A2Physics物理出國(guó)英語(yǔ)WavesonStringsMS

1.Awavetravelsalongastring.Theequationofthewaveisgivenby(y=0.02sin(2pi(5t2x)))where(y)and(x)areinmetersand(t)isinseconds.Determinethedirectionofthewave.

Answer:

Thegeneralformofawaveequationis(y=Asin(omegatpmkx)).Iftheequationisoftheform(y=Asin(omegatkx)),thewavetravelsinthepositive(x)direction.Inthegivenequation(y=0.02sin(2pi(5t2x))=0.02sin(10pit4pix)),sinceitisintheform(omegatkx)(where(omega=10pi)and(k=4pi)),thewavetravelsinthepositive(x)direction.

2.Astringhasalinearmassdensity(mu).Awaveonthisstringhasaspeed(v).Ifthetensioninthestringisincreasedbyafactorof4,whathappenstothespeedofthewave?

Answer:

Thespeedofawaveonastringisgivenbytheformula(v=sqrt{frac{T}{mu}}),where(T)isthetensioninthestringand(mu)isthelinearmassdensity.Lettheinitialtensionbe(T_1)andtheinitialspeedbe(v_1=sqrt{frac{T_1}{mu}}).Ifthetensionisincreasedto(T_2=4T_1),thenewspeed(v_2=sqrt{frac{T_2}{mu}}=sqrt{frac{4T_1}{mu}}=2sqrt{frac{T_1}{mu}}=2v_1).Sothespeedofthewavedoubles.

3.Twowavesonastringaregivenby(y_1=Asin(kxomegat))and(y_2=Asin(kx+omegat)).Whatistheresultantwavewhenthesetwowavessuperpose?

Answer:

Usingthetrigonometricidentity(sina+sinb=2sin(frac{a+b}{2})cos(frac{ab}{2})).Here(a=kxomegat)and(b=kx+omegat).

(y=y_1+y_2=Asin(kxomegat)+Asin(kx+omegat))

(y=2Asin(frac{(kxomegat)+(kx+omegat)}{2})cos(frac{(kxomegat)(kx+omegat)}{2}))

(y=2Asin(kx)cos(omegat)).Thisistheequationofastandingwave.

4.Awaveonastringhasafrequency(f)andawavelength(lambda).Ifthefrequencyisdoubledwhilethetensionandlinearmassdensityofthestringremainthesame,whathappenstothewavelength?

Answer:

Thespeedofawaveonastring(v=sqrt{frac{T}{mu}}),andalso(v=flambda).Since(T)and(mu)areconstant,(v)isconstant.Lettheinitialfrequencybe(f_1)andwavelengthbe(lambda_1),so(v=f_1lambda_1).Ifthenewfrequency(f_2=2f_1),and(v=f_2lambda_2).Since(v)isconstant,(f_1lambda_1=f_2lambda_2).Substituting(f_2=2f_1)gives(f_1lambda_1=2f_1lambda_2).Cancelingout(f_1)(since(f_1neq0)),weget(lambda_2=frac{lambda_1}{2}).Sothewavelengthishalved.

5.Astringoflength(L)isfixedatbothends.Thefundamentalfrequencyofvibrationofthestringis(f_1).Whatisthefrequencyofthesecondharmonic?

Answer:

Forastringfixedatbothends,thefundamentalfrequency(f_1=frac{v}{2L}),where(v)isthespeedofthewaveonthestring.Thefrequencyofthe(n)thharmonicisgivenby(f_n=nf_1).Forthesecondharmonic,(n=2).So(f_2=2f_1).

6.Thespeedofawaveonastringis(v).Thestringhasalength(L)andisfixedatbothends.Whatarethepossiblewavelengthsofstandingwavesonthisstring?

Answer:

Forastringfixedatbothends,theboundaryconditionsrequirethattherearenodesatthetwoends.Thepossiblewavelengths(lambda_n)aregivenby(lambda_n=frac{2L}{n}),where(n=1,2,3,cdots).When(n=1),itisthefundamentalmodewith(lambda_1=2L);when(n=2),(lambda_2=L);when(n=3),(lambda_3=frac{2L}{3})andsoon.

7.Awaveonastringhasanamplitude(A).Iftheamplitudeisdoubled,whathappenstotheenergyperunitlengthofthewave?

Answer:

Theenergyperunitlength(u)ofawaveonastringisproportionaltothesquareoftheamplitude,(uproptoA^{2}).Lettheinitialenergyperunitlengthbe(u_1)withamplitude(A_1),so(u_1=kA_1^{2})(where(k)isaconstant).Iftheamplitudeisdoubled,(A_2=2A_1),then(u_2=kA_2^{2}=k(2A_1)^{2}=4kA_1^{2}=4u_1).Sotheenergyperunitlengthisquadrupled.

8.Astringisvibratinginthethirdharmonic.Howmanynodesandantinodesarethere?

Answer:

Forastringfixedatbothendsvibratinginthe(n)thharmonic,thenumberofnodes(N=n+1)andthenumberofantinodes(A=n).When(n=3),thenumberofnodes(N=3+1=4)andthenumberofantinodes(A=3).

9.Awaveonastringisdescribedby(y(x,t)=0.1sin(3x4t))(inSIunits).Whatisthespeedofthewave?

Answer:

Thegeneralformofawaveequationis(y(x,t)=Asin(kxomegat)),wherethespeedofthewave(v=frac{omega}{k}).Inthegivenequation(y(x,t)=0.1sin(3x4t)),(k=3m^{1})and(omega=4s^{1}).So(v=frac{omega}{k}=frac{4}{3}m/s).

10.Twostringswiththesamelinearmassdensity(mu)arejoinedtogether.String1hasatension(T_1)andstring2hasatension(T_2).Awavetravelsfromstring1tostring2.If(T_1>T_2),whathappenstothewavelengthofthewaveatthejunction?

Answer:

Thespeedofawaveonastring(v=sqrt{frac{T}{mu}}).Letthespeedofthewaveinstring1be(v_1=sqrt{frac{T_1}{mu}})andinstring2be(v_2=sqrt{frac{T_2}{mu}}).Since(T_1>T_2),(v_1>v_2).Thefrequency(f)ofthewaveremainsthesamewhenitcrossesthejunction.Using(v=flambda),wehave(lambda_1=frac{v_1}{f})and(lambda_2=frac{v_2}{f}).Since(v_1>v_2),(lambda_1>lambda_2).Sothewavelengthdecreasesatthejunction.

11.Astringisfixedatoneendandfreeattheotherend.Whatarethepossiblewavelengthsofstandingwavesonthisstring?

Answer:

Forastringfixedatoneendandfreeattheotherend,theboundaryconditionsrequireanodeatthefixedendandanantinodeatthefreeend.Thepossiblewavelengths(lambda_n)aregivenby(lambda_n=frac{4L}{n}),where(n=1,3,5,cdots).When(n=1),(lambda_1=4L);when(n=3),(lambda_3=frac{4L}{3});when(n=5),(lambda_5=frac{4L}{5})andsoon.

12.Awaveonastringhasaphasevelocity(v_p).Thegroupvelocity(v_g)ofthewavepacketformedbysuperposingwavesofslightlydifferentfrequenciesisrelatedtothephasevelocity.Fornondispersivewavesonastring,whatistherelationshipbetween(v_p)and(v_g)?

Answer:

Fornondispersivewaves,thephasevelocity(v_p)isindependentofthefrequency.Thegroupvelocity(v_g)isgivenby(v_g=frac{domega}{dk}),andthephasevelocity(v_p=frac{omega}{k}).Inanondispersivemedium,(omega=vk)(where(v)isaconstant).Then(v_g=frac{d(vk)}{dk}=v)and(v_p=frac{vk}{k}=v).So(v_g=v_p).

13.Astringoflength(L)isvibratinginitsfundamentalmode.Themaximumdisplacementattheantinodeis(A).Whatisthedisplacementatapoint(x=frac{L}{4})fromoneofthefixedends?

Answer:

Theequationofastandingwaveonastringfixedatbothendsvibratinginthefundamentalmodeis(y(x,t)=Asin(frac{pix}{L})cos(omegat)).When(x=frac{L}{4}),(y(frac{L}{4},t)=Asin(frac{pi}{4})cos(omegat)=frac{A}{sqrt{2}}cos(omegat)).Themaximumvalueof(y)at(x=frac{L}{4})is(frac{A}{sqrt{2}}).

14.Awaveonastringisreflectedfromafixedend.Whathappenstothephaseofthewaveuponreflection?

Answer:

Whenawaveonastringisreflectedfromafixedend,thephaseofthewavechangesby(pi)radiansor180°.Thisisbecausethefixedendcannotmove,andthereflectedwavemustcancelouttheincomingwaveatthefixedend,whichrequiresaphasechangeof(pi).

15.Astringisunderatension(T)andhasalinearmassdensity(mu).Awaveissentalongthestring.Ifthetensionisincreasedby21%,whatisthepercentageincreaseinthespeedofthewave?

Answer:

Thespeedofawaveonastring(v=sqrt{frac{T}{mu}}).Lettheinitialtensionbe(T_1)andspeedbe(v_1=sqrt{frac{T_1}{mu}}).Thenewtension(T_2=T_1+0.21T_1=1.21T_1).Thenewspeed(v_2=sqrt{frac{T_2}{mu}}=sqrt{frac{1.21T_1}{mu}}=1.1sqrt{frac{T_1}{mu}}=1.1v_1).Thepercentageincreaseinspeedis(frac{v_2v_1}{v_1}times100%=frac{1.1v_1v_1}{v_1}times100%=10%).

16.Twowavesonastringhavethesameamplitude(A),frequency(f)andwavelength(lambda),butareoutofphaseby(frac{pi}{2})radians.Whatistheamplitudeoftheresultantwavewhentheysuperpose?

Answer:

Letthetwowavesbe(y_1=Asin(kxomegat))and(y_2=Asin(kxomegat+frac{pi}{2})=Acos(kxomegat)).Usingtheformulafortheresultantoftwowaves(y=y_1+y_2),andthemagnitudeoftheresultantamplitude(R)oftwowaveswithamplitudes(A_1=A)and(A_2=A)andphasedifference(phi=frac{pi}{2})isgivenby(R=sqrt{A_1^{2}+A_2^{2}+2A_1A_2cosphi}).Substituting(A_1=A_2=A)and(phi=frac{pi}{2})(so(cosphi=0)),weget(R=sqrt{A^{2}+A^{2}}=sqrt{2}A).

17.Awaveonastringisdescribedby(y=0.05sin(2pi(3t+4x))).Whatisthedirectionofthewave?

Answer:

Thegeneralformofawaveequationis(y=Asin(omegatpmkx)).Iftheequationisoftheform(y=Asin(omegat+kx)),thewavetravelsinthenegative(x)direction.Inthegivenequation(y=0.05sin(2pi(3t+4x))=0.05sin(6pit+8pix)),sinceitisintheform(omegat+kx)(where(omega=6pi)and(k=8pi)),thewavetravelsinthenegative(x)direction.

18.Astringoflength(L)isvibratinginthefourthharmonic.Whatisthedistancebetweentwoconsecutivenodes?

Answer:

Forastringfixedatbothendsvibratinginthe(n)thharmonic,thewavelength(lambda_n=frac{2L}{n}).When(n=4),(lambda_4=frac{2L}{4}=frac{L}{2}).Thedistancebetweentwoconsecutivenodesis(frac{lambda}{2}).Soforthefourthharmonic,thedistancebetweentwoconsecutivenodesis(frac{lambda_4}{2}=frac{L}{4}).

19.Awaveonastringhasanintensity(I).Iftheamplitudeofthewaveistripled,whathappenstotheintensity?

Answer:

Theintensity(I)ofawaveonastringisproportionaltothesquareoftheamplitude,(IproptoA^{2}).Lettheinitialintensitybe(I_1)withamplitude(A_1),so(I_1=kA_1^{2})(where(k)isaconstant).Iftheamplitudeistripled,(A_2=3A_1),then(I_2=kA_2^{2}=k(3A_1)^{2}=9kA_1^{2}=9I_1).Sotheintensityisincreasedbyafactorof9.

20.Astringisvibratinginastandingwavepattern.Thetimeaveragepowertransmittedbyastandingwaveonastringiszero.Why?

Answer:

Inastandingwave,theenergyisconfinedbetweentwoconsecutivenodes.Theenergyoscillatesbackandforthbetweenkineticenergy(associatedwiththemotionofthestringparticles)andpotentialenergy(associatedwiththestretchingofthestring).Thereisnonettransferofenergyinthedirectionofwavepropagation.Thepower(P)istherateofenergytransfer.Sincethereisnonetenergytransferinthedirectionofwavepropagation,thetimeaveragepowertransmittedbyastandingwaveonastringiszero.

21.Awaveonastringhasafrequency(f)andaspeed(v).Ifthefrequencyischangedto(2f)whilethetensioninthestringisadjustedsothatthespeedremainsthesame,whathappenstothewavelength?

Answer:

Weknowthat(v=flambda).Lettheinitialfrequencybe(f_1=f),wavelengthbe(lambda_1),andspeedbe(v_1=v),so(v_1=f_1lambda_1).Thenewfrequency(f_2=2f_1),andthenewspeed(v_2=v_1).Using(v_2=f_2lambda_2),wesubstitute(v_2=v_1)and(f_2=2f_1)into(v_2=f_2lambda_2)and(v_1=f_1lambda_1).Since(v_1=v_2),(f_1lambda_1=f_2lambda_2).Substituting(f_2=2f_1)gives(lambda_2=frac{lambda_1}{2}).Sothewavelengthishalved.

22.Astringoflength(L)isfixedatbothends.Thefirstovertonefrequency(f_2)andthefundamentalfrequency(f_1)arerelatedas(f_2=2f_1).Explainwhy.

Answer:

Thefundamentalfrequencyofastringfixedatbothendsisgivenby(f_1=frac{v}{2L}),where(v)isthespeedofthewaveonthestring.Thefirstovertonecorrespondstothesecondharmonic((n=2)).Thefrequencyofthe(n)thharmonicis(f_n=nf_1).When(n=2),(f_2=2f_1).Physically,inthefundamentalmode,thereisoneantinodeinthemiddleandtwonodesattheends.Inthesecondharmonic,therearetwoantinodesandthreenodes,andthewavelengthishalfofthefundamentalwavelength,whichresultsinthefrequencybeingtwicethefundamentalfrequency.

23.Awaveonastringisincidentonafreeend.Whathappenstothephaseofthewaveuponreflection?

Answer:

Whenawaveonastringisreflectedfromafreeend,thereisnophasechange.Thefreeendcanmovefreely,andthereflectedwavehasthesamephaseastheincidentwave.Thisisbecausetheboundaryconditionatthefreeendallowsthestringtomovewithoutanyrestraintthatwouldcauseaphaseshift.

24.Astringhasalinearmassdensity(mu=0.01kg/m)andisunderatension(T=4N).Whatisthespeedofawaveonthisstring?

Answer:

Thespeedofawaveonastringisgivenby(v=sqrt{frac{T}{mu}}).Substituting(T=4N)and(mu=0.01kg/m)intotheformula,weget(v=sqrt{frac{4}{0.01}}=sqrt{400}=20m/s).

25.Astandingwaveonastringisgivenby(y(x,t)=0.2sin(3x)cos(4t))(inSIunits).Whatarethewavelength(lambda)andtheangularfrequency(omega)?

Answer:

Thegeneralformofastandingwaveis(y(x,t)=Asin(kx)cos(omegat)).Comparingwith(y(x,t)=0.2sin(3x)cos(4t)),wehave(k=3m^{1})and(omega=4s^{1}).Therelationshipbetween(k)and(lambda)is(k=frac{2pi}{lambda}),so(lambda=frac{2pi}{k}=frac{2pi}{3}m).Theangularfrequency(omega=4s^{1}).

26.Awaveonastringhasaphaseconstant(phi).Whatdoesthephaseconstantrepresent?

Answer:

Thephaseconstant(phi)inthewaveequation(y(x,t)=Asin(kxomegat+phi))representstheinitialphaseofthewaveat(x=0)and(t=0).Itdeterminesthestartingpointofthewave'soscillation.Forexample,if(phi=0),thewavehasadisplacementof(y(0,0)=0).If(phi=frac{pi}{2}),then(y(0,0)=A),indicatingthatthewavestartsatitsmaximumdisplacementat(x=0)and(t=0).

27.Astringisvibratinginastandingwavepattern.Theenergyofthestringisstoredintwoforms.Whatarethey?

Answer:

Theenergyofavibratingstringinastandingwavepatternisstoredaskineticenergyandpotentialenergy.Thekineticenergyisassociatedwiththemotionofthestringparticles.Whenthestringparticlesaremovingattheirmaximumspeed(attheequilibriumpositionoftheoscillation),thekineticenergyismaximum.Thepotentialenergyisassociatedwiththestretchingofthestring.Whenthestringisatitsmaximumdisplacementfromtheequilibriumposition,thepotentialenergyismaximum.

28.Awaveonastringistravelingwithaspeed(v).Ifthetensioninthestringisquadrupledandthelinearmassdensityishalved,whathappenstothespeedofthewave?

Answer:

Thespeedofawaveonastringis(v=sqrt{frac{T}{mu}}).Lettheinitialtensionbe(T_1)andlinearmassdensitybe(mu_1),so(v_1=sqrt{frac{T_1}{mu_1}}).Thenewtension(T_2=4T_1)andthenewlinearmassdensity(mu_2=frac{mu_1}{2}).Thenewspeed(v_2=sqrt{frac{T_2}{mu_2}}=sqrt{frac{4T_1}{frac{mu_1}{2}}}=sqrt{8frac{T_1}{mu_1}}=2sqrt{2}v_1).Sothespeedofthewaveincreasesbyafactorof(2sqrt{2}).

29.Astringoflength(L)isfixedatbothends.Thefrequencyofthefifthharmonic(f_5)ismeasured.Ifthelengthofthestringisthenhalvedwhilekeepingthetensionandlinearmassdensitythesame,whatisthenewfrequencyofthefifthharmonic?

Answer:

Thefrequencyofthe(n)thharmonicofastringfixedatbothendsis(f_n=frac{nv}{2L}),where(v=sqrt{frac{T}{mu}}).Initially,(f_{5i}=frac{5v}{2L}).Whenthelengthishalved((L_f=frac{L}{2})),thenewfrequency(f_{5f}=frac{5v}{2L_f}).Substituting(L_f=frac{L}{2})gives(f_{5f}=frac{5v}{2timesfrac{L}{2}}=frac{5v}{L}=2timesfrac{5v}{2L}=2f_{5i}).Sothenewfrequencyofthefifthharmonicistwicetheoriginalfrequency.

30.Awaveonastringhasanamplitude(A)andafrequency(f).Ifthefrequencyisincreasedbyafactorof3whilekeepingtheamplitudethesame,whathappenstothemaximumspeedofapointonthestring?

Answer:

Thedisplacementofapointonthestringisgivenby(y=Asin(kxomegat)).Thevelocityofapointonthestringis(v_y=frac{partialy}{partialt}=Aomegacos(kxomegat)).Themaximumspeed(v_{max}=Aomega),and(omega=2pif).Initially,(v_{max1}=Aomega_1=2piAf_1).Ifthefrequencyisincreasedto(f_2=3f_1),then(omega_2=2pif_2=3times2pif_1=3omega_1).Thenewmaximumspeed(v_{max2}=Aomega_2=3Aomega_1=3v_{max1}).Sothemaximumspeedofapointonthestringistripled.

31.Astringisvibratinginastandingwavepattern.Thenodesarepointsofzerodisplacement.Whatisthephysicalsignificanceofthenodes?

Answer:

Nodesinastandingwaveonastringarepointswherethetwointerferingwaves(theincidentandthereflectedwaves)alwayscanceleachotherout.Physically,thesepointsdonotmove.Theyarefixedinpositionandrepresenttheboundariesbetweenregionsofthestringthatoscillateinoppositephases.Thepresenceofnodesisaconsequenceoftheboundaryconditions(e.g.,astringfixedatbothends)andtheinterferenceofthewavestravelinginoppositedirections.

32.Awaveonastringhasawavenumber(k).Whatistherelationshipbetweenthewavenumber(k)andthewavelength(lambda)?

Answer:

Thewavenumber(k)isdefinedas(k=frac{2pi}{lambda}).Thewavenumberrepresentsthenumberofwavelengthsper(2pi)radiansofthewave'sspatialvariation.Itisameasureofhowrapidlythephaseofthewavechangeswithposition.

33.Astringisundertension(T)andhasalinearmassdensity(mu).Awaveissentalongthestring.Ifthelinearmassdensityisincreasedbyafactorof4whilekeepingthetensionthesame,whathappenstothespeedofthewave?

Answer:

Thespeedofawaveonastringis(v=sqrt{frac{T}{mu}}).Lettheinitiallinearmassdensitybe(mu_1)andspeedbe(v_1=sqrt{frac{T}{mu_1}}).Ifthenewlinearmassdensity(mu_2=4mu_1),thenewspeed(v_2=sqrt{frac{T}{mu_2}}=sqrt{frac{T}{4mu_1}}=frac{1}{2}sqrt{frac{T}{mu_1}}=frac{v_1}{2}).Sothespeedofthewaveishalved.

34.Astandingwaveonastringisformedbythesuperpositionoftwowaves.Whataretheconditionsfortheformationofastandingwave?

Answer:

Thefollowingconditionsarerequiredfortheformationofastandingwaveonastring:

1.Twowavesmusthavethesameamplitude,frequency,andwavelength.

2.Thetwowavesmusttravelinoppositedirections.Thisusuallyoccurswhenawaveisreflectedbackonitself,suchasawaveonastringreflectingfromafixedorfreeend.

3.Thewavesmustinterferewitheachother.Whentheseconditionsaremet,theresultantwavepatternappearstobestationary,withnodesandantinodesatfixedpositions.

35.Awaveonastringhasafrequency(f)andawavelength(lambda).Ifthewavetravelsfromaregionoflowdensitystringtoaregionofhighdensitystring(whilethetensionremainsthesame),whathappenstothefrequencyandthewavelength?

Answer:

Thefrequencyofawaveremainsthesamewhenittravelsfromonemediumtoanother.Thisisbecausethefrequencyisdeterminedbythesourceofthewave.Thespeedofawaveonastring(v=sqrt{frac{T}{mu}}).Whenmovingfromalowdensity((mu_1))toahighdensity((mu_2>mu_1))stringwiththesametension(T),thespeed(v_2<v_1).Since(v=flambda)and(f)isconstant,if(v)decreases,thewavelength(lambda)alsodecreases.Sothefrequencyremainsthesameandthewavelengthdecreases.

36.Astringoflength(L)isvibratinginthethirdovertone.Whatisthenumberofantinodes?

Answer:

Thefirstovertonecorrespondstothesecondharmonic((n=2)),thesecondovertonecorrespondstothethirdharmonic((n=3)),andthethirdovertonecorrespondstothefourthharmonic((n=4)).Forastringfixedatbothends,thenumberofantinodesisequaltotheharmonicnumber(n).Sowhen(n=4),thenumberofantinodesis4.

37.Awaveonastringhasanangularfrequency(omega).Whatistherelationshipbetweentheangularfrequency(omega)andtheperiod(T)?

Answer:

Theangularfrequency(omega)andtheperiod(T)arerelatedbytheformula(omega=frac{2pi}{T}).Theperiod(T)isthetimetakenforonecompleteoscillationofthewave,andtheangularfrequency(omega)istherateofchangeofthephaseofthewavewithrespecttotime.

38.Astringisvibratinginastandingwavepattern.Theantinodesarepointsofmaximumdisplacement.Whatisthephysicalsignificanceoftheantinodes?

Answer:

Antinodesinastandingwaveonastringarepointswherethetwointerferingwaves(theincidentandthereflectedwaves)constructivelyinterfere.Atthesepoints,theamplitudeoftheresultantwaveismaximum.Theyrepresenttheregionsofmaximumenergytransferwithinthestandingwavepattern.Theantinodesoscillatewiththemaximumamplitudebetweenthepositiveandnegativemaximumdisplacements.

39.Awaveonastringisdescribedby(y(x,t)=Asin(kxomegat)).Whatisthevelocityofapointonthestringatposition(x)andtime(t)?

Answer:

Wefindthevelocityofapointonthestringbytakingthepartialderivativeofthedisplacementfunction(y(x,t))withrespecttotime.Given(y(x,t)=Asin(kxomegat)),then(v_y=frac{partialy}{partialt}=Aomegacos(kxomegat)).Themaximumvelocityofthepointoccurswhen(cos(kxomegat)=pm1),and(v_{max}=Aomega).

40.Astringisfixedatbothendsandisvibratinginthefundamentalmode.Thetensioninthestringis(T)andthelinearmassdensityis(mu).Whatisthefrequencyofvibration?

Answer:

Thespeedofawaveonastringis(v=sqrt{frac{T}{mu}}).Forastringfixedatbothendsvibratinginthefundamentalmode,thewavelength(lambda_1=2L),where(L)isthelengthofthestring.Usingtherelationship(v=flambda),andsubstituting(lambda=lambda_1=2L)and(v=sqrt{frac{T}{mu}}),weget(f_1=frac{v}{2L}=frac{1}{2L}sqrt{frac{T}{mu}}).

41.Awaveonastringhasagroupvelocity(v_g)andaphasevelocity(v_p).Inadispersivemedium,howdo(v_g)and(v_p)generallycompare?

Answer:

Inadispersivemedium,thephasevelocity(v_p=frac{omega}{k})andthegroupvelocity(v_g=frac{domega}{dk}).Inanormaldispersivemedium,thegroupvelocity(v_g)islessthanthephasevelocity(v_p)((v_g<v_p)).Inananomalousdispersivemedium,thegroupvelocitycanbegreaterthanthephasevelocity((v_g>v_p)).

42.Astringisvibratinginastandingwavepattern.Theenergyofthestringatanodeisalwayszero.Why?

Answer:

Atanodeinastandingwaveonastring,thedisplacementofthestringparticlesisalwayszero.Thekineticenergyofaparticleonthestringis(K=frac{1}{2}mv_y^{2}),andsincethevelocity(v_y=0)atthenodes(becausetheparticlesdonotmove),thekineticenergyiszero.Thepotentialenergyofthestringisrelatedtothestretchingofthestring.Atthenodes,thereisnostretchingofthestring(asthedisplacementiszero),sothepotentialenergyisalsozero.Therefore,thetotalenergyofthestringatanodeisalwayszero.

43.Awaveonastringisincidentonaboundarywherethestringchangesitslinearmassdensity.Partofthewaveisreflectedandpartistransmitted.Whatdeterminestheamountofreflectionandtransmission?

Answer:

Theamountofreflectionandtransmissionofawaveataboundarywherethelinearmassdensityofthestringchangesisdeterminedbytheimpedancemismatch.Theimpedance(Z)ofastringisgivenby(Z=sqrt{Tmu}),where(T)isthetensionand(mu)isthelinearmassdensity.Thereflectioncoefficient(R)andtransmissioncoefficient(T)arefunctionsoftheratiooftheimpedancesofthetwomedia.Iftheimpedancesofthetwopartsofthestringareverydifferent,therewillbealargeamountofreflectionandasmallamountoftransmission.Iftheimpedancesaresimilar,therewillbelessreflectionandmoretransmission.

44.Astringoflength(L)isvibratinginthefifthharmonic.Whatisthenumberofnodes?

Answer:

Forastringfixedatbothendsvibratinginthe(n)thharmonic,thenumberofnodes(N=n+1).When(n=5),thenumberofnodes(N=5+1=6).

45.Awaveonastringhasanamplitude(A)andawavenumber(k).Whatisthemaximumslopeofthestring?

Answer:

Thedisplacementofthestringisgivenby(y(x,t)=Asin(kxomegat)).Theslopeofthestringis(frac{partialy}{partialx}=Akcos(kxomegat)).Themaximumvalueoftheslopeoccurswhen(cos(kxomegat)=pm1).Sothemaximumslopeofthestringis(Ak).

46.Astringisvibratinginastandingwavepattern.Theenergyofthestringisconserved.Explainhow.

Answer:

Theenergyofavibratingstringinastandingwavepatternisconservedbecausethetotalenergyofthestringisthesumofitskineticandpotentialenergies.Asthestringvibrates,thekineticenergyismaximumwhenthestringparticlespassthroughtheirequilibriump