physics chapter 15 oscillations

' ..

Oscillations

CONCEPTS IN CONTEXT The body-mass measurement device shown is used aboard the International Space Station for the daily measureme nt of the masses of the astronaut s. The device consists of a spring coupled to a chair into which the astron aut is strapped. Pushed by the spring, the chair with the astronaut oscillates back and forth. We will see in thi s chapter that the frequ ency of oscillation of the mass-spring system depends on the mass, and therefore the frequency can serve as an indicator of the mass of the astronaut.

W hile learn ing about oscillating systems , we will consider such ques-tions as:

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

Cone - ;n -

Conte

15.1 Simple Harmonic Motion

? What is the total mechani cal energy of the astronaut- spring system? What are the kineti c and potential energies as the spring begins to push? At later times? (Example 5, page 482)

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

The motion ofa p art icle or ofa system ofparticles is periodic, or cyclic , if it repeats again an d again at regular intervals oftime. The orbital motion of a planet around the

Sun, th e uniform rotational motion of a carousel or of a circular saw blade, the back -rnd-forth mot ion of a pisto n in an auto mobile engine or in a wat er pump, the swing-:ng motion of a pendulum bob in a grandfather clock, and the vibration of a guitar string are example s ofperiodic motions.Iftheperiodic motion is a back-and-jOrth motion .dong a straight or curv ed lin e, it is called an oscillation. Thus, the motion of the piston :5 an oscillatio n, and so are the motio n of the pendulum and the moti on of the indi -vidual particles of the guitar 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 analogous equations also occur in the description of all other oscil-lating systems. We will also examine some of these other oscillating systems, such as :he pendulum.

15.1 SIMPLE HARMONIC MOTION Simple harm onic mot ion is a special kind ofone-dimensional periodic motion . In any

of one-dimensional periodic motion, the particle moves back and forth along J straigh t line, repeating th e same motion 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 harmonic (Fig. l5.l a), and so is the motion 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-down motion of th e blade of a saber saw Fig. 15.Ic), However, in th is first section we will merely deal with the mathematical

description of simple harmonic mo tion, and we will postpone until the next section (he question ofwhat causes the motion.

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 motion of a particle oscilla ting back and for th in response to the push and pu ll of a spring is simple 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 Concept

Tutorial

W hen motor rum, wheel. . .

. . .blade moves up and down.

470 CHAPTER 15 Oscillations

x

a

-0.8 Motion 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 motion up and down along th e x axis.

simple harmonic motion

period and angularfrequency

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 the mathematical form of a cosine 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 that the "angle" 1007Tt in the cosine r - -

tion is reckoned in radians. [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) how u;

factor multiplying tin Eq. (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 harmonic 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 motion and is ready to begin the next cycle.Thus, the period T,or the repeat ti .• the motion (the number of seconds for one complete cycle of the motion), is

T = 0.020 s

and the frequencyf 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 which the x coordinate attains its mac -mum and minimum 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 harmonic motion. More genera.' the motion of a particle is simple harmonic if the dependence of position on tim e 1-._ the form of a cosine or a sine function, such as

x = A cos(wt + 0) (15.-

The quantities A, co, and 0 are constants. The quantity A is called the amplitude .-the motion; it is simply the 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 relationship 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 the motion, that is, the peri od Tofthe motion, is relate i to the angular frequen cy by

27TT= 27T (15.5 or W = TW

The repetition rate, or the frequency of the motion, is l iT, so we may write

471 15.1 Simple Harmonic Motion

or w = 27rf (15 .6) frequency and angular frequency

Note th at th e angu lar frequency w and the frequencyf 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 , the label cycle in cyclels . an be omitted in the course of a calculation, and so can th e label radian in radian/so 3 ut it is useful to ret ain the se labels wh erever th ere is a chance of confus ion. The 51 urut offrequency is called the hertz (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, the frequency isf = l / T = 1/ (0.020 s) = SOls = 50 Hz, and .he angular frequency is

w = 27rf = 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

':i5tinguish th e angular frequ ency w from the ordin ary frequ encyf T he argu ment (wt + 8) of the cosine funct ion is called the phase of the oscilla-

. n, and the quantity 8 is called the phase constant.This cons tan t determi nes at what mes the part icle reaches the point of maxim um displacement, when cos(wt + 8) = 1. ne such instant is when

· at is, whe n

(15.8) phose constant and timeof

Hence the particle reaches the po in t of maximum displacem 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 time. If the phase constant is zero (8 = 0), then th e max-:::1Um displacem ent occurs at t = O.

Note that th e precedi ng equ ations con necting angularfrequency, period , and fre-cuency are formally the same as the equa tions connecting angular velocity, period, and :iequency of uniform rotational motion [see Eqs. (12.4) and (12.5)]. This coincidence

(b) (c) x

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

maximumdisplacement

x

0= - rr/ 4

Negative phase constant delays cosine peak to after! = O.

• GURE 15.3 Examples of cosine function s cos(wt + 8) for simple harm on ic moti on wi th different -hase cons tants. (a) 8 = O. The par ticle reac hes maximu m displacem ent at t = O. (b) 8 = 71'/4 (or 45°). Th e parti cle reaches maximum displaceme nt before t = O. (c) 8 = -71'/4 (or -45°).T he particle -eaches maxi mu m displacem en t af ter t = O.

472 CHAPTER 15 O scillation s

arises from a special geometrical relationship between simple harm onic and uniform circular motion . Suppose that a particle moves with si monic motion according to Eg . (15.4), with amplitude A and angular \ and consider a "satellite" particle that is constrained to move in un if :-lar motion with angular velocity w along a circle of radius A, centeremidpoint of the harm onic motion, that is, centered on x = thi s circle, called the reference circle. At time t = satellite are on the x axis at x = the x axis, so its position is

x = A cos(wt)

M eanwhil e, the satellite moves around the circle, and its angular posi

e = wtFIGURE 15.4 Par ticle oscilla ting along the x axis and satelli te particle mov ing Now note that the x coordinate of the satellite is the adjacent side of the triangl: around referen ce circle. T he particle and th e in Fig. 15.4: satellite are always aligned verti cally; that is, th ey have th e sam e x coordinat e. A co s e = A cos(wt) xsat =

Co mparing thi s with Eg. (15.9), we see th at the x coordinate of the satelli e coincides with the x coordinate of the particle; th at is, the particle and the : always have exactly the same x motion. This means that in Fig. 15.4 the S;1:

always on that point of the reference circle directly above or directly below the r _-This geomet rical relation ship between simple harmonic motion and unifo:-

cular mo tion can be used to generate simple harmonic motion from uniform motion . Figure 15.5 shows a simple mechanism 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 pcthe role of "satellite," and the midp oint of the slot in the arm plays the role or"; cle." The peg drags the arm left and right and makes it move with simple harrr motion. A mechanism of thi s kind is used in electr ic saber saws and other dcvi convert the rotational motion ofan electric motor int o the up-and-down motion saw blade or other moving component.

Finally, let us calculate th e instantaneous velocity and instant aneous acceler in simple harmonic motion. If the displacement is

x = Acos(wt + o)

then differentiation of thi s displacement gives the velocity

dx v = - = - w/l sin( wt + 0)

dt

Circular motion is converted into linear motion.

FIGURE 15.5 Rotating wheel with a peg driving a slotted arm Slotted arm is constrained to back and forth . move horizontally. _ I

(F _

(b)

O. Figure 1:,-0, both th e parricie _

A . Mter this time, the particle move

_

_ _

- -

_-

:'

-

(a)


MATH HELP DERIVATIVES OF TRIGONOME TRIC FUNCTIONS

Under the assump tion tha t the argumen t of each trigonometri c function is expressed in rad ians, the derivatives or the sine, cosine, and tangent 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

d 2 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 instantaneous 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 the midpoint (note that the max-unum magn itude of sinwt is 1).

Figure 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 motion: the particle moves at low speed (smallerdisplacements 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 the cosine is at its max imum (say, cos O = 1), the 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 Sequence of snapshots at uniform tim e in tervals of an oscillating mass on a spring (a- h). Note that the mass moves slowly at the extreme s of its moti on.

(a) (b) (c) (d) (e) (f) (g) (h)

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 harmonic mot ion as fun ction 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 has a small magnitude, and vice versa. Figure, -and b compare the velocity and the displacement for simple harmonic moti on a :.... ferent times. Graphically, the velocity is the slope of the position vs. time curve. ,

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

tion goes through zero, the magnitude of the 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 Fig. 15.7 c. This proportionality is a characteristic feature of si harmonic motion, 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 behavior occurs wh enever the second

tive of a quantity 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 moving up and dow:".EXAMPLE 1 simple harmonic moti on with a frequ ency of 50.0 H z, or

angular frequency of 314 radians/so Suppose that the amplitude 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 it ion 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 of0 such that cos 0 = O.The smallest values of0 that satisfy this 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 reaches 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 ment of t1:: cosine is initially negative, and remains negative until the motion reache s the fu:.: amplitude, x = 0.0120 m). So

7T(314/ s)t - - = -1052 .


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-halfof the ampli-tude is not one-half of the time taken to reach the full amplitude, 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 ofa 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 . 15.8a 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 discussed above, the largest acceleration occurs at the point of maximum displacement. From Eq . (15.13) this maximum acceleration is [since the maximum value of cos(wt + 8) is 1]

-a - w 2A (15.16) max

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 106 radians/s X 9.0 X 10-8 m = 0.19 m/smax

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 X 2 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 regard the mass as a par ticle, whose posi-tion coincides with the cen ter of the mass.

mCheckup 15.1

QUESTION 1: I s th e rotation al motion of the E arth about its axis periodi c m o ti -O scillatory moti on? QUESTION 2 : For a particle with sim ple harmonic motion, at what point of the m

does th e velocity attain maximum magnitude? Minimum magnitude? QUESTION 3 : For a particle with simple harmonic motion, at what point of the m -doe s the acceleration attain maximum magnitude? M inimum magnitude ? QUESTION 4 : Two particles execute simple harmoni c moti on with the same an:,. tude. One particle has twi ce the frequency of the other. Compare th eir maxim -velocities and accelerations . QUESTION 5 : Are the x coordinates of the particle and the satellite particle in t: -

15.4 always th e same? The y coo rd ina tes? The veloc ities? The x comp onents 0:' :: velocities? The accelerations? The x co mponents of the accelerations? QUEST ION 6 : Suppose that a particle with simple harm onic moti on passes th r the equ ilibrium point (x = 0) at t = O. In thi s case, which of the followin g is a - r -

ble value of the 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 consistsofaparticle coupled to an ideal, masslessspri ,-

obeys Hooke's Law, that is, a spring that provide s a force prop ortional to the elongazi or compression of the spring. One end of the spring is attached to the part icle. - _ the other is held fixed (see F ig. 15.9).W e will ign ore gravity and friction , so th e SFi - _ force is th e only force acting on the particle.The system has an equilibrium posi - . corresponding to the relaxed length of the spring. If th e particle is initially at s distance from this equilibrium position (see Fig. 15.10 ), then the stretched sori _ supplies a restoring force that pull s the particle toward the equilibrium position. -=---particle speeds up as it moves tow ard th e equilib riu m positi on , and it overs the equilibrium position. Then, the particle begins to compress the spring and ,,: down, coming to rest at the other side of the equilibrium positi on, at a distance ec to its initial distance.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 there is no frict ion .

The great importance of the simple harm onic oscillator is that many physical ,-tem s are mathematically 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 atom 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 ematical form in these systems as in the simple

moni c oscillator, and we can transcribe th e ge neral mathem atical results dir ectly fr -the latter to th e former.

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

F= -kx

15.2 The Simple Harmonic Oscillator 477

Here the displacement x is measured from the equilibrium position, which corresponds to x = O. The constant k is the spring constant. Note that 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 equation 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 ofa particle CQU-

?led to a spring must be simple 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 the 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. Thefrequency ofthe oscilla-

:?Tis unaffected by the amplitude with which it has been set in motion-if the 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; the force is negati ve. (b) Negative displ acement of th e particle; th e force is positive.


Spring scale oscillates about its shifted equilibrium.

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

Con<"pl. - I n -

Context

When you place a heavyencyclopedia, 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 position 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 chen scale? (N eglect other masse in the scale.)

SOLUTION : The mass of8 kg in conjunction with the internal spring of the scale forms a mass- and-spring system, to whi ch we can apply E q. (15.2 1). Ifwe square both sides of th is equat ion, we obtain

which gives us

k = 47T 2 -m (15 .23 T 2

With m = 8 kg and T = 0.4 s, th is becomes

8 kg k = 47T2 X - - - = 2 X 103 N /m

(0.4 S) 2

COMMENT: In thi s example, th ere is not only th e force of th e spring acting or. the mass, but also the force ofgravity 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 direct effect on the frequency of oscillation around equilibrium. T he fricti on forces cause the oscillations to stop aft er a few cycles , but only sligh tl:" redu ce the frequenc y (see Secti on 15.5). For negligible friction, th e frequ encv depends exclusively on the mass and th e spri ng constant.

Suppo se th at the astronaut in th e chapter ph oto has a mass 0 '-EXAMPLE 4 58 kg, including the chair device to whi ch she is attached . She

and the chair move und er the influence of th e force of a spring with k = 2.1 X 10"' N/m .There are no other forces acting. Cons ider th e motion to be along the 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 rough the equilibrium point ?

SOLUTION : Since the astronaut is initially at rest at x = 0.20 m, th is mus t be one of th e turning points of th e mo tion; thus, the amplitude of th e moti on must 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 onsequentl 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 evaluate thi s, we need the angu lar frequ ency of the oscillation. By E q. (15.19) thi s is

/ 2.1 X 103 N /m w = - = \ = 6.0 radian s/ s

m 58 kg

479 15.2 TheSimple 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 radians/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 ofvibra-tion and control the display on the face of the clock.

The advantage of the quartz crystal as a timekeeping element is that the vibra-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: Ifwe 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 ?

(A) 1/2 (B) 1/v2 (C) 1 (D) v2 (E) 2

CHAPTER 15 Oscillations480

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 which we can construct a potential energy. With this potential energy, we can :', ' -mulate a law of conservation of the mechanical energy: the sum of the kineti c en, _ and the potential energy is a constant; that 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 harmonic oscillator at each instant of time, and we will verify explici: that the sum of these energies is constant .

The kineti c energy of a moving particle is

K -- 21 mv 2 (15.:-

For simple harmonic motion, the speed is given by Eq. (15.12) , and the kinet ic ene: "-becomes

K = = sin(w t + 8)]2 1 2 ,, 2 . 2( ")= 2m w n sin wt + u

Since mw 2 = k [see Eq . (15.18)], 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

Eqs. (15.28) and (15.30), each oscillates between a minimum value ofzero and a max-imum value of kA 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 phase constant at 8 = ( At the initial time t = 0, the particle is at maximum distance from the equ ilibrium 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; through the equilibrium point and attains its maximum speed; thus, the kineti c energy is at its maximum value and the potential energy is zero . Thus energy is traded back and forth between potential energy and kinetic energy.

, Since the force F = - kx is conservative, the total mechanical energy E = K + C-is a constant of the motion. To verify this 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(= + 8) + cos wt + 8)]

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 1M 2

tM2

0

U

,T ,,, I I I I I

• e", "t f. f t : rJ)' ----!.-

I "' -,, I I

T/2 T

potential energy 1M 2

TI2 T

E=K+ U total energy

1M 2

tkA 2

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

T

Total energy remains constant.

W e can simplify thi s expression if we use the tr igon ometric identity sin2e + cos 2e = 1, which is valid for any angle e. With this identity, we find that the right side of Eq. (15.31) is simply ! kA2 :

(15.32)

This shows that the energy oJthe motion is constant and isp roportional to the square oJthe amplitude ojosciffat ion.

energy of simple harmonic oscillator

0 TI2

CHAPTER 15 Oscillati ons 482

Concepts - In-

Context

By means of Eq . (15.32), we can express th e maximum displacement in terms or the energy. For this, we need only solve Eq. (15.32) for A:

xmax = A = V2E/k (15.33'1

Likewise, we can express the maximum speed in terms 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,

Ifwe solve this for V max ' we find

V rnax = V2E/m (15.35)

These equations tell us th at both the maximum displacement and the maximum speed increase with the energy-they both increase in proporti on to the square root of the 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 the total mechanical energy? What is the kinetic energy and what is the potential energy at t = O? What is the 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 spring con-stant is k = 2.1 X 103 N/m .The total mechani cal energy is

E = = X 2.1 X 103N/m X (0.20 m) 2 = 42J

At t = 0, the astronaut is at rest at x = 0.20 m.The kineti c energy is zero and the potential energy is at its maximum,

At t = 0.20 s, the astronaut has nonzero speed, and the kinetic energy is given by Eq. (15.28 ). With 8 = 0 (see Example 4), we find

K = =

= X 2.1 X 103 N/m X (0.20 m) 2

X sin2(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 = kA 2cos\wt)

= X 2.1 X 103 N /m X (0.20 m) 2

X cos2(6.0 radian s/s X 0.20 s)

= 6 J (15.37)

COMMENT: Note that th e sum of the kineti c and potential energi es is K + U = 36 J + 6 J = 42 J, which agrees with our result for the total mechanical energy.

483 15.3 Kinetic Energy and Potential Energy

The hydrogen molecule (H2) may be regarded as two particle s EXAMPLE 6 joined by a spring (see Fig. 15.13). The center of the spring is

the center of mass of th e mole cule . This point can be assumed to rem ain fixed, so this molecul e consists of two identical simple harm oni c oscillators vibr ating in opp osite directions. The spring constant for each of the se oscillators is 1.13 X 103 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 that the tot al vibrational energy of the mol-ecule is 1.3 X 10- 19J. Find th e correspo nding 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): spring . The particle- move svrn me tri callv relative to th e center 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.

Each atom has half the total energy of the molecule; thus, the energy per atom is

E = X 1.3 X 10 - 19J = 6.5 X 10 - 20 J

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

J¥ 20 JE 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 1.67 X 10- 27 kg

half of

H '

Checkup 15.3

QUESTION 1: Two harmonic oscillators have equal 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 attained 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 some tim e the energy is purely kinetic. At what later time will it be purely potential? At wh at later time again purely kineti c? QUESTION 4 : If the particle in a simple harmonic oscillator experi ences a frictional force (say, air resist ance), is the energy constant? Is the amplitude A constant? QUESTION 5: The mass, frequency, and amplitude of one oscillator are each twice that of a second oscillator. What is the ratio of their stored energies, E1/ Ez?

(A) 2 (B) 4 (C) 8 (D) 16 (E) 32


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 . .:. some 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.

When 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 ofcircle (see Fig. 15.15). The motion is two-dimensional; how. the position of the pendulum can be completely described by a single paramett :-: .• angle ebetween 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 fixed suspension 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 mass m at a distance 1from the axis of rotation:

\5 Suspension force 5 exerts FIGURE 15.15 Stroboscop ic ph otograph no torque. ,

r rII rI

of a swinging pendulum.The pendulum moves slowly at the extremes of its mot ion .

Ie I rI r r rI rI rI

w

m FIGURE 15.16 "Free-body" diagram for the string-bobYComponent 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.

I

485 15.4 The Simple Pendulum

Hence the equation of rotational moti on (15.38) becomes

(15.40)

J r

g . ea = -- 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 eis 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 Tangent; and see Fig. 15.17).

With this approximation, the equation of mot ion become s

g a = - -e (15.43)

1

or, since the angular acceleration is a = d 2e/dt 2,

g- -e (15.44) 1

This equation has the same mathematical form 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, 1replaces 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

-- QR! FIGURE 15.17 lf the angle 13 is length of the straight line PQ is app roxiraa te:v the same as the length of the circular .'..1' R.

SMALL-ANGLE APPROXIMATIONS FOR SINE ,MATH HELP COSINE , AND TANGENT

With the assumption that an angle eis expressed in radians and that this angle is small, the trigonometric functions have the simple approximations

sine = e cose = 1 - e 2/2 tane = e

To und erstand how thes e approxim at ions come about, consider the small angle eshown in Fig. 15.17. The sine of

this angle is sin e = PQ/ I. Ife is small, the length of (he st raight line 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 = PRJI. But the ratio PR / ! is the definition of the angle eexpressed in radian s, so sin e =e. Similar arguments give the above approximations for the cosine and the tan gent. These approxim ati o ns are usually satisfactory 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 Oscillatio ns

The frequen cy and the period of the pendulum 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 the pendulum bob or on th e amplitude of oscillation (but, of course, our calculac depends on the assumption that the 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 property can be easily verified by swinging two pendulums of equal lengths ; -by side , wit h different ampli tudes. The pendulums will continue to swing in stt ._ ;-a long while.

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 ler-z -3.0 m. From Eq. (15.47) we then fin d

T = 21T -: = 21T 2 = 3.5 sJl g 9.81 m / s

The "second s" pendulum in a pendulum clock built for an a:;::- _EXAMPLE 8 nomi cal observatory has a period of exactly 2.0 s, so each

way mo tion of the pen dulum takes exactly 1.0 s. What is th e length of such , "secon ds" pendulum at a place whe re the 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 both sides of E q. (15.47) and th en solve for th e len ":'o f, we find

W ithg = 9.81 m/ s2 and th e kn own per iod T = 2.0 s, this gives

2.0 s)2 21 = 21T X 9.81 m/s = 0.994 m (

W ith g = 9.79 m/s' , it gives

2.0 S) 2 21= - X 9.79 m/ s = 0.992 m ( 21T

487 15.4 The Simple Pendulum

FIGURE 15.19 T his electromec ha nical clock, regul ated by a pendu lum, served as the U.S. frequency sta ndard 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 construction 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 them from dis-rurbances caused by variat ions of th e atmospheric pressure and temperatur 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 Section 1.3).

Ano ther impor tant application of pendulums is the measurement of the acceler-ation of gravity g. For th is purpose it is necessary only to time the swings of a pendu-lum of kn own len gth; the value of g can then be calculated from Eq . (15.4 7). The pendulums used for precise determin ations ofg 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 swinging rigid body is called a physical pendulum; its period is related to its size and shape.

A physical pendulum has a moment of inertia I about its point EXAMPLE 9 ofsuspension, and its center of mass is at a distance d from this

point (see Fig. 15.20a). Find the period of this pendulum.

SOLUTION : Figure 15.20b shows the "free-body" diagram for the pendulum.The suspension force S has zero moment arm about the pivot, and so exert s no torque. The weight acts at the center of mass, at a distance of d from the 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 2e/dt 2 is the angular acceleration for the rotational mo tion . With the usual small-angle approximation 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 derivat ive of e is proportional to the negative of e, the motion will again be simple harmonic. Hence the 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 physical pe ndulum consisting of a rigid bodv swing ing about a point of suspe nsion . (b) "Free-bodv' diagr am for the physical pe ndulum .The weigh t acts at th e cen ter o f mass.

488 CHAPTER 15 O scillatio ns

(15..1"-"

and the period is

T = 27T = J (15.':' w mgd

COMMENT: Note that for a simple pendulum, the moment of inertia about

point of suspension is J = mPand the distance of the center of mass from this poir., is d = f. Accordingly, Eq. (15.4 9) yields T = 27T V mf 2jmgf = 27TV/fi , whi '--shows that the formula for the period of the simple pendulum is a special case 0: the general formula for th e physical pendulum.

Fi nally, we must emp hasize th at the approximation con tained in Eq. (15.43 ' -valid only for small angles. If the amplitude of oscillation of a pendulum is more tha; a few degrees-say, more than lOO-the approximation (15.43) begins to fail, and t:-.-: motion of the pendulum begins to deviate from simple harm oni c motion . At

amplitudes, the period of th e pendulum depen ds on the ampl itud e-th e larger .;- -:-amplitud e, the larger the period. For instance, a pendulum oscillating with an arnpii-tude of30° has a period 1.7% longer than the value given by Eq. (15.47).

mCheckup 15.4

QUESTION 1: Ifwe 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 times the mass of rc other. Ifwe want 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 amplitude . 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 simple harmonic oscillator or a pendulum 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, there is alwavs some 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 gradually brake the pendulum, reducing its amplitud e of oscillation. Although good oscillators have low friction, sometimes more friction is desirable for damping out unwanted oscillations, as with the kitchen scale of Example 3, so that a steady, equilibrium position can be att ained.

15.5 Damped Oscillationsand Forced Oscillations 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 function of tim e for a harmonic oscillato r with fairly strong fric-tion. The amplitude of oscillation suffers a noticeable decrease from one 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 the dashed line in Fig. 15.21. Increasing the friction shortens the time it takes for the amplitude to decrease, and slows the frequency ofoscillation 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 examine the damped harmonic oscillator in detail.

Since the oscillator must do work against the friction , the mechanical energy grad-ually decreases. The energy loss per cycle is a constant fraction of the energy E that the oscillator has at the beginn ing of the cycle. Ifwe represent the energy loss per cycle by tiE , then !i.E is proportional to E:

( 27f ) , (15.51) Q of oscillator !i.E = Q E

H ere, the constant of proportionality has been written in the somewhat compli cated form 27f/ Q, which is the form usually adopted in engineering.The quantity Q is called the quality factor of the oscillator. In terms of the damping constant b,

Q= (15.52)b

An oscillator with low friction has 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 roughly coincides with the number of cycles the oscillator completes before the oscillations damp away significantly.M echanical oscillators oflow 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 oscillations 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

- In - mass measurement device described in Examples 4 <1• •': ; Context 0.200 m. Suppose that, because offriction, the 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 rearranging Eq. (1.5 ..i :

EQ = 2'1T-tJ.E

At maximum displacement, the total energy is all potential energy, so E = . The spring constant k = 2.1 X 103 N/m was given in Example 4. We to .. .:. Example 5 that the when the amplitude was A = 0.200 m, the energy storec

E = = X 2.1 X 103 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 103N/m X [(0.200m)2 - (0.185 m)2] = 6.1 J

Hence the quality factor is

E 42JQ = 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. Any 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 periodicdriving force,

_ Q = 10 With the addition of a harm onic driving force of amplitude Foa . " angular frequency to , the equation of motion (15.50) becomes

d 2x dx m - = - kx - b - + F. cos wt (15.53

dt 2 dt 0

If the frequency w of the driving force coincides with the frequencv Woof the natural oscillations of the oscillator, then even a quite small 00 \'-ing force can gradually build up large amplitudes. Under these condi .. tions the driving force steadily feeds energy into the oscillations, and the amplitude of th ese grows until the friction becomes so large that it inhibits further growth. The ultimate amplitude reached depends on the amount of friction; in an oscillator oflow friction, or high Q, this ulti-

Naturalfrequency Wa

yt------------------w

Fa3-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 the final amplitude of oscillation att ained as a function of the frequency of

15.5 Damped Oscillations and ForcedOscillations 491 I the harmonic driving force for two mass-and-spring system s with th e same natural angular frequency W o = but different values of Q, th e quality factor. Notice that large amplitudes occur over a range ofdriving frequen cies, and that some enh ance-ment over the stati c Hooke's-Law displ acement x = - Fo/ k occurs for any frequ ency of forced oscillation near or below th e natural frequ ency wo.lf the oscillator is forced precisely at reson ance, the amplitude can be shown to take th e value

(15 .54) amplitude at resonance of damped, driven harmonic oscillator

This is simply the magnitude of the static displa cement x = - Fo/ k multiplied by Q; thus th e quality factor is equi valent to an amplitude enhancement facto r for a system at reson ance.

The phenomenon of reson ance plays a cruci al role in many piece s of industrial machinery-if on e vibrating part of a machine is driven at resonance by a perturbing force origi nating from some other part , th en the amplitude of oscillation can build up to violent levels and shake the machine apart. Such dangerous resonance effects can occur not only in moving pieces of machinery, but also in structures that are normally regarded as static. In a famou s accident that took place in 1850 in Angers, Fran ce, the stomp-ing of 487 soldiers marching over a suspension bridge excit ed a reson ant swin gin g motion of the brid ge; the moti on quickly rose to a disastrous level and broke the bridge apart, causing the death of 226 of the soldiers (Fig. 15.23).

FIGURE 15.23 Resonance disaster: the collapse of the bridge at An gers, as illustrated in a contemporary newspap er.

mCheckup 15.5

QU ESTION 1: Suppose that the driving force has a frequen cy half as large as th e frequ ency of th e oscillator. Would you expect a buildup of oscillations by resonan ce? QUESTION 2 : Suppose that th e driving force has a frequency twice as large as th e

:l frequency of the oscillator. Would you expect a bu ildup of oscillations by resonance? It

QUESTION 3: Suppose th at a bell has a high Q (it continues to ring for a long tim e e after you strike it) . Ifyou rest your hand against the bell after striking it, how does thi s alter th e Q?1-

QUESTION 4 : An oscillator begin s with 1.00] of mechanical energy. After 10 oscilla-:y tions, the energy stored has dropped to 0.90]. What is the approximate Q of the system?

(A) 6.3 (B) 10 (C) 63 (D ) 100 (E) 630

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SUMMARY MATH HELP D erivatives of trigonometric functions (_0MATH HELP Small-a ngle approxima tions for sine, cosine, and tangent

PHYSICS IN PRACTIC E C haos (p . -

SIMPLE HARMONIC MOTION x = A cos(wt + 8) where A is the amplitude (the maximum displacement from x = 0); w is the angular frequency, and 8 is the phase constant.

PERIOD (time for one cycle) T = 27T/W

FREQUENCY (number of cycles per second) f= l /T = W/ 27T

PHASE CONSTANT AND TIME OF MAXIMUM DISPLACEMENT

MAX IMUM VELOCITY {E o

MAXIMUM ACCELERATION

EQUATION OF MOTION OF SIMPLE HARMONIC OSCILLATOR wh ere k is the spring constant .

ANGULAR FREQUENCY AND PERIOD OF T = 27T k (15.19, 15._: SIMPLE HARMONIC OSCILLATOR w= [fm ..

ENERGY OF SIMPLE HARMONIC OSCILLATOR E = =

(15.: ;

(15.: .

-

-

(15.3_

ANGULAR FREQUENCY AND PERIOD OF SIMPLE PENDULUM T = 27T J} ,

", l.-:",,, :

I

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(15.46, 15.47

ANGULAR FREQU ENCY AND PERIOD OF PHYSICAL PENDULUM wh ere I is the mom ent of inerti a of the pendulum.

T = Imgd < ;,'.t

,, ,.

(15.48, 15.49)

Ima. ...--=.-

ENERGY LOSS PER CYCLE OF DAMPED OSCILLATOR w h ere Q is th e qu ali ty factor.

AMPLITUDE AT RESONANCE OF DAMPED HARMONIC OSCillATOR w here Fa is the am pli tu de of a h armo n ic driving force .

QUESTIONS FOR DISCUSSION 1. Is the motion of th e piston of an au tomobile engine simple

harmonic motion- H ow does it differ from simple harm oni c motion ?

2. In our calculatio n of the frequency of the simple harm onic oscillato r, we igno red the mass of the spring. Qu alitatively, how does th e mass of the spring affect the freque ncy?

3. A gra ndfa th er clock is regulated by a pendulum . If the clock is runni ng late, how mu st we adjust th e length of the pend ulum?

4. F igure 15.24 shows the escapement of a pendulum clock, i.e., the linkage th at pe rmits the pendulum to contro l the rota tion of the whee ls of the clock. Explain how the wheel turns as th e pen dulum swings .

gear of clock

pendulu m bob

FIGURE 15.24 Escapement mechani sm of a pendu lum clock. At the instant shown, the tooth at the left has escaped from the left arm, and the tooth on the right is pushing against the right arm .

5. Wou ld a pen d ulum clock keep goo d time on a ship?

6. G alileo claim ed th at th e oscillato rs of a pend ulum are isoch ro-nous, even for an amplitud e of oscillation as large as 30°. W hat is your opinion of th is claim?

21Tf::,E= - - E (15.51)Q

F. (15.54)

k

7. W hy would you expect a pendu lum oscillatir g with an am pii-tude of nearly (but no t quite) 180° ro have a very , = -

8. Can a pend ulum oscillate with an amp litude of n r chan l

9. F igu re 15.25 shows a "tilted pendu lum" des igned bv ill' - ' 1'1

H uygens in the seventeenth century. W hen the pc idulum is til ted , it s period is longer than when the pcn du urn i , vertic Expl ain.

FIGURE 15.25 H uygens' tilted pendulum.

10. M ost grandfather clocks have a lenticular pendulum bob which supp osedly minimizes friction by "slicing" through the air. H ow-ever, expe rience has shown that a cylindr ical pendulum bo b expe riences less air friction . C an you suggest an explanation?

11. Gali leo descr ibed an expe rime nt to compare the acceleration of gravity of lead and ofcork:

I took two balls, one of lead and one of cork , the former bei ng more than a hund red times as heavy as th e latt er, and suspend ed th em from two equal th in strings, each four or five bracchi a long. P ulling each ball aside from the ver-tical, I released them at th e same instant , and the" . falling along th e circumferences of the circles having the ' trings as radii , passed th rough the vertical and retu rned along the same pa th .T his free oscillation, repea ted more than a hu n-dred times, showed clearly that the heavv bod ,' kep t time with the ligh t body so well th at neither in a hundred oscil-lati on s, nor in a thousand , will the former anticipate the latt er by even an ins tan t, so pcrfecrlv do th ev keep step.

Since air friction affects the cork bali mu ch more than the lead ball, do you think G alilee 's result s are credible-

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. ' . • Newton reported a more careful experiment th at avoided

the inequality of friction:

I tried the thing in gold, silver, lead, glass, sand, common salt, wood, water, and wheat. I provided two equalwooden boxes. I fIlled th e one with wood, and suspended an equal weight of gold (as exactly as I could) in the centre of oscillation of the other.The boxes, hung by equal threads of 11 feet, made a cou ple of pendulums perfectly eq ual in wei ght and figure . . . and, placing the one by the other, I observed them to play together forwards and backwards for a long while, with equal vibrations. . . . And by th ese experiments , in bodies of the same weight , one could have discovered a dif-ference of matter less than the thousandth part of the whole.

Explain how Newton's experiment was bett er th an Galil ee's.

12. A simple pendulum hangs below a table, with its strin g through a small hole in th e tabletop. Suppose you gradually pull the string while the pend ulum is swinging. W hat happens to the frequency ofoscillation - To the (angular) amplitude?

PROBLEMS

15 .1 Simple Harmonic Motion t

1. A particle moves as follows as a funct ion of time:

x = 3.0 cos (2.0t)

where distan ce is measured in meters and time in seconds.

(a) What is the amplitude of this simple harm onic motion? T he frequency?The angular frequency?The period ?

(b) At wha t time does the par ticle reach the midpoint , x = O? T he tur ning point?

2. A particl e is performing simple harmonic motion along the x axis according to the equation

x = 0.6 cos ( where the distance is measured in meters and the time in seconds.

(a) C alculate the position x of th e particl e at t = 0, t = 0.50 s, and t = 1.00 s.

(b) C alculate the instant aneous velocity of the particle at th ese time s.

(c) Cal culate the instantan eous acceleration of the particle at these tim es.

3. A part icle moves back and forth along the x axis betwe en th e points x = 0.20 m and x = -0.20 m. T he period of the motion is 1.2 s, and it is simple harmonic. At the time t = 0, the particle is at x = 0.20 m and its velocity is zero.

For help. see Online Concept Tutorial 16 at www.wwnorron.comlphysics

13. Shor ter people have a shorter length ofstride, but a

_

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rate of step when walking "naturally." Explain.

14. A girl sits on a swing who se ropes are 1.5 m long. Is this . simple pendulum or a physical pendulum?

15. A simple pendulum consists of a part icle of mass m att acz to a string oflength I. A physical pendulum consists of _ of mass In attach ed to a string in such a way that the ce

mass is at a distance 1from the point of support. Which re dulum has th e sho rte r period?

16. Suppose that the spring in the front-wheel suspension 0:-automobile has a natural frequency of oscillation equal to " frequency of rotation of the wheel at, say, 80 km/h. W hy i. th is bad?

17. When marching soldie rs are about to cross a bridge , the:' break step. Why?

(a) What is the frequen cy of the motion ?The angular frequency?

(b) What is the amplitude of th e mot ion ?

(c) At what time will th e part icle reach the point x = 0; At what time will it reach the point x = -0.10 m?

(d) What is the speed of the particle when it is at x = O? What is the speed of the parti cle when it reaches the point x = -0.10 m?

4. Suppose that the peg on the rotating wheel illustr ated in Fig. 15.5 is located at a radius of 4.0 cm. The wheel turns at :. rate of 600 rev/m in. What is the amplitude of the simple harmonic moti on of the slot ted arm? What are th e period , the frequency, and the angula r frequency?

5. Co nsider that the particle in Fig. 15.4 is executing simple harmonic motion according to Eq, (15.1).

(a) What is the speed of th e satellite for this case?

(b) At t = 0.050 s, th e particle is at the midp oint and its instant aneous velocity is parallel to tha t of the satellite. What is the speed of th e particl e? How does it compare with the speed of th e satellite?

6. A given point on a guitar string executes simple harm onic motion with a frequency of 440 Hz and an amplitude of 1.2 mm. What is the maximum speed of thi s mo tion? The maximum acceleration?

7. A piston in a windmill-dr iven water pump is in simple harmonic motion. The motion has an amplitude of 50 ern and the mass of th e piston is 6.0 kg. Find th e maximum net force on

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the piston when it oscillates 80 times per min ute. Find the maximum velocity.

8. A particl e moves in simple harmonic mot ion according to x = A COS (WI + 8). At I = 0, the particl e is at x = 0 with initial velocity Vo > O. W hat is the phase constant 8 ?

9. T he position of a body can be described by x = A cos (WI + 8). The angular frequency w, the initial position xo, and the initial velocity Vo are known. Find the amplitude A and the pha se constan t 8 in terms of to, xo' and vo'

10. The central par t of a piano string oscillates at 261.7 H z with an amplitude of 3.0 mm . What is the angular frequency of the motion?The period? W hat is the maximum velocity? What is the maximum acceleration?

11. In a mod ern nonlinear dynamics experiment, small beads (spheres) are vibrated on a plate; when the beads start to move, interesting patterns form (see Fig . 15.26). If the plate vibrates at 250 H z, for what amplitud e of motion will the beads star t to lift off? (Hi nt: This will occur when the maximum accelera-tion of the plate equals g = 9.81 m /s2

.)

FIGURE 15.26 Oscillating beads.

12. A mass moves in a circle of radius 10 ern , centered on the origin in the x-y plane, with an angular velocity of 7T/ 4 radian / soAt 1= 0, the mass is on the positive x axis. What are the x compone nts of the position, velocity, and acceleration of the mass at I = 1.0 s? At I = 2.0 s?

13. A particle executes simple harmonic moti on. Its displacement is given by x = A cos (WI + 8), where as usual, the amplitude A is a positive constant. A t I = 0, the particle is in the origin and moving in the positive x direction. What is the appropriate choice of the phase constant 0 in thi s case?

*14. Experience shows that from one- third to one- half of the pas-sengers in an airliner can be expected to suffer motion sickness if the airliner bounces up and down with a peak acceleration of 0.4 g and a frequency of about 0.3 Hz. Assume that thi s up-and-down motion is simple harmonic. W hat is the amplitude of the mot ion?

*15. The frequency of a mass attached to a spring is 3.0 H z. At time 1= 0, the mass has an ini tial displacement of0.20 m and an initial velocity of 4.0 m /s.

(a) W hat is the position of the mass as a 'unction of time:

(b) W hen will the mass first reach a turning: point? \ Vhar will be its acceleration at that time?

15.2 The Simple Harmonic Oscillator

16. A man of mass 70 kg is boun cing up and down on a t :,0 s6c k (see Fig. 15.27). H e finds that ifhe holds himself rigi -1 lr,,:i lets the stick do the bouncing (after getting it started). me period of the up- and- down moti on is 0.70 S. What is the spring constan t of the spring in the pogo stick? Assume that the bottom of the stick remains in touch with the floor an :l igno re the mass of the stick.

FIGURE 15.27 M an on pogo stick.

17. T he cable described in Example 8 in Chapter 14 can be regarded as a spring. What is the effective spring constant of this spring? What is the frequency of oscillation whe n a mass of7 .1 X 103 kg is attached to the lower end of the cable and allowed to oscillate up and down? Neglect the mass of the cable in your calculation.

18. A simple harm onic oscillator consists of a mass sliding on a frictionless surface under the influence of a force exerted by a spring connected to the mass. T he frequency of this harmonic oscillator is 8.0 Hz. Ifwe connec t a second, identical spring to the mass, parallel to the first spring, what will be the new fre-quency of oscillation?

19. T he body of an automobile of mass 1100 kg is supported by four vertical springs attac hed to the axles of the wheels. In order to test the suspension, a man pushes down on the body of the automobile and then suddenly releases it.T he body rocks up and down with a period of 0.75 s. What is the spring constant of each of the springs? Assume that all the springs are ident ical and that the compressional force on each spring

, I : • is the same; also assume that the shoc k absorbers of the automobile are completely worn out so that they do not affect th e oscillation freque ncy.

20 . Deuterium (D ) is an isoto pe of hydrogen. The mass of the deuterium atom is 1.998 tim es larger th an the mass of th e hydrogen atom. G iven th at the freq uency ofvibration of th e H 2 molecule is 1.31 X 1014 Hz (see Example 6), calcul ate th e frequency of vibrat ion of the D 2 molecule. Assume the "spring" con necting the atoms is the same in H 2 and D 2.

*21. Calculate the frequency of vibration of the HD molecule con-sisting of one atom of hydrogen and one of deuterium. See Problem 20 for necessary data. (Hint:The center of mass is stationary.)

22. A mass attached to a spring oscillates with an amplitude of 15 cm; the spring constant is k = 20 N /m. When the position is half the maximum value, the mass moves with velocity v = 25 cm /s. Determine the period of the moti on. Find the value of the mass.

23 . A mass of 150 g is attached to a spring of con stant k = 8.0 N /m and oscillates without fricti on . The mass is dis-placed 20 cm from equilibrium and, at t = 0, is released from rest. If the position as a function of time is written x = A cos (wt + 8 ), determine the values of A, w, and 8. What is the maximum velocity of the mass? Its maximum accelerati on? ..

24. The equilibrium position of the bottom end of a light, hang-ing spring shifts downward by 15 cm wh en a 200-g mass is hu ng from it. The mass is then displaced an addi tional 5.0 cm and released . What is the period of moti on ?

25. A th ickness monitor is a laboratory instrument used to deter-min e the thickness of a thin film that is deposited on the sur -face of a qu art z crystal. We may treat the crystal as a spring-and -mass system with k = 6.0 X 105 N /m and m = 0.50 g. What is the frequenc y of oscillation of thi s system? This frequ ency changes slightly as mass is added to the crystal. If the frequency decreases 0 .010%, how much mass was dep osited' If the area of the crystal is 2.0 cm2 and the mass density of the film mat erial is 7.5 g/cm 2, how thick was the dep osit ed film ?

*26. A thin metal rod is attached to the ceiling and a mass M = 15 kg is attached to the bottom of the rod. The rod is 2 .0 m long and has a 9.0-mm2 cross-sectional area. Regard the rod as a (stiff) spring. I f the Youn g's modulus of the rod material is 22 X 1010

N /m 2, what is its spring constant (for small elongations and compressions)? If the mass is dis placed verti cally, wh at is its frequen cy of oscillation (in H z)? Neglect the mass of the rod.

*27. A mass III = 2.5 kg han gs from the ceiling by a spring with k = 90 N /m . Initially, the spring is in its unstretched configu-ration and the mass is held at rest by your hand . If, at time t = 0, you release the mass, what will be its posit ion as a function of time ?

*28. The wheel of a sports car is suspended below the body of the car by a vertical spring with a spring con stant 1.1 X 104 N /m . The mass of th e wheel is 14 kg, and the diameter of the wheel is 61 cm .

(a) What is the frequency of up-and-down oscillation; wheel? Regard the wh eel as a mass on one end ofa ; c

and regard the body of the car as a fixed suPPOft for :.::. oth er end of the spring.

(b) Suppose th at the whe el is slightly out of round, havi r-. bump on one side. As the wheel rolls on the street, it receives a periodic push each tim e th e bump come s ir, contact with the street. At wh at speed of the rransla : motion of the car will the frequency of this push coi with the natu ral frequency of the up-and-down oscili tion s of the wheel? What will happen to the car at rhi: speed ' (N ote: This problem is not qui te realist ic becae « the elasticity of the tire also contributes a restoring forc

to the up-and-down motion of the wheel.)

*29. A mass m slides on a frictionless plane inclined at an ang e with the horizontal. The mass is attached to a spring, parai.e to the plane (F ig. 15.28); th e spring constant is k. H ow rnu is the spring stretched at equilibrium? What is the freque n of the oscillations of the mass up and down on the plane ?

m

e

FIGURE 15.28 Mass sliding on a frictionless inclined plane.

**30 . Two identical masses slide with one-dimensional motion on a frictionless plane under the influence of thr ee identical springs attached as show n in Fig. 15.29 .The magn itude of each mass is m, and the spring constant of each spring is k.

.J - \ .\ ':. ,II., '."'" ml I. , i t. m2

FIGURE 15.29 Two masses sliding on a frictionles s plane.

(a) Suppose that at time t = 0, the masses are at their equilib-rium positions and thei r instantaneous velociti es are VI = - v 2. Find the position of each mass as a function of time. What is the frequen cy of the motio n?

(b) Suppose that at time t = 0, the masses are at their equilib-rium positions and their instan taneous velocit ies are V I = v2• Find the position of each mass as a fun ction of time . What is the frequency of the motion?

:

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" 3 1. A cart co nsists of a body and four wheels on frictionless axles . The bod y has a mass m . The wheels are un ifo rm di sks of mass M and radius R.The cart rolls, without slip ping, ba ck an d forth on a horizontal plane under th e influen ce of a spri ng attache d to one end of the cart (F ig. 15.30 ). The spr ing con-stan t is k.Taking in to account the moment of inertia of the wh eels, find a formula for th e frequ ency of th e back-and-forth mo tion of the cart .

FIGURE 15.30 A cart attached to a spring.

15.3 Kinetic Energy and Potentia l Energy

32 . Suppose that a part icle of mass 0.24 kg act ed upon by a spring undergoes simple harmonic motion with th e paramet ers given in Problem 3.

(a) What is th e total en ergy of thi s motion ?

(b) At what tim e is th e kin eti c ene rgy zero? At what time is th e potential energy ze ro?

(c) At what time is th e kin etic energy equ al to th e po te ntial energy?

33 . A ma ss of 8.0 kg is attached to a spri ng and oscillates with an amplitude of 0 .25 m and a frequ ency o f 0.60 H z . What is th e energy of the motion?

34. A simple harmonic oscill at or co nsists of a mass of 2.0 kg slid-ing back and fort h along a hori zontal frictio nless track while pushed and pull ed by a spr ing with k = 8.0 X 102 N /m. Suppose th at wh en the mass is at th e equilibrium point, it has an inst antaneou s speed of 3.0 m / s. What is the ene rgy of th is harmoni c oscill at or? What is the am plitu de of oscill ation?

35 . A simple harmon ic oscillato r of mass 0.60 kg oscillates with a frequ en cy of3.0 Hz and an amplitude of 0.15 m. Suppose that, while th e ma ss is ins tant aneously at rest at it s turning poi nt, we quickly attach ano ther mass of 0.60 kg to it . H ow does thi s change the amp Iirude of the motion ?The frequency? The energy? The maximum sp eed?The maximum accelera tio n?

*36. The separation be tween the equilibrium positions of th e two atoms of a hyd rogen m olecule is 1.0 X 10- 10 m. Using the dat a given in Ex ample 6, calculate the value of the vibr a tional energy that corresponds to an am plitude of vibration of 0.5 X 10- 10 m for each atom. Is it valid to treat th e motion as small oscillatio n if the ene rgy has thi s value ?

37. A 500-g mas s is co nnected to a sp ring and execu tes sim ple harmon ic motion . The period of th e motion is 1.5 s, and the total mec h anical energy of the system is 0 .50 J. Find the amplitude of mot ion.

38. A ma ss oscillate s on a ; pO lg. :\ t the point s in a cvcle wh en the kin et ic is on e- half of the po tenti al cn ergv, the dis-placement from eq ilib rium is 15 em anti the ins rantaneous velocity is ::':25 cm/ s. \ \ hat i, the riod o r ' he motion '

*39. O ne end of a horizontal 'pri!:'::0 " constant k tixed an d th e other end is attached to a m ass »: L . 1 niess surface . T he spring is in itially in its eq uilibrium i 'iri ; . , ; : = J. a :' rce F, constant th ereafte r, is applied in ' di rectio 0;' e' of the spri ng. Sometime later, the mas;

in the d irecti on of the force. What is he V. eti e . T . 2'

time?

'40 . A mass of 3.0 kg slid ing along a fr ict ionless , om T ':: . m , strikes and com presses a spring o f co nstant k = .300 . . . . The spring stops the ma ss. How far does th e mas, rrav : wh o being slowed by th e spring ? H ow long does th e mas; [ ;1,, :: r' stop?

*41. Two masses m j and 1112 are joined by a spri ng of spr ing con-stant k. Sh ow th at the frequ ency of vibratio n of th ese masse, along th e line co nnecting them is

w =

(H int: The center of ma ss remains at rest. )

"42 . Although it is usually a goo d app roximation to negl ect the mass of a spring , so me times thi s mass mu st be tak en in to account. Suppose that a un iform spring has a relaxed length I and a mass m'; a mass m is attach ed to th e en d of th e spring . The m ass m ' is uniformly distributed along the spring . Suppose that if th e mov ing end of the spring has a spee d v, all othe r points of th e spring have speed dir ectly propo rtional to their di stanc e from th e fixed end ; for in stance, a point midway between the moving and the fixed end ha s a speed

(a) Sh ow that th e kin et ic energy in the spring is t, m' v 2 and that the kinetic ene rgy of th e mass m and the sp ring is

C onsequently, the effective mass of th e com bina tio n . I ' Is m ' un-

(b) Sh ow that th e frequ ency of oscillation is

co = Vk/ (m + tm' ).

(c) S up pose th at a spr ing has a mas s of 0 .05 kg. The fre-que ncy of oscillatio n of a 4.0-kg mas s attached to this spring will th en be so mewhat smaller than calcula ted for a massless spr ing. H ow much smaller? Ex press your ans wer as a percent age of th e value obtained for a ma ssless spring .

15 .4 The Simple Pendulum" 43 . The longest pendulum in existence is a 27-m Foucault pendu-

lum in Portland, Oregon. What is th e peri od of this pendulum?

t For help, see O nline Concept Tu torial 17 at www.wwno rton.corn/physics

• 44. At a construction site, a bucket full of concrete hangs from a

crane . You observe that the bucket slowly swings back and forth, 8.0 tim es per minute. What is th e length of the cable from which the bucket hangs?

45. The elevator cage of a skyscrape r hangs from a 300- m-long steel cable. The elevator cage is guided with in the elevator shaft by railings. Ifwe remove these railings and we let the elevator cage swing from side to side (with small amplitude), what is its period of oscillation?

46. On the Earth, a pendulum oflength 0.994 m has a period of 2.00 s (compare Example 8). Ifwe take this pendulum to the surface of Jup iter, where g = 24.8 m/ s2, what will be its period?

47. A mass suspended from a parachute descendin g at constant velocity can be regarded as a pendulum. What is the frequency of the pendulum oscillations of a hum an body suspended 7.0 m below a parachute?

48. A "seconds" pendulum is a pendulum that has a period of exactly 2.0 s; each one-way swing of th e pendulum th erefore takes exactly 1.0 s. What is the length of the seconds pendu-

(a) Find th e angul ar frequency of oscillation in terms or' .' R, and K.

(b) If the disk is turned through an initial angle of eoanc released, what is the maximum rotational angular ve of the subsequent motion ?

(c) For wh at value of eo do the answers to (a) and (b) hav- '. same value?

*56. The balance wheel in a clock is a torsional oscillator with a pe:-of0.50 s.The restoring torque of the wheel spring is T = - K-

where K is the torsional constant. If the wheel is essenriallv

hoop, that is, all of its mass m = 8.0 g is concentrated at its radius R = 1.0 ern, what is the value of K (in N un/radian)?

*57. A pendulu m hangs from an inclined wall (see Fig. 15.31). Suppose that this pendulum is released at an initial angle of , . and it boun ces off the wall elastically when it reaches an ang c

- 5°. What is the period of this pendulum?

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lum in Paris (g = 9.809 m/s 2) , Buenos Aires (g = 9.797 m/s''), and W ashington , D .C. (g = 9.801 m/ s2)?

49. A gran dfather clock controlled by a pendulum oflength 0.9932 m keeps good time in New York (g = 9.803 m/ s").

(a) Ifwe take thi s clock to Austin, Texas (g = 9.793 m/s"), how man y minutes per day will it fall beh ind?

(b) In order to adjust the clock, by how many millim eters must we short en the pendulum?

50. The pendulum of a grand fath er clock has a length of 0.994 m. If th e clock run s late by 1.0 minute per day, how much must you short en th e pendulum to make it run on time?

51. A small model of a 10- story construction crane used on a Hollywood movie set should appear realistic in motion .T o make it look large, th e mass hanging from the crane "cable" (actually,a rod) is constrained to oscillate with a period of 10 s. H ow long does this make the cable seem?

52. An astronaut land s on an asteroid and sets up a pendulum that has a period of 1.0 s on Earth. She finds that th e pendulum has a period of8 9 s on the aste roid. What is the local value of the accelerati on due to gravity on the asteroid?

53 . A circular painting is 2.00 m in diameter and has uniform thickness. It hangs on a wall, suspended by a nail 10 cm from the top edge. If it is pushed slightly, wh at is the period of small oscillations of the painting?

54. A hula hoop (a thin, uniform toy hoop ) of radius 1.0 m hangs over a nail. If it is set to swinging with small amplitude, what is the period of motion?

*55. A torsional oscillator consists of a horizontal uniform disk of mass M and radius R attached at its center to the end of a mass-less vertical fiber. Some such oscillators can execute simple har-monic (twisting) motion with very large amplitudes (amplitudes greater than one rotation are possible).The restoring torque of

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hanging from an inclined wall.

*58. The pendulum of a pendulum clock consists of a rod of lengtl. 0.99 m with a bob of mass 0.40 kg. The pendulum bob swing; back and forth along an arc oflength 20 cm.

(a) What are the maximum velocity and the maximum accel-eration of the pendulum bob along the arc?

(b) What is the force that the pendulum exerts on its suppor when it is at the midpoint of its swing? A t the endpoint? Neglect the mass of the rod in your calculation s.

*59. The pendulum of a regular clock consists of a mass of 120 g at the end of a (massless) wooden stick of length 44 cm.

(a) What is the tot al energy (kinetic plus potential) of this pendulum when oscillating with an amplitude of 4°?

(b) W hat is th e speed of th e mass wh en at its lowest point?

60. At the National Institute of Standards and Techn ology in G aith ersburg, M aryland, the value of the accelerati on of grav-ity is 9.800 95 m/s 2. Suppose th at at th is locat ion a very precise physical pendulum, designed for measurem ents of the acceler-ation of gravity, has a period of 2.103 56 s. Ifwe take this pen-dulum to a new locat ion at th e U.S. Coas t and Geode tic Survey,in nearby W ashington , D .C., it has a period of2.10354 s. What is the value of th e acceleration of gravity at thi s new

FIGURE 15.31 Pendulu m

the fiber is proportional to the angular rotation; that is, T = location ? What is the percentage change of the acceleratio n - Ke, where K is called the torsional constant of the system. between the two locat ions?

I111

61. Consider a meterstick swing ing about a pivot through its upper end . What is the period of oscillation of th is physical pen dulum ?

' 62 . A pendulum consists of a brass rod with a brass cylinder attached to the end (Fig. 15.32).The diameter of the rod is 1.00 ern and its length is 90 .00 cm; th e diam ete r of the cylinder is 6.00 ern and its length is 20.00 cm. What is the period of thi s pendulum?

90.00 em .

/>20.00 em I

.J ';./ FIGURE 15.32

/"'6.00 em A physical pendulum.

*63. To test that th e acceleration of gravity is the same for a piece of iron and a piece of brass, an expe rimenter take s a pendulum oflength 1.800 m with an iron bob and ano ther pendulum of th e same length with a brass bob and starts them swinging in un ison. After swing ing for 12.00 min, th e two pendulums are no more than one- quarte r of a (one-way) swi ng out of step. What is the largest difference between the values ofg for iron and br ass consis tent with the se dat a? Exp ress your answer as a fractional di fferen ce.

*64. C alculate the natural period of the swinging motion of a human leg. T reat the leg as a rigid physical pendulum with an axis at the hip jo in t. Pre tend th at the mass distribution of the leg can be approximated as two rod s joined rigidly end to end. The upp er rod (thigh) has a ma ss of 6.8 kg and a length of 43 ern; th e lower rod (shin plus foot) has a mass of 4.1 kg and a length of 46 ern. Using a watch, measure the period of th e natural swingi ng motion of your leg when you are standing on on e leg and letting the other dangl e freely. Alternatively, measure the peri od of th e swing ing motion of your leg when you walk at a nor mal rate (this approxim ates the natural swingi ng moti on ). Compare with th e calculated number.

*65. A hole has been drilled through a metersti ck at the 30-cm mark and th e meterstick has been hung on a wall by a nai l passing through thi s hole. If the metersti ck is given a push so th at it swings about the nail, what is the period of th e moti on?

*66. A physical pe ndulum has th e shape of a disk of radius R.The pe ndulum swings abo ut an axis pe rpendicular to th e plane of th e disk at a distan ce / from the center of the disk.

(a) Sh ow th at th e frequ ency of th e oscilla tions of th is pendu-lum is

*67. A physical pen dulum consis ts of a massless rod of length 2/ rot ating abo ut an axis through its cen ter. A mass mI is attac hed at th e lower end of th e rod , and a sm aller mass m2 at th e upp er end (see Fig. 15.33). What is th e peri od of th is pen-dulum ?

•

FIGURE 15.33 A physical pendulum with two bobs.

*68. Suppose that a physi cal pendulum consists of a thin . . ..:cr of mass m suspended at one end . Suppose that this roc h ' 5 an initial position e= 20° and an initial angu lar velocity tu = O. C alculate the force F that the supp ort exerts on th e pen iulu:r, at thi s initial instant (give horizontal and verti cal com nents) .

*69. The door ofa house is made of wood of un iform thic kness. T .e door has a mass 'Of 27 kg and measures 1.90 m X 0.91 m. T he door is held shut by a tor sional spring with K = 30 N -rn/radian arr anged so tha t it exerts a torque of 54 N 'rn when the do or is fully open (at right angles to th e wall of the hou se). What angular speed doe s the door att ain if it slams shut from th e fully ope n position ? What linear spee d does th e edge of th e door att ain>

" 70. G alilee claime d to have verified experim entally that a pendu -lum oscillating with an amplitude as large as 30° has the same period as a pendulum of identical length oscilla ting with a mu ch smaller amplitude . Suppose th at you let two pendulums of length 1.5 m oscillat e for 10 min. Initially, the pendulums oscilla te in step. If the amplitude of one of th em is 30° and the amplitude of the other is 5°, by wh at fraction of a (one-way) swi ng will the pendulums be out of step at th e end of th e 10-min interval ?What can you conclude about Galilee 's claim?

"*71. A thin vertical rod of steel is clamped at its lower end. When you pu sh th e upp er end to one side , be nding the rod, the upper end moves (approxima tely) along an arc of circle2 of radiu s R and the rod opposes your push with a res toring force F = -KO, where e is th e angu lar d isplacement and K is a con-sta nt. Ifyou att ach a mass m to th e upper end , wha t will be th e frequency of smal l oscillations? For wh at value of m doe s the rod beco me un stable; th at is, for wh at value of m is w = O? Treat the rod as massless in your calcula tions . (H in t:T hi nk of th e rod as an inverted pendulum of length R, with an ext ra restoring force - Ke.)

2 The radius R of the approximating (osculating) circle is somewh at shorter (b) For wha t value of / is th e frequ ency a maximum? than the leng th of the rod.

**72. According to a proposal describe d in Pro blem 83 of Chapter 1, very fast trains could travel from one city to anoth er in stra ight subter ranean tunnels (see F ig. 15.34). For the following calcu-lations, assume that the density of the Earth is constant, so the acceleration of gravity as a functi on of the radial distanc e r from the center of the Earth is g = (GM/R3)r.

FIGURE 15.34 A straigh t runnel connecting rwo points on the surface of the Earth .

(a) Show that the compo nen t of the acceleration of gravity along the track of the train is

gx = -(GM/ R3)x

wh ere x is measured from the midpoint of the track (see F ig. 15.34).

(b) Neglectin g friction, show that the motion of the train along the track is simple harmonic mot ion with a period independent of the leng th of the track,

7T = 27T GM

(c) Starting fro m rest, how long wo uld a train take to roll freely along its track from San Francisco to W ashingt on, D.C.? W hat wo uld be its maxi mum speed (at the mid-point )? Use the numbers you calculated in Problem 83 of Chapter 1 for the length and depth of the track.

"'73. A physical pendulum consists of a long, thin cone suspended at its apex (Fig. 15.35). The heigh t of the cone is t. What is the period of thi s pendulum?

FIGURE 15.35 A long, thin con e.

**74. The net gr avitati onal force on a particl e placed midwav between two equal spherical bodies is zero. H owever, if

part icle is placed som e distance away from th is equilibriuz point, then the gr avitati onal force is not zero.

(a) Show th at if the par ticle is at a distance x from the e

rium point in a direct ion toward one of the bodies, th force is approxima tely 4GMmx/r3 , where M is the n a;-each spherical body, m is the mass of the particle, and ::. the distance betwee n the sphe rical bodi es. Assume x«

(b) Show that if the parti cle is at a distance x from the equi. rium point in a direction perpendicular to the line connec the bodies, then the force is approxima tely - 2GMm x/ r where the negative sign indicates that the direction of 1-.:: force is toward the equilibrium point.

(c) What is the frequency of small oscillations of the mass about the equilibriu m point wh en moving in a direct: perpend icular to the line connecti ng the bodies? Assu: that the bodies remain stationary.

(a) Find the tension in the string of th is pendulum; assume that (J « 1.The mass of the suspended particle is m.

(b) T he tension is a function of time. At what t ime is the rer.-sia n maximum ? What is the value of th is maximum tens:

15.5 Damped Oscillations and Forced Oscillations

76. Roughl y, what is the frequency of stomping of soldiers on the mar ch? What must have been the resonan t freq uency of the brid ge at Angers th at bro ke when soldiers marched across it:

77. A pendulum ofleng th 1.50 m is set swingi ng with an initial amplitude of 10°. After 12 min, friction has redu ced the ampl itude to 4°. What is the value of Q for this pendulum ?

78. The pendulum of a g randfa ther clock has a length of 0.994 111 and a mass of 1.2 kg.

(a) If the pendulum is set swinging, the friction of the air reduce s its amplitude of oscillation by a factor of 2 in 13.0 min. What is the value of Q for thi s pendulu m?

(b) Ifwe want to keep thi s pendulum swinging at a con stant amplitude of 8°, we must supply mechanic al energy to it at a rate sufficient to mak e up for the frictional loss. W hat is the required mechanical power?

79. When a swing in motion is not being "pumped," the angular ampl itude of oscillation decreases because of air and othe r fricti on. The motion of a 3.0-m-long swing decreases in amplitude from 12° to 10° after 5 complete cycles. W ha t is the Q of the system? If the rider and seat are treated as a point mass with m = 25 kg, at wh at average rat e is mechanic al energy bei ng dissipated)

-

-

80 . A horizontal spri ng of constant k is attached to a ma ss m that slides on a sligh tly frictional floor. After the mass is di splaced a di stan ce A from equilibrium and released, the am plitude of oscillatio n decrea ses to 0.95A after 10 cycles. W hat is the Q of th is system?

81. A har monic for ce F= Facos wt, where Fa = 0.20 N, is applied to a damped harmonic oscillator of spring con stant k = 15 N 1m and ma ss m, wh ere w = T he am plitude of osc illation increases rapidly at first, and then settles to a constant value, A = 40 ern. W hat is the Q of the system? What wou ld the am plitude be if the angul ar frequency of th e force F had been mu ch less

82 . A microclec tromechanical system (M EM S) consists of a microsc opic silicon mechanical oscillator (see Fig. 15.36) with

FIGURE 15.36 A microelectromechanical system (MEMS) oscillator, the silicon memb rane structure suspended above the faceted silicon tren ch .

a spring co nstant k = 5 X 10-3 N/m . When it oscillates in a vacuum-sealed device (to remove air fric tion) , the Q of such an oscill ator is large: Q = 5 X 106. What amplitude of mo tion will th e oscillator attain if an oscilla ting for ce of amplitude 1 X 10- 18 N (near the curren t limits offorce detection) is applied?

83 . Using electro n-bea m lith ogra ph-.., engine ers are attempting to fabrica te nanoclectro rn cchar ical svsre rn (N E M S) osc illators wi th frequenci es as high as 100 G H z (for communications and h igher-speed com puti ng). If rhe equivalen t mass of such

Fan oscillato r is 1.0 X lO- g and. m inimurnarnplirude of 0.10 nm is needed to detect an app lied harmonic force o f

10am plitu de 1.0 X 10 - N , wha t must the minimum Q of such an oscillator be>

*84. Consider the motion of the damped harm on ic oscillator plotted in Fig . 15.2 1.

(a) According to thi s plot , what fractio n of its am pli tude does th e oscillator lose in its first osc illa tion>

(b) W hat fracti on of its ene rgy does the oscillator lose in irs first oscillation?

(c) According to Eq . (15 .5 1), what is th e value ofQ for thi s oscillato r>

*85 . If you stand on one leg and let the ot her dangle freely back and forth starting at an init ial am plitude of, say, 20° or 30°, the amplitude will decay to one -half of the initial amplitu de after about four swings . Regarding the d angling leg as a d amped oscillator, what value of Q can you deduce from thi s?

REVIEW PROBLEMS 86. A particle per for ms simpl e harmonic motion along th e x axis

with an am plitude of 0.20 m and a period of 0.80 s. At t = 0, the particle is at max im um di stan ce from th e origin; that is, x = 0 .20 m.

(a) What is th e equa tion tha t describes the position of the part icle as a func t ion of time ?

(b) C alculate th e positi on of the particle at t = 0.10 s, 0.2 0 s, 0.30 s, and 0040 s.

87 . I n an electric saber saw, the ro tational moti on of the electric mo tor is converted into a back-and-forth motion of the saw blade by a mech anism similar to th at shown in Fig. 15.5. Suppose th e peg of the rot ati ng wheel moves aro und a circle of diameter 3.0 cm at 4000 revlmin and thereby moves the slo tted arm to whi ch the saw blade is bolted. What are the amp litude and the frequency of th e bac k-and- forth simple harm on ic mo tion of the blade ?

88. In response to a sound wave, the middle of you r eardrum oscillates back and for th with a frequen cy of4000 H z and an

5amplitud e of 1.0 X 10 - m . What is th e m aximum speed of the eardr um >

89 . Suppose that two particles are performing sim ple harm onic mot ion along the x axis wi th a period of 8.0 s. The first parti-cle moves acco rd ing to the equation

x = 0.3 0 cos ( :t) and the second parti cle according to th e equation

x' = 0.30 sin ( :t ) where th e distance is measured in m ete rs and the ti me in seconds .

(a) When doe s th e first par ticle reach th e midpoint? T he turning point? Draw a diagram showing rhc pan icle and its sa telli te particle at these tim es.

(b) When do es the second particle reach the mid po int?The turning poin t? Draw a diagram showin g the part icle and its satellite particle at these times.

(c) By some argum ent, establish that wh enever the first parti-cle passes through a point on the x axis, the seco nd pani -cle passes th rough this same point 2.0 s later.

. , .. • 90. A part icle of 6.0 kg is executing simple har monic motion

along the x axis under the influence of a spring.T he part icle moves according to the equation

x = 0.20 cos (3.0t)

where x is measured in meters and t in seconds .

(a) What is the frequency of the motion? W hat is the spring constant of th e spring? What is the maximum speed of the motion?

(b) Suppose we replace the parti cle by a new part icle of2.0 kg (but we keep the same spr ing), and suppose we start the mo tion with th e same amp litu de of 0.20 m. What will be the new frequency of the motion? What will be the new maximum speed?

91. The mo tion of the piston in an automobile eng ine is approxi-mately simple har monic. Suppose that the piston travels back and forth over a distance of 8.50 em and has a mass of 1.2 kg. W hat are its maximum accele ration and maximu m speed if the eng ine is turni ng at its high est safe rate of 6000 rev/min? What is the maxim um force on th e piston?

92 . A Small Mass M easurement In strument (SMMI) was used in Skylab to measure the masses of biological samples, small ani -mals, chemicals, and other such items used in life-sciences experiment s whil e in orbit (see F ig. 15.3 7).The sample to be measured is strapped to a tray supported by leaf springs, and the mass is determined from the observed period of oscillation of the tray-and-mass.To calibr ate th is instrument, a test mass of 1.00 kg is first placed on the tray; the peri od of oscillation is then 1.08 s. Suppose that when the test mass is removed and an unknown sample is placed on the tray, the per iod becomes 1.78 s. What is the mass of the sample? Assume that the mass of the tray (and the st raps) is 0.400 kg.

FIGURE 15.37 Small Mass M easurement Instru ment.

93. A simple harmonic oscillator has a frequency of 1.5 H z. W hat will happen to the frequency if we cut the spring in half and attach both halves to the mass so that both springs push jointly?

94. A physicist of 55 kg stands on a bathro om scale (a spring scale, with an intern al spring). She observes that when she mount s the scale suddenly, the poin ter of th e scale first oscil-lates back and fort h a few time s with a frequen cy of 2.4 Hz.

(a) What value of the spring constant can she de -these da ta?

(b) If she then takes a child of20 kg in her arms ar-; _: stands on the scale, what will be the new frequer: oscillation of the point er?

95 . Ropes used by mountain climbers are quite elastic, ::. :: :..

-

_

.r:

_

- behave like springs. A rope of 10 m has a spring COil " --

k = 4.9 X 103 N/ m. Supp ose th at a mountain climber hangs on thi s rope, whic h is stretched vert ically down. . the frequency of up-and-down oscillations of the rno clim ber?

96. C onsider a particle of mass In moving along the x axis

the influence of a spring ofspring constan t k.T he equii": po int is at x = 0, and the amplitude of th e motion is "i. (a) At what point x is the kinetic energy of the particle ec

to its potent ial energy?

(b) W hen th e particle reaches the point x =!A, what r: :.. -:-of its energy is potential, and what fraction is kinetic-

97. A simple har monic oscilla tor consists of a mass of 3.0 kg' ing back and forth along a hor izontal frict ionless track wi-pushed and pulled by a spring with k = 6.0 X 102 N/ m. Suppose that in itially the mass is released from rest at a 6:;-ranee of 0.25 m from the equ ilibrium point. W hat is the energy of this harmon ic oscillator? W ha t is the maximum speed it attains when passing th rough th e equilibrium poir.: :

98. A simple har monic oscillator of mass 0.80 kg oscillates wi frequency of 2.0 H z and an amplitude of 0.12 m. Supp ose th at , while the mass is instantaneously at rest at its turn ing po int, we quickly shi ft th e fixed end of the spring to a new fixed position, 0.12 m farth er away from the mass. How doe; thi s change the amplitude of the motion? The frequen cy? energy?The maximum speed? The maximum accelera tion?

99. A pendulum has a length of 1.5 m. What is th e perio d of th is pendulum? Ifyou wanted to construct a pendulum with exactly half this period , how long would it have to be?

100. A n "inte rrupted" pendulum consists of a simple pendulum of length ! that encounters a nail placed at a d istance below the point of supp ort. If this pendulu m is released from one side, it will begin to wrap around the nail as soon as it passes through the vertical position (F ig. 15.38). What is the period of th is pendulum?

.._--------

•II \ I\

\ I I

I

FIGURE 15.38 An II

"int errupted" pendulum.

11111.1 1 _

505

101. A physical pendulum consists of a uniform spherical bob of mass M and rad ius R suspended from a massless string of length L (see F ig. 15.39). Taking into account the size of the bob, show th at the period of small oscillat ions of th is pendu-lum is

f R2 + (R + L)2T = 27T ::...5_--,----_ --,-_) g(R + L )

I I I L I I

M

FIGURE 15.39 A physical pendulum with a large bob.

Answers to Checkups

102. A un iform rod ofleng th L is swinging abou t a pivot at a dis-tan ce x from its cen ter (see Fig. 15.40). Fi nd the period of oscillat ion of this physical pendulum as a functi on of x. For wh at choice of x is the period sho rtest?

<: _j./'<

L

.l"

/ -

FIGURE 15.40 A swinging rod.

103. A swing oflengt h 2.0 m han gs from a horizontal branch or' a tree . With what freq uency should you rock the branch to bi il(: up oscillat ion s of the pendulum by reson ance?

Answers to Checkups

Checkup 15 .1 1. The E arth's rot ational motion is periodic; it repeats with each

daily cycle. It is not a back-and-forth mot ion along a line or are, so it is not an oscillation.

2. The velocity attains its maximum magni tude at x = 0, that is, wh ere the displacement is zero; the velocity attains its mi ni-mum magni tude at x = ±A, that is, at the points of maximum displacement. This is because the displacement and velocity are 90° out of phase; if one is a cosine func tion, the other is a sine function [see Eqs . (15 .11) and (15.12) ].

3. T he acceleration attains its maxi mum magnitude at x = ±A, that is, at the point of maximum displ acement from the origin; the acceleratio n attains its minimum magnitude at x = 0, that is, where th e displacement is zero .This is becau se the dis-placement and accelera tion are 180° out of pha se; if one is a cosine function , the other is a negati ve cosine fun ction [see Eq s. (15.11 ) and (15 .13)].

4. ]f the maxim um displacem ent is A, the maximum velocity is wA [com pare Eqs. (15.11) and (15.12)]. Thus, for the same am plitude, the particle with twice the frequency has twice the maximum veloc ity. Similarly, the maximum acceleration is w2A, so the particle with twice the frequency has 4 times the maximum acceleration.

5. As described in Section 15.1, the x coor dinates of the particl e and satellite are identical. Obviously, the y coordinates are not, since the particl e is always at y = 0, whil e the satellite executes circu lar mo tion.The velocitie s are no t the same , since the par-ticle has zero y velocity, unlike the satellite.The x com pon ent s of the velocities and accelera tions are the same, since they are derived from the identical time dep endence of the x coordi-na te. Sin ce the parti cle is always at y = 0, th e y compo nent s of the veloci ty and acceleration are not the same .

6. (C) 7T/2. Ifwe in sert x = 0 and t= 0 in x =A cos(cot + 8) [Eq. (15.4)], then we see that 0 = cos 0, which is tru e if o= 7T/2 or if 0 = -7T / 2. Of thes e two, only 0 = 7T/ 2 is listed .

_ _

Checkup 15.4 1. The period of the simple pendulum is proport ion,

to the square root of th e length (T = 27T V'!'i;; ), so ' ,. of the shorter pend ulum will be decreased by a factor ' 1/ \/2. The frequency is the inverse of the perio d ( .'= and so will increase by \/2.

2 . Two pendulum s of the same length have the same 0

quency of oscillation, since w = 'Viii,But the pendulu m is (th is formula is equ ally valid to" . pendulu m and the simple harm oni c oscillater).Thus : - the same energy of oscillation , a mass 3 times smaller move with an amplitude that is v3 times larger.

3 . (B) V3g/2 1.The angular frequency of such a physical , lum is given by Eq. (15.48), w = Vrngd/ I .T he disranc measured from the point of suspension to the center or' I thus is half of the length of the rod ; that is, d = 1/2. Insc this and the given mom ent of inertia yields v = (mgl/ 2)/ (mI2/ 3) = V 3g/21.

Checkup 15.5 1. Yes, at least some sligh t buildup always occurs at frequen

below resonance. Figur e 15.22 indicates th at forced oscili - tions far below th e reson an t frequency approac h an arnplic A = Fa/ k, the magnitud e of th e static-force spring displa• . ment.This occurs because a slowly varying force allows the- (faster) mass-spring system to follow the force over time.

2 . No. In this case the slowly responding oscillator cannot fo • the oscillating force; it is as if the response is averaged nearlv equally over the positive and negati ve force con tributions.

3. Your hand prov ides friction; you remov e energy from the be '. When 6.E, the energy lost per cycle, increases, th e Q must decreas e.

4. (E) 630 . W e can solve Eq. (15.51) for Q and obtain Q = 27TE/6. E. Since 0 .10 J is lost in 10 cycles, abou t 0.010 J is 10,: each cycle.Thus Q = 27T(1.0]) / (0.010 J) = 2007T = 630 .

Co

Cc

Checkup 15.2 1. The force on th e particle attains m axi mum magnitude at the

extreme displacem ents (the turn ing points) of the motion, x = :i:.A. T he force on the part icle atta ins the minimum mag-nitude ofzero when the particle passes through the equ ilib-rium point, x = O.

2 . Since the frequency is given by w = doub ling the mass decreases the frequency by a factor of 1/ \/2.

3. If the spring is cut when the particle is at the equ ilibrium point, the parti cle will con tin ue moving wit h the constant velocity it h ad there, 'Vm.v. = wA. I f we cut it whe n the particle is at x = A. where the par ticle is instan tan eously at rest , the particle will remain at rest there.

4 . (B) 1/ v2.A stronger spring causes oscillation s with a high er frequ ency, and so a sho rter period . The period varies inversely wi th the square root of the spring constant [Eq . (15.2 1)].

Checkup 15.3 1. Since energy is proport ion al to the square of the ampli tude

[Eq. (15 .32)], th e oscillator wit h twice the amp litude has 4 tim es th e energy. Since th e maximum speed is proport iona l to the ampl itude, 'V"'"" = wA [Eq . (15 .34)], the oscillator with twice the amplitude also ha s twice the maximum speed.

2 . Both oscillators have the same energy, since E = But th e maximum speed is inversely pro portional to the square root of the mass [Eq . (15 .35)], so the particle wit h twice th e mass has a smaller maximum speed by a factor of 1/\/2.

3. T he energy is purely kinetic whe n th e oscillator passes through equilibrium. The energy will be pur ely pot ential at maximum amplitude, which is on e-q uarter of a cycle later, or 2.0 s later. An oscillator passes th rough equilibrium twice each cycle (once in each direction ), so th e energy will be purely kinetic 4.0 s after th e initial tim e, or another 2.0 s after th e energy is purely pot enti al.

4 . Friction removes energy from th e system, so the energy will decrease wh enever th e particle is moving, and will not remain constan t. Since E = the amp litude A will also decrease each cycle due to friction .

S. (E) 32.T he sto red energy is E = But from Eq. (15.18), 2 2A 2k = mw , so E = mw ; if each of In, tv, and A increases by a

factor of 2, then the energy increases by a facto r of 25 = 32.

IIII IIWI ;.

physics chapter 15 oscillations

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