Physics Chapter 15 Oscillations

'

..

Oscillations

CONCEPTS

IN

CONTEXT

The body-mass measurement device shown is used aboard the International Space Station for th e daily measureme nt of th e masses of th e astron aut s. The device consists of a spring coupled to a chair into which the astron aut is strapped. Pu shed by the spring, the chair with the astronaut oscillates back and forth. W e will see in thi s chapter that the frequ ency of oscillation of the mass-spring system dep end s on th e mass, and th erefore the frequency can serve as an indi cator of the ma ss of th e astrona ut.

W hile learn ing abo ut oscillating systems , we will consider such ques-tion s as:

?

W hen the spring pushes and pulls the astronaut, wh at is the position of the astronaut as a functio n of time ? The velocity of the astronaut? (Example 4, page 478)

Cone

-

; n

15.1 Simple Harmonic Motion

?

What is th e total mechani cal ener gy of the astronaut- spring system? What are the kineti c and potential energies as th e spring begins to push? At later times? (E xampl e 5, page 482)

?

Good oscillato rs have low friction. H ow do we measure the quality of an oscilla-tor? (Ex ample 10, page 490)

T

he motion

of

a p art icle or

of

a system

of

p articles is periodic, or cyclic ,

if

it repeats again an d agai n at regular interv als oftime . The orbital motion of a planet aro und the Sun, th e uniform rotational motion of a carousel or of a circular saw blade, th e back -rn d- forth mot ion of a pisto n in an auto mobile engine or in a wat er pump, th e swing-:ng motion of a pendulum bob in a grandfather clock, and the vibration of a guitar string are example s of periodic motion s.

If

the periodic motion is a back-and-jOrth motion .dong a straight or curv ed lin e, it is called an oscillation. Thus, the motion of th e piston

:5an oscillatio n, and so are th e mo tio n of the pendulum and th e moti on of th e indi

-vidual particles of the gu itar string.

In this chapter we will examine in some detail the motion of a mass oscillating back and forth under the push and pull exerted by an ideal, massless spring. The equa -rions that we will develop for the description of thi s mass-spring system are of great im portance because analogou s equations also occur in the description of

all

other oscil-lating systems. We will also examine some of these other osciloscil-lating systems, such as :he pendulum .

15.1 SIMPLE HARMONIC MOTION

Simple harm onic mot ion is a special kind of one-dimensional periodic motion . In any of one-dimensional periodic motion, th e particle moves back and forth alon g

J straigh t line, rep eating th e same moti on again and again . In the special case of

simple harmonic motion, thep art icle'sposition can be exp ressed as a cosine or a sinef unc-.lon ofti me. As we will see later, the motion of a mass oscillating back and forth under :he push and pull of a spring is simple harm onic (Fig. l5.l a), and so is the moti on of

J pendulum bob swinging back and forth (provided the amplitude of swing is small;

see Fi g. l5 .lb), and so is th e up -and-d own moti on of th e blad e of a sa be r saw Fig. 15

.Ic),

However, in th is first section we will merely deal with th e mathematical description of simple harm onic mo tion, and we will postpone until th e next section (he question ofw hat causes the moti on.

As a num erical example of simple harmonic motion, suppose th at the tip of th e blade in Fig. l5.lc moves up and down between x

=

-0.8 cm and x

=

+0.8 cm (where

.a) (bl (el

-FIGURE 15 .1 (a) T he motio n of a particle oscilla ting back an d for th in response to the push and pu ll of a spring is sim ple harmonic. (b) The mot ion of a pendulum bob is approximately simple ha rmo nic. (e) T he mo tion of a saber saw blade is simple harmonic.

469

Online

Concep t

Tutorial

W hen motor rum, wheel. . . ...blade moves up and down.

470 CHAPTER 15 Oscillations

x

a

-0.8

M otion is simple harmonic if position is a cosine (or sine) function of time. FIGURE 15 .2 P lot of positionvs. time for a case of sim ple harmonic m otion up and down along th e x axis.

simple harmonic motion

period and angular frequency

the x axis is assumed to be vertical); furth er suppose that the blade c.=. pletes 50 up-and-down cycles each second. Figure 15.2 gives a plot :---posi tion of the tip of the blade as a fun ction of tim e. The plot in 15.2 has th e mathematical form of a cosin e function of the time t.

x = 0.8COS(1007Tt)

wh ere it is assumed th at dist ance is measured in centimeters and --in seconds, and it is assumed th at th e "angle"

1007Tt

in the cosine r -

-tion is reckoned in radi ans. [The factor 1007T multiplying t in Eq. (": • has been selected so as to obtain exactly 50 complete cycles each sec -which is typical for saber saws; we will see below in Eq. (15 .5) ho w u;

factor multiplying tin E q. (15.1) is related to the period of the mo ues

Cosines and sines are called harmonic functions, which is why we _

the motion harmonic. For the harm onic motion plotted in Fig. 15.2,

ar .

=

0, the blade tip is at its maximum upward displacement [evaluating Eq. (15.1) at t

=

we have cos 0

=

1, so x

=

0.8 ern] and is just starting to move; at t

=

0.005 S, it

P"

through the midpoint [since coS(1007T X 0.005)

=

cos

(7T/ 2)

=

0, Eq . (15.1) gives x = 0]; at t

=

0.010 S, it reaches maximum downward displacement [COS(7T)

=

-1, so x

=

-em]; at t

=

0.015 S, it again passes through midpoint . Finally,at t

=

0.020 s, the tip rercrr, to its maximum upward displacement, exactly as at t = O-it has completed one

the mot ion and is ready to begin the next cycle.Thus, the

period

T,

or the repeat ti .• the moti on (th e number of seconds for one complete cycle of the motion), is

T

= 0.020 s

and the frequency

f

of the motion, or the rate of repetiti on of the motion (the nu ru:

of cycles per second), is

f = -

1

= - -

1

=

50/s

T

0.020 s

The points x = 0.8 em and x = -0.8 em, at whi ch the x coordinate attains its mac

-mum and mini-mum valu es, are the

turning points

of the motion ; and the po.r.'

x = 0 is the midpoint.

Equation (15.1) is a special example of simple harm onic motion. More genera.' the mot ion of a particle is simple harmonic if the dependence of position on tim e

1-._

th e form of a cosine or a sine fun ction, such as

x

= A

cos(wt

+

0)

(15.-The quantities

A,

co,

and

0

are consta nts. The quantity

A

is called the

amplitude

.-the motion; it is simply th e distance between .-the midpoint

(x

=

0) and either of '._ turning points

(x

=

+

A

or

x

=

-A) .

The quantity w is called the

angular

its value is related to the period

T

To establish the relation ship between wand

T,

nor-that if we increase the time by T (from t to t

+

T), the argument of the cosine in Eq. increases by w

T.

For this to be one cycle of the cosine function, we must require w

T

=

2-;-T hus, the repetition tim e of th e motion, that is, the peri od 2-;-Tofthe motion, is relate

i

to the angular frequen cy by

27T

T=

27T _{or (15.5}

W =

T

W

471 15.1 Simple Harmonic Motion

or w

=

27rf

(15 .6) frequency and angular frequency

N ote th at th e angu lar frequency w and th e frequency

f

differ by a facto r of 27r, -.vhich corresponds to 27r rad ians = 1 cycle. The units of angu lar frequ ency are radi -m s per second (radia ns/s) ,The units of frequency are cycles per secon d (cycles/s) . Like the label revoluti on th at we used in

revl s

in rotational mot ion , th e label cyclein

cyclels

. an

be o mitted in th e course of a calculation, and so can th e label radian in radian/so 3 ut it is useful to ret ain the se lab els wh erever th ere is a chance of confus ion. The

51

urut

of

frequency is called th e hert z (Hz):

1 hert z

=

1 H z

=

1

cyclels

=

l i s

(15 .7)

hertz

(Hz)

For ins tance, in the example of the motion of th e saber saw blade , the per iod of th e 'lotion is

T

=

0.020 s, th e frequ en cy is

f

=

l / T

=

1/ (0.020 s)

=

SOls

=

50 Hz, and .he angular frequency is

w

=

27r

f

=

27r X

SOl s

=

314 radiansls

ere, in th e last step of the calculati on, th e label radians has been insert ed, so as to

':i5tingu ish th e angul ar frequ ency w fro m the ordin ary frequ ency

f

T he argu ment

(wt

+

8 ) of the cosine funct ion is called th e phase of the

oscilla-. n, and th e quantity 8 is called th e phase constant.This cons tan t determi nes at what

mes th e part icle reaches th e point of maxim um displacement, wh en

cos(wt

+

8)

=

1. ne such inst ant is when

· at is, whe n

(15.8) phose constant and time of H ence th e particle reach es th e po in t of maximum di splacem ent at a time

0/

w before

· = 0 (see Fig. 15 .3) . O f course, the particl e also passes th rough thi s point at periodic ntervals before and after this tim e. If the pha se constant is zero (8 = 0), th en th e max-:::1Um displacem ent occurs at t

=

O.

N ote that th e precedi ng equ ations con necting angularfrequency, peri od , and fre-cu ency are formally th e same as the equa tions connecting angular velocity, period, and :iequency of u niform rot ati onal motion [see E qs. (12.4) and (12.5)]. This coincidence

(b) (c)

x

Positive phase constant advances cosine peak to before ! = O.

maximum displacement

x

0= - rr/ 4

Negative phase constant delays cosine peak to after! = O. • GURE 15.3 Examples of cosin e fu nction s cos(wt

+

8) for sim ple harm on ic moti on wi th d ifferent

-hase cons tants. (a) 8

=

O. The par ticl e reac hes maximu m displacem ent at t = O. (b) 8

= 71'/4

(or 45°).

Th e parti cle reach es maximu m displaceme nt befor e t

=

O. (c) 8 = -71'/4 (or -45°).T he particle -eaches maxi mu m dis placem en t af ter t = O.

472 CHAPTER 15 O scillation s

arises from a special geometrical relation ship between simple harm onic and un iform circular mot ion . Sup pose that a particle moves with si monic motion according to Eg .

(15.4),

with amplitude A and angular \ and consider a "satellite" particle th at is constrained to move in un if :-lar motion with angu:-lar velocity w along a circle of radius A, centere midp oint of the harm onic motion, that is, centered on x

=

thi s circle, called th e reference circle. At time t =

satellite are on the x axis at x = th e x axis, so its position is

x = A cos(wt)

M eanwhil e,

th e satellite moves around the circle, and its angular posi

e

= wt

FIGURE 15.4

Par ticle oscilla ting along

th e x axis and satelli te p article mov in g _{Now note that the x coordinate of the satellite is the adjacent side of the tria ngl:} around referen ce circle. T he particle and th e _{in Fi g.}

_15.4:

satellite are always aligned verti cally; that is,

th ey h ave th e sam e x coordinat e. _xsat

=

A co s

e

=

A cos(wt)

Co m paring thi s with E g.

(15.9),

we see th at th e x coordinate of the satelli e coincides with the x coordi nate of th e particle; th at is, the particle and th e : always have exactly the same x motion. This mean s that in Fi g.

15.4

the S;1:

always on that point of th e referen ce circle directly above or directly below the r

_-This geomet rical relation ship between simple harmonic motion and uni fo:-cular mo tion can be used to generate simple harm onic moti on from uniform moti on . Figur e

15.5

shows a simple mech anism for accomplishing this by mea slotted arm placed over a peg that is attached to a wheel in uniform circular rn The slot is vert ical, and the arm is constrained to move horizontally.The pc the role of "satellite," and th e midp oint of the slot in th e arm plays th e role

or";

cle." The peg drags the ar m left and right and makes it move with simple harrr moti on. A mechanism of thi s kind is used in electr ic saber saws and other

dcvi

convert the rotational motion of an electric motor int o the up-and-down motion saw blade or other moving component.

Finally, let us calculate th e instanta neous velocity and instant ane ous acceler in simp le harm onic motion. If the displ acement is

x = Acos(wt + o )

th en differentiation of thi s displacement gives th e velocity

dx

v

= - =

- w/l

sin( wt

+

0)

dt

Circular motion is converted into linear motion.

FIGURE 15.5

Rotating w heel

with a peg driving a slotted arm Slotted arm is constrained to

back and forth . move horizontally.

_{_ I}

( F _

(b)

O. F igure

1:,-0, both th e par ri

cie _

A .

Mter this time, the particle move

_ _ _

-

_-:' -(a)

473 15.1 Simple Harmonic Motion

MATH HELP

DERIVATIVES OF TRIGONOME TRIC FUNCTIONS

Under the assump tion tha t the argumen t of each trigo nometri c function is expressed in rad ians, the derivatives or the sine, cosine, and tan gent are

d . d d 2 b

- srn bu = b cos bu - cos bu = - b sin bu - tan bu = b sec bu = -

-2-du du du cos bu

and differentiation of th is velocity gives the accelerat ion

d2 x dv 2

a

=

- 2

= -

=

- w

A

cos(wt

+

0)

(15.13) dt dt

Here we have used the standa rd form ulas for the derivatives of the sine function and the cosine func tion (see M ath H elp: D erivatives of Tr igonometric Function s). Bear in mind that the argu ments of the sine and cosine func tions in this chapter (and also the next) are always expressed in radians, as required for the validity of the standard for-:nulas for derivatives.

As expected, the instantaneou s velocity calculated from Eq. (15.12) is zero for

wt

+

0 = 0, when the particle is at the turn ing point. Furthermore, the ins tantaneous velocity attains a maximum magnitude of

(15 .14) maximum velocity

.or

co:

+

0 = 7T/ 2, when the particle passes thr ough th e midpoi nt (note th at the

max-unum magn itude of sinwt is 1).

F igure 15.6 shows a multiple-exposure photograph of the oscillations of a particle .n simple harm onic motion .The picture illustrates the variationso f speed in simple har-:nonic motio n: the particle moves at low speed (smaller displacements between snapshots) :-.ear the turn ing points, and at high speed (larger displacements) near the midpoint .

The velocity (15.12) is a sine function, whereas the displacemen t (15.11) is a cosine .u nction ,When t he co sine is at its max im um (say, cos O

=

1), th e sine is small sin 0

=

0); whe n the cosine is smal l (say, cos 7T/ 2 = 0) , the sine is at it s maxim um

fiGURE 15 .6 Seq uence of snapsho ts at

uniform tim e in tervals of an oscillating mass on a spring (a- h). N ote t hat the mass moves slowly at the extreme s of its moti on.

1

x

474

acceleration in simple harmonic motion

(a)

(b) v

When displacement has a large magnit ude, velocity has a small magnitude.

(c)

1:""

""'i

:

t

!

•

t

If

W hen displacemen t is large and positive, acceleration is large and negative.

FIGURE 15 .7 (a) Position , (b) veloci ty, and (c) acceler ation of a particle in simple harmo nic m ot ion as fun cti on s of tim e.

CHAPTER 15 Oscillati ons

(sin 7T/ 2 = 1). Hence the displacement and the velocity are out of step- whe-has a large magnitude, th e other h as a small magnitude, and vice versa. Figure, -and b compare the velocity -and the displacem ent for simple harmonic moti on a :.... ferent tim es. Graphically, th e velocity is the slope of the position vs. time curve. ,

th e position goes through a maximum or minimum, the slope is zero; when the

tion goes through zero , th e magnitude of th e slope is a maximum . Comparison

of

Eqs. (15.11) and (15.1 3) shows that

Thus, the acceleration is always prop ortional to the displacement x, but is in th e

o:r

site direction; see Fi g. 15.7 c. This proportio nality is a characteristic feature of

si

harmonic motio n, a fact th at will be useful in the next section. Even when a

nomenon does not involve motion along a line (for example, rotational motion c -behavior of electric circuits), harmonic beh avior occurs wh enever the second tive of a qu antity is proportional to the negative of that quantity, as in Eq.

(15.

The sine and cosine functions (or a combination of them) are the

only

fun ctio ns . . have thi s property.

Consider the blade of a saber saw m oving up and dow:".

EXAMPLE 1

simple harmonic moti on with a frequ enc y of 50.0 H z, or angular frequen cy of 314 radians/so Suppose that the amplitu de of the

morioc

1.20 em and th at at time t

=

0, the tip of the blade is at

x

=

0 and its

velocin

positive. What is the equation describing the position of the tip of the blade function of time? How long does the blade take to travel from x = 0 to x = 0.60 To 1.20 em?

SOLUTION : The pos ition as function of time is given by Eq. (15.4):

x

=A

cos(wt

+

0)

with to

=

314 radians/s and

A

=

0.0120 m. Since x

=

0 at t= 0, we must adop; • value of

0

such that cos

0

=

O.

The smallest values of

0

th at satisfy thi s cond itio , _

0=

7T/2 and

0

= -7T / 2 (other possible values of

0

differ from the se by :±:27T, = ,:-etc.). From Eq . (15.12), we see th at to obtain a positive value of vat t

=

0, we nee; a negat iv e value of 0; that is, 0 = - 7T/ 2. So the equation describing the motion "

x

=

(0.0120 m) cos [

(314/s)t - ;]

The tip of the blade reache s x

=

0.00 60 m when

0.0060 m = (0.0120 m) cos [ (314/s) t - ; ]

that is, when

cos[(314/ s)t -

7T/2] =

m.

With our calculator we obtain cos- 1 =

- 1.05 radians (here, we have to select a negative sign, since th e argu me nt oft1::

cosine is initially negative, and remains negative until the motion reache s the

fu:.:

amplitu de, x = 0.0120 m). So

7T

475

15.1 Simple Harmonic Motion

from which

-1.05

+

(1T/2)

t = =

0.0017 s

314 /s

To find when the tip of the blade reaches

x =

0.0120 m, we can use Eq. (15.8),

which gives

8

(-1T /2)

t = - - = - =

0.0050 s

w

314 / s

COMMENT:

Note that the time taken to reach a distance of one-half of the

ampli-tude is not one-half of the time taken to reach the full ampliampli-tude, because the

motion does not proceed at constant speed.

In an atomic-force microscope (AFM), a cantilever beam with

EXAMPLE 2

a sharp tip (Fig . 15.8a) oscillates near a surface. We can map

the topog raphy of a surface (see Fig . 15.8b) by slowly moving the tip laterally as it

oscillates vertically, much like a blind person tapping a cane on the ground. The AFM

tip shown in Fig . 15.8 a oscillates with a period of3.0 X

10-

6 s.

The tip

moves up and down with amplitude 9.0 X

10-

8

m. What is the maximum

vertical acceleration of the tip? It s maximum vertical velocity?

(a) (b)

SOLUTION :

As discu ssed above, the largest acceleration occurs at the point of

maximum displacem ent. From Eq . (15.13) thi s maximum acceleration is [since

the maximum value of

cos(wt

+

8)

is 1]

-a _max - w 2A

(15.16)

From Eq . (15.5) and the period T

=

3.0 X

10-

6

s, we obtain the angular frequency

21T

21T

6

w

= -

=

6

=

2.1

X

10 radians/s

T

3.0

X

10- s

Thus, with

A

=

9.0 X

10-

8

m, the maximum acceleration is

This is more than 40000 standard

g's,

an enormous acceleration.

The maximum velocity is, from Eq. (15.12),

v =

wA

=

2.1 X

10

6

_radians/s

_X

_{9.0 X}

_10-

8

_m

₌

_0.19

_m/s

max

FIGURE 15 .8 (a) Atomic-force

micro-scope (AFM) cantilever and tip . (b) AFM image of the surface of a crystal, obt ained by scanning the vibrating tip across the surface. The area shown is 2 fLm X2 us«. The ragged terr aces are single atomic "steps."

CHAPTER 15 O scillation s

476

I _{When displaced and}

«

released, the mass will oscillate about equilibrium.

• x

FIGURE 15 .9 A mass atta ched to a spring slides back and forth on a frict ionless surface . W e reg ard the m ass as a par ticle, whose posi-tion coincides with th e cen ter of th e ma ss.

m

Checkup 15.1

QUESTION 1: I s th e rotation al motion of the E arth abou t its axis periodi c m o ti -O scillator y moti on?

QUESTION 2 : For a particle with sim ple harmoni c motion, at what point of the m does th e velocity attain maximum magnitude? Minimum magnitude?

QUESTION 3 : For a particle with sim ple harmonic moti on, at what point of th e m -doe s the acceler ati on attain maximum magnitude? M inimu m magnitude ? QUESTION 4 : Two particles execute simple harmoni c moti on with the same an:,.

tude. One particle ha s twi ce the frequenc y of the oth er. Comp are th eir maxim -velocities and accelerations .

QUESTION 5 : Are the x coordin at es of the particle and the satellite particle in t:

-15.4

alwa ys th e same? The

y

coo rd ina tes? The veloc ities ? The x comp onents 0:' :: velocities? The accelerations? The x co mpo nents of the acceleratio ns?

QUEST ION 6 : Suppose that a particle with sim ple harm onic moti on passes th r the equ ilibrium point

(x

=

0)

at t =

O.

In thi s case, wh ich of the followin g is a r

-ble valu e of th e phase constant 8 in

x

= A

cos(wt

+

8)?

(A) 0

(B)

7T/4

(C)

7T/2

(D)

37T/4

(E

-15.2 THE SIMPLE HARMONIC OSCILLATOR

The simple harmonic oscillator consists

of

aparticle coupled to an ideal, masslessspri

,-obeys Hooke's Law,

that is, a spring th at provide s a forc e prop ortional to the

elongazi

or compression of the spring . One end of the spring is attached to th e part icle. - _ the other is held

fixed

(see F ig.

15.9).

W e will ign ore gravity and friction , so th e

SF

i - _

force is th e only force acting on the particle. The system has an equilibrium posi - . corresponding to the relaxed len gth of the spring. If th e particle is initially at s distance from this equilibrium position (see Fig.

15.10 ),

then the stretc he d

sori _

supplies a restoring force that pull s th e particle toward th e equilibrium positio n.

-=---particle speeds up as it moves tow ard th e equ ilib riu m p ositi on , and it overs the equilibrium positi on. Then, th e particle begins to com press the spring and ,,: down, coming to rest at the other side of the equilibriu m p ositi on, at a distan ce ec

to its initial distanc e. The compressed spring then pushes the particle back towar " equilibrium position .The particle again speeds up, overshoot s the equilibrium posio and so on. The result is that the particle oscillates back and forth about the equili b ' position-forever if th ere is no frict ion .

The great importance of the simple harm onic oscillato r is that many physical ,-tem s are mathe matically equivalent to simple harmonic oscillators; that is, th ese ' . . ' tems have an equation of motion of the same mathematical form as the simple harmo oscillator.

A

pendulum, the balance wh eel of a watch, a tuning fork, the air in an or", pipe, and the at om s in a diatomic molecule are systems of thi s kind ; the restor ing fo r

and the inerti a are of the same math em atical form in these systems as in the simple

moni c oscillato r, and we can tran scribe th e ge neral mathem atic al results dir ectly fr -the latter to th e fo rmer.

To obtain the equation of motion of the simple harmonic oscillator, we begi n \,.i' · Hooke's Law for th e restoring force exerted by the spring on the particle [co mpare Eq. (6.11)]:

15.2 The Simple Harmonic Oscillator 477

Here the displacement x is measured from the equilibrium position, which corresponds

to x = O. The con stant k is the spring constant. Note th at the force is negative if x is positive (stretched spring; see Fig . 15. LOa) ; and the force is positive if the displace-ment is negative (compressed spring; see Fig. 15. lOb).

With the force as given by Eq. (15.16), the equ ation of motion of the particle is

(15.18)

This equation says that the acceleration of the particle is always proportional to the distance x, but is in the opposite direction. We now recall, from Eq. (15.15), that such a proportionality of acceleration and distance is characteri stic of simple harmonic motion, and we therefore can immediately conclude that the motion of a particle CQU-?led to a spring must be sim ple harmonic motion. By comparing Eqs. (15 .18) and

15.15), we see that these equations become identical if

2 k

w=-m

and we therefore see that the angular frequency w of th e oscillation of the particle on

J spnng IS

(15 .19)

Con sequently, the frequency and the period are

(15.20)

"-TId

1

{;z

T=

f

=

27T\j1;

(15.21)

With the value (15.19) for the angular frequency, the expression (15.4) for the posi-20n as a function of time becomes

(15.22) According to Eq. (15.20) the frequency of the 'motion of the simple harm onic oscil-tor depends only on the spring constant and on the mass. The frequency

of

the

oscilla-:?Tis unaffected by the amplitude with which it has been set in motion-ifth e oscillator has -' frequency of, say,2 Hz when oscillating with a small amplitude, then it also has a fre--=Juency of 2 Hz when oscillating with a large amplitude. This property of the oscilla-:or is called isochronism.

Note that the period is long if the mass is large and the spring constant is small. This is as expected, since in each period the spring must accelerate and decelerate the :nass, and a weak spring will give a large mass only little acceleration.

equation of motion

for simple harmonic oscillator

angular frequency, frequency, and period for simple harmonic oscillator

(a)

-A

+A x

Spring force always acts toward equilibrium position. (b)

I'

jF"

I I .. -A I +A x I

x=o

FIGURE 15.10 (a) Positive d isplacement of the particle; th e force is negati ve. (b) Negativ e displ acement of th e particle; th e for ce is positive.

478 CHAPTER 15 Oscillations

Spring scale oscillates about its shifted equilibrium.

FIGURE 15.11 A h eavy book on a spring scale osc illates up and dow n.

Con<"pl.

- I n

-Context

When you place a heavy encyclopedia, of mass 8

kg,

on a kitchen

EXAMPLE 3

scale (a spring scale; see F ig. 15.11), you notice th at before comi ng to equilibrium , the pointer of the scale oscillates back and forth aroun d the equilibrium positi on a few time s with a period of 0.4 s. What is the effective spring constant of th e internal spring of th e kit ch en scale? (N eglect other m asse in the scale.)

SOLUTION : The mass of8

kg

in conjunction with th e internal spring of th e scale forms a ma ss- and-spring system, to whi ch we can apply E q. (15.2 1). Ifwe square both side s of th is equat ion, we obtain

wh ich gives us

k

= 47T 2 -

m

(15 .23

T

2 With m

=

8

kg

and

T

=

0.4 s, th is becom es

8

kg

k

=

47T2X - - -

=

2

X 103 N /m

(0.4 S) 2

COMMENT: In thi s example, th ere is not only th e for ce of th e spring acting or. the mass, but also the force of gravity on the mass (the weight) and friction forces. The force of gr avity determines whe re the spring will reach equilibrium, but this force has no dir ect effect on the frequency of oscillation arou nd equilibrium. T he fricti on forces cause the oscillations to stop aft er a few cycles , but only slig h tl:" redu ce the freq uenc y (see Secti on 15. 5). For negli gible friction, th e frequ en cv depe nds exclusively on the m ass and th e spri ng constant.

Suppo se th at the astro naut in th e chapter ph ot o has a mass 0

'-EXAMPLE 4

58

kg,

including the ch air device to whi ch she is attached . She and the chair move und er the influence of th e force of a spri ng with

k

= 2.1 X

10"'

N/m .There are no other forces acting. C ons ider th e motion to be along th e

x axis.

with the equilibrium point at x

=

O.

Suppose th at at t

=

0, she is (instantaneously at rest at x

=

0.20 m . Where will she be at t

=

0.10 s? At t

=

0.20 s? What will her veloci ty be when she passes th rou gh the equilib rium point ?

SOLUTION : Since the astronaut is initially at rest at x

=

0.20 m , th is mus t be one of th e turni ng points of th e mo tion; thus, the amplitude of th e moti on mu st be

A

=

0.20 m . Furth erm ore, since at t

=

0 the astronau t is at the turning po int, the phase cons tan t [) = 0 [see E q. (15.8)]. C onsequ entl y, at tim e t = 0.10

s,

th e posi-tion of th e astro naut will be

x =

A

cos wt = 0.20 m X cos( w X 0.10 s)

To evalu ate thi s, we need the angu lar frequ en cy of th e oscillation. By E q. (15.19) thi s is

/ 2. 1 X 103 N /m

w

= - = \

=

6.0 radian s/ s

479 15.2 The Simple Harmonic O scillator

With this value of

w,

x =

0.20 m

X

cos(6 .0 radian s/s

X

0.10 s)

=

0.20 m

X

cos(0.60 radian)

=

0.20 m

X

0.83

=

0.17 m

Likewise, at time

I =

0.20 s, the position will be

x

=

A cos (WI)

=

0.20 m

X

cos (6.0 radian s/s

X

0.20

s)

=

0.20 m

X

cos(1.2 radian)

=

0.20 m

X

0.36

=

0.072 m

The astronaut passes through the equilibrium point when

wI =

ni2 (which

makes cos

wI =

0). To find her velocity when she passes through the equilibrium

point, we take the derivative of

x

with respect to

I,

and then evaluate the resultin g

expression at

wI

=

tt

/2 :

dx d

v

= - =

-(Acoswl)

=

-wAsinwI

dl dt

(15.24)

=

-6.0 radians/s

X

0.20 m

X

sin (n / 2)

=

- 1.2 m/s

Simple harmonic oscillatorsare used as the timekeeping element in modern watches.

These watches use a quartz crystal as a spring-and-mass system. The crystal is elastic,

with a high Young's modulus, and it therefore acts as a very stiff spring.The mass is not

attached as a lump to the end of thi s spring, but it is uniformly distributed over the

volume of the crystal (hence this spring-mass system is said to be "distributed," in

contrast to a "lumped" system with separate springs and masses). The crystal is set

into vibration by electric impulse s, instead of mechanical pushes .The electric circuits

attached to the crystal not only keep it vibrating, but also sense the frequency of

vibra-tion and control the display on the face of the clock.

The advantage of the quartz crystal as a timekeeping element is that the vibr

a-rions of the crystal are extremely stable , because any accelerations from bumping the

watch are completely negligible compared with the immense accelerations of the

oscil-lating masses in the crystal. Ordinary quartz clocks are accurate to within a few

sec-onds per month; high-precision clocks are accurate to within 10-

5

s per month.

Checkup

15.2

QUESTION 1:

For a particle with simple harmonic motion, at what point of the motion

does the force on the particle attain maximum magnitude? Minimum magnitude?

QUESTION

2: Suppose we replace the particle in a simple harmonic oscillator by a

particle of twice the mass. How does this alter the frequency of oscillation?

QUESTION 3 :

If we suddenly cut the spring of a simple harmonic oscillator when the

particle is at the equilibrium point (x

=

0), what is the subsequent motion of the

par-ticle? If we suddenly cut the spring when the particle is at maximum displacement

(x

=

A)?

QUESTION 4 :

Suppose we replace the spring in a simple harmonic oscillator by a

stronger spring, with twice the spring constant. What is the ratio of the new period of

oscillation to the original period ?

CHAPTER 15 Oscillations

480

15.3 KINETIC ENERGY AND

POTENTIAL ENERGY

We know from Section 8.1 that the force exerted by a spr ing is a conservative f

.

for whi ch we can construct a potential energy. With thi s potential energy, we can :', '

-mulate a law of conservation of th e mech anical energy: the sum of the kineti c en, _

and the potential energy is a constant; th at is,

E= K+ U= [constant]

(15.: :

In this section we will see how to calculate the kineti c energy and the potential ener

of the simple harmoni c oscillator at each instant of time, and we will verify explici:

that the sum of the se energies is constant .

The kineti c energy of a moving particle is

K

-

-

21 mv 2

(15.:-For simple harmonic moti on, the speed is given by Eq. (15.12) , and the kinet ic ene:

"-become s

K

= =

+ 8)]2

1 2 ,, 2 . 2( ") = 2m w n sin wt

+

u

Since

mw 2 =

k [see Eq . (15.1 8)], we can also write this as

The potential energy associated with the force

F

=

-kx is [see Eq. (8.6)]

(1L

For simple harmonic motion, with

x = A cos(wt

+

8), this becomes

(15,3

The kineti c energy and the potenti al energy both depend on time. Accordin g

Eq s. (15.28) and (15.30), each oscillates between a minimum value ofzero and a

max-imum value of

2.

Figure 15.12 plots the oscillations of the kinetic energy and

c' . _

potent ial energy as functions of time; for simplicity,we set the pha se constant at 8

= (

At the initi al time

t =

0, the particle is at maximum distance from the equ ilibriu m

point and its instantaneous speed is zero; thu s, the potential energy is at its maxirnurr,

value, and the kinetic energy is zero. A quarter of a cycle lat er, th e particle passe;

throu gh the equilibrium point and attains its maximum speed; thu s, the kineti c energy

is at its maxim um value and the potential en ergy is zero . Thus energy is traded back

and forth betw een potential energy and kinetic energy.

, Since the force

F

= -

kx is conservative, the total mech anical energy E

=

K

+

C-is a constant of the motion. To verify thC-is conservation law for the energy explicitly,we

take the sum of Eqs. (15.28) and (15.30),

E=K+U

=

+

8)

+

+

8)

(15.31)

2(

15.3 Kinetic Energy and Potential Energy 481

I

zero K.E . intermediate K.E . maximum K.E. zero K.E. max.imum K.E. zero K.E.

maximum P.E . intermediate P.E. zero P.E . maximum P.E. zero P .E. maximum P.E.

,,

,,

,

_{..' I'!) y\'l;}

_{,_ .}

,,

I I K _{kinetic energy}

1

M 2

tM

2 0 U

,

T

,

,

,

I I I I I • e ", "t f. f t : rJ)' ----!.-I

"'

-,,

I I T/2 T potential energy

1

M 2 TI2 T E=K+ U total energy

1

M 2 tkA 2

FIGURE 15.12 Kinetic energy and potential energy of a sim ple harm onic oscillator as a function of time.

T

Total energy remains constant.

W e can simplify thi s expression if we use the trigon ometric identity sin

2

e

+

cos 2

e

=

1, which is valid for any angle

e.

With this identity, we find that the right side of E q.

(15.31) is simply

! kA

2 :

(15.32)

Th is shows th at

the energy oJth e motion is constant and isp roportio nal to th e square oJthe amplitude

oj

osciffat ion .

energy of simple harmonic oscillator

CHAPTER 15 Osc illati ons

482

Concepts

- In

-Context

By means of Eq . (15.32), we can express th e maximum displacement in terms or

th e energy. For this, we need only solve Eq. (15.3 2) for

A:

x

_max

=

A

=

V2E/k

(15.33'1

Lik ewise, we can express the maximum speed in term s of the energy. For this, we note

that when the particle passes through the equilibrium point , the energy is purely kinetic:

E -

1 2

- 2 m V m ax

(15.34,

If

we solve this for

V _{m ax '}

we find

V _{rn ax} =

V2E/m

(15.35)

These equati ons tell us th at both the maximum displacement and the maximum speed

increase with the energy-they both increase in prop orti on to th e square root of th e

energy.

For the 58-kg astronaut (with chair) moving under the influence

EXAMPLE 5

of the spring in the body-m ass measurement device described in

Example 4, what is th e total mechanical energy? What is the kinetic energy and

what is the potential energy at

t

=

O? What is th e kinetic energy and what is the

potential energy at

t =

0.20 s?

SOLUTION:

From Example 4, the amplitude is A

=

0.20 m and the sprin g

con-stant is

k

=

2.1

X

10

3

N /m .The total mechani cal energy is

E

=

=

X

2.1

X

10

3N

/m

X

(0.20

m) 2

=

42J

At

t

= 0, the astronaut is at rest at

x

=

0.20 m.The kineti c energ y is zero and

the potential energy is at its maximum,

At

t =

0.20 s, the astro naut has nonzero speed, and the kinetic energy is given

by Eq. (15.28 ). With 8

=

0 (see E xample 4), we find

K

= =

= X

2.1

X 103

N/m

X

(0.20

m) 2

X

sin

2

(6.0 radians/s

X

0.20 s)

=

36J

(15.36)

The potential energy is given by Eq . (15.30) , again with 8 =

0:

U

=

=

X

2.1

X

10

3

N /m

X

(0.20

m) 2 X

cos

2(6.0

radian s/s

X

0.20 s)

=

6 J

(15.37)

COMMENT:

Note th at th e sum of th e kineti c and potential energi es is K

+

U

=

483 15.3 Kinetic Energy and Potential Energy

The hydrogen molecule (H₂₎ may be regarded as two particle s

EXAMPLE 6

join ed by a spring (see Fi g. 15.13). The center of the spring is the center of m ass of th e mole cule . This point can be assumed to rem ain fixed , so this molecul e co nsists of two identical simple harm oni c oscillators vibr ating in opp osite dir ection s. The spring constant for each of the se oscillators is 1.13 X

10

3

N/m ,

and th e mass of each hydrogen atom is 1.67 X 10-27

kg.

Find the fre-

- H

-,

,

,

quency of vibration in hertz. Suppose th at the tot al vibrational energy of the

mol-ecule is 1.3 X 10 - 19

J.

Find th e correspo nd ing amplitude of oscillation and th e

FIGURE 15 .13 A hnirogen molecule, maximum speed.

rep resenred a; two ?:J:ci.: s jo i cd bv a

SOLUTION: The frequ ency is given by Eq. (15.20): spri ng . Th e parti cle- move svrn me tri callv relative to th e ce nter 0 :' :: 1l;;.

3

=

.L

{i

=

X 10

N /m)

112

=

1.31 X 1014Hz

f

2'Tf

\j;;

2'Tf 1.67 X 10-27 kg

Thus molecular vibrational frequ encie s can be quite high, about a hundred thou-sand billion cycles per second.

E ach atom has half the total energy of the molecule; thu s, the energy per atom is

E

=

1.3 X 10 - 19

J =

6.5 X 10 -20

J

According to Eqs. (15 .33) and (15.35), th e amplitude of oscillation and th e max-im um speed of each atom are then

J¥

_E

20

_J

X 6.5 X 10 - - 11 x = - = = l.lX lO m max k 1.13 X 103

N /m

and

J¥

E

2 X 6.5 X 10 20

J

3

v

= -

=

=

8.8 X 10

m / s

max m 1.6 7 X 10-27 kg half of H '

Checkup 15.3

QUESTION 1: Two harmonic oscillators have eq ual masses and spring con stants. One of them oscillates with twice the amplitude of the other. C ompare the energies and compare the maximum speeds attain ed by th e particles.

QUESTION 2: Two harmonic oscillators have equal spring constants and amplitudes of oscillation . One has twice th e mass of the other. C ompare the energies and the max-imum speeds attained by the particles.

QUESTION 3: The period of a simple harmonic oscillator is 8.0 s. Suppose that at som e tim e the energy is purely kinetic. At what later time will it be purely potential? At wh at later time again purel y kineti c?

QUESTION 4 : If the particle in a simple harmonic oscillat or experi en ces a frictional force (say, air resist ance), is the ener gy con stant? Is the amplitude

A

constant?

QUESTION 5 : The mass, frequency, and amplitude of one oscillator are each twice that of a second oscillato r. What is the ratio of their stor ed energies,

E

1/

Ez?

484

CHAPTER 15 Oscillations

Online

15 .4 THE SIMPLE PENDULUM

Concept

Tutorial

A

simple pendulum

consi sts of a bob (a mass) suspended by a string or a ro . .:. som e fixed point (see Fig. 15.14). The bob is assumed to behave like a particle oi rn, and the string is assumed to be massless. Gravity acting on the bob provides a res; ing force . When in equilibrium, the pendulum hangs vertically, just like a plu mb ' --When released at some angle with the vertical, the pendulum will swing back Swinging mass m is

assumed concentrated at a distance I.

Wh en released, mass will swing down toward equilibrium. I I r I I I I I I

ie

I I I I I I I I I

forth along an arc of circle (see Fig. 15.15). The motion is two-dimensional; how. the position of the pendulum can be completely described by a single paramett :-: .• angle

e

between the string and the vertical (see Fig. 15.14) . We will reckon thi s _:: _ as positive on the right side of the vertical, and as negative on the left side .

Since the bob and the string swing as a rigid unit, the motion can be regard - _ rotation about a horizontal axis through the point of suspension, and the equati -motion is that of a rigid body [see Eq. (13.19)):

fa = T (E ..=.

-Here the moment of inertia L and the torque T are reckoned about the horizontal s

through the point of suspension, and

a

is the angular acceleration.

Figure 15.16 shows the "free-body" diagram for the string-bob system w itr;

the external forces. These external forces are the weightw of magnitude w

=

mg

acti . on the mass m and the suspension force

S

acting on the string at the point of sup.

-FIGURE 15.14

A pendulum swing ing

about a fixedsuspension point. The angle

e

The suspension force exerts no torque, since its point of application is on the axis is reckoned as positive if the deflection of rotation (its moment arm is zero). The weight exerts a torque [see Eq. (13.3)] the pendulum is toward the right, as in this

-mglsin

e

(15.

T =

figur e.

where 1is the length of the pendulum, measured from the point of suspension to c. -center of the bob. The minus sign in Eg. (15.39) indicates that this is a restoring whi ch tends to pull the pendulum toward its equilibrium position.

The moment of inertia

1

of the string-bob system is simply that of a particle ma ss m at a distance

1

from the axis of rotation:

\5

Suspension force 5 exerts

FIGURE 15.15

Stroboscop ic ph otograph _{no torque.}

,

r r I I r I of a sw inging pendulum . The pendulum

moves slowly at the extremes of its mot ion .

Ie

I r I r r r I r I r I w m

FIGURE 15.16

"Free-body"

di agram for the string-bob

Y

Component of weight

perpendicular to string

_el

system. The torque exerted by the exerts a torque wi sin

e,

_{weight w has magnitude wi sin}

_e,

or mglsin

e.

or mglsin

e.

485

15.4 The Simple Pendulum

Hence th e equation of rotational moti on (15 .38) becom es

(15.40) J r

g .

e

a

= - - Sin (15.41) 1

We will solve thi s equation of motion only in the special case of small oscillations about the equilib rium positi on. If

e

is small, we can make the approximation

sine

= e

(15.42)

where the angle is measured in radians (see M ath Help: Small-Angle Approximation s ror Sine , Cosine, and Tan gent; and see Fig. 15.17).

With this approximation, th e equation of mot ion become s

g

a =

- -e

(15.43)

1

or, since th e angular acceleration is a =

d

2

e/

dt

2,

g

- -e

(15.44)

1

This equation has the same mathematical for m as Eq. (15.17) . Comparin g the se two equations, we see th at th e angle

e

replace s the distance x, the angular acceleration replaces the linear acceleration,

1

replaces m, and

g

replaces

k.

H ence the angular motion is simple harmonic. M aking the appropriate replacements in Eq. (15.4), we find that (he moti on is described by the equ ation

e

=

A

cos(wt

+ 8)

(15.45 )

with an angular frequency [comp are Eq s. (15.19) and (15.44)]

(15.46)

w

=17

For small anz les. sinfJ =fJ.

e

- - Q

R!

FIGURE 15.17

lf the angle 13 is

length of the straigh t line PQ is app roxiraa te:v the same as the length of th e circular .'..1' R.

SMALL-ANGLE APPROXIMAT IONS FOR SINE ,

MATH HELP

COSINE , AND TANGENT

With the assumption th at an angle

e

is expressed in radians and that this angle is small, th e trigonometric functions have th e simple approximations

sine =

e

cose

=

1 -

e

2

/2

tane =

e

To und erstand how thes e approxim at ion s come about, consider the small angle

e

shown in Fig. 15.17. The sine of

this angle is sin

e

=

PQ/

I.

If

e

is small, the length of (he st raight lin e

PQ

is appr oxim ately the same as the length of the curved circular arc PR (for small angles, the curved arc is almost a straight line). Thus, sin e = PR JI. But the ratio PR / ! is the definition of the angle

e

expressed in radian s, so sin

e

=

e.

Simil ar arguments give the above approximations for the cosine an d the t an gent. Th ese approxim ati o ns are usually sati sfactory if

e

is less th an ab out 0 .2 rad ians, or about 10°.

486

angular frequency, frequency, and period for simple pendulum

For simplicity, we assume all of mass is concentrated at one point .

FIGURE 15 .18 W oman on a swing .

CHAPTER 15 Osc illatio ns

The frequen cy and the period of the pe ndu lum are the n

and

(1

-No te that these expressions for th e freque ncy and th e period depen d only 0 •. - .

length of the pendulu m and on the acceleration ofgravity; they do not depend on th of th e pendulum bob or on th e amplitude of oscillation (but, of course, our calculac depends on the assumption th at th e angle 0, and thu s the amplitude of motion, is S::'__

Like the simple harmonic oscillator, the pendulum has the property of

isochronis-a-e-itsfreque ncy is (approximately) independent

of

the amplitude w ith which it is suiins:

This prope rty can be easily verified by swinging two pe nd ulu ms of equal lengths ; -by side , wit h different ampli tudes. The pe ndulums will continue to swing in stt ._ ;-a lon g wh ile.

A woman sits in a swing oflength 3.0 m (see Fig. 15.18) . \ \

:-EXAMPLE 7

is the period of oscillatio n of thi s swing?

SOLUTION: W e can regard the swing as a pendulum of an approx ima te lerz -3.0 m. From Eq . (15.47) we then fin d

T

=

21T

Jl

-:

=

21T

2 = 3.5 s

g

9.81 m / s

The "second s" pend ulum in a pend ulum clock built for an a:;::- _

EXAMPLE 8

nomi cal ob servatory has a pe riod of exactly 2.0 s, so each way mo tion of the pen dulum takes exac tly 1.0 s. What is th e len gth of such , "secon ds" pendulum at a place whe re th e acceleratio n of gravity is g

=

9.81 m :;: . At a place wh ere the acceleration of gravity is 9.79 m / s2?

SOLUTION : If we squa re bo th sides of E q. (15 .47 ) and th en solve for th e len_o ":' f, we find

W ithg

= 9.81 m / s2 an d th e kn own per iod

T

= 2.0 s, this gives

2.0

s)2

2

1 =

₍

21T

X 9.81 m/s = 0.994 m

W ith g

= 9.79 m /s' , it gives

2.0 S) 2

2

1= -

X 9.79 m / s = 0.992 m

( 21T

487

15.4 The Simple Pendulum

FIGURE 15.19 T h is elec tro mec ha n ical clo ck, regul at ed by a pendu lum, served as the U.S. frequency sta nd ard in th e 1920s. It s mas ter pendulum is encl osed in the can ister at right .

The most familiar application of pendulums is the construc tion of pendulum clocks.

Up to about 1950, the most accurate clocks were pendulu m clocks of a special design ,

which were kept inside airt ight flasks placed in deep cellars to protect th em from

dis-rurbances caused by variations of th e atmospheric pressure and temp eratur e (see Fig .

15.19). The best of these high-precision pendulu m clocks were accurate to within a

tew thousandths of a second per day. Later, such pendulu ms were superseded by quartz

clocks (see Section 15.2) and then by atomic clocks (see Sectio n 1.3).

Ano ther impor tant application of pendulums is th e measureme nt of th e

acceler-ation of gravity

g.

For th is pur pose it is necessary only to time th e swings of a

pendu-lum of kn own len gt h; the value of g can then be calculated from Eq . (15.4 7). The

pendulums used for precise determin ations of

g

usually consist of a solid bar swinging

about a kn ife edge at one end, instead of a bob on a string. Such a pendulu m

consist-ing of a swconsist-ingconsist-ing rigid body is called a

physical pendulum;

its period is related to its

size and shape.

A physical pendulu m has a mome nt of inertia I about its point

EXAMPLE 9

of suspension, and its cent er of mass is at a distance

d

from this

point (see Fig. 15.20a). Fi nd th e period of thi s pendulum.

SOLUTION :

Figure 15.20b shows the "free-body" diagram for th e pendulum .The

suspension force S has zero mo me nt arm about the pivot, and so exert s no torqu e.

The weight acts at th e cent er of mass, at a distance of

d

from th e point of suspe

n-sion, and it exerts a torque [see Eq. (13.3)]

T =

- mgdsin e

H ence the equation of rot ational motion (15.38) is

fa

=

- mgd sin e

where

a

=

d

2

e/

dt

2

is the angu lar acceleration for the rotatio nal mo tion . With the

usual small-angle appr oxima tion sin e""

e,

this becomes

mgd

-e

t

As in the case of the simple pendulum, we compare this with Eq. (15.17). Since the

second time derivative of

e

is proportional to the negative of

e,

the mo tion will

again be simple harmonic. Hence th e angular frequency of oscillation is

(a)

Rigid body hangs from pivot. I '

1---'

\ 1

" >.

"

,

I ,

1'\ , -,

When displaced from \ \ equilibrium and released, ' '" body swings back and forth. ,¥ / ;;.",,/

(b)

d is distance from pivot to center of mass.

FIGURE 15.20 (a) A ph ysical pe ndulum consisting of a rigi d bodv swing ing about a point of suspe nsion . (b) "Free-bodv' diagr am for the p hysical pe ndulum . The weigh t acts at th e cen ter o f ma ss.

488 CHAPTER 15 O scillatio ns

(15.

.1"-"

and th e period is

T =

27T

=

J

_(15.':'

w mgd

COMMENT:

N ote that for a simple pendulum, th e mom ent of inertia about

point of suspension is

J =

mP

and the distance of the center of mass from this poir.,

is

d

=

f.

Accordin gly, Eq. (15.4 9) yields

T

=

27T V

mf 2jmgf

=

27TV/fi,

whi

'--shows th at the formul a for the period of the simple pendul um is a special case

0:

the gen eral formula for th e physical pendulum.

Fi nally, we must emp hasize th at th e appr oximation con tained in Eq. (15.43 '

-valid only for small angles. If the amplitud e of oscillation of a pendulum is more

tha;

a few degrees-say, more than lOO- the approximation (15.4 3) begins to fail, and t:-.-:

motion of th e pendulum begins to deviate from simpl e harm oni c motion . A t

amplitudes, th e perio d of th e pendulu m depen ds on th e ampl itud e-th e larger .;-

-:-amplitud e, th e larger th e period. For instance, a pendulum oscillating with an

arnpii-tude of 30° has a period 1.7% longer th an th e value given by E q. (15.47).

m

Checkup 15.4

QUESTION 1:

If we shor ten the string of a pendulum to half its original length , wh,

-is the altera tion of the period?The frequency?

QUESTION 2 :

Two pendulums have equal length s, but one has 3 tim es the mass of rc

other. Ifwe wa nt the energies of oscillation to be the same, how much larger must

v, _

make the amplitude of oscillation of the less massive pendulum?

QUESTION 3 :

A un iform metal rod oflength fhangs from one end and oscillates wit...

small am plitude . Such a rod, rotating about one end, has momen t of inertia

J

=

t

m. :

(Table 12.3). What is

io,

the angular frequency of oscillation?

(A)

ViiI

(B) V3g/2f

(C)

V3ii!

(D)

V6ifi

15.5 DAMPED OSCILLATIONS AND

FORCED OSCILLATIONS

So far we have proceeded on the assumption th at th e only force act ing on a simp le

harmonic oscillator or a pe ndulum is th e restori ng force

F = - kx

or the restor ing

torque

T = - mgf

sin

e.

H owever, in a real oscillator or a real pendulum, th ere is alwavs

som e extra force caused by friction. For instance, if the pendulum start s its swingio",

mot ion with some initial ampli tude , the friction against the air and against the point

of suppo rt will grad ually brake th e pendulum, redu cing its amplitud e of oscillation.

Although good oscillators have low friction, some times more friction is desirable for

damping out unwanted oscillations, as with th e kitchen scale of Exa mple 3, so that a

steady, equilibrium positio n can be att ained.

15.5 Damped Oscillations and Forced O scillations 489

I

x decaying envelope.

7/

....

-

...

-

...

_Ol+r\+ "t,r

-FIGURE 15.21

Plot ofposition vs.time

for a particle with damped harmon ic mot ion.

If the friction force is proportional to the velocity, the equation of motion becomes

d

2x

dx

m -

=

-kx-b - (15.50)

dt

2

dt

where b is called the friction constant , or the damping constant . Figure 15.21 is a plot of the position as a fun ction of tim e for a harmonic oscillato r with fairly strong fric-tion. The amplitude of oscillation suffers a noticeable decrease from on e cycle to the next. Such a gradually decreasing oscillation is called damped harmonic motion. The oscillation amplitude decreases exponentially with time, as indic ated by th e dashed line in Fig. 15.21. Increasing the friction shortens the time it take s for the amplitude to decrease, and slows the frequency of oscillation somewhat. If the damping is very large, a displaced "oscillator" merely moves back to its equilibrium position, without oscillating. In Section 32.4, we will exami ne the damped harmonic oscillator in detail.

Since the oscillator must do work against the friction , the mech anical energy grad-ually decreases. The energy loss per cycle is a constant fraction of the energy

E

that th e oscillator has at the beginn ing of the cycle. Ifwe represent the energy loss per cycle by

ti E ,

then !i.E is proportional to E :

(

27f ) ,

(15.51) Q of oscillator

!i.E =

Q

E

H ere, the con stant of proportionality has been written in the somewhat compli cated form

27f

/ Q,

which is th e form usually adopted in engineering.The quantity Q is called th e quality factor of the oscillator. In terms of the damping constant b,

Q=

_(15.52)

b

An oscillator with low friction ha s a high value of Q, and a small energy loss per cycle; an oscillator with high friction has a low value of Q, and a large energy loss per cycle. The value ofQ rou ghly coincides with the number of cycles the oscillator completes before the oscillations damp away sig nificantly. M echanical oscillators ofl ow friction, such as tuning forks or pian o strings, have Q values of a few th ousand; th at is, they "ring" for a few thousand cycles before their oscillation s fade not iceably.

I

I

490 A

_k Amplitude at natural &equencyi s enhanced by quality factor Q.

CHAPTER 15 Osc illations

The maximu m displacement from equilibrium of the ..

Con cepts I

EXAMPLE 10

- I n -

mass measurement device described in Examples

4 <1• •': ; Context

0.200 m. Suppose that, because of friction, th e amplitude one cycle later is

O.

. :

What is the quality factor for this damped harmonic oscillator?

SOLUTION :

We can solve for the quality factor

Q

by rearrangin g Eq. (1.5 ..

i :

E

Q

=

2'1T-tJ.E

A t maximum displacement, the tot al energy is all potential energy, so E

=

The spring constant

k =

2.1 X

10

3

N/m was given in Example 4. We

to .. .:.

Example 5 th at the when th e amplitude was A

=

0.200 m, the energy

storec

E

=

=

X

2.1

X

10

3

N/ m

X

(0.200 m)2

=

42J

The energy lost dur ing the cycle is the difference between the energy whe;. ..

amplitude was

A =

0.200 m and the energy one cycle later, when the arnplire ' ..

A'= 0.185 m:

=

X

2.1

X

10

3N

/m

X

[(0.200m)2 - (0.185 m)2]

=

6.1 J

Hence the quality factor is

E

42J

Q

=

2'1T -

=

2'1T

X - -

=

43

tJ.E

6.1 J

To maintain th e oscillations of a damped harmonic oscillator at a constant le. ..

it is necessary to exert a periodic force on the oscillator, so the energy fed into the

oscx-lator by th is extra force compensates for the energy lost to friction.

.."

extra force is also needed to start the oscillations of any oscillator,

dampe.;

or not , by supplying the initial energy for the motion. A ny such exrrc

force exerted on an oscillator is called a driving force. A familiar exam

-pie of a driving force is the "pumping" force that you must exert on a pia:"

ground swing (a pendulum) to start it moving and to keep it

movi r.z

at a constant amplitude. This is an example of a

periodicdrivin g

force,

_ Q= 10

With th e addition of a harm onic driving force of amplitude F

_o

a . "

angular frequency

to ,

the equation of motion (15.50) become s

d

2

x

dx

m - = - kx - b -

+

F. cos

wt

(15.53

dt

2

_dt

₀

If the frequency

w

of the driving force coincides with the frequen cv

Woof the natural oscillations of the oscillator, then even a quite small 00

\'-ing force can gradually build up large amplitudes. Under thes e condi ..

tions the driving force steadily feeds energy int o the oscillations, and

th e amplitude of th ese grows until the friction becomes so large that it

inhibits further growth. The ultimate amplitude reached depend s on the

amount of friction; in an oscillator oflow friction, or high

Q,

this

ulti-Naturalfrequency Wa

y t - - - w Fa

3-_k

FIGURE

15 .22 Am plitude of a forced damped harmonic oscillator as a function of the frequency of the oscillating force.

mate amplitude can be extremely large. The buildup of a large

ampli-tude by the action of a driving force in tune with the natural frequency

of an oscillator is called resonance. Figu re 15.22 shows the value of th e

final amplitude of oscillation att ained as a function of the frequency of