answers to test yourself questions

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 = = =v

f

330

2561 29. m.

b λ = =×

= × −v

f

330

25 101 32 10

32. m.


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


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.)


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.Theangleofrefractionwillbelargerthanθandsoasθ

increasestheangleofrefractionwillbecome90°andsowillnotenterwater.Thishappenswhen

sin sinsin .

θ θ340

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.


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 = = =c

f

340

2251 51. m.

39

40 Thewavelengthofsoundisλ = = =c

f

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 =2L

n.Thefrequenciesoftwoconsecutive

harmonicsarethen fc cn

L= =

λ 2,300

2=cn

Land360

1

2=

+c n

L

( ).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

= ⇒ = .


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 wededucethatthefrequencyofoscillationofpointPandhencealsoofthewaveis

45

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

= =×

=λ

1 3 10

2 45 6

4.

.. kHz.

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