Expression for motion

  • Thread starter Reshma
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  • #1
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I need your guidance in this problem.

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?
 

Answers and Replies

  • #2
747
4
Never mind, I found the solution :biggrin: !
 

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