How Do You Solve the Wave Equation Using Coefficient Equations?

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Homework Statement



[PLAIN]http://img33.imageshack.us/img33/8236/waveeq.jpg



The Attempt at a Solution



We calculate second differential with respect to x, and t, substitute into the wave equation.

We then equate the coefficients: [A''(x) + (w/v)^2A(x)]sin(wt)=0

We know from SHM equation that: A''(x) = -(w/v)^2A(x), and hence A''(x) = -k^2 A(x)

But where do we go from here? Any hints?

Also, what about part b?
 
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From A''(x) = -k^2 A(x), we seek a solution of the form A(x) = Csin(kx + psi)

Apply our boundary conditions of y(0,t) and y(L,t) both = 0.

We end up with sin(kL) = 0, where kL varies from 0 to 2PI, this implies that kL=nPI where n=1,2,3...

Because it's quantised, we can say k(n) = nPI/L, where n=1,2,3...

Since k = w/v, w(n) =nPI/L . vWhere w(n) are the normal mode frequencies.

Could someone verify this is correct?
 
Also, any clues for b)?
 
Looks good for part (a).
For (b), I'm not quite sure what they are getting at. In a sense, you already showed this in your derivation for part (a). Maybe they want you to think in terms of the wavelength λ and how it relates to the string length L.