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Math Help - Question on damping string vibration.

  1. #1
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    Question on damping string vibration.

    I cannot derivate the formula of b_n . Please read the question at the bottom and look at the way how I derivate the coeficients. You can see the formula don't match.
    The original question is:
    Find the solution of the equation for damping vibration string stretched from x=0 to x=L where

    \frac{\partial ^2 u}{\partial t^2} \;+\; 2k\frac{\partial u}{\partial t} \;=\; c^2\frac{\partial ^2 u}{\partial x^2} \;\;\;\;\; (1)

    With boundary and initial condition given:

     u(0,t) = u(L,t)= 0 \;\;\; (2)
     u(x,0) = f(x)\; ,\; \frac{\partial u}{\partial t}(x,0) = g(x) \;\;\;\;(3)

    We assume u(x,t)=X_{(x)}T_{(t)} \;\Rightarrow\; X''+\mu^2X=0 \;\;and\;\; T'' \;+\; 2kT' \;+\; (\mu c)^2 T \;=\; 0

    Attempted steps:
    1)\;\; X''+\mu^2X=0 \Rightarrow \mu=\mu_n = \frac{n\pi}{L} \;\;\;\Rightarrow\;\;\; X \;=\; X_n \;=\; sin(\frac{n\pi}{L})x \;\;\;\;n=1,2,3......(4)


    2)\;\; T'' \;+\; 2kT' \;+\; (\frac{n\pi}{L}c)T \;=\; 0\;\; \;\;(5)

    Using constant coeficients ODE, m=\frac{-2k^+_-\sqrt{4k^2-4(\frac{n\pi}{L}c)^2}}{2}

    Three cases of k^2-(\frac{n\pi}{L}c)^2 \;\;\;\;Let\;\lambda_n =\sqrt{|k^2-(\frac{n\pi}{L}c)^2|}

    Case \;1\;\;\;\;k^2-(\frac{n\pi}{L}c)^2\;< 0 \Rightarrow\; n > (\frac{kL}{\pi c}) \;\;\;\Rightarrow\;\; \; T= e^{-kt}[c_n cos(\lambda_n t) \;+\; d_n sin(\lambda_n t)]\;\;(6)

    Case \;2\;\;\;k^2-(\frac{n\pi}{L}c)^2\;= 0 \Rightarrow\; n = (\frac{kL}{\pi c}) \;\;\;\Rightarrow\;\; \; T= h_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; j_{n=\frac{kL}{\pi c}}te^{-kt} \;\;\;\;(7)

    Case \;3\;\;\;\;k^2-(\frac{n\pi}{L}c)^2\;> 0 \Rightarrow\; n < (\frac{kL}{\pi c}) \Rightarrow T= e^{-kt}[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] (8)
    For n=1,2,3.......

    1) For\; \frac{kL}{\pi c} \;\; is\; not\; integer

    u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)]

    \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n cos(\lambda_n t) \;+\; b_n sin(\lambda_n t)] \;\;\;

    \Rightarrow \; u(x,t) = \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n \; e^{-kt} \; cosh(\lambda_n t)]

     \;+\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x) [b_n\; e^{-kt}\; sinh(\lambda_n t)] \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n \; e^{-kt} \; cos(\lambda_n t)]

    \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x) [b_n\; e^{-kt}\; sin(\lambda_n t)]

    u(x,0) \;=\; f(x) \;=\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n\; sin(\frac{n\pi}{L}x) \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n \; sin(\frac{n\pi}{L}x)

     \;=\; \sum_{n=1} ^{\infty} \; a_n\; sin(\frac{n\pi}{L}x)

     \frac{du(x,t)}{dt} = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} a_n sin(\frac{n\pi}{L}x )[-k\;cosh(\lambda_n t) + \lambda_n sinh(\lambda_n t)]

    \;\;\;\;\;\;\;\;\; \;+ \;\;e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} b_n sin(\frac{n\pi}{L}x )[-k\;sinh(\lambda_n t) + \lambda_n cosh(\lambda_n t)]

    .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;a_n sin(\frac{n\pi}{L}x )[-k\;cos(\lambda_n t) - \lambda_n sin(\lambda_n t)]

    \;\;\;\;\;\;\;\;\;\; \;+\;\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n sin(\frac{n\pi}{L}x )[-k\;sin(\lambda_n t) + \lambda_n cos(\lambda_n t)]

    \Rightarrow \;\;\frac{d(u}{dt}(x,0) \;=\;g(x) \;=\; -\;k\sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

    .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-\;\;k \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n \lambda_n sin(\frac{n\pi}{L}x )

    \Rightarrow \;\;\frac{d(u}{dt}(x,0) \;=\;g(x) \;=\; -\;k\sum_{n=1} ^{\infty} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

    \;\;\;\;\;\;\;\;\;\;\;=\; -\; kf(x) \;\;+\;\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

     Therefore \;\; g(x)+kf(x) \;=\; \sum_{n=1}^{\infty}b_n \lambda_n sin(\frac{n \pi x}{L})\;\;.

    ((g(x)+kf(x)) \;,\; sin(\frac{m \pi x}{L}) ) \;=\; ( sin(\frac{m \pi x}{L}) \;,\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x ))

    \Rightarrow\; \int_0^L [g(x)+kf(x)]sin(\frac{m\pi}{L}x ) dx \;=\; b_n\lambda_m \int_0^L sin^2(\frac{m \pi x}{L}) dx

     a_n=\frac{2}{L}\int_0^L f(x)]sin(\frac{n\pi}{L}x) dx \;\;\;and\;\;\; b_n=\frac{2}{\lambda_n L}\int_0^L [g(x)+kf(x)]sin(\frac{n\pi}{L}x ) dx (10)


    2) For\; \frac{kL}{\pi c} \;\; is\; integer

    u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)]

    \;\;\;\;\;\;\;\;\;\; \;+\; sin(\frac{k}{c}x)[ a_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; b_{n=\frac{kL}{\pi c}}te^{-kt} ] \;

    .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n cos(\lambda_n t) \;+\; b_n sin(\lambda_n t)]

    u(x,0) \;=\; f(x) \;=\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n\; sin(\frac{n\pi}{L}x) \;+\; a_{n=\frac{kL}{\pi c}} sin(\frac{k x}{c})

    \;\;\;\;\;\;\;\;\;\;\;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n \; sin(\frac{n\pi}{L}x)

    u(x,0) \;=\; f(x) \;=\;\sum_{n=1} ^{\infty} \; a_n\; sin(\frac{n\pi}{L}x)

     \frac{du(x,t)}{dt} = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} a_n sin(\frac{n\pi}{L}x )[-k\;cosh(\lambda_n t) + \lambda_n sinh(\lambda_n t)]

    \;\;\;\;\;\;\;\;\;\;\;+ \;\;e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} b_n sin(\frac{n\pi}{L}x )[-k\;sinh(\lambda_n t) + \lambda_n cosh(\lambda_n t)]

    .\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;+\; sin(\frac{k}{c}x)[ -a_{n=\frac{kL}{\pi c}}ke^{-kt} \;+\; b_{n=\frac{kL}{\pi c}}(e^{-kt}-kte^{-kt}) ] \;

    .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;a_n sin(\frac{n\pi}{L}x )[-k\;cos(\lambda_n t) - \lambda_n sin(\lambda_n t)]

    \;\;\;\;\;\;\;\;\;\; \;+\;\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n sin(\frac{n\pi}{L}x )[-k\;sin(\lambda_n t) + \lambda_n cos(\lambda_n t)]


     \Rightarrow \;\;\frac{du}{dt}(x,0) \;=\;g(x) \;=\; -k\sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n sin(\frac{n\pi}{L}x ) \;+\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

    .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -\;\;k a_{n=\frac{kL}{\pi c}} sin(\frac{k}{c}x)\;+\; b_{n=\frac{kL}{\pi c}} sin(\frac{k}{c}x)\;\;\;-\;\;\;k \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n sin(\frac{n\pi}{L}x )

    \;\;\;\;\;\;\;\;\;\;\;+\;\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n \lambda_n sin(\frac{n\pi}{L}x )

    \int_0^L g(x)sin(\frac{n\pi}{L}x ) dx \;=\; (-ka_n +b_n\lambda_n)\int_0^L g(x)sin^2(\frac{n\pi}{L}x ) dx n=\frac{kL}{\pi c}

    \Rightarrow\;\; -\;ka_n\;+\;b_n\lambda_n \;=\; \frac{2}{L}\int_0^L g(x)sin^2(\frac{n\pi x}{L}) dx =\frac{kL}{\pi c}

     b_n \;=\; \frac{2}{\lambda_n L}\int_0^L g(x) sin(\frac{n\pi x}{L}) dx + \frac{ka_n}{\lambda_n}
    except\;\; n=\frac{kL}{\pi c} \;where\; b_{n=\frac{kL}{\pi c}} = \frac{2}{L}\int_0^L g(x)sin(\frac{k}{c}x) dx + ka_{n=\frac{kL}{\pi c}}


     a_n\; =\; \frac{2}{L}\int_0^L f(x)sin(\frac{n\pi}{L}x) dx

    b_n=\frac{2}{\lambda_n L}\int_0^L [g(x)+kf(x)]sin(\frac{n\pi}{L}x ) dx \;\;\;+\;\;\; \frac{2k}{\lambda_n L}\int_0^L f(x)sin(\frac{n\pi}{L}x) dx

    With\;\;\; b_{n=\frac{kL}{\pi c}} \;=\; \frac{2}{L}\int_0^Lg(x)sin(\frac{k}{c}x) dx \;+\; ka_{n=\frac{kL}{\pi c}}


    Question: For \frac{kL}{\pi c} \;is\;integer

    The text book gave this equation:
    -\;k a_n\;+\;\lambda_n b_n\;=\;\frac{2}{L}\int_0^L f(x)sin(\frac{n\pi x}{L} dx

    I tried and I cannot arrive to this formula no matter what. Please look at (10) above.

    It look very similar to the formula for b_{n=\frac{kL}{\pi c}}\;\; when \;\;\frac{kL}{\pi c}\;\;is \; an\; integer.

    But the book seldom make mistake!!! What did I do wrong?
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  2. #2
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    Anyone please? Read just the last part, you might have the answer. I posted a long drawn formulas just in case anyone need to verify how I got to this point.
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