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Expression for motion

  1. Jan 6, 2006 #1
    I need your guidance in this problem.

    The midpoint of a stretched string of length 'L' is pulled a distance U = L/10 from its equilibrium position so that the string forms 2 sides of an isosceles triangle. The string is then released. Find the expression for the motion by the Fourier series method.

    My problem is I have no idea of how Fourier analysis is done cause my course never included this topic. But the formula hinted in my text is:
    [tex]U(x,t) = \sum_n\left[A_n\sin \frac{n\pi x}{L}\cos \frac{n\pi ct}{L} + B_n\sin \frac{n\pi x}{L}\sin \frac{n\pi ct}{L}\right][/tex]
    Can someone help me out here?
  2. jcsd
  3. Jan 10, 2006 #2
    Never mind, I found the solution :biggrin: !
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