INTRODUCTION

SCILL I 5

In this chapter, we'll concentrate on a kind of periodic motion that's straightforward and that,fortunately, actually describes many real-life systems. This type of motion is called simple harmonicmotion. The prototypical example of simple harmonic motion is a block that's oscillating on the endof a spring, and what we learn about this simple system, we can apply to many other oscillatingsystems.

SIMPLE HARMONIC MOTION (SHM):THE SPRING....BLOCK OSCILLATORWhen a spring is compressed or stretched from its natural length, a force is created. If the spring isdisplaced by x from its natural length, the force it exerts in response is given by the equation

Fs =-kx

143

Thisismown as Hooke's Law. Theproportionality constant, k, is a positive numbercalledthespring(or force) constant that indicates how stiffthe spring is. Thestiffer the spring,the greaterthe valueof k. The minus sign in Hooke's Lawtells us that Fs and x always point in oppositedirections. Forexample, referring to the figure below, when thespring is stretched(x is to the right), thespringpullsback (Fis to the left); when the spring is compressed (xis to the left), the springpushes outward (Fis to the right). In allcases, the springwants to return to itsoriginallength. Asa result,thespringtriesto restorethe attachedblockto the equilibrium position, whichis the positionat whichthenet forceon the blockis zero. For this reason,we say that the spring provides a restoring force.

spring

Ix=o

equilibriumposition

springstretched

Ix>O

springcompressed

Ix<o

144 II CRACKING THE AP PHYSICS EXAM

Solution. The displacement of the spring has a magnitude of 14 -12 =2 em =0.02 m so, accordingto Hooke's Law, the spring exerts a force of magnitude F = kx = (400N/m)(0.02 m) =8 N. Therefore,we'd have to exert this much force to keep the spring in this stretched state.

Springs that obey Hooke's Law (called ideal or linear springs) provide an ideal mechanism fordefining the most important kind of vibrational motion: simple harmonic motion.

Consider a spring with force constant k, attached to a vertical wall, with a block of mass m on africtionless table attached to the other end.

km

Ix=O

equilibriumposition

Grab the block, pull it some distance from its original position, and release it. The spring will pullthe block back toward equilibrium. Of course, because of its momentum, the block will pass throughthe equilibrium position and compress the spring. At some point, the block will stop, and thecompressed spring will push the block back. In other words, the block will oscillate.

During the oscillation, the force on the block is zero when the block is at equilibrium (the point wedesignate as x =0). This is because Hooke's Law says that the strength of the spring's restoring forceis given by the equation Fs =kx, so Fs =0 at equilibrium. The acceleration of the block is also equalto zero at x =0, since Fs =0 at x =0 and a F/m. At the endpoints of the oscillation region, wherethe block's displacement, x, has the greatest magnitude, the restoring force and the magnitude of theacceleration are both at their maximum.

OSCIllATIONS III 145

Fsis maxais max

Fs =0a=O

Fsis maxaismax

Ix =Xmax

,,,'fr---------,, ', ', ', ', ', ', ', ', '

,,,'f

m

Ix=O

equilibriumposition

,,,,,,,,,,,,:4<-m-- oscillation region _n .jIo-:

,,,'f

IX =-Xmax

j---------,, ', ', ', '

SHM IN TERMS OF ENERGYAnother way to describe the block's motion is in terms of energy transfers. A stretched or compressedspringstores elastic potential energy, which is transformed into kinetic energy (andbackagain); thisshuttling ofenergy between potential andkinetic causes theoscillations. Fora springwithspring constantk, the elastic potential energy it possesses-relative to its equilibrium position-is given by theequation

Us =t kx2

Notice that the farther you stretch or compress a spring, the more work you have to do, and, asa result, the more potential energy that's stored.

In termsofenergytransfers, we candescribe theblock'soscillations asfollows. Whenyou initiallypull the blockout, you increase the elastic potentialenergyof the system. Upon releasing the block,this potential energy turns into kinetic energy, and the blockmoves. As it passes through equilib­rium, Us =0,so all the potentialenergyis kinetic. Then, as the blockcontinues through equilibrium,it compresses the spring and the kinetic energyis transformed back into elastic potential energy.

ByConservation of Mechanical Energy, the sum K+ Us is a constant. Therefore, when the blockreaches the endpoints of the oscillation region (that is, when x =±Xm.), Us is maximized, so K mustbe minimized; in fact, K =°at the endpoints. As the blockis passing through equilibrium, x =0, soUs =°and K is maximized.

146 II CRACKING THE AP PHYSICS EXAM

Us is max Us=O Usis maxK=O Kismax K=Ov=O vis max v=O, , ,, t ,, , t

V t Vyr---------, r ... --------~, ', t ,, t ,, ' I

m ,II,,

I I Ix =-Xmax x=O X = X max

equilibriumposition

The maximum displacement from equilibrium is called the amplitude of oscillation, and isdenoted by A. So instead of writing x = xmax' we write x = A (and x =-xmax will be written as x =-A).

Solution. First let's get an expression for the maximum elastic potential energy of the system:

U - 1 k 2 U - 1 k 2 - 1 kA2S -"2 x => S,max -"2 Xmax - "2

When all this energy has been transformed into kinetic energy-which, as we discussed earlier,occurs just as the block is passing through equilibrium-the block will have maximum kinetic energyand maximum speed.

U K 1 kA2 - 1 2S,max~ max ::::}"2 - "2 m Vmax

Vmax = ~~2r--------

= (500 N I m)(0.04 m)21.5 kg

= 0.73 mls

OSCillATIONS III 147

Solution. By definition, flUs =-Ws' and, since F is not constant, the work done by F must becalculated using a definite integral. As the end of the spring moves from position x = Xl to X = x

2' the

work it does is equal to

=[-t k X 2J::=-(tkx~ -tkxi)

Therefore,

So if we designate USI = 0 at Xl = 0, the equation above yields Us = tkx2, as desired. (Even if aspring does not obey Hooke's Law, this method can still be used to find the work done and thepotential energy stored.)

Solution. The total energy of the oscillator is the sum of its kinetic and potential energies. ByConservation of Mechanical Energy, the sum K + Us is a constant, so if we can determine what thissum is at some point in the oscillation region, we'll know the sum at every point. When the block isat its amplitude position, X =8 em, its speed is zero; so at this position, E is easy to figure out:

E = K+ Us = 0 + tkN = t(500 N/m)(0.08 m)? = 3.2 J

This gives the total energy of the oscillator at every position. At any position x, we have

1 m2

so when we substitute in the numbers, we get

v = E- tkx2 _ (3.2 J)- t(500 N I m)(0.04 m)2tm - t(2.0 kg)

=1.7 mls

148 CRACKING THE AP PHYSICS EXAM

Solution. Theblockwill come to rest when all of its initialkinetic energyhas been transformedintothe spring's potential energy. At this point, the blockis at its maximumdisplacementfrom equilib­rium, that is, it's at one of its amplitude positions, and

K, +Uj = K, +Uf

tmv~ +0 =0+tkA2

A=~m:~~-----

_ (2.0 kg)(2.0 m/ S)2

500N/m=0.13m

THE KINEMATICS OF SHMNow that we've explored the dynamicsof the block's oscillations in terms of force and energy,let'stalk about motion-or kinematics. As you watch the blockoscillate, you should noticethat it repeatseachcycleofoscillation in the sameamount of time.Acycle is a round-trip: forexample,frompositionx =A over to x = -A and backagain to x =A. Theamount of timeit takes to completea cycle is calledthe period of the oscillations, or T. If T is short, the blockis oscillating rapidly, and if T is long, theblock is oscillating slowly.

Anotherway of indicatingthe rapidity of the oscillations is to count the number ofcycles that canbe completedin a giventimeinterval;the morecompletedcycles, the more rapid the oscillations. Thenumber of cycles that can be completedper unit time is calledthe frequency of the oscillations, orf,and frequency is expressedin cycles per second. One cycle per second is one hertz (abbreviated Hz).

One of the most basic equations of oscillatory motion expresses the fact that the period andfrequency are reciprocals of each other:

Therefore,

. d # secondspeno =

cyclewhile

# cyclesfrequency = d

secon

OSCILLATIONS III 149

Solution. Theperiod is definedas the timerequired for one full cycle. Moving from oneend of theoscillation region to the other is only half a cycle. Therefore, if the block moves from its positionofmaximum spring stretchto maximumspringcompression in 0.25 s, the timerequiredfora fullcycleis twice as much; T =0.5 s. Because frequency is the reciprocal of period, the frequency of theoscillations is1=liT =1/(0.5 s) =2 Hz.

Solution. The frequency of the oscillations, in hertz (which is the number of cycles per second), is

1=45.5 cycles x 1min =0.758 cycles 0.758 Hzmin 60s s

Therefore,

'1 1T=-= =1.32 sf 0.758 Hz

Oneof the defining properties of the spring-block oscillator is that the frequency and period canbedetermined from themassoftheblock and theforce constant ofthespring. Theequations areasfollows:

1=1- (k and T=21t rm21t~;; ~~

Let's analyze these equations. Suppose wehad a small massona verystiff spring; thenintuitively, wewouldexpect that this strong springwouldmake the small massoscillate rapidly, with highfrequencyandshortperiod. Both ofthese predictions aresubstantiated by theequations above, because ifmis smalland k is large, then the ratiokim is large (high frequency) and the ratiomlk is small (short period).

150 CRACKING THE AP PHYSICS EXAM

Solution. According to the equationsabove,

f=.!- {k =.!- 300N/m =1.9 Hz2n ~;; 2n 2.0kg

R 2.0kgT =2n - =2n =0.51 s

k 300N/m

Notice thatf ~ 2 Hz and T ~ 0.5 s, and that thesevalues satisfy the basic equationT =l/f

Solution. Since the samespring is used,k remainsthe same. According to the equationgivenabove,f is inversely proportional to the square root of the mass of the block: f oc 1/-fiii .Therefore, if mdecreases by a factor of 4, thenf increases by a factor of ..J4 = 2.

Theequationswe saw above for the frequency and period of the spring-blockoscillator do notcontainA, the amplitude of the motion. In simpleharmonicmotion, both the frequency and the periodare independent of the amplitude. The reason that the frequency and period of the spring-blockoscillator are independent of amplitude is that F, the strength of the restoringforce, is proportionalto x, the displacement from equilibrium, as given by Hooke's Law: Fs =kx.

Solution. If the systemexhibits simpleharmonic motion, then the period and frequency are indepen­dent ofamplitude. Thisisbecausethesamespringand blockwereused in the two trials, so theperiodand frequency willhave the samevaluesin the secondtrial as theyhad in the first. Butthe maximumspeed of the blockwill be greater in the secondtrial than in the first. Since the amplitude is greaterin the secondtrial, the systempossesses more total energy (E=tkN). Sowhen the blockis passing

. through equilibrium (itspositionofgreatestspeed),the secondsystemhas moreenergyto converttokinetic, meaningthat theblockwillhavea greaterspeed.In fact, fromExample 8.2, we knowthat vmax

=A.jk / m so, since A is twiceas great in the secondtrial than in the first, vmax willbe twiceas greatin the second trial than in the first.

OSCILLATIONS l1li 151

Solution.

(a) Imagine that the block was displaced a distance x to therighfof its'equilibriuri]position. Then the force exerted by the first spring would be FI :: ....kIxarid the forceexerted by the second spring wouldbe F2 =-k2x. The net force exerted by thesprings wouldbe

Since Felf =-(kl +k2)x, we see that kelf =kl +k2,

(b) Imagine that the block was displaced a distance x to the right of its equilibriumposition, Then the force exerted by the first spring would be FI =-kIx and the forceexerted by the second spring would be F2 =-k

2x. The net force exerted by the

springs would be

FI +F2 =-kIx + -k2x =-(kl + k2)x

As in part (a), we see that, since Feff =-(kl +k2)x,

we get keff =k; +k2,

CRACKINGTHEAP PHYSICS EXAM

(c) Imagine that the block was displaced a distance x to the right of its equilibriumposition. Let Xl be the distance that the first spring is stretched, and let x

2be the

distance that the second spring is stretched. Then X =Xl + x2• But Xl =-F/kl

andx2 =-F /k2, so

Therefore,

(d) If the two springs have the same force constant, that is, if kl=kz =k, then in the first

two cases, the pairs of springs are equivalent to one spring that has twice their forceconstant: keff =kl +kz=k +k =2k. In (c), the pair of springs is equivalent to a singlespring with half their force constant:

kk e k--=-=-k+k 2k 2

THE SPRING.....BLOCK OSCILLATOR:VERTICAL MOTIONSo far we've looked at a block sliding back and forth on a horizontal table, but the block could alsooscillate vertically. The only difference would be that gravity would cause the block to movedownward, to an equilibrium position at which, in contrast with the horizontal 5HM we've exam­ined, the spring would not be at its natural length.

Consider a spring of negligible mass hanging from a stationary support. A block of mass m isattached to its end and allowed to come to rest, stretching the spring a distance d. At this point, theblock is in equilibrium; the upward force of the spring is balanced by the downward force of gravity.Therefore,

kd d -- mg=mg:::::} k

OSCILLATIONS III 153

k k k

-------------------11I1

d:II

eqUilibriumPOSition-n-~-0 nnn y=O _n n -

Next, imagine that the block is pulled down a distance A and released. The spring force increases(because the spring was stretched farther); it's stronger than the block's weight, and, as a result, theblock accelerates upward.AB theblock's momentum carries it up, throughtheequilibrium position, thespringbecomes lessstretched thanitwasatequilibrium, soF5 islessthan theblock's weight. Asa result,the block decelerates, stops, and accelerates downward again, and the up-and-down motion repeats.

When the block is at a distance y below its equilibriumposition, the spring is stretched a totaldistance ofd +y, so the upward spring force is equal to k(d +y), while the downward force stays thesame,mg. The net force on the blockis

F =k(d +y) - mg

but this equationbecomes F=ky, because kd =mg (aswe saw above). Since the resultingforce on theblock, F =ky,has the formofHooke'sLaw,we know that the vertical simpleharmonic oscillations ofthe blockhave the same characteristics as do horizontaloscillations, with the equilibrium position,y =0,not at the spring's natural length,but at the point where the hangmg blockis in equilibrium.

154 II CRACKING THE AP PHYSICS EXAM

Solution.

(a) The frequency is givenby

f 1 rr 1 300 N / m == 2.3 Hz== 2n~; == 2n 1.5 kg

(b) Before the blockis pulled down, to begin the oscillations, it stretches the spring by adistance

d == mg == (1.5 kg)(9.8 N / kg) 0.049 m == 4.9 emk 300N/m

Since the amplitude of the motion is 2.0 em,the spring is stretcheda maximumof4.9cm +2.0 em == 6.9 em when the blockis at the lowestpositionin its cycle, and aminimum of 4.9 cm- 2.0 em == 2.9 em when the blockis at its highestposition.

k k k

minimum stretch4.9 - 2 =2.9 em

------------~--

t1

, 1

Y =-2 cm - • - : 1- - __ I,

- - - -- y =0 -- - --:-- - ----1I ,- - --

y, 'y=+2cm---, 1

', - - _:

111

d =4.9 em :111

...equilibrium position - - - - - - m

maximum stretch4.9 + 2 =6.9 em

OSCILLATIONS 155

THE SINUSOIDAL DESCRIPTION OF SHMThe position of the block during its oscillation can be written as a function of time. Take a look at theexperimental set-up below.

constant speed

block undergoes :simple harmonic :oscillations :

III

'If

A small pen is attached to the oscillating block, and it makes a mark on the paper as the paper ispulled along by the roller on the right. Clearly, the simple harmonic motion of the block is sinusoidal.

The basic mathematical equation for describing simple harmonic motion is:

y =A sin ( wt)

where y is the position of the oscillator, A is the amplitude, to is the angular frequency (defined as2 ref, wherefis the frequency of the oscillations), and tis time. Since sin( wt) oscillates between -1 and+1, the quantity A sin( w t) oscillates between -A and +A; this describes the oscillation region.

If t =0, then the quantity A sin (w t) is also equal to zero. This means that y =0 at time t =O.However, what if y 1= 0 at time t =O? For example, if the oscillator is pulled to one of its amplitudepositions, say y = A, and released at time t = 0, then y =A at t =O. To account for the fact that theoscillator can begin anywhere in the oscillation region, the basic equation for the position of theoscillator given above is generalized as follows:

y=Asin(wt+ ¢o)

where ¢ 0 is called the initialphase. The argument of the sine function, to t + ¢ 0' is called the phase

(or phase angle). By carefully choosing ¢o' we can be sure that the equation correctly specifies the

oscillator's position no matter where it may have been at time t =o. The value of ¢ 0 can be calculatedfrom the equation

no _ • -l(Yatt=o)'l'o-smA

156 III CRACKING THE AP PHYSICS EXAM

Solution. First,A = 3 emand ro = 21tf = 21t(4.0 S-l) = 81t S-l. The value of the initial phase is

ill • -1(Yatt=o) . _1 3 cm . -11 1'1'0 =sm -- =sm --=sm ="21tA 3cm

Therefore, the position of the oscillator at any time t is given by the equation

Y= (3 cm)·sin[(81t S-I)t +t1tl

So, at time t = 0.3 s, we find that y=(3 cm)·sin[(81ts-1)(0.3 s)+t1tl = 0.93 em. (Make sure yourcalculatoris in radian mode when you evaluate this expression!)

Solution.

(a) To determineYat time t = 0, we simply substitute t = 0 into the given equation andevaluate:

Yatt=O =(4 cm)·sin(-t1t) =-4 em

(b) The amplitude of the motion,A, is the coefficient in front of the sin(ro t + f/J 0)expression. In this case, we read directly from the given equation that A = 4 em,

(c) The coefficient of t is the angular frequency, oi. In this case, we see that to = 61t S-l.

Since to = 21tfby definition, we have f = ro 1(2 1t) = 3 Hz.

(d) Since T = 1If, we calculate that T = 11f = 1/(3 Hz) = 0.33 s.

[C] INSTANTANEOUS VELOCITY AND ACCELERATIONIf the positionof a simpleharmonicoscillator is givenby the equation Y=A sin (co t + f/J 0)' its velocityand acceleration can be found by differentiation:

v(t) = y(t) = !£[Asin(cot + f/Jo)] = Arocos(cot + f/Jo)dt

and

a(t)=iJ(t) =!£[Arocos(cot +</>0)] =-Aro2 sin(cot+ f/Jo)dt

Notice that both the velocity and acceleration vary with time.

OSCILLATIONS III 157

Solution. Since the maximum value of cos(cot + cfJo) is I, the maximum value of v is vmax =A co;similarly, sincethe minimumvalueof sin(cot + cfJo) is-I, the maximumvalueofa isamax =A co 2. Fromthe equation for y,we see that A =4 em =0.04 m and co =61t S-l. Therefore, vmax =A CO =(0.04 m)(61t S-l) =0.75 m/s and amax =Aco 2 =(0.04 m)(61t S-1)2=14m/s2•

PENDULUMSA simple pendulum consists of a weight of mass mattached to a massless rod that swings, withoutfriction, about the vertical equilibrium position. Therestoringforce is provided by gravityand, as thefigurebelowshows,the magnitudeof the restoringforce when the bob is at an angle e to theverticalis given by the equation:

Frestoring =mg sin e

, , , ,

158 CRACKING THE AP PHYSICS EXAM

IIII

AIIII

equilibriumposition

L =length of pendulum\\\

\\\\

\\

\\\

IIIII

t ..e "

Although the displacement of the pendulum is measured by the angle that it makes with thevertical, rather than by its linear distance from the equilibrium position (as was the case for thespring-block oscillator), the simple pendulum shares many of the important features of the spring­block oscillator. For example,

" Displacement is zero at the equilibrium position.

.. At the endpoints of the oscillation region (where e =±ernaJ, the restoring force andthe tangential acceleration (at) have their greatest magnitudes, the speed of thependulum is zero, and the potential energy is maximized.

" As the pendulum passes through the equilibrium position, its kinetic energy andspeed are maximized.

Frestoring = maxat=maxU=mgh=maxK=Ov=O

.>- ... _.

:h___ .J __

Frestoring = maxat =max

U=mgh=max ~K=Ov=O

I~_...JA,-- _____IF - 0 \restoring -

at =0U=OK=maxv=max

Despite these similarities, there is one important difference. Simple harmonic motion results froma restoring force that has a strength that's proportional to the displacement. The magnitude of therestoring force on a pendulum is mgsin e, which is notproportional to the displacement e. Strictlyspeaking, then, the motion of a simple pendulum is not really simple harmonic. However, if e issmall, then sin e ::::: e(measured in radians) so, in this case, the magnitude of the restoring force isapproximately mge, which is proportional to e. So if emax is small, the motion can be treated assimple harmonic.

If the restoring force is given by mg e, rather than mg sin e, then the frequency and period of theoscillations depend only on the length of the pendulum and the value of the gravitational accelera­tion, according to the following equations:

f=.l rg21tfL and T=21tJf

OSCILLATIONS 159

Notice that neither frequency nor period depends on the amplitude (the maximum angulardisplacement, erna); this is a characteristic feature of simple harmonic motion. Also notice thatneither depends on the mass of the weight.

Solution. The equation T = 21t~L / g shows that Tis inversely proportional to -Ii, so ifg decreases

by a factor of 6, then T increases by a factor of .J6. That is,

TonMoon =.J6 x Ton Earth =.J6(1 s) =2.4 s

160 II CRACKING THE AP PHYSICS EXAM

CHAPTER 8 REVIEW QUESTIONS

SEGION I: MULTIPLE CHOICE

1. Which of the following is/are characteris­tics of simple harmonic motion?

I. The acceleration is constant.II. The restoring force is

proportional to thedisplacement.

III. The frequency is independent ofthe amplitude.

(A) II only(B) I and II only(C) I and III only(0) II and III only(E) I, II, and III

2. A block attached to an ideal springundergoes simple harmonic motion. Theacceleration of the block has its maximummagnitude at the point where

(A) the speed is the maximum.(B) the potential energy is the minimum.(C) the speed is the minimum.(0) the restoring force is the minimum.(E) the kinetic energy is the maximum.

3. A block attached to an ideal springundergoes simple harmonic motion aboutits equilibrium position (x = 0) withamplitude A. What fraction of the totalenergy is in the form of kinetic energywhen the block is at position x = t A?

(A) 1/3(B) 3/8(C) 1/2(0) 2/3(E) 3/4

4. A student measures the maximum speedof a block undergoing simple harmonicoscillations of amplitude A on the end ofan ideal spring. If the block is replaced byone with twice the mass but the amplitudeof its oscillations remains the same, thenthe maximum speed of the block will

(A) decrease by a factor of 4.

(B) decrease by a factor of 2.(C) decrease by a factor of .fi .(0) remain the same.

(E) increase by a factor of 2.

5. A spring-block simple harmonic oscillatoris set up so that the oscillations arevertical. The period of the motion is T. Ifthe spring and block are taken to thesurface of the Moon, where the gravita­tional acceleration is 1/6 of its value here,then the vertical oscillations will have aperiod of

(A) T/6

(B) T/3

(C) T/.J6(0) T(E) T.J6

6. A linear spring of force constant k is usedin a physics lab experiment. A block ofmass m is attached to the spring and theresulting frequency, f, of the simpleharmonic oscillations is measured. Blocksof various masses are used in differenttrials, and in each case, the correspondingfrequency is measured and recorded. Iff2is plotted versus l/m, the graph will be astraight line with slope

(A) 4rt2/F

(B) 4rt 2/k

(C) 4rt 2k

(0) k/(4rt 2)

(E) k2/(41t 2)

OSCILLATIONS II 161

7. A block of mass m= 4 kg on a frictionless,horizontal table is attached to one end of aspring of force constant k = 400 N I m andundergoes simple harmonic oscillationsabout its equilibrium position (x =0) withamplitude A =6 em. If the block is at x =6cm at time t =0, then which of the follow­ing equations (with x in centimeters and tin seconds) gives the block's position as afunction of time?

(A) x =6 sin(10t + tn)

(B) x =6 sin(10 1Ct + t1C)

(C) x=6sin(101Ct- t 1C )

(0) x = 6 sin(10t)

(E) x = 6 sin(10t - tn)

8. [C] A block attached to an ideal springundergoes simple harmonic motion aboutits equilibrium position with amplitude Aand angular frequency m. What is themaximum magnitude of the block'svelocity?

(A) Am(B) Nm(C) Am 2

(0) AI to(E) AI m2

162 III CRACKING THE AP PHYSICS EXAM

9. A simple pendulum swings about thevertical equilibrium position with amaximum angular displacement of 5° andperiod T. If the same pendulum is given amaximum angular displacement of 10°,then which of the following best gives theperiod of the oscillations?

(A) TI2(B) TI-J2(C) T(0) T-J2(E) 2T

10. A simple pendulum of length L and massm swings about the vertical equilibriumposition ( ()= 0) with a maximum angulardisplacement of ()max' What is the tensionin the connecting rod when thependulum's angular displacement is()= erna.?(A) mg sin e(B) mg cos (}:::(C) mgL sin ()max

(0) mgL cos ()max

(E) mgL(l - cos ()rna)

SEalON II: FREE RESPONSE

1. The figure below shows a block of mass m (Block 1) that's attached to one end of an idealspring of force constant k and natural length 1. The block is pushed so that it compresses thespring to 3/4 of its natural length and then released from rest. Just as the spring has extendedto its natural length L, the attached block collides with another block (also of mass m) at reston the edge of the frictionless table. When Block 1 collides with Block 2, half of its kineticenergy is lost to heat; the other half of Block l's kinetic energy at impact is divided betweenBlock 1 and Block 2. The collision sends Block 2 over the edge of the table, where it falls avertical distance H, landing at a horizontal distance R from the edge.

----------)0-iL

L

k 1- - --,, ,

Blot 1 rfBlock 2

: m Im ~ n -,

JJ 'J ,J ,J ,H: \, \, ,, ,J ,J ,J ,

: ~,----------------->

R

(a) What is the acceleration of Block 1 at the moment it's released from rest from its initialposition? Write your answer in terms of k, L, and m.

(b) If v1 is the velocity of Block 1 just before impact, show that the velocity of Block 1 just

f. . 1

a ter Impact IS '2V1'

(c) Determine the amplitude of the oscillations of Block 1 after Block 2 has left the table.Write your answer in terms of L only.

(d) Determine the period of the oscillations of Block 1 after the collision, writing youranswer in terms of To, the period of the oscillations that Block 1 would have had if it didnot collide with Block 2.

(e) Find an expression for R in terms of H, k, L, m, and g.

OSCILLATIONS III 163

2. A bullet of mass m is fired horizontally with speed v into a block of mass M initially at rest, atthe end of an ideal spring on a frictionless table. At the moment the bullet hits, the spring is atits natural length, 1. The bullet becomes embedded in the block, and simple harmonic oscilla­tions result.

kM

I

x=O

(a) Determine the speed of the block immediately after the impact by the bullet.

(b) Determine the amplitude of the resulting oscillations of the block.

(c) Compute the frequency of the resulting oscillations.

(d) Derive an equation which gives the position of the block as a function of time(relative to x =0 at time t =0).

3. A block of mass M oscillates with amplitude A on a frictionless horizontal table, connected toan ideal spring of force constant k. The period of its oscillations is T. At the moment when theblock is at position x = tA and moving to the right, a ball of clay of mass m dropped from

- above lands on the block.

r-------,, ,, ,, ,, ,, ,, ,, ,,x=-A

I

x=O

ma

M

;-------1: : k

,x=A

(a) What is the velocity of the block just before the clay hits?

(b) What is the velocity of the block just after the clay hits?

(c) What is the new period of the oscillations of the block?

(d) What is the new amplitude of the oscillations? Write your answer in terms of A, k, M,andm.

(e) Would the answer to part (c) be different if the clay had landed on the block when itwas at a different position? Support your answer briefly.

(f) Would the answer to part (d) be different if the clay had landed on the block when itwas at a different position? Support your answer briefly.

164 CRACKING THE AP PHYSICS EXAM

object of total mass M is allowed to swing around a fixed suspension point P. The"h;n~t-'~ moment of inertia with respect to the rotation axis perpendicular to the page throughP is denoted by I. The distance between P and the object's center of mass, C, is d.

P

(a) Compute the torque T produced by the weight of the object when the line PC makes anangle ewith the vertical. (Take the counterclockwise direction as positive for both eand T.)

(b) If e is small, so that sin emay be replaced bye,write the restoring torque Tcomputed in part (a) in the form T =-ke.

A simple harmonic oscillator whose displacement from equilibrium,z, satisfies an equation ofthe form

has a period of oscillation given by the formula T = 2n / .Jb .

(c) Taking z equal to e in Equation (*), use the result of part (b) to derive an expression forthe period of small oscillations of the object shown above.

(d) Answer the question posed in part (c) if the object were a uniform bar of mass M andlength L (whose moment of inertia about one of its ends is given by the equation

1= iMP.)

OSCILLATIONS 165