# Solving the Wave Equation

1. May 20, 2010

### Lavace

1. The problem statement, all variables and given/known data

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

3. 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?

Last edited by a moderator: May 4, 2017
2. May 20, 2010

### Lavace

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 . v

Where w(n) are the normal mode frequencies.

Could someone verify this is correct?

3. May 20, 2010

### Lavace

Also, any clues for b)?

4. May 20, 2010

### Redbelly98

Staff Emeritus
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.