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Thread: Application of One-Dimensional Wave Equation

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Application of One-Dimensional Wave Equation

    I'm working on my final project in this independent study I'm in. I worked out this problem, but I want to verify that I'm right.

    A string of length L with both endpoints fixed, such as the string of a piano, is struck at its midportion, the impact thus imparting to the string an initial velocity $\displaystyle \psi(x)$ which is defined by:

    $\displaystyle 0, 0<= x<\frac{L-a}{2}$
    $\displaystyle 1, \frac{L-a}{2}<x<\frac{L+a}{2}$
    $\displaystyle 0, \frac{L+a}{2}<x<=L $

    Find the motion of the string.

    I set up the following boundary value problem:

    $\displaystyle y_{tt}=a^2y_{xx}$

    $\displaystyle y(0,t)=0$

    $\displaystyle y(L,t)=0$

    $\displaystyle y(x,0)=0$

    $\displaystyle y_{t}(x,0)=\psi(x)$

    and I got the following Fourier Series Solution:

    $\displaystyle y(x,t)=\frac{4L}{\pi^2a}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}sin{\left[\frac{(2n-1)\pi a}{2L}\right]}sin{\left[\frac{n\pi a}{L}t\right]}sin{\left[\frac{n\pi}{L}x\right]}$

    Does this look right? I would appreciate any help!!!
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    Quote Originally Posted by Chris L T521 View Post
    $\displaystyle 0, 0<= x<\frac{L-a}{2}$
    $\displaystyle 1, \frac{L-a}{2}<x<\frac{L+a}{2}$
    $\displaystyle 0, \frac{L+a}{2}<x<=L $
    The solution is given by,
    $\displaystyle y(x,t) = \sum_{n=1}^{\infty} \left( A_n \cos \frac{\pi n at}{L} + B_n \sin \frac{\pi n at}{L} \right) \sin \frac{\pi n x}{L}$.

    Where, $\displaystyle A_n = \frac{2}{L} \int_0^L 0\cdot \frac{\pi n x}{L} dx = 0$ because the string is streched.

    And, $\displaystyle B_n = \frac{2}{\pi n a}\int_0^L \phi (x) \frac{\pi n x}{L} dx$.

    All you need to do now is compute $\displaystyle B_n$ coefficient and you got your answer.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    I understand how the Coefficients are found, but would you briefly explain how you got that series?
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    Quote Originally Posted by Chris L T521 View Post
    I understand how the Coefficients are found, but would you briefly explain how you got that series?
    Are you familar with how to work with Boundary Value Problems? I would post out all the steps but it will take far to much typing. It would be better if you write your work and I tell you if I agree with it or disagree. Try reading this, see if it helps you.
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    Rhymes with Orange Chris L T521's Avatar
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    I was thinking it over today, and I realized where that series came from (Because I had derived it a couple weeks back...but I was tired, and didn't catch that.)

    There is one question I still have: Are you sure that your $\displaystyle B_n$ is correct? I got this for $\displaystyle B_n$

    $\displaystyle B_n=\frac{2}{n\pi a}\int_{0}^{L}\psi(x)sin{\left(\frac{n\pi}{L}x\rig ht)}\,dx$.

    When I simplify the integral, this is what I'm left with:


    $\displaystyle B_n=\frac{2}{n\pi a}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}\psi(x)sin{\left(\frac{n\pi} {L}x\right)}\,dx$.

    Does this seem right?
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Are you sure that your $\displaystyle B_n$ is correct?
    It is possible I got the $\displaystyle B_n$ wrong. Because maybed I mis-copied this solution directly out of my book because I am not going to rederive it each time. The link I gave you goes into more detail. There is a full solution there.

    This is Mine 96th Post!!!
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