new physics at work workbook answer ch 8.docx
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new physics at work workbook answer ch 8.docx
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newphysicsatworkworkbookanswerch8
8Sound
Practice8.1(p.152)
1C
2D
v=
=0.21ms1
Byv=fl,
frequencyofthewave=
=
=2.0Hz
3D
Amplitudeofthewave
=displacementofbeadc
=3cm
Beadseandmaretwosuccessivecompressionsinthewave.
Separationbetweenthemisequaltothewavelengthlofthewave.
l=8´5=40cm
4A
Periodofbeadc=4´0.1=0.4s
Frequency=
=
2.5Hz
5(a)
(b)
6(a)Longitudinalwave
(b)
(c)(i)Wavelength=3.4m
(ii)Frequency=
=100Hz
(iii)Speed=f=100´3.4=340ms1
Practice8.2(p.161)
1C
2C
Toproducethesecondecho,soundtravelsfromtheboytocliffA,thentocliffBandfinallytotheboyagain.Thatis,thesoundhastravelledadistanceof2AB.
Bys=vt,
2AB=340´1.68
AB=286m
3Byv=fl,
l=
=
=1.30m
Therefore,thedistancebetweentwosuccessivecompressionsis1.30m.
4Lengthofthelake=vt=333´
=250m
5(a)Byv=fl,
l=
=
=0.34m
Thetwoloudspeakersare1mapart.
=2.94
Therefore,theyare2.94wavelengthsapart.
(b)Thepositionsofconstructiveinterference(loudsounds)intheinterferencepatternbecomemorewidelyspaced.
6(a)Byv=fl,
f=
=
=256Hz
Speedofsoundinthemedium
=fl=256´5.47=1400ms–1
(b)Refraction
7(a)Byv=fl,
l=
=
=0.68m
=2.21
Thewidthofthedoorwayis2.21timesthewavelengths.
(b)Sincethewidthofthedoorwayiscomparabletothewavelengthofthesoundwaves,soundwavesdiffractaroundthedoorwaywhentheypassthroughit.
8(a)
(b)
9Thenodallineswouldmovefurtherapart.
Practice8.3(p.171)
1D
2B
3(a)Byv=fl,
wavelength=
=
=0.833m
(b)Yes.Thefrequencyofsoundwavedoesnotchangewhiletravellinginironandair,soitisstill6000Hz,withintheaudiblefrequencyrange.
4s=vt=340´2.5=850m
Thefireworkis850mfromtheaudiencewhenitexplodes.
5
–
=1.4
s=511m
Theimpactoccurredat511maway.
6(a)Byv=fl,
wavelength=
=
=0.01m
(b)Asmallwavelengthresultsinlessdiffractionwhentheultrasonicwavesencounteranyobstacles.Therefore,ultrasoundcanbeusedtodetectthespecifictargetsmoreaccurately.
(c)Timeelapsedbetweenthetwopulses
=20´103´5=0.1s
Distanceoftheshoaloffishbelowtheship=1500´
=75m
7Distancebetweenthebatandthebird
=340´
=34m
8Timeperiodbetweenwingbeats
=
s
s=vt=340´
=0.567m
Thesoundwavestravel0.567mbetweenwingbeats.
9Speedofsound=
=
=333ms–1
Practice8.4(p.181)
1B
2D
3D
4(a)
(b)
5(a)Period=
=
=3.82´10–3s
(b)Byv=fl,
wavelength=
=
=1.30m
6(a)Noisesat130dBproduceearpain.
(b)Students’conversation,loudspeakers
(Orotherreasonableanswers)
(c)Studentskeepquiet,lowerthevolumeofloudspeakers
(Orotherreasonableanswers)
7(a)
(b)Sincelowfrequencynoisehaslongerwavelength,itbendsmorewhenpassingthebarrier.Therefore,flatsintheshadowofthebarrierarestillaffectedbythelowfrequencynoise.
Revisionexercise8
Multiple-choice(p.184)
1D
2C
3D
4B
5C
Pathdifference
=S2P–S1P=4.2–3.4=0.8m
Constructiveinterferencetakesplaceatapositionwherethepathdifferenceisequaltowholenumberofwavelengths.Therefore,itisimpossiblethatthewavelengthofthesoundequals0.6m.
6B
7B
8A
9C
10D
11D
12(HKCEE2005PaperIIQ13)
13(HKCEE2006PaperIIQ17)
14(HKCEE2006PaperIIQ18)
15(HKCEE2006PaperIIQ34)
16(HKCEE2007PaperIIQ36)
17(HKCEE2007PaperIIQ37)
18(HKCEE2007PaperIIQ39)
Conventional(p.186)
1(a)Assumethetimeoftravelbylightisnegligible.(1A)
Speedofsound=
(1M)
=
=345ms–1(1A)
(b)Heightofthecliff=vt(1M)
=345´0.2
=69m(1A)
2(a)Theelasticstringsetstheairparticlesaroundittovibrate.(1A)
ThevibrationsofairparticlesaretransmittedtoJoe’searandsohecanhearthesound.(1A)
(b)(i)Frequency=
(1M)
=
=500Hz(1A)
(ii)Applyv=fl.(1M)
wavelength=
=
=0.68m(1A)
3(a)Thefrequencyrangeofthehummingsoundisfrom15Hzto80Hz.(1A)
(b)Byv=fl,(1M)
Forthe15Hzsound,
l=
=
=22.7m(1A)
Forthe80Hzsound,
l=
=
=4.25m(1A)
Thelongestandshortestwavelengthsofthehummingsoundarerespectively22.7mand4.25m.
(c)No,humancannothearthewholerangeofthehummingsound.(1A)
Thisisbecausesoundoffrequencybelow20Hzisoutoftheaudiblefrequencyrange.(1A)
4(a)(i)Whentheloudspeakerproducessounds,theloudspeakerconemovesinandoutrapidly.(1A)
Thisstretchesandcompressestheairinfront.(1A)
Asaresult,aseriesofrarefactionsandcompressionstravelsthroughtheairandtheflamefollowsthemotionofair.(1A)
(ii)Byv=fl,(1M)
wavelengthofthesound
=
=
=0.85m(1A)
(b)Theboxabsorbsalltheenergycarriedbythesound.(1A)
Thereisavacuuminthebox.(1A)
(Orotherreasonableanswers)
5(a)Getapartnertohitalongironrailfromadistance.(1A)
Thenhearthesoundstravellingthroughfirsttherailandthentheair.(1A)
(b)Timeneededtotravelthroughtheair
=
=29.4s(1M)
Timeneededtotravelthroughtheground
=
=2.5s(1M)
Timedifference
=29.4–2.5=26.9s(1A)
6(a)(i)S1Q=
(1M)
=8.14m(1A)
(ii)S2Q=
(1M)
=8.38m(1A)
(b)PathdifferenceofQ
=S2Q–S1Q(1M)
=8.38m–8.14m
=0.24m(1A)
(c)PathdifferenceofQ=l
Thewavelengthofthesoundemittedis0.24m.(1A)
(d)Speedofsoundinair
=fl(1M)
=1.4´1000´0.24
=336ms–1(1A)
7(a)Yes,theyhavethesamepitch.(1A)
ThisisbecausetheyhavethesamenumberofwaveformsontheCROscreen.(1A)
(b)NotesX,YandZhavethesameloudness.(1A)
Thisisbecausetheamplitudesoftheirwaveformsarethesame.(1A)
(c)Theysounddifferentlybecausetheyhavedifferentqualities(waveforms)(1A)
whichdependonthenumber(1A)
andamplitudeofovertonesaddedtothefundamentalfrequency.(1A)
(d)Ifthenotehasasoundintensitylevelof140dB,itwillcausepermanentdamagetotheear.(1A)
8(a)Bothsoundandlightarewaves.(1A)
Soundwavesarelongitudinalwaveswhilelightwavesaretransversewaves.(1A)
Thetravellingspeedofsoundwavesintheairis330ms–1whilethetravellingspeedoflightwavesintheairis
3´108ms–1.(1A)
(Orotherreasonableanswers)
(b)
Forlightwaves:
(Decreasedwavelengthinwater)(1A)
(Thewavebendstowardsthenormalinwater.)(1A)
Forsoundwaves:
(Increasedwavelengthinwater)(1A)
(Thewavebendsawayfromthenormalinwater.)(1A)
9(a)
(Correctamplitude)(1A)
(Correctperiod)(1A)
(b)ThewaveformofMichael’svoicehasalargeramplitude.(1A)
ThewaveformofMichael’svoiceshowsthesamefundamentalfrequencybutdifferentnumberandamplitudeofovertonesfromJulia’s.(1A)
(c)
(Sameamplitudeandperiod)(1A)
(Correctquality)(1A)
10(a)Interference(1A)
(b)(i)Period
=
=
=2.22´103s(1M)
f=
(1M)
=
=450Hz
Byv=fl,(1M)
wavelengthofthesound
=
=
=0.756m(1A)
(ii)
(Samefrequency)(1A)
(Smalleramplitude)(1A)
(c)(i)Diffractionoccurs(1A)
andsoundwavesspreadoutafterpassingthegap.(1A)
(ii)Increasingthepitchofthesoundmeansdecreasingthewavelength.(1A)
Soundwaveswouldspreadatasmallerdegreeafterpassingthegap.(1A)
11(a)(i)Anultrasoundsignalemittedbytheshipisreflectedwhenithitsanobstacle.(1A)
Thedistancebetweentheshipandtheobstacleisequalto
wherevisthespeedofsoundintheseaandtisthetimeelapsedfromemittingthesignaltoreceivingtheecho.(1A)
(ii)Depthoftheseabed
=
(1M)
=
=300m(1A)
(iii)Thewavelengthofultrasoundissmallerthanthatofaudiblesound,soultrasounddiffractslesswhenitmeetsanobstacle.Thismakesultrasounddetectthelocationsofobstaclesmoreaccurately.(1A)
(b)(i)X-rayisakindofelectromagneticwavewhileultrasoundisnot.(1A)
X-rayisatransversewavewhileultrasoundisalongitudinalwave.(1A)
ThetravellingspeedofX-rayinair(3108ms–1)ismuchhigherthanthetravellingspeedofultrasoundinair(340ms–1).(1A)
(ii)X-raymayharmthefoetus.(1A)
12(a)Alongitudinalwaveisoneinwhichthevibrationsofparticlesarealongthedirectionoftravelofthewave.(2A)
(b)(i)EorM(1A)
(ii)AorI(1A)
(c)(i)EandM(1A)
(Orotherreasonableanswers)
(ii)AandE(1A)
(Orotherreasonableanswers)
(d)(i)Anyoneofthefollowing:
(1A)
A,B,H,IorJ
(ii)Anyoneofthefollowing:
(1A)
D,E,F,LorM
(iii)Anyoneofthefollowing:
(1A)
C,GorK
13(a)Takedirectiontotherighttobepositive.
(Correctcurve)(1A)
(Correctlabellingofparticles)(1A)
(Correctdisplacement)(1A)
(b)Amplitude=8cm(1A)
Wavelength=80cm(1A)
(c)(i)Frequency=5Hz(1A)
Period=
=
=0.2s(1A)
Speed=fl(1M)
=5´0.8
=4ms1(1A)
(ii)
(Correctlabelledaxes)(1A)
(Correctcurve)(1A)
(d)(i)
(Correctcurve)(1A)
(ii)
(Correctpositionsofparticles)
(1A)
14(a)(i)SoundwavesarereflectedbythewindowsintoroomB.(1A)
(Correctdiagram)(1A)
(ii)Opentherightwindowinsteadoftheleftone.(1A)
(Correctdiagram)(1A)
(b)(i)Interference(1A)
(ii)LoudandsoftsoundsareheardalongPQ.(1A)
Atsomepositions,soundwavesfromtheloudspeakersreinforceeachother(constr
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