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  • ANSWERS TO TEST YOURSELF QUESTIONS 4 1physics for the iB Diploma © camBriDge University press 2015

    answers to test yourself questionstopic 44.1 oscilliations

    1 a Anoscillationisanymotioninwhichthedisplacementofaparticlefromafixedpointkeepschangingdirectionandthereisaperiodicityinthemotioni.e.themotionrepeatsinsomeway.

    b Insimpleharmonicmotion,thedisplacementfromanequilibriumpositionandtheaccelerationareproportionalandoppositeeachother.

    2 Itisanoscillationsincewemaydefinethedisplacementoftheparticlefromthemiddlepointandinthatcasethedisplacementchangesdirectionandthemotionrepeats.Themotionisnotsimpleharmonichoweversincethereisnoaccelerationthatisproportional(andopposite)tothedisplacement.

    3 Itisanoscillationsincethemotionrepeats.Themotionisnotsimpleharmonichoweversincetheaccelerationisconstantandisnotproportional(andopposite)tothedisplacement.

    4 a Theaccelerationisoppositetothedisplacementsoeverytimetheparticleisdisplacedthereisaforcetowardstheequilibriumposition.

    b Theaccelerationisnotproportionaltothedisplacement;ifitwerethegraphwouldbeastraightlinethroughtheorigin.

    5 a i Itwasnotintendedtoaskaboutthemass–apologies! ii Theperiodis8.0s;theparticleisatoneextremepositionatt=0andagainatt=4.0s.Thisishalfaperiod.

    b EP / J

    t /s

    0.0

    0.5

    1.0

    1.5

    2.0

    6543210 7

    4.2 travelling waves

    6 Thedelaytimebetweenyouseeingthepersonnexttoyoustandupandyoustandingupandthenumberdensityofthepeoplei.e.howmanypeopleperunitmeter.Forafixeddelaytime,thecloserthepeoplearethefasterthewave.

    7 Thereisadisturbancethattravelsthroughthelieofdominoesjustasadisturbancetravelsthroughamediumwhenawaveispresent.Youcanincreasethespeedbyplacingthemclosertogether.Anexperimenttoinvestigatethismightbetoplaceanumberofdominoesonalineoffixedlengthsuchthatthedominoesareafixeddistancedapart.Wemustgivethesameinitialpushtothefirstdomino(forexampleusingapendulumthatisreleasedfromafixedheightandstrikesthedominoatthesameplace.Wethenmeasuretimeformwhenthefirstdominoishituntilthelastoneishit.Dividingthefixeddistancebythetimetakengivesthespeedofthepulse.Wecanthenrepeatwithadifferentdominoseparationandseehowthespeeddependsontheseparationd.

  • 2 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 4

    8 a Wavelength–thelengthofafullwave;thedistancebetweentwoconsecutivecrestsortroughs b Period–thetimeneededtoproduceonefulloscillationorwave c Amplitude–thelargestvalueofthedisplacementfromequilibriumofanoscillation d Crest–apointonawaveofmaximumdisplacement e Trough–apointonawaveofminimumdisplacement

    Distance /m

    Dis

    plac

    emen

    t /cm 2

    4

    –4

    –2

    00.5 1.0 1.5 2.0

    λ amplitude A

    Time / ms

    Dis

    plac

    emen

    t /cm 2

    4

    –4

    –2

    0

    period T

    642 108

    9 a Inwavemotiondisplacementreferstothedifferenceinthevalueofaquantitysuchasposition,pressure,densityetcwhenthewaveispresentandwhenthewaveisabsent.

    b Inatransversewavethedisplacementisatrightanglestothedirectionofenergytransfer,inalongitudinalitisparalleltotheenergytransferdirection.

    c Thefallingstoneimpartskineticenergytothewateratthepointofimpactandsothatwatermoves.Itwillcontinuemoving(creatingmanyripples)untiltheenergyisdissipated.

    d Wemustrecallthattheintensityofawaveisproportionaltothesquareoftheamplitude.Theamplitudewilldecreasefortworeasons:first,someenergyisboundtobedissipatedasthewavemovesawayandsotheamplitudehastodecrease.Second,evenintheabsenceofanyenergylosses,theamplitudewillstilldecreasebecausethewavefrontsgetbiggerastheymoveawayfromthepointofimpactoftheripple.Theenergycarriedbythewaveisnowdistributedonalongerwavefrontandsotheenergyperunitwavefrontlengthdecreases.Theamplitudemustthendecreaseaswell.

    10 a Fromlefttoright:down,down,up. b Fromlefttoright:up,up,down.

    11

    12 a λ = = =vf

    330

    2561 29. m.

    b λ = =×

    = × −v

    f

    330

    25 101 32 10

    32. m.

  • ANSWERS TO TEST YOURSELF QUESTIONS 4 3physics for the iB Diploma © camBriDge University press 2015

    13 a Awaveinwhichthedisplacementisparalleltothedirectionofenergytransferredbythewave. b i

    20 4 6 8 x /cm

    ii Atx=4.0cm

    c i 92 3 5 710 4 86 x/cm

    ii Thecompressionisnowatx=5.0cm.

    14 a fv

    = = =λ

    340

    0 40850

    .Hz

    b i Acompressionoccursatx=0.30m.Moleculesjusttotheleftofthispointhavepositivedisplacementandsomovetotheright.Moleculesjusttotherightmovetotheleftcreatingthecompressionatx=0.30m.

    ii Bysimilarreasoningx=0.10misapointwhereararefactionoccurs.

    4.3 Wave characteristics

    15 Addingthepulsespointbypointgivesthefollowingdiagram.

    16

    17 Addingthepulsespointbypointgivesthefollowingdiagram.

    1 unit

    2 units

    2 cm

    t = 0.5 s t = 1.0 s t = 1.5 s

    1 cm

    1 cm1 cm

    1 unit

    1 unit

    18 Weaddthepulsespointbypoint.Forexampleatx=0bothwaveshavezerodisplacementandsowegetzerodisplacementforthesum.Atx=10cm,thebluepulsehasy=0.50cmandtheredpulsehasy=0.75cm.Thesumis1.25cm.Atx=20cm,thebluepulsehasy=0andtheredpulsehasy=1.0cm.Thesumis1.0cm.Atx=30cm,thebluepulsehasy=–0.50cmandtheredpulsehasy=0.70cm.Thesumis0.20cmandsoon.

    19 a Awavefrontisasurfaceonwhichallpointshavethesamephase.

    rayx

    y

    z

    wavefronts

    λ λ

  • 4 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 4

    b Arayisthedirectionnormaltowavefrontsthatcorrespondstothedirectionofenergytransfer.

    a b

    source of disturbance

    point source

    20 a Polarisedlightislightinwhichtheelectricfieldoscillatesonthesameplane. b Lightcanbepolarisedbypassagethroughapolariserandbyreflectionoffanon-metallicsurface.

    21 Inapolarisedwavethedisplacementmustbeonthesameplane.Inalongitudinalwavethedisplacementisalongthedirectionofenergytransferandsobelongstoaninfinityofplanesatthesametime.Henceitcannotbepolarised.

    22 a Thelightisnotpolarised.Inthecaseofunpolarisedlightincidentonananalyser,theintensityofthetransmittedlightwouldbehalftheincidentintensityandsoconstantasrequiredinthequestion.

    b Sincethereisanorientation(callitX)oftheanalyserthatmakesthetransmittedintensityzero,itfollowsthattheincidentlightwaspolarisedinadirectionatrightanglestothedirectionX.

    c Sincetheintensityneverbecomeszerothelightwasnotpolarised.Sincetheintensityvarieshowever,itfollowsthattheincidentlighthasunequalcomponentsinvariousdirectionssoitispartiallypolarised.

    23 a ThisrelatesthetransmittedintensityItotheincidentintensityI0whenpolarisedlightisincidentandthentransmittedthroughananalyser.TherelationisI=I0cos2θwhereθistheanglebetweenthetransmissionaxisandthedirectionoftheincidentelectricfield.

    bI

    I 0

    2 2 25 0 82= = ° =cos cos .θ

    24 a Thelighttransmittedthroughthefirstpolariserwillbepolarisedinagivendirection.Thesecondpolariser’saxisisatrightanglestothisdirectionsotheelectricfieldhaszerocomponentalongtheaxisofthesecondpolariser.Hencenolightgetstransmitted.

    b Lightwillbetransmittedsincenowtherewillbeacomponentoftheelectricfieldalongthesecondpolariser’saxis.

    c Thesituationisnowidenticaltoaandsonolightgoesthrough.

    4.4 Wave behaviour

    25 a From1 00 38 1 583 2. sin . sin× ° = × θ wefindsin. sin

    .sin . .θ θ2 2

    11 00 38

    1 5830 3889 22 9=

    × °⇒ = = °− .

    b nc

    cc

    c

    ngg= ⇒ = =

    ×= × −

    3 0 10

    1 5831 9 10

    88 1.

    .. m s

    c Thefrequencyinwateristhesameasthatinairandsoλλ

    ga

    n= =

    ×= ×

    −−6 8 10

    1 5834 3 10

    77.

    .. m.

    26 a ts

    c= =

    ×= × −

    3 0

    3 0 101 0 10

    88.

    .. s

    b Inthistime,1 0 10 6 0 10 6 0 108 14 6. . .× × × = ×− fullwaveshavebeenemitted.(Or,thewavelengthis

    λ = ××

    = × −3 0 10

    6 0 105 0 10

    8

    147.

    .. mandinalengthof3.0mwecanfit

    3 0

    5 0 106 0 10

    76.

    ..

    ×= ×− fullwaves.)

  • ANSWERS TO TEST YOURSELF QUESTIONS 4 5physics for the iB Diploma © camBriDge University press 2015

    27 Firstwefindtheangleofrefraction(angleθinthediagram).

    4.0 cm

    x

    θφ

    d

    1.00×sin40°=1.450×sinθ,henceθ=26.3°.Thismeansthatx =°

    =4 0

    26 34 46

    .

    cos .. cm .

    Nowϕ = ° − ° = °40 2 26 3 13 7. . andsod = × ° =4 46 13 7 1 06. sin . . cm .

    28 Letθbetheangleofincidencefromair.Theangleofrefractionwillbelargerthanθandsoasθincreasestheangleofrefractionwillbecome90°andsowillnotenterwater.Thishappenswhensin sin

    sin .θ θ

    340

    90

    1500

    340

    150013 11=

    °⇒ = = °− .

    29 Thediagrammustbesimilartotheonebelow.

    wavelength λ

    a b

    30 Thereisnoappreciablediffractionhere;thewavecontinuesstraightthroughtheopening.

    31 Thereispoorreceptionbecauseofdestructiveinterferencebetweenthewavesreachingtheantennadirectlyandthosereflectingoffthemountain.Thepathdifferenceisdoublethedistancebetweenthehouseandthemountain.Thewavereflectingoffthemountainwillsufferaphasedifferenceofπandsotheconditionfordestructiveinterferenceis2d n= λ.Thesmallestd(otherthanzero)correspondston = 1andsod = 800 m.

    4.5 standing waves

    32 Astandingwaveisaspecialwaveformedwhentwoidenticaltravelingwavesmovinginoppositedirectionsmeetandthensuperpose.Thiswave,unlikeatravelingwave,hasnodesi.e.pointswherethedisplacementisalwayszero.Theantinodes,pointswherethedisplacementisthelargestdonotappeartobemoving.Astandingwavediffersfromatravelingwaveinthatitdoesnottransferenergyandthattheamplitudeisvariable.Inastandingwavepointsinbetweenconsecutivenodeshavethesamephasewhereasinatravellingwavethephasechangesfromzeroto2πafteradistanceofonewavelength.

    33 Astandingwaveisformedwhentwoidenticaltravelingwavesmovinginoppositedirectionsmeetandthensuperpose.

    34 a Anodeisapointinthemediumwherethedisplacementisalwayszero. b Anantinodeisapointinthemediumwherethedisplacement,atsomeinstant,willassumeitsmaximumvalue. c Speedreferstothespeedofthetravellingwaveswhosesuperpositiongivesthestandingwave.

  • 6 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 4

    35 a Wemustdisturbthestringwithafrequencythatisequaltothefrequencyofthesecondharmonic. b

    36 Thewavelengthofthewavewillremainthesame(andequaltotwicethelengthofthestring).Sincethespeedincreasesby 2 thefrequencymustdothesameandsois354Hz.

    37 Thefirstharmonichaswavelength2L(Listhelengthofthestring)andthesecondawavelengthL.Theratioofthefrequenciesisthen2sincethespeedisthesame.

    38 a Thewavelengthofthefundamentalis2L=1.00m.Thefrequencyisthen fv

    L= =2

    225 Hz

    b Thesoundproducedbythevibrationsofthestringwillhavethesamefrequencyi.e.225Hzandsothe

    wavelengthofsoundwillbeλ = = =cf

    340

    2251 51. m.

    39

    40 Thewavelengthofsoundisλ = = =cf

    340

    3061 11. m.Standingwaveshavewavelengthgivenbyλ = 4L

    nwith

    n=1,3,5,….Therefore4

    1 111 11

    4

    L

    nL

    n= ⇒ =

    ×.

    .m .Thisgives0.28mand1.4mforn=3andn=5.

    41 a Thewavelengthisgivenbyλ = =4 0 800L

    n n

    .andalsobyλ = =

    c

    f

    c

    427.Hence

    c

    nc

    n n427

    0 800 427 0 800 342 1= ⇒ =×

    = −. .

    m s .Theanswermakesphysicalsenseonlyifn=1(thefirstharmonic

    isestablished)inwhichcase c = −342 1m s .

    b Thenextharmonicwillhavewavelength4

    0 8000 800

    40 200

    ′= ⇒ ′ = =

    L

    nL

    nn.

    .. .Withn=3weget

    ′ =L 0 600. m.

    42 a Thewavelengthsintheopentubearegivenby λ = 2Ln

    .Thefrequenciesoftwoconsecutive

    harmonicsarethen fc cn

    L= =

    λ 2,300

    2=cn

    Land360

    1

    2=

    +c nL

    ( ).Thismeansthat

    360

    300

    12

    2

    11 2 1 1 2 0 2=

    +

    ⇒+

    = ⇒ + = ⇒ =

    c nLcnL

    n

    nn n n

    ( )

    . . . 11 5⇒ =n ;wehavethefifthandsixthharmonics.

    b Weget300340 5

    22 833 2 8=

    ××

    ⇒ = ≈L

    L . . m.

    43 Thetwoharmonicshavethesamefrequencyandhencethesamewavelength.Thewavelengthofthefirst

    harmonicintheopen-openpipeisλ = 2LX.Thewavelengthofthefirstharmonicintheclosed-openpipeis

    λ = 4LY.Hence2 4 2L LL

    LX YX

    Y

    = ⇒ = .

  • ANSWERS TO TEST YOURSELF QUESTIONS 4 7physics for the iB Diploma © camBriDge University press 2015

    44 Withonesteppersecondyoushakethecupwithafrequencyofabout1 Hz.Inthefirstharmonicmodethewavelengthwouldbeabouttwicethediameterofthecupi.e.16 cm(wehaveantinodesateachend).Thisgivesaspeedofv = × = −1 16 16 1cm s .

    45 a Astandingwaveismadeupoftwotravelingwaves.Thespeedofenergytransferofthetravelingwavesistakentobethespeedofthestandingwave.

    b Fromy t= 5 0 45. cos( )π wededucethatthefrequencyofoscillationofpointPandhencealsoofthewaveis45

    222 5

    ππ

    = . Hz.Thewavelengthisthenλ = = =v

    f

    180

    22 58 0

    .. m.Sincethediagramshowsasecondharmonic

    thisisalsothelengthofthestring. c Thephasedifferenceisπandsoy t t= + = −5 0 45 5 0 45. cos( ) . cos( )π π π .

    46 a Thehitcreatesalongitudinalwavethattravelsdownthelengthoftherodandreflectsoftheend.Thereflectedwavespushesthehammerback.

    b vs

    t= =

    ×= ×−

    −2 4

    0 18 101 3 103

    4 1.

    .. m s

    c Weassumefree-freeendpointsandsothewavelengthisgivenby2.4m.Thefrequencyisthen

    fc

    = =×

    1 3 10

    2 45 6

    4.

    .. kHz.