A2Physics物理出國(guó)英語(yǔ)5.5Oscillations(2)

1.Inasimpleharmonicmotion,theaccelerationoftheobjectisdirectlyproportionaltoitsdisplacementfromtheequilibriumpositionandisalwaysdirectedtowardstheequilibriumposition.Mathematically,(a=omega^{2}x),where(a)istheacceleration,(x)isthedisplacement,and(omega)istheangularfrequency.IfanobjectinSHMhasadisplacement(x=0.2m)andanangularfrequency(omega=5rad/s),whatistheacceleration?

Usingtheformula(a=omega^{2}x),substitute(omega=5rad/s)and(x=0.2m).Then(a=(5)^{2}times0.2=5m/s^{2}).Thenegativesignindicatesthattheaccelerationistowardstheequilibriumposition.

2.Theperiod(T)ofasimplependulumisgivenbytheformula(T=2pisqrt{frac{l}{g}}),where(l)isthelengthofthependulumand(g)istheaccelerationduetogravity.Ifthelengthofapendulumisincreasedbyafactorof4,whathappenstotheperiod?

Lettheoriginallengthbe(l_1)andthenewlengthbe(l_2=4l_1).Theoriginalperiod(T_1=2pisqrt{frac{l_1}{g}})andthenewperiod(T_2=2pisqrt{frac{l_2}{g}}=2pisqrt{frac{4l_1}{g}}).Since(sqrt{4}=2),(T_2=2T_1).Sotheperioddoubles.

3.ForamassspringsysteminSHM,thetotalenergy(E)isthesumofkineticenergy(K=frac{1}{2}mv^{2})andpotentialenergy(U=frac{1}{2}kx^{2}),where(m)isthemass,(v)isthevelocity,(k)isthespringconstant,and(x)isthedisplacement.Atthemaximumdisplacement(x=A)(amplitude),thevelocity(v=0).Whatisthetotalenergyatthispoint?

Substitute(v=0)intothetotalenergyformula(E=K+U).Since(K=frac{1}{2}mv^{2}=0)at(x=A),thetotalenergy(E=frac{1}{2}kA^{2}).

4.AparticleisinSHM.Thedisplacement(x)asafunctionoftime(t)isgivenby(x=Acos(omegat+varphi)),where(A)istheamplitude,(omega)istheangularfrequency,and(varphi)isthephaseconstant.If(A=0.3m),(omega=2rad/s),and(varphi=frac{pi}{6}),whatisthedisplacementat(t=pis)?

Substitutethegivenvaluesintotheformula:(x=0.3cos(2pi+frac{pi}{6})).Usingthepropertyofcosinefunction(cos(a+2pi)=cos(a)),wehave(x=0.3cos(frac{pi}{6})=0.3timesfrac{sqrt{3}}{2}approx0.26m).

5.Thefrequency(f)ofasimpleharmonicoscillatorisrelatedtotheangularfrequency(omega)bytheformula(omega=2pif).Ifthefrequencyofanoscillatoris(5Hz),whatistheangularfrequency?

Substitute(f=5Hz)intotheformula(omega=2pif).Then(omega=2pitimes5=10pirad/sapprox31.4rad/s).

6.Inadampedharmonicoscillator,theamplitudeofoscillationdecreasesovertime.Theamplitude(A(t))attime(t)isgivenby(A(t)=A_0e^{frac{bt}{2m}}),where(A_0)istheinitialamplitude,(b)isthedampingconstant,and(m)isthemass.If(A_0=0.5m),(b=1kg/s),and(m=2kg),whatistheamplitudeat(t=4s)?

First,calculatetheexponent:(frac{bt}{2m}=frac{1times4}{2times2}=1).Then(A(4)=0.5e^{1}approx0.5times0.368=0.184m).

7.Thephasedifference(Deltavarphi)betweentwosimpleharmonicmotions(x_1=A_1cos(omegat+varphi_1))and(x_2=A_2cos(omegat+varphi_2))is(Deltavarphi=varphi_2varphi_1).If(x_1=0.2cos(3t+frac{pi}{3}))and(x_2=0.3cos(3t+frac{pi}{6})),whatisthephasedifference?

Usingtheformula(Deltavarphi=varphi_2varphi_1),substitute(varphi_1=frac{pi}{3})and(varphi_2=frac{pi}{6}).Then(Deltavarphi=frac{pi}{6}frac{pi}{3}=frac{pi}{6}).Thenegativesignindicatesthat(x_2)lagsbehind(x_1).

8.Forasimplependulum,thetimeperiod(T)onthesurfaceoftheEarthis(T_E=2pisqrt{frac{l}{g_E}}),where(g_E)istheaccelerationduetogravityonEarth((g_Eapprox9.8m/s^{2})).OntheMoon,(g_M=frac{1}{6}g_E).WhatisthetimeperiodofthesamependulumontheMoon?

ThetimeperiodontheMoon(T_M=2pisqrt{frac{l}{g_M}}).Substitute(g_M=frac{1}{6}g_E)intotheformula:(T_M=2pisqrt{frac{l}{frac{1}{6}g_E}}=sqrt{6}times2pisqrt{frac{l}{g_E}}=sqrt{6}T_Eapprox2.45T_E).

9.Thevelocity(v)ofaparticleinSHMisgivenby(v=omegaAsin(omegat+varphi)).If(A=0.4m),(omega=3rad/s),and(varphi=0),whatisthemaximumvelocity?

Themaximumvalueofthesinefunctionis1.So,themaximumvelocity(v_{max}=omegaA).Substitute(A=0.4m)and(omega=3rad/s),then(v_{max}=3times0.4=1.2m/s).

10.Inaforcedharmonicoscillator,theamplitudeofoscillationdependsonthedrivingfrequency(omega_d)andthenaturalfrequency(omega_0)oftheoscillator.When(omega_d=omega_0),resonanceoccurs.Foramassspringsystemwith(m=1kg)and(k=4N/m),whatisthenaturalfrequency(omega_0)?

Thenaturalfrequency(omega_0=sqrt{frac{k}{m}}).Substitute(m=1kg)and(k=4N/m),then(omega_0=sqrt{frac{4}{1}}=2rad/s).

11.AparticleinSHMhasadisplacementtimegraph.Thedisplacement(x)isgivenby(x=Asin(omegat)).Iftheparticlestartsfromtheequilibriumposition((x=0))at(t=0),andreachesthemaximumdisplacement(x=A)at(t=frac{T}{4})(where(T)istheperiod),whatistheangularfrequency(omega)intermsof(T)?

Weknowthat(omega=frac{2pi}{T}).At(t=frac{T}{4}),(x=Asin(omegatimesfrac{T}{4})).Since(x=A)at(t=frac{T}{4}),then(A=Asin(omegatimesfrac{T}{4})),whichmeans(sin(omegatimesfrac{T}{4})=1).So(omegatimesfrac{T}{4}=frac{pi}{2}),andsolvingfor(omega)gives(omega=frac{2pi}{T}).

12.Theacceleration(a)ofaparticleinSHMisrelatedtothedisplacement(x)by(a=omega^{2}x).Ifthemaximumacceleration(a_{max})is(20m/s^{2})andtheamplitude(A)is(0.5m),whatistheangularfrequency(omega)?

Themaximumaccelerationoccursatthemaximumdisplacement((x=A)).So(a_{max}=omega^{2}A).Rearrangingfor(omega),weget(omega=sqrt{frac{a_{max}}{A}}).Substitute(a_{max}=20m/s^{2})and(A=0.5m),then(omega=sqrt{frac{20}{0.5}}=sqrt{40}approx6.32rad/s).

13.Forasimplependulum,therestoringforce(F=mgsintheta),where(m)isthemassofthependulumbob,(g)istheaccelerationduetogravity,and(theta)istheangulardisplacementfromthevertical.Forsmallangles(theta)(inradians),(sinthetaapproxtheta).If(m=0.2kg),(g=9.8m/s^{2}),and(theta=0.1rad),whatistherestoringforce?

Usingtheapproximation(F=mgtheta),substitute(m=0.2kg),(g=9.8m/s^{2}),and(theta=0.1rad).Then(F=0.2times9.8times0.1=0.196N).Thenegativesignindicatesthattheforceistowardstheequilibriumposition.

14.Theenergyofasimpleharmonicoscillatorisconservedintheabsenceofdamping.Ifthetotalenergy(E=frac{1}{2}kA^{2})andthespringconstant(k)isdoubledwhilekeepingtheamplitude(A)constant,whathappenstothetotalenergy?

Lettheoriginalenergybe(E_1=frac{1}{2}k_1A^{2})andthenewenergybe(E_2=frac{1}{2}k_2A^{2}).Given(k_2=2k_1),then(E_2=frac{1}{2}(2k_1)A^{2}=2timesfrac{1}{2}k_1A^{2}=2E_1).Sothetotalenergydoubles.

15.ThedisplacementofaparticleinSHMis(x=0.3cos(4t+frac{pi}{4})).Whatisthephaseofthemotionat(t=0)?

Substitute(t=0)intothephase(omegat+varphi).Here,(omega=4rad/s)and(varphi=frac{pi}{4}),sothephaseat(t=0)is(frac{pi}{4})radians.

16.Amassspringsystemhasaspringconstant(k=5N/m)andamass(m=0.5kg).Whatisthefrequencyofoscillation?

First,findtheangularfrequency(omega=sqrt{frac{k}{m}}=sqrt{frac{5}{0.5}}=sqrt{10}approx3.16rad/s).Then,usingtherelation(omega=2pif),wecanfindthefrequency(f=frac{omega}{2pi}=frac{sqrt{10}}{2pi}approx0.5Hz).

17.Inadampedoscillator,thequalityfactor(Q)isdefinedas(Q=frac{omega_0m}),where(omega_0)isthenaturalfrequency,(m)isthemass,and(b)isthedampingconstant.If(omega_0=4rad/s),(m=2kg),and(b=1kg/s),whatisthequalityfactor?

Substitutethegivenvaluesintotheformula:(Q=frac{4times2}{1}=8).

18.ThedisplacementoftwoparticlesinSHMare(x_1=0.2cos(3t))and(x_2=0.2cos(3t+frac{pi}{2})).Aretheyinphaseoroutofphase?

Thephasedifference(Deltavarphi=frac{pi}{2}).Since(Deltavarphineq0)and(Deltavarphineq2npi)((n=0,1,2,cdots)),thetwoparticlesareoutofphase.

19.Asimplependulumhasalength(l=1m).WhatisitsperiodonEarth?

Usingtheformula(T=2pisqrt{frac{l}{g}})with(l=1m)and(gapprox9.8m/s^{2}),wehave(T=2pisqrt{frac{1}{9.8}}approx2s).

20.Thekineticenergy(K)ofaparticleinSHMis(K=frac{1}{2}mv^{2}=frac{1}{2}momega^{2}A^{2}sin^{2}(omegat+varphi)).Atwhatdisplacementisthekineticenergyhalfofthetotalenergy?

Thetotalenergy(E=frac{1}{2}kA^{2}=frac{1}{2}momega^{2}A^{2}).Wewant(K=frac{1}{2}E).So(frac{1}{2}momega^{2}A^{2}sin^{2}(omegat+varphi)=frac{1}{2}timesfrac{1}{2}momega^{2}A^{2}).Then(sin^{2}(omegat+varphi)=frac{1}{2}),(sin(omegat+varphi)=pmfrac{1}{sqrt{2}}).And(x=Acos(omegat+varphi)).Using(sin^{2}alpha+cos^{2}alpha=1),when(sin(omegat+varphi)=pmfrac{1}{sqrt{2}}),(cos(omegat+varphi)=pmfrac{1}{sqrt{2}}),so(x=pmfrac{A}{sqrt{2}}).

21.TheaccelerationofaparticleinSHMis(a=9x).Whatistheangularfrequency?

Comparingwith(a=omega^{2}x),wehave(omega^{2}=9),so(omega=3rad/s).

22.AparticleinSHMhasadisplacement(x=0.5sin(2tfrac{pi}{3})).Whatisthedirectionofmotionat(t=0)?

First,findthevelocity(v=frac{dx}{dt}=0.5times2cos(2tfrac{pi}{3})).At(t=0),(v=0.5times2cos(frac{pi}{3})=0.5times2timesfrac{1}{2}=0.5m/s).Since(v>0),theparticleismovinginthepositive(x)direction.

23.Theperiodofamassspringsystemis(T=1s).Whatisthespringconstantifthemass(m=0.2kg)?

Using(T=2pisqrt{frac{m}{k}}),wecansolvefor(k).Rearranginggives(k=frac{4pi^{2}m}{T^{2}}).Substitute(m=0.2kg)and(T=1s),then(k=4pi^{2}times0.2approx7.9N/m).

24.Inadampedharmonicmotion,iftheamplitudereducestohalfofitsinitialvaluein(t=5s),whatisthedampingconstant(b)intermsofthemass(m)?

Using(A(t)=A_0e^{frac{bt}{2m}}),when(A(t)=frac{A_0}{2})and(t=5s),wehave(frac{A_0}{2}=A_0e^{frac{5b}{2m}}).Then(frac{1}{2}=e^{frac{5b}{2m}}).Takingthenaturallogarithmofbothsides,(ln(frac{1}{2})=frac{5b}{2m}).So(b=frac{2mln2}{5}approx0.28m).

25.Thephasedifferencebetweentwowaves(y_1=Asin(omegat))and(y_2=Acos(omegat))is:

Weknowthat(cos(omegat)=sin(omegat+frac{pi}{2})).Sothephasedifference(Deltavarphi=frac{pi}{2})radians.

26.Asimplependulumistakentoahighaltitudelocationwheretheaccelerationduetogravity(g)isreduced.Whathappenstotheperiodofthependulum?

Using(T=2pisqrt{frac{l}{g}}),if(g)decreases,thevalueof(sqrt{frac{l}{g}})increases.Sotheperiod(T)increases.

27.ThemaximumaccelerationofaparticleinSHMoccursat:

Theacceleration(a=omega^{2}x).Themaximumvalueof(|a|)occurswhen(|x|)ismaximum,i.e.,atthemaximumdisplacement(x=pmA).

28.Theenergyofasimpleharmonicoscillatoris(E=frac{1}{2}kA^{2}).Iftheamplitudeistripled,whathappenstotheenergy?

Lettheoriginalenergybe(E_1=frac{1}{2}kA_1^{2})andthenewenergybe(E_2=frac{1}{2}kA_2^{2}).Given(A_2=3A_1),then(E_2=frac{1}{2}k(3A_1)^{2}=9timesfrac{1}{2}kA_1^{2}=9E_1).Sotheenergyincreasesbyafactorof9.

29.ThedisplacementofaparticleinSHMis(x=Asin(omegat)).Thevelocityismaximumwhen:

Thevelocity(v=omegaAcos(omegat)).Themaximumvalueof(v)occurswhen(cos(omegat)=pm1),whichmeans(omegat=npi)((n=0,1,2,cdots)),andatthesepoints(x=0)(equilibriumposition).

30.Amassspringsystemisoscillatingwithanamplitude(A).Thepotentialenergyisequaltothekineticenergywhen:

Thetotalenergy(E=frac{1}{2}kA^{2}),(U=frac{1}{2}kx^{2}),and(K=frac{1}{2}mv^{2}).When(U=K),then(U=frac{1}{2}E).So(frac{1}{2}kx^{2}=frac{1}{2}timesfrac{1}{2}kA^{2}),whichgives(x=pmfrac{A}{sqrt{2}}).

31.Theangularfrequencyofasimplependulumis(omega=sqrt{frac{g}{l}}).Ifthelength(l)isquadrupled,whathappenstotheangularfrequency?

Lettheoriginalangularfrequencybe(omega_1=sqrt{frac{g}{l_1}})andthenewangularfrequencybe(omega_2=sqrt{frac{g}{l_2}})with(l_2=4l_1).Then(omega_2=sqrt{frac{g}{4l_1}}=frac{1}{2}sqrt{frac{g}{l_1}}=frac{1}{2}omega_1).Sotheangularfrequencyishalved.

32.Inaforcedoscillationsystem,resonanceoccurswhen:

Resonanceoccurswhenthedrivingfrequency(omega_d)isequaltothenaturalfrequency(omega_0)oftheoscillator,i.e.,(omega_d=omega_0).

33.ThedisplacementofaparticleinSHMisgivenby(x=0.1cos(5t+frac{pi}{4})).Thetimetakenfortheparticletomovefrom(x=0)to(x=frac{0.1}{sqrt{2}})is:

When(x=0),(0.1cos(5t+frac{pi}{4})=0),so(5t+frac{pi}{4}=frac{pi}{2}+npi).When(x=frac{0.1}{sqrt{2}}),(0.1cos(5t+frac{pi}{4})=frac{0.1}{sqrt{2}}),so(5t+frac{pi}{4}=frac{pi}{4}+2npi)or(5t+frac{pi}{4}=frac{pi}{4}+2npi).Forthefirstpositivetimeinterval,from(x=0)((5t_1+frac{pi}{4}=frac{pi}{2}))to(x=frac{0.1}{sqrt{2}})((5t_2+frac{pi}{4}=frac{pi}{4})),wesolvefor(t_1=frac{pi}{20})and(t_2=0).Thetimetaken(Deltat=frac{pi}{20}s).

34.Amassspringsystemhasanaturalfrequency(f_0=2Hz).Whenadampingforceisapplied,thenewfrequency(f):

Foralightlydampedoscillator,thefrequency(f)isapproximatelyequaltothenaturalfrequency(f_0).Inthecaseofadampedoscillator,thefrequency(f=sqrt{omega_0^{2}frac{b^{2}}{4m^{2}}}/2pi).Forlightdamping((b)issmall),(fapproxf_0).

35.Therestoringforceinasimplependulumisproportionalto:

Forsmallangles,therestoringforce(F=mgsinthetaapproxmgtheta).Since(theta=frac{x}{l})(where(x)isthearcdisplacement),(Fapproxfrac{mg}{l}x).Sotherestoringforceisproportionaltothedisplacement(x)fromtheequilibriumposition.

36.Thetotalenergyofasimpleharmonicoscillatoris(E).Atadisplacement(x=frac{A}{2})fromtheequilibriumposition,theratioofkineticenergy(K)topotentialenergy(U)is:

Thepotentialenergy(U=frac{1}{2}kx^{2}=frac{1}{2}k(frac{A}{2})^{2}=frac{1}{8}kA^{2}).Thetotalenergy(E=frac{1}{2}kA^{2}),sothekineticenergy(K=EU=frac{1}{2}kA^{2}frac{1}{8}kA^{2}=frac{3}{8}kA^{2}).Theratio(frac{K}{U}=frac{frac{3}{8}kA^{2}}{frac{1}{8}kA^{2}}=3).

37.AparticleinSHMhasadisplacementtimeequation(x=0.2sin(4tfrac{pi}{6})).Theinitialdisplacementis:

Substitute(t=0)intotheequation:(x(0)=0.2sin(frac{pi}{6})=0.1m).

38.ThetimeperiodofasimplependulumonthesurfaceoftheEarthis(T).Ifthependulumistakentoaplanetwheretheaccelerationduetogravityis(4g),thenewtimeperiod(T')is:

Using(T=2pisqrt{frac{l}{g}})and(T'=2pisqrt{frac{l}{4g}}),wehave(T'=frac{1}{2}T).

39.Inadampedharmonicoscillator,theamplitudedecaysexponentially.Thetimeconstant(tau)isdefinedas(tau=frac{2m}).Ifthemass(m)isdoubledandthedampingconstant(b)ishalved,whathappenstothetimeconstant?

Theoriginaltimeconstant(tau_1=frac{2m_1}{b_1}).Thenewtimeconstant(tau_2=frac{2m_2}{b_2}),with(m_2=2m_1)and(b_2=frac{1}{2}b_1).Then(tau_2=frac{2times2m_1}{frac{1}{2}b_1}=4timesfrac{2m_1}{b_1}=4tau_1).Sothetimeconstantisquadrupled.

40.ThephaseofaparticleinSHMis(frac{pi}{3})at(t=0).Iftheangularfrequency(omega=2rad/s),whatisthephaseat(t=2s)?

Thephase(varphi(t)=omegat+varphi_0).Substitute(omega=2rad/s),(t=2s),and(varphi_0=frac{pi}{3}).Then(varphi(2)=2times2+frac{pi}{3}=4+frac{pi}{3})radians.

41.Amassspringsystemoscillateswithanamplitude(A).Theaveragekineticenergyoveronecompletecycleis:

Thetotalenergy(E=frac{1}{2}kA^{2}).Thekineticenergy(K)andpotentialenergy(U)varyoverthecycle,buttheaveragekineticenergyoveronecycleis(frac{1}{2}E=frac{1}{4}kA^{2})becausetheaveragevalueofthekineticenergyandpotentialenergyoveracycleisthesameinasimpleharmonicoscillator.

42.ThevelocityofaparticleinSHMis(v=0.3cos(2t)).Themaximumaccelerationis:

First,findtheacceleration(a=frac{dv}{dt}=0.3times2sin(2t)).Themaximumvalueof(|a|)occurswhen(|sin(2t)|=1).So(a_{max}=0.6m/s^{2}).

43.Asimplependulumhasalength(l).Ifthemassofthependulumbobisdoubled,theperiodofthependulum:

Using(T=2pisqrt{frac{l}{g}}),theperiodofasimplependulumdoesnotdependonthemassofthebob.Sotheperiodremainsthesame.

44.Inaforcedoscillationsystem,theamplitudeofoscillationatresonanceis:

Atresonance((omega_d=omega_0)),theamplitudeofoscillationismaximum.Foralightlydampedforcedoscillator,theamplitude(A=frac{F_0}{bomega_0}),where