# Normal modes of square membrane

bon

## The Attempt at a Solution

Ok so I've been able to do the first few parts and have derived that Wm,n = c pi / L (m^2 + n^2)

I've thus been able to show that the second lowest freq is a factor of root(5/2) times larger than the first and that there are two modes that have this frequency. But how do you do the next two parts?

How do I show that by combining them it is possible to have a node along either diagonal of the square? I guess I have to add the two solutions and subtract, but then how do I show that these nodes are along the diagonal?

Also what about the last part? How am i meant to deduce the fundamental mode for a triangular membrane?

Thanks!

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## The Attempt at a Solution

OK so I've been able to do the first few parts and have derived that Wm,n = c pi / L (m^2 + n^2)

I've thus been able to show that the second lowest freq is a factor of root(5/2) times larger than the first and that there are two modes that have this frequency. But how do you do the next two parts?

How do I show that by combining them it is possible to have a node along either diagonal of the square? I guess I have to add the two solutions and subtract, but then how do I show that these nodes are along the diagonal?

Also what about the last part? How am i meant to deduce the fundamental mode for a triangular membrane?

Thanks!

It looks like what you have so far is:

um,n(x,y,t) = Am,n·sin(mπx/L)·sin(nπy/L)·cos(ωm,nt), where ωm,n = (cπ/L)·√(m2 + n2).

You actually had ωm,n = (cπ/L)·(m2 + n2), but since you got the correct ratio, I assume you inadvertently left out the radical. I also took the liberty to include subscripts on u(x,y,t) to indicate what values of m & n are used.

Notice, that if um,n(x,y,t) is a solution to the equation, then, of course, so is un,m(x,y,t).

Not only that, um,n(x,y,t) ± un,m(x,y,t) are also solutions.

In particular, look at u1,2(x,y,t) - u2,1(x,y,t).

Use the double angle identity for sine. Do some factoring, then use the identity:

cos(θ) - cos(φ) = -2·sin((θ+φ)/2)·sin((θ-φ)/2)

That will get the node along the y = x diagonal.

As for the last question: What does it mean for there to be a node along a diagonal?