p260 gupta hw1 solutions (oscillations) due at the ...ayush.pbworks.com/f/solutions+hw1.pdfp260...

P260 Gupta HW1 Solutions (Oscillations) Due at the beginning of lecture on Thursday, 09/10/09

This first assignment covers the chapter of oscillations. Relevant section numbers are posted at ayush.pbworks.com/Schedule . This first assignment might seem a bit long – but only because many problems walk you through a series of mini-steps. Some advice:

1. Don’t start by hunting for equations. Rather, think about the physical situation, analyze the causal relations that are relevant to the question, and then try and express that in mathematics.

2. Please explain your reasoning. 3. Always think about the plausibility of your answer before handing it in. 4. Please write legibly. If we cannot read it, we will not grade it.

P1. (Adapted from Birkett and Elby The solution to this was also in Birkett and Elby) When hanging from the ceiling, Jim’s very light spring is 15 cm long. When a 0.50 kg cart is attached to the bottom of the spring and allowed to settle (first figure) the spring is 20 cm long. Suppose the cart-and-spring system is removed from the ceiling and arranged horizontally (second figure). The cart is free to roll with negligible friction. At t = 0, the cart is released from rest, 12 centimeters from the wall.

(a) How far from the wall is the cart when it reaches its maximum speed? 15 cm. The spring with nothing attached is 15 cm long. So, 15 cm is its

equilibrium length, the length at which it is neither stretched nor compressed (assuming it’s very light) In general, a simple harmonic oscillator reaches its top speed at equilibrium. (Think: why?)

Note: If the coils were heavy, their weight stretches out the spring when it hangs vertically. In that case, the spring is longer when hanging vertical than it is when sitting horizontal on a table. Here, however, the coils are very light and so the length does not change appreciably from the equilibrium length when the spring is hanging vertically (without the cart).

(b) Find the greatest distance from the wall that the cart reaches during its oscillations.

18 cm. The cart was released 3 cm to the left of equilibrium. Because it starts from rest, it never gets more than 3 cm from equilibrium. So, the rightmost point of its oscillations is 3 cm to the right of equilibrium, which is 18 cm from the wall.

(c) (i) Find the cart’s acceleration (magnitude and direction) when it’s 17 cm from the wall and heading rightward. (ii) Is the acceleration the same or different when the cart is 17 cm from the wall but heading leftward? Explain.

We can use Newton's 2nd law. When 17 cm from the wall, the cart is x = 2 cm to the right of equilibrium (Remember in using Fspring = -kx, x is measured from point of equilibrium). So, Fnet = Fspring = –kx, and hence,

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a =Fnetm

=−kxm

.

P260 Gupta HW1 (Oscillations) Due at the beginning of lecture on Thursday, 09/10/09

We know that x = 2 cm and m = 0.50 kg. So, to finish this problem, we just need to know k.

Subproblem: Find the spring constant The key is to think about the forces acting on the cart when it’s hanging from the

vertical spring. According to the problem statement, the spring in that case is 20 cm long, which is xhang = 5.0 cm longer than its equilibrium length. Since the cart hangs motionless, the upward force from the spring, |Fspring| = kxhang, balances the downward force of gravity, mg:

kxhang = mg, [For you to think: Why did we lose the minus in front of kx?] and hence, k = mg/xhang = (0.50 kg)(9.8 m/s2)/(0.050 m) = 98 N/m. Note that we

converted xhang from centimeters to meters, to make the units work out. Returning now to the problem at hand, we can use the above equation for the

cart’s acceleration to find a when the cart is 17 cm from the wall, i.e., x = 2 cm from equilibrium. We get

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a =Fnetm =

−kxm = −

(98 N/m)(0.020 m)0.50 kg = −3.9 m /s2 .

This answer doesn’t depend on which way the cart is heading. The minus sign means leftward, the direction of the net force. So, when the cart passes that point and heading rightward, it’s slowing down; and when it passes that point moving leftward, it’s speeding up.

(d) Taking rightward as positive, sketch rough, non-numerical graphs of the

cart’s position and velocity vs. time, for one full oscillation. Mark the following times on your graph: when the spring first reaches equilibrium (t1), when the cart reaches its rightmost point (t2), when the spring next reaches equilibrium (t3), and when the cart reaches its leftmost point (t4).

(Ignore the numbers on the graph!) Note, position starts out below zero, because the cart starts to the left of the

equilibrium position. The velocity starts at zero and grows positive because the cart starts from rest, but heads to the right.


(e) Find the maximum speed reached by the cart during its oscillations. Answer: vmax = .42 m/s. In part (a), we found that the cart reaches its fastest speed at equilibrium. Energy

conservation gives us one way to think about this issue (we can also resolve this in terms of forces and acceleration). As the spring decompresses after the cart is released, the spring’s elastic potential energy gradually converts into the cart’s kinetic energy. At equilibrium, the spring is neither stretched nor compressed; it has no elastic potential energy. So, at that moment, all of the system’s energy “lives” in the cart as kinetic energy. Therefore, that’s when the cart has its maximum kinetic energy, and hence, it maximum speed.

To find vmax, just “translate” that explanation into an equation: Choosing “initial” as the moment the cart is released from x = –3.0 cm (measuring x from equilibrium position), and “final” as the moment the cart reaches equilibrium (x = 0), we can say that the initial energy, which is all potential, equals the “final” energy, which is all kinetic:

E0 = Ef

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K0

0+U0 = K f +U0

0

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12 kx2 =

12 mvmax

2

. Plug in x = –0.030 meters as the initial position, and k = 98 N/m from part (c),

then solve for vmax to get 0.42 m/s (f) Find the cart’s distance from the wall 5.0 seconds after it’s released. Since the cart feels a proportional restoring force, a force whose strength is

proportional to the cart’s distance from equilibrium, we can solve Newton's 2nd law to find the cart’s position as a function of time. After finding that function, we’ll plug in t = 5.0 s.

Review of the connection between proportional restoring forces and simple harmonic motion. Here’s a summary of the reasoning. The spring causes the cart to feels a net force Fnet = –kx. So,

according to Newton's 2nd law, the cart’s acceleration is a = Fnet/m = –kx/m. Since acceleration is the second derivative of position with respect to time, we can rewrite this acceleration equation as

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d2 xdt2 = −

km x.

To simplify the ensuing math, define the angular frequency as

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ω = k /m . Now we can rewrite the differential equation as

d2xdt2 = −ω 2x.

The solutions to this differential equation are sinusoidal functions such as x = Acos wt and x = Asin

wt. End of review.


In part (d), we found (by graphing!) that the position vs. time is a negative cosine. So, the particular sinusoidal function describing this cart is

x = –Acos wt. To find the cart’s position at t = 5.0 s, we just need to know the constants A and

w, and then we can plug in t = 5.0 s. Since the cart oscillates between x = –3 cm and x = +3 cm. So, its amplitude—its maximum distance from equilibrium—is A = 3.0 cm = 0.030 m. And as mentioned above, the angular frequency is defined as

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ω =km

=98 N /m0.50 kg = 14 s -1 .

So, the cart’s position at t = 5.0 s is

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x = −Acosωt = −(0.030 m)cos 14 s-1( ) 5.0 s( )[ ] = −0.019 m, which means 1.9 centimeters to the left of equilibrium. (Remember: Angle

expressions in these equations are in radians, not degrees!) (g) At what time does the cart first reach the rightmost point of its motion? Answer:0.22seconds.Weeseetwoclosely‐relatedwaystoaddressthis. Method1:Useoscillationkinematics.Asdiscussedinquestion2,

therightmostpointoftheoscillationsisx=3.0cm=0.030meters.Andwejustfiguredoutthecart’spositionasafunctionoftime:

x=–Acosωt.So,tofindthetimeatwhichthecartreachesitsrightmostpoint,justplugx=

0.030mintothatequationandsolveforttoget0.22s. Method2:Reasonintermsoftheperiod.Afulloscillationtakesthe

cartfromLeftmostpoint‐>Rightmostpoint‐>Leftmostpoint.So,thisquestionasksforthetimeneededtocompletehalfanoscillation(Leftmostpoint‐>Rightmostpoint).Well,thetimeneededtocompleteawholeoscillationistheperiod,T.Hence,wejusthavetofindTandcutitinhalf.

Angularfrequencyandperiodarerelatedby

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T =2πω

.(Seenote)

Sinceω=14s–1,asfoundinpart(b)above,theperiodworksouttobeT=0.45seconds.Therefore,thecartreachesitsrightmostpointatt=T/2=0.22s.

NOTE:Here’showtofigureouttherelationshipbetweenangularfrequencyandperiod.

Eachtimetheargumentofasineorcosineincreasesby2πradians(360°),thefunction“startsover,”Forinstance,cos0°=cos360°=cos720°,andsoon,becauseinasense,0°and360°and720°areallthesameangle.Physically,thismeansthecartreturnstoitsinitialpositionandbeginsanewoscillationeachtimeωtincreasesbyanadditional2πradians.Insummary,whenthetime


increasesbyoneperiod(T),theargumentofthecosine,ωt,increasesby2π.TranslatingthatboldfacesentenceintoanequationgivesusωT=2π.

Tosomestudents,thismakesmoreintuitivesenseifyourepresentωtbyadialspinninginacircle.Afulloscillationofthecartcorrespondstoafullrotationofthedial.Inotherwords,afullrotationofthedialthrough2πradiansoccurswhenthetimeincreasesbyoneperiod.Again,translatingthatlastsentenceintoanequationgivesωT=2π.

P2.(BirkettandElby)Att=0,anessentiallyfrictionlesschunkofdryice(frozencarbondioxide)ofmassm=0.10kgisreleasedfromrestfromD=0.020meterstotherightofthebottomofahemisphericalbowlofradiusR=0.25meters.(Seethediagram.)Setx=0atthebottomofthebowl.Wechooserightwardasthepositivex‐direction.

(a)Sketchrough,non‐numericalgraphsofthechunk’shorizontalposition,

velocity,andaccelerationvs.time.

(don’tpayattentiontothenumericalvaluesontheaxis!)(b)Whendoesthechunkfirstreachthebottom?Solvethisusingforces

ratherthantorques.Hint2b.Answer: 0.25 seconds.

You can address this by writing the chunk’s position as a function of time (see part (c)) and then solving for the first time at which x = 0. Or, you can reason in terms of period. A full oscillation is

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Rightmost point →1

Equilibirum→2

Leftmost point →3

Equilibirum →4

Rightmost point.

‐1.5

‐1

‐0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9

Position

Velocity

Acceleration


So, this question asks about the first quarter of a complete oscillation, the first of four “legs.” Therefore, we’re solving for a quarter of the time needed to complete a full oscillation—a quarter of a period. Let’s solve for T and divide by 4.

Tofindtheperiodletuslookatthe

natureoftherestoringforcefortheice.Theforcesontheiceare,thenormal

forcefromthebowlpushingitup(atandangle)andtheweightdown.Theweightisseparatedintothecomponentsalongthenormalforce(perpendiculartothebowlsurfaceatthatpoint)andthecomponentparalleltothesurface.Fortherestoringforce,weareonlyconcernedwiththeforceparalleltothebowlsurface–theperpendicularforcesdon’tchangethespeedoftheice.

So,thenetforceonthechunkalongthedirectionofmotion(i.e.,alongthesurfaceofthebowl)isgivenbyFnet=–mgsinθ.Andsinθ=opposite/hypotenuse=x/R.So,wecanrewritethisnetforceas

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Fnet = −mgsinθ = −mg xR

= −mgR

x.

Thisisindeedaproportionalrestoringforce:ittakestheformFnet=–keffx,withkeff=(mg/R).

[Note:inrealityθ=s/Rwheresisthearclengthalongthesurfaceofthebowl;butforthetinydisplacementoftheicechunk,sisequaltoxbecauseforsmallanglesthechunk’smotionisalmostentirelyhorizontal)

Thex‐coordinateoftheicechunkcouldbesaidtobegovernedbythe

differentialequation

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d2 xdt2 = −ω2 x. where ωistheangularfrequency

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ω =keff

m =mg/R( )

m =gR ,

Aswe’veseen,thesolutionstothisdifferentialequationaresinusoidal.Our

graphofxvs.tinpart(a)showsthattheparticularsinusoidalfunctiondescribingthechunk’spositionisacosine:

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x = Acosωt . HereAistheamplitudeoftheoscillation(0.02m)Andthen,


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T =2πω

=2πg/R

= 2π Rg = 2π 0.25 m

9.8 m /s2 = 1.0 s.

Sincethechunkcompletesafulloscillationinonesecond,itcompletesaquarteroscillation(Rightmostpointequilibrium)inaquartersecond.

(c)Findthechunk’shorizontalposition3.0secondsaftergettingreleased.

Hint2c.Justpluggingint=3.0secondsintotheequationforthepositionofthe

chunk,weget

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x = Acosωt = 0.020 m( ) cos 39.2 s -1( ) 3.0 s( )[ ] = 0.0042 m, Answer:about4millimeterstotherightofequilibriumpositionatthe

bottomofbowl.(d)Thechunkiscaughtandthere‐releasedfromrest,fromx=0.01meters

(insteadof0.02meters)fromthebottomofthebowl.Doesitreachthebottominmoretime,lesstime,orthesametimeasitdidbefore?Explain.

Sametime.Althoughthechunknowneedstocoveronlyhalfasmuchdistancetoreachequilibrium,itslidesonlyhalfasfast,onaverage,asitdidbefore.Thesloweraveragespeedexactlycompensatesforthesmallerdistanceitmustcover.So,inbothcases,thechunkreachesequilibriumatthesametime.

Thisreasoningmaylookmorefamiliarwhenphrasedintermsofamplitudeandperiod.Thetimeneededforthechunktoreachequilibriumisaquarterofaperiod.Butbecauseofthecompensationbetweendistanceandaveragespeed,theperiodofasimpleharmonicoscillatorisindependentoftheamplitude;biggerorsmalleroscillationstakethesametime.

(e)Inreality,thechunkisn’tcompletelyfrictionless.Sketchxvs.t,showing

manyoscillations.(Includetheeffectoffrictiononthemotion.)Howdoesyournewgraphdifferfromtheidealizedpositiongraphyoudrewinpart(a)?

Frictionshouldbesmallenoughthattheoscillationsareunderdamped,i.e.,thechunkoscillatesbackandforthmuchasitwouldintheabsenceoffriction,


excepttheoscillationsgraduallygetsmallerinamplitudeuntildyingoutcompletely.

P3. (Adapted from Birkett and Elby) In the figure the two identical springs haveequilibrium length L0 and spring constant k0. A puck ofmassM, shown here atequilibrium, can slide with negligible friction on the table/surface. The puck isattached on each side to the spring. The other end of the spring is nailed to thetable. In the diagram, the springs are stretched to a length L (So, the length thestringisstretchedbeyonditsrelaxedlengthisL‐L0).

(a) The puck is displaced a small distance in positive x‐direction, and released.

Whatistheperiodofoscillation?AnswerintermsofM,k0,L,andL0.(b) Nowthepuckisdisplacedfromequilibriumasmalldistanceinthepositivey‐

direction,andreleased.Itoscillatesbackandforthinthey‐direction.Doyouexpect thenewperiodofoscillation tobegreater than, less than,orequal toyourpart (a)answer? Givean intuitive, conceptual answer. Nopointsoff ifyouturnouttobeincorrect.

(c) Findtheperiodofoscillationforthepuckoscillatinginthey‐direction.(d) Revisityourpart(b)reasoning.Wasitcorrect?Ifnot,seeifyoucanexplainin

common‐sense termswhy the periodsworked out thewaytheydid.P3.Solutions(ThissolutioniscontributedbyEricKuo)

‐1.5

‐1

‐0.5

0

0.5

1

1.5

0 10 20 30 40

Position

Velocity

Acceleration


a) Whenthepuckisatequilibrium,bothspringsarepullingwithaforcek0(L‐

L0).Sincethey’repullinginoppositedirections,thenetforceonthepuckis0,soit’snotaccelerating.Intuitively,Iknowthatifyoudisplacethepuckalittlebitinthehorizontaldirection,it’sgoingtostartoscillatingabouttheequilibriumpointbecauseofthesprings.InordertocalculatetheperiodIwanttowritethisintuitionoutusingmath.

Ifthepuckis,forexample,

adistance“x”totherightoftheequilibriumposition,thenthespringontheleftisstretchedoutmorethanLandthespringontherightisstretchedoutlessthanL.Icanwriteouttheforceofthetwospringsandfindthenetforceonthepuck.

Nowthatwehavethenetforce,Newton’s2ndlawtellsushowthepuckmust

beaccelerating.

Sincethepuckisoscillating,weknowthatthepositionofthepucklooks

somethinglikex~cos(at),whereaissomeconstant.Inordertosatisfythe

equationwegetfromNewton’s2ndlaw,wewantx~cos

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2k0M

t

.Ifyoudon’t

believethis,trytoplugxandd2x/dt2intotheequationtoshowthatthissatisfiestheequation.

Oneperiodiswhenthepuckgoesthroughonecycle.Cos

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2k0M

t

goes

throughonecyclewhen

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2k0MT = 2π

.Therefore,theperiodofthepuck’soscillation


is

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T = 2π M2k0 .Note:theanswerdoesn’tinvolveLorL0.Isthisbelievable?Well,

thepuckstartsatanequilibriumpositionwheretheforcesare0.Whenwetalkaboutoscillations,weonlycareaboutthechangeinnetforceasit’sgettingdisplaced,soitdoesn’tmatterifLis1cmor1000m;thechangeinforceofeachspringdependsonhowmuchmoretheygetstretched/compressedfromthepuck’sequilibrium.That’swhyL(andsimilarly,L0)getcancelledoutfromthestartwhenwecalculatedthenetforceonthepuck.

b) Here’soneintuitiveanswerthatIheardfromastudent:Iexpecttheperiodofoscillationinthey‐directiontobeshorterthanthe

periodinthex‐direction.Whenyoumovethepuckupinthey‐directionbothspringsarepullingthepuckinthesamedirection(down),whereaswhenyoumovethepuckinthex‐directionthespringsarepullinginoppositedirections;theleftspringpullsthepucktotheleftandtherightspringpullsthepucktotheright.Asaresult,fortheverticaloscillation,thetwospringsareworkingtogether,sotheymustpullharderandmaketheoscillationfaster,andthereforetheperiodshorter,thaninthecaseofhorizontaloscillation.

Thereisnowronganswerhereaslongasyouranswermakesintuitivesense

toyou!Makesureyouexplainyourreasoningclearly.c) Ifyoumovethepuckinthe

y‐direction,thenthespringsaregoingtogetstretchedoutatanangleθ.Thenetforceofthetwospringsisgoingtobeinthey‐direction,becausethehorizontalcomponentsofthepullofthetwospringswillcancelout,soallwehavetodoisfigureoutthecomponentofthesprings’forceinthey‐direction.

Sincethisproblemisdealingwithsmalldisplacements(comparedto

thelengthofthesprings),wecanestimatethatthelengthofthespringwhenstretchedisL.Thismightseemtotallycrazy,solet’sdoanexamplebyplugginginsomevalues:L=20cm,y=1.5cm.UsingthePythagoreantheorem,weseethatthe

lengthofthespringwhenstretchedis

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202 +1.52 = 20.06 ≈ 20cm. Sothemagnitudeofthetotalforceexertedbyonespringisk(L‐L0)

andthey‐componentofthisforce(ifwesaythatthepuckisabovetheequilibriumposition)is‐k(L‐L0)sinθ≈‐k(L‐L0)θforsmallangles(itisminussincethespringsmustbepullingdown).Fromourapproximationfrombefore(thelengthofthe


stretchedspringisL),wenowhaveanisoscelestrianglewithareallysmallangle,andweknowhowtorelatethesidesandtheangleinthiscase.

Sopluginθ=y/LtogetthenetforceofBOTHspringsinthey‐

direction:‐2k0y(1‐L0/L).Nowthatwehavethenetforce,justlikeinparta),wecan

againuseNewton’s2ndLawtofindtheperiod.Ifyourepeatthereasoninginpart

a),theperiodofverticaloscillationis

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2π M

2k0(1−L0L).

d) Since(1‐L0/L)isalwayslessthan1,theperiodofverticaloscillationis

alwayslongerthanforhorizontaloscillation.Onewaytomakesenseofthisisthatbecausetheangleθmadewhenyoumovethepuckinthey‐directionisreallysmall,they‐componentoftheforceisalsoreallysmall.Anotherwaytosaythisisthis:inthex‐direction,alloftheforceofthespringcontributestothenetforce,butinthey‐directiononlysinθ(whichisverysmall)oftheforcecontributestothenetforce.

ThisansweristheoppositeofwhatIthoughtwhenIintuitivelyreasonedin

partb)whenIsaidthatthespringsinhorizontaloscillationarepullingagainsteachotherwhereasinverticaloscillationsthespringsarebothworkingtogethertopullharder.HowcanIresolvemyanswerinc)withmyreasoninginpartb)?Well,first,inx‐oscillation,eventhoughthespringsarepullinginoppositedirections,whenthepuckisdisplaced,onespringpullsmoreandtheotherpullsless.Inthisway,thespringsareworkingtogether.Imagineinatug‐of‐warwherebothteamsarepullingequally,onewaytowinistopullmorethanyouropponent,butonewaytheotherteamcan“helpyou”winisbypullingless.

Yourreasoninginpartb)wasprobablydifferent,soyouwillhavetowork

outyourownresolution.Themostimportantthinghereistolookbackatyourownthinkingandtrytotweakyourownintuitiontomakeintuitivesenseoftheresultsofpartsa)andc).Ifyourreasoninginb)wasalreadycorrect,trytothinkofdifferentreasonswhysomeonemightchooseadifferentanswer,andtheytrytorespondtothatperson’sclaims.Whatistheflawinthatlineofreasoningandhowcouldyoumakethatcleartoanotherperson?

P4.(BirkettandElby)Considerapendulumconsistingofaverysmallballofmassm1attachedtotheendofarodofmassm2andlengthL(seefigure).


(a) Thependulumisdisplaced2°fromequilibriumandreleased.Thenit’sdisplacedby4°andreleased.Inwhichcase,ifeither,doesthependulumtakemoretimetocompleteonefulloscillation?Explain.

(b) Forthe2°release,findthetimeittakesthependulumtocompleteafulloscillation,expressingyouranswerintermsofconstantsandthegivenquantities.

(c) Theperiodofasimplependulum(ballhangingfrommasslessstring)is .Howcanyouusethatpreviously‐derivedresulttodouble‐checkyourpart(b)answer?

Solution to P4: Also contributed by Eric Kuo P4.a) They should have the same period! The reasoning is exactly the same as

in problem 2d). Although the 4° oscillation has to cover a greater distance, the average speed of the pendulum is greater (one way to see this is that the speed of the pendulum at the bottom of its swing is greater the larger the initial displacement). These two factors balance out and the periods are the same.

b) If the pendulum was only the ball at the end, we might have done a similar

strategy as in problem 2 by approximating the small circular swing of the pendulum as horizontal oscillation and solving the problem with forces. The problem is that now we have a heavy bar that is rotating, so we have to take that rotation into account. Now, instead of talking about forces, we’re going to consider torques and the rotational analog of Newton’s 2nd Law: τ = Iα.

The total torque is sum of the torque on the bar and the torque on the ball.

The way to find the torque is τ = r F sinθ, where r is the distance from the pivot point to where the force is acting. One thing to remember is that the force on the bar effectively acts on the center of mass of the bar, which is in the middle. We also need to know the moments of inertia of the ball and the bar.

Now, we plug this all into the rotational analog of Newton’s 2nd Law.

!

2" L/g


This equation is exactly the type of equation that we get in problem 3a).

We follow the exact same logic as in that problem and find that the period here is:

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T = 2πm1L

2 +m2L

2

3m1gL +

m2gL2

Notice that at no point did we use the fact that it was a 2° release. The

only point where the angle is relevant is where I say sinθ ≈ θ, which is only true for small angles. This calculation is equally valid for an initial displacement of 2° and 4°, as well as any other angle small enough for the small angle approximation to be valid.

c) Our system would be just like a simple pendulum if the mass of the bar m2

= 0. If we plug in m2 = 0 into our expression for the period of our pendulum, we get

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T = 2π Lg , which is exactly what we’d expect for a simple pendulum. Although this

doesn’t guarantee that we got the correct answer in part b), we are a little more confident because it at least gives us the answer we would expect in one particular situation.

P5. (Adapted from “General Problems in Physics” by I. E. Irodov) A uniform rod is placed on two spinning wheels as shown. The axes of the wheels are separated by a distance L = 20 cm, the coefficient of friction between the rod and the wheels is k = 0.18 N/s2. The initial position of the rod is such that the center of the rod was a tiny little bit to the right of the point mid-way from the center of the two wheels. (a) Qualitatively explain why the rod would oscillate back and forth about the point

midway from the center of the wheels? Thinking about the forces on the rod, we have the weight (Mg) acting down, the normal force from each wheel on the rod acting up (Nl and Nr) {using Nl for the left wheel and Nr for the right side wheel}. And you have the forces of friction from the left and the right wheels, fr and fl. Also note that the friction between the right wheel and


the rod is to the left (since the wheel is rotating to the left); and the friction between the leftwheel and the rod is to the right. The rod does not move vertically – so the forces in the vertical direction must balance. Now, if the center of the rod is to the right of the midpoint between the wheels, then the right side wheel carries more of the weight of the rod. So, the normal force on that wheel has to be more than the normal force on the left side wheel. This means that the friction between the right wheel and the rod is more than that between left wheel and rod. And this difference in the friction leads to a net force on the rod to the left! And the situation is reversed if the center of the rod is to the left of the midpoint between the wheels. So, if the center of the rod is displaced to the right, the net force tries pushes it to the left. By the time the rod’s center is aligned with the point midway between the wheels, it is already moving. So right when the centers are aligned, even if the forces on the rod are balanced, it keeps moving to the left. As soon as the rod overshoots to the left, the net friction on it is to the right, which slows down, and turns it around. The story repeats and the rod oscillates. (b) Consider the moment when the center of the rod is exactly midway from the center of the two wheels. This is the equilibrium point, when the forces and torques on the rod are balanced. Why doesn’t the rod just come to rest at that equilibrium point? Right when the center of the rod is midway between the wheels, the forces on the rod are balanced (The friction force from the left and right wheels are exactly equal and they cancel) But, by the time the rod reaches that point, it was already moving! And since the force at that point is zero, it just keeps moving. There is nothing to stop it! And so the rod does not come to rest at the equilibrium point. (Parts C & D are required only for the Honors section; optional for others)

(c) In this situation, show that the net force on the rod can act as a restoring force for the rod to perform simple harmonic motion in the horizontal direction?

Vertical forces balance: Nr+Nl = Mg Horizontal forces provide the restoring force. Net force in horizontal direction: Fnet = fr -fl and this net force is to the left

(assuming the initial position of the rod as offset to the right). Some of you might object that fr is to the left and so must be negative – but see here, just use the common sense with signs. fr is greater in size than fl, so I get fr -fl and the result of this subtraction points to the left. I will just work with the size of these forces until almost the final step, when I write the force as a vector.

Now since k is the coefficient of friction between the rod and the wheels, then we can write Fnet = k (Nr – Nl)

Intuitively, I expect that Nr-Nl should be proportional to the amount that the rod is offset to the right, because that is what makes the right wheel support a bit of extra weight making Nr greater than Nl. And the larger the offset distance between center of rod and midpoint between wheels, the larger this difference.

To do so mathematically, think about how the two normal forces are opposing one another (We want to find their difference, so it would make sense to think how they

oppose). Acting together they balance the weight of the rod. But the torque due to Nr is opposite the direction of torque due to Nl.


At equilibrium, when the center of rod is midway between wheels, if I take the torque about that center point (marked “O” in the picture), all the torques would cancel (the contribution of weight in that case is 0, and the torque from the left normal force balances that due to the right normal force; Also, the friction forces do not contribute to the torque since they pass through that center point.

If I take the torque about the point O when the rod is offset by a distance x, then Torques in counter-clockwise direction = Nr (L/2 + x) Torques in clockwise direction = Mg(x) +Nl(L/2 – x) Balancing the equations, we get Nr – Nl = (2Mg/L)x .... proportional to x, as we

intuitively expect! Putting that back into the force equation Fnet = (2Mgk/L)x Finally, since the net force is to the left, expressing this as vectors, we get Fnet = - (2Mgk/L)x, (Fnet and x being vectors in this equation.) Since the net force has the general form of being proportional to the deviation

from equilibrium and in the opposite direction, it acts as a restoring force. (d) Find the period of oscillation of the rod. Almost done. Writing the net force equation in terms of acceleration of the rod, Ma = - (2Mgk/L)x, or a = - (2gk/L)x, which gives us x(t) = x0 cos (√(2gk/L)

t) And so the period is T = 2π√(L/2gk) = π√(2L/gk) We are not done yet. We need to evaluate if our math expression is supported by

causal dependencies in the physical system. Well, I should expect that if the coefficient of friction between rod and wheels was more, then the restoring force would be more, making the rod move faster. And so, it makes sense that the period is inversely proportional to k. As for L, the further apart the wheels are the less is the proportional dis-balance between the normal forces, for the same deviation of the rod from equilibrium. So it is good that the period is proportional to L P6. (Courtesy: Prof. Redish) The Damped Oscillator: A class looked at the oscillation of a mass on a spring. They observed for 10 seconds and found its oscillation was well fit by assuming the mass' motion was governed by Newton's second law with the static-spring force Fspring = -k Δs where Δs represents the stretch or squeeze of the spring and k is the spring constant. However, it was also clear that this was not an adequate representation for times on the order of 10 minutes, since by that time the mass had stopped oscillating.

(a) Suppose the mass was started at an initial position x0 with a velocity 0 m/s. Write down the solution, x(t) and v(t), for the equations of motion of the mass using only the static-spring force. What is the total energy of the oscillating mass?

This is simple: x(t) = x0 cos (ωt) where ω = √(k/m) and v(t) = - x0ω sin(ω t) We have talked about this in class lots of times. The solutions to the

equations of motion of mass on spring are sinusoidal. Assuming that at t=0, the


mass is at the rightmost extremity, it would be a positive cosine function. The velocity is negative – of course, the mass would rush to the left initially!

The total energy of the mass is ½ kx02 – equal to the total potential energy

put into the spring in pulling it through x0 distance. Assume that there is also a velocity dependent force (the damping force) that the

surrounding medium exerts on an object when it is moving. Let's make the simplest assumption, that the damping force is linear in the velocity, Fdamping = -γv. Assume further that over one period of oscillation the velocity dependent piece is small. Therefore, take x(t) and v(t) to be the same as if the system weren’t damped.

(b) Calculate the work done by the damping force over one period in terms of the parameters k, m, and γ .

Think about the mass at a point x to the right of the equilibrium position and moving towards the left.

So, The force at a point x(t) is given by F = −γv = γx0ω sin(ω t) (Is it okay that there is no minus sign in front of that sin(ω t)? )

In moving through a distance dx more to the left, the work done by the force would be dW = F dx = F dx sin(θ). Here θ is the angle between the Force vector and dx vector. In our case, dx is to the left. And since the mass is moving to the left, the damping force is to the right. So θ = 180°

So, dW = − γx0ω sin(ω t) dx (Here, the negative sign is because of the angle between force and dx; the negative sign is good – because we know that the damping force is taking energy away from the mass, slowing it down. So the work done should be negative.)

Now dx = |v|dt ... I take the absolute value of v because I am looking here for just the size of the displacement – so I just take the speed, not the velocity.

So, dx = x0ω sin(ω t) dt Putting that back into the equation for Work done, dW = − γx0

2 ω2 sin2(ω t) dt

To get the total work done by this damping force over one cycle, we need

to integrate this expression over the entire period.

€

W = −γx02ω 2 sin2(ωt) dt

0

T

∫

Working out the integral, you get W = −γx02

ω2 T/2 where T is the period of the oscillation -- T = 2π√(m/k) and ω2= k/m

Putting that back into the equation, W = −πγx02√(k/m)

(Stepping back, does this make sense? A higher γ would mean a larger

damping force – so the work done by that force would be more. A larger x0 or a larger ω (=√(k/m) would lead to a larger maximum speed and that also would imply a larger average damping force – so the work done would be more. In


other words, the mathematical dependencies in our expression align well with what we expect from the causal relations in our physical system)

(c) The find the fraction of the energy lost in one period. The work that we got in the earlier part is just the amount of energy lost to

the damping in one cycle. The fraction of energy lost is just

€

πγx02 km

12kx

02

= 2 πγkm

...

Note, we just take the positive value of the work done – because it is the size of this work that equals the amount of energy lost. The sign just tells us that it is a “loss” of energy from our spring-mass system.

Note, k , the stiffness of the spring has a larger effect on the energy stored in the spring than on the energy lost to the damping force ... the energy lost to the damping force depends on √k, but the energy stored in spring depends on k. It is this reason that k appears in the denominator in the above expression.

p260 gupta hw1 solutions (oscillations) due at the ...ayush.pbworks.com/f/solutions+hw1.pdfp260...

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