quadrant, the x component of the velocity should be positive, as I found.

15.12 (a) In Example 15.3 is the initial condition that = 0 satisfied by Eq.15.7? (b) Is it correct to say that the x component of the velocity given by Eq.15.7 must be positive whenever x is negative so as to make sure the cart movesback to its equilibrium position?

! Example 15.4: Cart struck with spring already compressed

Cart 1 of mass m = 0.50 kg fastened to a spring of spring constant k = 14 N/m ispushed 15 mm in from its equilibrium position and held in place by a ratchet (Fig-ure 15.27). An identical cart 2 is launched at a speed of 0.10 m/s toward cart 1. Thecarts collide elastically, releasing the ratchet and setting cart 1 in motion. After thecollision cart 2 is immediately removed from the track. What is the maximum com-pression of the spring, and how many seconds elapse between the instant the cartscollide and the instant the spring reaches maximum compression?

v x

(t )

3

2

o 2

km (14 Nm)(0.50 kg) 28 s2

x(2.0 s) (30 mm) sin (5.3 s1)(2.0 s)

2 12 mm

i 2

v x(2.0 s) (5.3 s1)(30 mm) cos (5.3 s1) (2.0 s)

2 0.15 ms

4

(a)

(b)

(c)

Figure 15.27

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Getting started: If I ignore the effect of the spring during the collision, I can saythat the elastic collision interchanges the velocities of the two carts. I make a sketchof the initial condition of the cart, choosing the positive x axis to the right, the ini-tial displacement of the cart to the left, and the equilibrium position at x = 0. I alsodraw a reference circle and sketch the resulting oscillation (Figure 15.28). Withthis choice of axis, the x component of the initial velocity of the cart is = –0.10

m/s. Because m and k are the same as in Example 15.3, the angular frequency isagain = 5.3 s–1.

In contrast to the situation in Example 15.3, the initial displacement of cart 1 isnot equal to the amplitude of the oscillation. The collision increases the cart’s dis-placement from the equilibrium position. In other words, cart 1 moves leftwardimmediately after the collision. It continues moving in this direction until the elas-tic restoring force building in the compressing spring causes it to stop. This posi-tion gives the amplitude of the oscillation.

Devise plan: I can determine the value of this amplitude from the cart’s mechan-ical energy, which is the sum of the initial potential energy in the spring and the ini-tial kinetic energy of the cart. The potential energy in the spring is given by Eq. 9.23;

because the equilibrium position xo is at the origin, this equation reduces to. The kinetic energy is given by . At the position of max-

imum compression all of the mechanical energy is stored in the spring and x = – A,so .

Once I know A, I can determine the initial phase from Eq. 15.6. I can then usethat same equation to solve for t at the position of maximum compression where

x = – A.

Execute plan: (a) The initial kinetic and potential energies are

K = (0.50 kg)(–0.10 m/s)2 = 0.0025 J

(14 N/m)(–15 mm)2

and so Emech = K + U = (0.0025 J) + (0.0016 J) = 0.0041 J.

At the position of maximum compression, all of this energy is stored in thespring, and so 0.0041 J, or with the k value given:

."

(b) Substituting the value for A determined in part a and the initial condition xi 15 mm at t 0 into Eq. 15.6, I obtain

(–15 mm)/(24 mm) = –0.63 or sin–1(–0.63).

Two initial phases satisfy this relation, –0.68 and – + 0.68 = –2.5,but only the latter gives a negative x component of the velocity (see Eq. 15.7) asrequired by the initial condition.

sin i i

x(0) A sin(0 i) (24 mm) sin i 15 mm

A 2(0.0041 J)

14 Nm 0.024 m 24 mm

12 kA2

U 12 kx2

12 1.0 m

1000 mm 2

0.0016 J

12

3

Emech U spring 12kA2

K 12mv2

v x,i

1

2

U spring 12kx2

i i

!15 0

x (mm

x x

eq"

v

igure 15.28

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15.7 Restoring torques

At the first instant of maximum compression , which means. Solving for t yields [ –(–2.5)]/(5.3 s–1)

= 0.17 s."

Evaluate result: The amplitude is larger than the initial displacement from theequilibrium position, as I would expect. From my sketch I see that it takes about

one-eighth of a cycle to go from the position of impact to the position of maxi-mum compression. From Eq. 15.1 I see that the time it takes to complete one cycleis = 2 /(5.3 s–1) = 1.2 s, and so the value I obtained is indeed close toone-eighth of a cycle. The assumption that the influence of the spring can be ignoredduring the collision is justified because the force exerted by the spring is smallcompared to the force of the impact: at a compression of 15 mm, the magnitudeof the force exerted by the spring is (14 Nm–1)(0.015 m) = 0.020 N. The force of im-pact is given by the rate of change in the cart’s momentum, .The magnitudeof the cart’s momentum change is = (0.50 kg)(0.10 m/s) = 0.050 kgm/s. If the collision takes place in 20 ms, then the magnitude of the force of impactis (0.05 kg m/s)/(0.020 s) = 2.5 N, which is more than 100 times as large as the mag-nitude of the force of the spring.

15.13 Suppose cart 2 were not removed from the track of Figure 15.27 im-mediately after the collision but instead were left stationary at the collisionpoint. (a) At what instant do the two carts collide again? (b) Describe the mo-tion of the two carts after this second collision.

! Example 15.5: Vertical oscillations

A block of mass m = 0.50 kg is suspended from a spring of spring constant k = 100

N/m. (a) How far below the end of the relaxed spring at xo is the equilibrium po-sition xeq of the suspended block (Figure 15.29a)? (b) Is the frequency with whichthe suspended block oscillates about this equilibrium position xeq larger than,smaller than, or equal to that of an identical system that oscillates horizontallyabout xo on a surface for which friction can be ignored (Figure 15.29b)?

Getting started: I begin by making a free-body diagram for the suspendedblock, choosing the x axis pointing down. Two forces are exerted on the block: anupward force exerted by the spring and a downward gravitational forceexerted by Earth (Figure 15.30a). When the suspended block is in translationalequilibrium at xeq (which lies below x o) the vector sum of these forces must bezero. When the block is below xeq the spring stretches further causing the magni-

tude of the force exerted by the spring to increase, and so the vector sum of theforces exerted on the block points up. When the block is above xeq (but below theposition x oof the end of the spring when it is relaxed), the spring stretches lesscausing the magnitude of the force exerted by the spring to decrease, and so thevector sum of the forces exerted on the block points down. The vector sum of and thus serves as a restoring force.

I also make a free-body diagram for the horizontal arrangement, showing onlythe horizontal forces (the force of gravity and the normal force cancel out). I let

F c

sb F G

Eb

1

sin( t i) 1

T 2

4

t i 12 t (

12 i)

12

p

t

p mv

F

c sb

F G

Eb

(a)

xo

(b)

Figure 15.29 Example 15.

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the x axis point to the right. In this case the restoring force is the force exerted bythe spring on the block (Figure 15.30b).

Devise plan: In translational equilibrium, the vector sum of the forces exertedon the suspended block is zero, so I can determine the magnitude of the force ex-erted by the spring. I can then use Hooke’s law (Eq. 8.20) to determine the distance

between the equilibrium position and xo. To compare the oscillation frequenciesof the two systems, I should write down the simple harmonic oscillator equation foreach system in the same form as Eq. 15.12.

Execute plan: (a) In translational equilibrium, the vector sum of the forces ex-erted on the block is zero, so

,

where is the displacement of the spring’s end from its relaxed position.Solving for , I obtain

= mg/k = (0.50 kg)(9.8 m/s2)/(100 N/m) = 0.049 m."

(b) For the horizontal arrangement, I have in the situation depicted in my sketch(Figure 15.30b)

, (1)

and so, if I let the origin be at the position of the end of the relaxed spring, xo = 0,the rightmost factor in Eq. 1 reduces to kx and the simple harmonic oscillatorequation becomes

.

Next I turn to the vertical arrangement. In the position illustrated in Figure15.30a, the x component of the upward force exerted by the spring is

. (2)

From part a I know that k( xeq xo) is equal to mg. The x component of the vectorsum of the forces exerted on the block at position x is thus

.

If, as usual, I let the origin be at the equilibrium position, xeq = 0, the equation of motion is identical to Eq. 1, and so the oscillation frequencies f of the two systemsare the same. "

Evaluate result: The two oscillations take place about different equilibrium

F x F GEbx F c

sbx mg k( x xeq) mg k( x xeq)

F c sbx k ( x xo) k ( x xeq) k( xeq xo)

m d2 x

dt 2 kx

F x F c sbx k( x xo)

xeq xo

xeq xo

xeq xo

F x F c sbx F GEbx k ( xeq xo) mg 0

3

2

4

x

(a)

(b)

xo

→

a→

F sb

c

xo

xeq

x

x

x

→

a

→

F Eb

G

→

F sb

c

Figure 15.30

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15.8 Damped oscillations

positions, but the effect of the combined gravitational and elastic force in the ver-tical arrangement is the same as that of just the elastic force in the horizontalarrangement because the force exerted by the spring is linear in the displacement.

15.14 Convince yourself that the argument presented in the solution to Example 15.5 is also valid for displacements of the block above xeq.

15.7 Restoring torques

Some simple harmonic oscillators involve rotational rather than translational displacements. As an example, consider the disk suspended from a thin fiber shownin Figure 15.31. Such a setup is called a torsional oscillator . If the disk is rotated inthe horizontal plane over an angle , the fiber is twisted, and elastic potential en-ergy is stored in the fiber. When the disk is released, it oscillates as energy is con-verted back and forth between elastic potential energy in the fiber and rotational

kinetic energy of the disk. Translational oscillations, as in our simple pendulumand block-spring setup, are due to the interplay between a restoring force and in-ertia; rotational oscillations are due to an interplay between a restoring torqueand rotational inertia.

For a quantitative description of the disk’s oscillation, we use the rotationalform of the equation of motion for the disk, which relates the sum of the torquescaused by the forces acting on the disk to its rotational acceleration (Eq. 12.8)

, (15.24)

where I is the rotational inertia of the disk.

To complete the description, we must relate the torque to the rotational dis-placement — just as we related force to translational displacement for translation-al oscillations. In the case of the rotating disk, the torque is caused by the twistedfiber: a clockwise rotation of the fiber causes a counterclockwise torque and viceversa; larger rotational displacements of the disk cause larger torques. For small ro-tational displacements, we might expect that the relationship between rotationaldisplacement and torque is linear:

. (15.25)

The minus sign indicates that the direction of the torque is opposite the directionof the rotational displacement: the torque tends to rotate the disk back toward its

equilibrium position. This rotational form of Hooke’s law is found to hold for a widevariety of materials. The constant (the Greek letter kappa), called the torsional

constant , depends on the properties of the material being twisted. Just as stiff springs have large spring constants, so do stiff materials have large torsional constants.

If we let the rotational position at equilibrium be zero, , we can combineEqs. 15.24 and 15.25, yielding

( o)

I

m

o 0

0

fiber

support

! m

Figure 15.31 Torsional osci

You can skim section 15.7.

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, (15.26)

or . (15.27)

Notice the similarity between this equation and two earlier ones: Eq. 15.9 for ageneric harmonic oscillator and Eq. 15.21 for a translational oscillator. The mean-ing of the symbols is different, but the mathematical form is identical. The disktherefore executes simple harmonic motion, and the rotational position is asinusoidal function of time.

To find the position as a function of time, we exploit the correspondence be-tween Eqs. 15.21 and 15.27, substituting rotational displacement for translation-al displacement x in Eq. 15.23 and rotational inertia I for mass m and the torsionalconstant for the spring constant k in Eq. 15.22:

, (15.28)

(torsional oscillator), (15.29)

where in Eq. 15.28 is the maximum rotational displacement (the amplitude of the rotational oscillation).

15.15 For the torsional oscillator shown in Figure 15.31, what effect, if any,does a decrease in the radius of the disk have on the oscillation frequency f ?Assume the disk’s mass is kept the same.

The pendulum is another example of a rotational oscillator. For a pendulum, how-ever, the restoring torque is not caused by an elastic force, and so we cannot use Eq.15.25. Instead, we must write an expression for the restoring torque due to gravity andrelate the magnitude of this torque to the rotational displacement of the pendulum.Consider an object suspended about a rotation axis located a distance l

cm from the ob-

ject’s center of mass, as illustrated in Figure 15.32a. The restoring torque is providedby the component of the force of gravity perpendicular to l

cm :

(Figure 15.32b). The lever arm of this force relative to the rotation axis is l cm

, and sofor a rotational displacement , the torque caused by the force of gravity about theaxis is

, (15.30)

where m is the object’s mass. For small rotational displacements, , andso once again we obtain a restoring torque that is linearly proportional to the rotational displacement:

. (15.31)

sin

l cm(mg sin )

F GEo mg sin

m

I

m sin( t i)

(t )

I I d2

dt 2

d2

dt 2

I

(ml cm g)

CM

(a)

(b)

l cm

axis

CM

l cm

→

F Eo

G F Eo//

G

F Eo⊥

G

→

→

→

F so

c

Figure 15.32 (a) Oscillating pen-dulum. (b) Extended free-bodydiagram for the pendulum.

ey point: once again, the

scillation frequency has

he form

sqrt( (stiffness) / (inertia) )

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After is substituted for the left side of Eq. 15.31, we get

. (15.32)

and so the angular frequency of oscillation is

(pendulum). (15.33)

! Example 15.6: The simple pendulum

Suppose a simple pendulum consisting of a bob of mass m suspended from a stringof length l is pulled back and released. What is the period of oscillation of the bob?

Getting started: I begin by making a sketch of the simple pendulum (Figure15.33). I indicate the equilibrium position by a vertical dashed line.

Devise plan: The period of the pendulum is related to the angular frequency(Eq. 15.1). To calculate the angular frequency, I can use Eq. 15.33 with l cm

= l . If Itreat the bob as a particle, I can use Eq. 11.30 to calculate the bob’s rotational in-ertia about the point of suspension, and so I = ml 2.

Execute plan: Substituting the bob’s rotational inertia into Eq. 15.33 I get

.

and so, from Eq. 15.1, 2 /T , I obtain:

."

Evaluate result: My result is independent of the mass m of the bob, in agree-ment with what is stated in Section 15.4. Increasing g decreases the period as itshould: a larger g means a larger restoring force and so the bob is pulled back tothe equilibrium position faster. It also makes sense that increasing l increases theperiod: as my sketch shows, for larger l the bob has to move a larger distance to re-turn to the equilibrium position.

! Example 15.7: Measuring g

The oscillations of a thin rod can be used to determine the value of the accelera-tion due to gravity. A rod that is 0.800-m long and suspended from one end is ob-served to complete 100 oscillations in 147 s. What is the value of g at the locationof this experiment?

Getting started: I begin by making a sketch of the situation. To simplify the

4

T 2

2 l

g

mlg

ml 2 g

l

3

2

1

ml cm g

I

d2

dt 2

ml cm g

I

I I d2 dt 2

1

l

m

Figure 15.33

l

Figure 15.34

The frequency and period for a

"simple pendulum" are

important, and worth remembering.

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problem, I shall assume that the rod pivots about the fixed end (Figure 15.34).

Devise plan: I calculated the rotational inertia of a thin rod about one end inExample 11.11: I = , where m is the rod’s mass and l its length. I can use

this result and Eq. 15.33 to solve for g, noting that l cm

in Eq. 15.33 is half the lengthof the rod ( ).

Execute plan: Substituting the rod’s rotational inertia and the center-of-massdistance into Eq. 15.33, I get

.

Solving for g yields g = .The period of the rod is (147 s)/(100) = 1.47 s, and the angular frequency is =

2 f = 2 (1/T ) = 2 /(1.04 s) = 4.27 s–1. Substituting this information into the ex-pression for g yields g = (4.27 s–1)2 (0.800 m) = 9.74 m/s2."

Evaluate result: Again, the mass of the rod does not appear in the problem, be-cause the restoring force is proportional to the rod’s mass. The value I obtained,9.74 m/s2, is reasonably close to the acceleration due to gravity near Earth’s sur-face, giving me confidence in my solution.

15.16 Can you use the experiment described in Example 15.7 to measure thegravitational constant G?

! Example 15.8: The simple pendulum as a translational oscillator

The angular frequency of a simple pendulum can be calculated by treating thependulum as a translational oscillator and, as we did in Section 15.4, consideringthe effect of the force of gravity on the horizontal displacement of the bob and ig-noring, for small displacements, the slight up and down motion. Show that thistreatment yields the same result as you obtained in Example 15.6.

Getting started: I make a sketch of the pendulum, showing it displaced fromits equilibrium position (Figure 15.35a). In this problem we are going to concen-trate on the horizontal displacement, labeled x. I also draw a free-body diagram forthe pendulum bob (Figure 15.35b).

Devise plan: The component of the gravitational force perpendicular to thestring supporting the pendulum bob, , provides the restoring force. For smalldisplacements, this component is nearly parallel to the x axis and so I can take

to be the restoring force in the x direction. This will permit me to write downthe simple harmonic oscillator equation for the pendulum.

Execute plan: From my free-body diagram, I see that . In

l cm 12l

l cm 12l

1

4

23

23

2l

m12lg

13ml 2

3 g

2l

3

13 ml 2

2

2

F G

Eb

F GEb

mg sin F GEb 3


(a)

(b)

l

s x

x

y→

→

F sbc

→

F EbG

G

F F⊥

F EbG

→F Eb//

Figure 15.35

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the small-angle approximation, this becomes . From the definitionof rotational coordinate, I have where s is the arc length over which thebob is displaced and x the bob’s horizontal displacement (Figure 15.35a). For smalldisplacement and so the restoring force is

,

The simple harmonic oscillator equation for the pendulum bob is thus

,

or .

Comparing this result with Eq. 15.11, d 2 x/dt 2 2 x, yields the same angular fre-quency of oscillation as in Example 15.6: ."

Evaluate result: Given that my result is the same as that obtained via a differ-ent method, I can feel confident that it is correct.

15.17 Imagine placing a pendulum in an elevator. While the elevator accel-erates upward, is the frequency f of the pendulum larger than, smaller than,or equal to that when the elevator is at rest?


Up to this point, we have ignored any dissipation of energy in simple harmonic mo-tion. Mechanical oscillations always involve some friction, however, and so the en-ergy of the oscillation is slowly converted to thermal energy. As this conversiontakes place, the oscillating object slows down. The general term for this slowingdown because of energy dissipation is damping, and the system is said to executea damped oscillation.

Consider, for example, the two systems in Figure 15.36. Figure 15.36a shows ablock oscillating under the action of a spring. In Figure 15.36b the block is replacedby a cart with a vane on top. As the block oscillates, it rubs against the floor, andits motion is slowed down by friction with the floor. The motion of the cart is hin-dered by air drag caused by the vane.

The figure also shows the free-body diagrams for the block and the cart. Eachobject is subject to two forces in the horizontal direction: a restoring force exert-ed by the spring and a damping force — a force due to friction or drag that dissi-pates the energy of the oscillation. Damping forces can take on many forms. Forthe oscillating block rubbing on the floor, the magnitude of the force of kineticfriction is , with k the coefficient of kinetic friction; the direction of this force is always opposite the direction of motion (that is, in the direction op-posite the direction of the velocity vector).

Drag forces exerted by air or liquids at low speeds tend to be proportional to

F GEb mg

4

gl

a x d2 x

dt 2

g

l x

ma x mg

l x

F G

Eb mg

s

l mg

x

l

mg

l x

s x

sl

F k fb k(mg)

[(H1F)]

(a)

(b)

→

v

→

F sbc

→

F scc →

F acd

→

F

→

v

→a

→

a

Figure 15.36 Damped osing systems.

Read text and graphs in section 15.8, but gloss over the equations.

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the velocity of the object: the faster you ride your bicycle, for instance, the largerthe drag force exerted on you by the air. So

(low speed), (15.34)

where the coefficient b, called the damping coefficient , depends on the shape of theobject on which the drag force acts and on the properties of the gas or liquid ex-

erting the force. The SI units of the damping coefficient are kg/s. The minus sign tellsyou that the direction of the drag force is opposite the direction of the velocity.

For most oscillators, velocity-dependent drag forces are the main cause of en-ergy dissipation. In the presence of drag, the equation of motion for the cart inFigure 15.36 becomes

. (15.35)

where k is the spring constant and m the mass of the cart. When we write the x com-ponents of the velocity and the acceleration as time derivatives of x, Equation15.35 can be written in the form

. (15.36)

As long as the damping is not too large, the solution to this differential equationis a sinusoidally varying function whose amplitude decreases with time. The solu-tion, given here without proof, is

, (15.37)

F x F c scx F d

acx kx bv x ma x

F

dao bv

m d2 x

dt 2 b

dx

dt kx 0

x(t ) Ae bt 2m sin( dt i)


x

x x x

x

t

b = 0

t t

t

(a)

(c)

Ae−bt /2m

(d) (e)

(b)

b = 0.025 mω

b = 0.1 mω b = 0.25 mω b = 0.5 mω

Figure 15.37 Damped simpleharmonic motion for variousvalues of the damping coeffi-cient b not so large as to pre-vent oscillations.

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For mechanical oscillators with low friction (piano strings, tuning forks), Q can beas large as a few thousand, meaning the oscillator executes thousands of oscilla-tions before its energy decreases by a factor e–1. A bell with a large Q “rings” longerthan one with a small Q.

15.18 A tuning fork that sounds the tone musicians call middle C oscillatesat a frequency f = 263 Hz. If the amplitude of the fork’s oscillation decreas-es by a factor 3 in 4.0 s, what are (a) the time constant of the oscillation and(b) the quality factor?


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Chapter glossary

Chapter glossary( SI units of physical quantities are given in parentheses)

Angular frequency (s–1): A positive scalar equal tothe rotational speed of a rotating object whose projec-tion produces an oscillation of frequency f , so that

. (15.4)

Amplitude A (m): A positive scalar equal to the mag-nitude of the maximum displacement in a periodic mo-tion.

Damped oscillation: Oscillation whose amplitude de-creases over time due to dissipation of energy.

Fourier’s theorem: Any periodic function with periodT can be written as a sum of sinusoidal functions of frequency , where n is a positive integer.

Frequency f (Hz): A positive scalar equal to the num-ber of cycles per second of a periodic motion. The fre-quency is the inverse of the period

. (15.2)

Fundamental frequency f 1 (Hz): Frequency of the firstterm (n = 1) in the sum of sinusoidal functions of fre-quency that add up to describe a periodicfunction.

Harmonic: Term in a sum of sinusoidal functions usedto describe periodic functions. The lowest frequencyterm is called the first harmonic. The other terms arecalled higher harmonics.

Hertz (Hz): Derived SI unit of frequency: 1 Hz 1s–1.

Oscillation: Periodic back-and-forth motion. Alsocalled vibration.

Periodic motion: Any motion that repeats itself at reg-

ular time intervals, called the period T of the motion.

Phase (unitless): A scalar specifying the time-de-pendent argument of the sine function describing sim-ple harmonic motion:

f n nT

2 f

f n nT

f 1T

(

The constant , the initial phase, specifies the pha

t = 0.

Phasor: Rotating arrow whose component on a veaxis traces out a simple harmonic motion.

Reference circle: Circle traced out by the tip of asor.

Simple harmonic motion: Motion for which theplacement is a sinusoidally varying function of given by

. (

where A is the amplitude, the angular frequency

the initial phase of the motion. The period T

simple harmonic oscillator is independent of the amtude A.

Simple harmonic oscillator equation: Equation sfied by any system that undergoes simple harmmotion:

(15

where x represents some sort of displacement andthe angular frequency of the oscillation. Any suchtem is called a simple harmonic oscillator .

Spectrum: Graph showing the squares of the ampli A of the simple harmonic components of a perfunction versus frequency f . The spectrum provivisual display of the frequency content of the func

Time constant (s): A scalar giving the interval which the energy of a damped simple harmonic o

lator is reduced by a factor 1/e.

Vibration: See oscillation.

i

i

d2 x

dt 2 2 x,

( t ) t i.

x(t ) A sin( t i)

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t

t

screen

x

x1

x2ω →

v

Figure S15.5

34 CHAPTER 15 Periodic motion Solutions

→

F psc

→

F tcc

→

F scc

→

F EcG

→

F csc

Figure S15.1

Solutions

15.1 (a) Two forces are exerted on the spring — one bythe cart, the other by the post — and three forces are

exerted on the cart: one by the spring, one by Earth, andone by the track. Figure S15.1 shows the free-body dia-grams during compression (when the spring is stretched,the direction is reversed for the two horizontal forces inthe diagram). (b) None; the only forces that have anonzero displacement along their line of action are thecontact forces between the spring and the cart, but theseare internal forces and so do no work on the system. (c)The contact force exerted by the cart on the springpoints in the same direction as the force displacement,and so the work done by this force on the spring is pos-itive. This makes sense because, as the spring is com-

pressed, its elastic potential energy increases. (d) Theforce exerted by the spring on the cart is directed in thedirection opposite the force displacement, and so thework done by this force on the cart is negative. Because

and the force displacement is the same forboth forces, the work done by the spring on the cart isthe negative of that done by the cart on the spring.

15.2 (a) Negative. Just before reaching the positionwhere the spring is maximally stretched, the velocityis positive (points left) and just afterwards, it is nega-tive (points right). So the change in velocity and the

acceleration are both negative. (b) It is largest wherethe curvature of the x(t ) curve is largest, when thespring is maximally stretched or compressed at the in-stants represented in Figures 15.2a, 15.2e, and 15.2i. Itis smallest (zero), when the cart is at the equilibriumposition at the instants represented in Figures 15.2c

and 15.2 g.

15.3 (a) a: tangential component of gravitational force;

F c

cs F c

sc

b: vertical component of elastic force in ruler, c: tan-gential component of gravitational force, d:verticalcomponent of elastic force in string. (b) gravitational

potential energy in a and c; elastic potential energy inb and d.

15.4 (a) Factor of 4. From Chapter 9 you know thatthe potential energy of a spring is proportional to thesquare of the displacement from the equilibrium posi-tion. So if the spring is compressed twice as much theinitial potential energy is four times as much. The ini-tial kinetic energy is zero and so the energy is fourtimes as large. (b) The initial compression determinesthe amplitude, and so the energy in the oscillator is pro-portional to the square of the amplitude.

15.5 (a) Upward, see Figure S15.5. (b) Downward.The two shaded regions in Figure S15.5 indicate thatthe shadow moves over increasingly smaller distances x during a given time interval t until it reaches thetop. In other words, the shadow slows down. Thismeans that the direction of the acceleration vector isopposite to that of the velocity vector.

15.6 (a) A pure sinusoidal function requires just a sin-gle term in the Fourier series, and so the spectrum con-sists of a single peak at frequency f = 1/T . The peakheight A, is the square of the amplitude of the func-tion. (b) As T increases, f = 1/T decreases, and the sin-gle peak in the spectrum shifts to a lower frequency.

15.7 (a) At x = xo and for large x, F x = 0, and so theseare equilibrium positions. Only the position at xo is astable equilibrium position. (At very large x, the equi-librium is unstable as the object will tend to accelerate

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Solutions

TABLE S15.8Small angle approximation

polar angle polar angle in degrees in radians sin error (%

1 0.0174533 0.0174524 0.0051

5 0.0872665 0.0871557 0.127010 0.1745329 0.1736482 0.509520 0.3490659 0.3420201 2.0600

in the negative x direction.) (b) The shape of the curvetells you that the magnitude of the restoring force islarger for a negative displacement from xo than for aequal positive displacement.

15.8 If you did not set your calculator to work with

radians before getting the five values, you might haveconcluded that the small-angle approximationis not correct! Because angles are really ra-

tios of arc lengths and radii (Section 11.4),applies only if is expressed in radians. Thus your firststep is to convert each angle from degrees to radians,then (with your calculator set at “rad”) get the sine val-ues. As Table S15.8 shows, the approximation is correctto better than 1% up to rotational positions corre-sponding to polar angles of 10°.

15.9 (a) If the length of the pendulum is increased, the

displacement of the pendulum bob for a given angleincreases, but the restoring force remains the same. Sothe restoring force for a given displacement is smallerand thus the period is longer (for a mathematical proof see Example 15.6). (b) The lower acceleration due togravity on the Moon decreases the restoring force ex-erted on the pendulum, and so the period is increased.(c) Neither mass nor the spring constant is affected bythe smaller gravitational attraction on the Moon, andso the object’s period is the same on the Moon and onEarth.

15.10 (a) x positive, v x positive, a x negative. (b) x neg-ative, v x negative, a x positive.

15.11 (a) Substituting the maximum displacement x = A into Eq. 15.15, you get U = . (b) Yes; at max-1

2 m 2 A2

sin

sin

imum displacement, K = 0 so and E = U .

15.12 (a) Yes. The velocity is given by the derivatiEq. 15.6, as Eq. 15.7 shows. Because , sututing t = 0 into Eq. 15.25 gives = 0No. In Figure 15.26, for example, the cart moves tleft (negative x component of the velocity) even

crossing the position x = 0. Only after reaching xdoes the cart turn around and the x component o velocity become positive.

15.13 (a) Because cart 2 remains in place after thecollision, the carts collide again when cart 1 returits initial position at x = –15 mm. Because the sine tion is symmetric about the maximum, this occurs a time interval twice as long as that required to rmaximum compression. So t = 2(0.17 s) = 0.34 s. (bmediately after the second collision, cart 1 has zerlocity and then begins a new oscillation wit

amplitude of 15 mm. Cart 2 moves away to the riga constant speed of 0.10 m/s.

15.14 If the block is lifted above xeq, Eq. 2 remvalid — the only difference now is that x – xeq is ntive because x is on the other side of xeq. The vesum of the forces is then positive, reflecting thethat the restoring force is now downward.

15.15 Decreasing the radius of the disk reduces ittational inertia (which means it rotates more eaDecreasing I increases (Eq. 15.29) and hence

creases (Eq. 15.4).

15.16 No. The oscillating rod experiment determthe acceleration due to gravity g, not G. The twrelated (Eq. 13.4, ), but the relationcontains Earth’s mass, which cannot be determinedependently. [As you may remember, Earth’s mdetermined from G, which is measured in

g GmER2E

v x cos 2 i 2

st r i ng

displacement

equilibrium positio

part 2

y

12 l

y

l'12

F F 12xc F F 32x

c

F F 12yc F F 32y

c →

→ →

→→

F 12c →

F

Figure S15.9

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36 CHAPTER 15 Periodic motion Solutions

(a) (b)

a = 0

→

F sb

c

→

F sb

c

→

F Eb

G →

F Eb

G

→→ →

a

Figure S15.17

Cavendish experiment (Section 13.5), not the otherway around, see Example 13.3.]

15.17 Larger. The upward acceleration effectively in-creases g, which means an object in your hand feelsheavier. If g increases, the frequency of the pendulum

increases too (Example 15.6). An alternative way tosee that f increases is to look at the free-body diagramsfor the pendulum bob in an elevator at rest (FigureS15.17a) and in an elevator that is accelerating up (Fig-ure S15.17b). For the elevator at rest (or moving at con-stant velocity), the upward vertical component of thetensile force exerted by the string on the bob and thedownward force of gravity add approximately to zero(neglecting the small vertical acceleration due to thependulum motion). When the elevator accelerates up-ward, the tensile force exerted by the string on the bobmust increase so that the vertical component of

becomes larger than the gravitational force. This caus-es the pendulum bob to accelerate along with the ele-vator. If becomes larger, however, the horizontalcomponent also increases, and so the restoring force

F c

sb

F c

sb

for a given displacement increases, which increases f .

15.18 (a) The time-varying amplitude is given by, where A is the initial amplitude. If theamplitude decreases by a factor 3, xmax/ A = 1/3 and so

. Taking the natural logarithm (Ap-pendix ••) of both sides, you get =ln(1/3) = –1.1, = (–2.0 s)/(–1.1) = 1.8 s. (b)

(263 Hz)(1.8 s) = 3.0

xmax Ae t 2

2 f 2 Q

(4.0 s)2 e(4.0 s)2 13

mazur ch15

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