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

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snickersnee
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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.

Textbook sketch of a string displaced to a square-wave shape: height A from x = 0 to L/2, then -A from L/2 to L, with the piecewise definition of the initial displacement written beside it.
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Homework Equations



Equation for Fourier sine coefficients: a_n = (2/L) times the integral from 0 to L of f(x) sin(nπx/L) dx.

b_n = 0 because released from rest[/B]

The Attempt at a Solution



Handwritten integral of -A sin(n pi x / L) dx, followed by the result A L cos(n pi x / L) over pi n, plus C.


Equation a_n = -(A / (2nπ)) (-cos(nπ) + 2 cos(nπ/2) + 1), written as plain mathematics showing the Fourier coefficient a_n in terms of A and n.

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Attachments
  • Equation for a_n: a_n = (2/L)[ -AL cos(nπx/L)/(nπ) evaluated from 0 to L/2, plus AL cos(nπx/L)/(nπ) evaluated from L/2 to L ].
    upload_2015-11-22_13-11-10.png
    1.6 KB · Views: 456
  • Equation a_n = -(AL/(nπ))(-cos(nπ) + 2cos(nπ/2) + 1)
    upload_2015-11-22_13-12-38.png
    1 KB · Views: 483
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)##