physics1课后习题答案6.pdf

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1、A-PDFSplitDEMOP18-22(a)Thefrequencyofvibrationfisthesameforboththealuminumandsteelwires;theydonnot,however,needtovibrateinthesamemode.Thespeedofwavesinthealuminumisv1,thatinthesteelisv2.Thealuminumvibratesinamodegivenbyn1=2L1f=v1,thesteelvibratesinamodegivenbyn2=2L2f=v2.Bothn1an

2、dn2needbeintegers,sotheratiomustbearationalfraction.Notethattheratioisindependentoff,sothatL1andL2mustbechosencorrectlyforthisproblemtoworkatall!Thisratioissrn2L22(0:866m)(7800kg=m3)5===2:50n1L11(0:600m)(2600kg=m3)2Notethatsincethewireshavethesametensionandthesamecrosssection

3、alareaitisacceptabletousethevolumedensityinsteadofthelineardensityintheproblem.Thesmallestintegersolutionisthenn1=2andn2=5.Thefrequencyofvibrationisthenssn1vn1T(2)(10:0kg)(9:81m=s2)f====323Hz:2L12L11A2(0:600m)(2600kg=m3)(1:00106m2)(b)Therearethreenodesinthealuminumandsixinthe

4、steel.Butoneofthosenodesisshared,andtwoareontheendsofthewire.Theansweristhensix.239A-PDFSplitDEMOE19-1(a)v=f=(25Hz)(0:24m)=6:0m=s.(b)k=(2rad)=(0:24m)=26rad/m;!=(2rad)(25Hz)=160rad/s.Thewaveequationiss=(3:0103m)sin[(26rad/m)x+(160rad/s)t]E19-2(a)[
P]m=1:48Pa:(b)f=(334rad/s)

5、=(2rad)=167Hz.(c)=(2rad)=(1:07rad/m)=1:87m.(d)v=(167Hz)(1:87m)=312m=s:E19-3(a)Thewavelengthisgivenby=v=f=(343m=s)=(4:50106Hz=7:62105m:(b)Thewavelengthisgivenby=v=f=(1500m=s)=(4:50106Hz=3:33104m:E19-4Note:Thereisatypo;themeanfreepathshouldhavebeenmeasuredinm"instead

6、ofpm".=1:0106m;f=(343m=s)=(1:0106m)=3:4108Hz:minmaxE19-5(a)=(240m=s)=(4:2109Hz)=5:7108m.E19-6(a)Thespeedofsoundisv=(331m=s)(6:21104mi/m)=0:206mi/s:Invesecondsthesoundtravels(0:206mi/s)(5:0s)=1:03mi,whichis3%toolarge.(b)Countsecondsanddivideby3.E19-7Marchingat120pac

7、esperminutemeansthatyoumoveafooteveryhalfasecond.Thesoldiersinthebackaremovingthewrongfoot,whichmeanstheyaremovingthecorrectfoothalfasecondlaterthantheyshould.Ifthespeedofsoundis343m=s,thenthecolumnofsoldiersmustbe(343m=s)(0:5s)=172mlong.E19-8Ittakes(300m)=(343m=s)=0:87sfortheco

8、ncertgoertohearthemusicafterithaspassedthemicro