For a string with fixed ends, which normal modes are missing?

AI Thread Summary
For a string with fixed ends, the odd nature of the function indicates that even modes will not be present. However, the presence of cos(n*pi) complicates this, as it does not vanish and can yield values of +1 or -1. It is crucial to confirm that the ends are fixed at ξ(0,t) = ξ(L,t) = 0, which allows for analysis at the midpoint ξ(L/2,t). To find the missing modes, one should split the integral for a_n into two segments and apply matching conditions at the midpoint. Understanding these principles is essential for identifying which normal modes are absent.
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Homework Statement



Here's the problem. I was able to find the a_n and b_n values, my question is mainly on part (c), how do I find which modes are missing? The function is odd, so even modes should disappear, but cos(n*pi) doesn't disappear, it's either +1 or -1. I'd greatly appreciate any help.

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Homework Equations



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b_n = 0 because released from rest[/B]

The Attempt at a Solution



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When it says fixed ends, are we to assume that at x = 0 and x = L, the ends are fixed at ##\xi(0,t) = \xi(L,t) = 0##?
If so, then you should be able to say something about ##\xi( L/2,t)##.

Using your equation for a_n, break that into two integrals, one from 0 to L/2 and one from L/2 to L. Then enforce an appropriate matching condition for ##\xi(L/2,t) ## and ##\xi_t(L/2,t)##
 
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