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The-incident-wave-set-up-on-a-string-of-length-fixed-at-each-end-is-given-by-y-1-Asin-kx-wt-i-what-is-the-equation-of-motion-of-the-reflected-wave-y-2-ii-obtain-the-resultant-y-y-1-y-2-of-the




Question Number 38822 by NECx last updated on 30/Jun/18
The incident wave set up on a string  of length fixed at each end is given  by:   y_1 =Asin(kx−wt)  i)what is the equation of motion  of the reflected wave,y_2 .  ii)obtain the resultant,y=y_1 +y_2   of the two waves.  iii)what type of resultant wave is  this?  iv)for what values of x will the  amplitud of the resultant wave   become zero?  v)for what values of x will y be  maximum?
Theincidentwavesetuponastringoflengthfixedateachendisgivenby:y1=Asin(kxwt)i)whatistheequationofmotionofthereflectedwave,y2.ii)obtaintheresultant,y=y1+y2ofthetwowaves.iii)whattypeofresultantwaveisthis?iv)forwhatvaluesofxwilltheamplitudoftheresultantwavebecomezero?v)forwhatvaluesofxwillybemaximum?
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18
i)y_2 =Asin(kx+wt)  ii)y_1 +y_2 =Asin(kx−wt)+Asin(kx+wt)  =2Asin(kx)coswt  iii)standing wave  iv)amplitude=2Asinkx  max amplitude=2A   min amplitude=0  when sinkx=0 =sinnΠ  kx=nΠ    x=((nΠ)/k)=((nΠ)/(2Π))×λ=n(λ/2)  v)max amplitude=2A   when sinkx=±1    sinkx=sin(n+(1/2))Π  kx=(n+(1/2))Π  κ=((2Π)/λ)  2(Π/λ)x=(n+(1/2))Π  x=(n+(1/2))(λ/2)
i)y2=Asin(kx+wt)ii)y1+y2=Asin(kxwt)+Asin(kx+wt)=2Asin(kx)coswtiii)standingwaveiv)amplitude=2Asinkxmaxamplitude=2Aminamplitude=0whensinkx=0=sinnΠkx=nΠx=nΠk=nΠ2Π×λ=nλ2v)maxamplitude=2Awhensinkx=±1sinkx=sin(n+12)Πkx=(n+12)Πκ=2Πλ2Πλx=(n+12)Πx=(n+12)λ2

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