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Power transported by a wave problem Please Help

  1. Nov 23, 2007 #1
    Power transported by a wave problem...Please Help!!!

    1. On a real string, some of the energy of a wave dissipates as the wave travels down the string. Such a situation can be described by a wave function whose amplitude A(x) depends on x: y=A(x)sin(kx-[tex]\omega[/tex]t), where A(x)=A[tex]_{o}[/tex]e[tex]^{-bx}[/tex]. What is the power transported by the wave as a function of x, where x>0?



    2. P=[tex]\mu[/tex]v[tex]\omega[/tex][tex]^{2}[/tex]cos[tex]^{2}[/tex](kx-[tex]\omega[/tex]t)
    [tex]\mu[/tex]=[tex]\frac{m}{L}[/tex]
    v=[tex]\sqrt{\frac{F_{T}}{\mu}}[/tex]
    [tex]\omega[/tex]=2[tex]\pi[/tex]f
    v=f[tex]\lambda[/tex]-->f=[tex]\frac{v}{\lambda}[/tex]




    3. I tried two approaches, neither of which I am sure if it is the right way to solve the problem...not sure what the question is really asking for.
    P(x)=[tex]\mu[/tex]v[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex]cos[tex]^{2}[/tex](kx-[tex]\omega[/tex]t)
    so, by substituting to "simplify": P(x)=[tex]\mu[/tex][tex]\frac{v}{\lambda}[/tex][tex]\lambda[/tex](2[tex]\pi[/tex][tex]\frac{v}{\lambda}[/tex])[tex]^{2}[/tex](A(x))[tex]^{2}[/tex]cos[tex]^{2}[/tex](kx-[tex]\omega[/tex]t)
    P(x)=[tex]\mu[/tex]4[tex]\pi[/tex][tex]^{2}[/tex]v(A(x))[tex]^{2}[/tex]cos[tex]^{2}[/tex](kx-[tex]\omega[/tex]t)
    P(x)=[tex]\mu[/tex][tex]^{2}[/tex](4[tex]\pi[/tex])[tex]^{2}[/tex][tex]\frac{F_{T}}{\mu}[/tex](A(x))[tex]^{4}[/tex]cos[tex]^{4}[/tex](kx-[tex]\omega[/tex]t)
    P(x)=16[tex]\mu[/tex][tex]\pi[/tex][tex]^{4}[/tex]F[tex]_{T}[/tex]A[tex]^{2}_{o}[/tex]e[tex]^{-4bx}[/tex]cos[tex]^{4}[/tex][/tex](kx-[tex]\omega[/tex]t)
    and I'm not sure where to go from there...if it's simplified enough for the answer or if it's the completely wrong approach.

    Or:
    y=A[tex]_{o}[/tex]e[tex]^{-bx}[/tex]sin(kx-[tex]\omega[/tex]t)
    v_{y}=\frac{d}{dt}\left[A[tex]_{o}[/tex]e[tex]^{-bx}[/tex]sin(kx-[tex]\omega[/tex]t)\right]=-\omegaA[tex]_{o}[/tex]e[tex]^{-bx}[/tex]cos(kx-[tex]\omega[/tex]t)
    P=F_{Ty}v_{y}\approxF_{T}v_{y}tan\theta=-F_{T}\frac{\gamma}{dt}\frac{\gamma}{dx}
    P=-F_{T}\left[-\omegaA[tex]_{o}[/tex]e[tex]^{-bx}[/tex]cos(kx-[tex]\omega[/tex]t)\right]\left[kA[tex]_{o}[/tex]e[tex]^{-bx}[/tex]cos(kx-[tex]\omega[/tex]t)\right]=F_{T}\omegakA[tex]_{o}[/tex]^{2}e[tex]^{-bx}[/tex]cos^{2}(kx-[tex]\omega[/tex]t)
    So those are my two answers worked out completely showing all steps. Please help and thank you.
    Also note that the greek letters look like superscripts but they aren't supposed to be.
     
  2. jcsd
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