answers to test yourself questions
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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 = = =v
f
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.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.
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 = = =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
= ⇒ = .
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 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.