# A standing wave with different linear mass densities

I have a question about a standing wave with different linear mass densities throughout the string. Suppose that we had a string of linear mass density $$\mu$$ joined at $$x = L$$ to a string with linear mass density $$\mu/9$$ and length $$3L$$ to form a composite string of total length $$4L$$. The first end of this composite string is fixed at $$x = 0$$ and the last end at $$x = 4L$$ is free to oscillate in the $$±y$$ direction. How would I show that $$\sin(k_1L)$$ = ± $$\frac{\sqrt3}{2}$$?

This question seemed very weird to me at first - since a standing wave isn't a travelling wave, I can't find reflected/transmitted amplitudes the standard way. However, am I right in thinking that since a standing wave can be decomposed into 2 left/right travelling waves of the same amplitudes, I can do this?

I have seen a sort of solution here : Standing waves on string with different densities, but this is of a string of length 2L, tied down at both ends, whereas my string is of length 4L, with one end fixed and the other end free, and I'm not quite sure how to apply the same technique here.

If anyone could point me to the right direction - it would be very much appreciated.

I suppose that free-to-oscillate means the following

$$\frac{\partial u}{\partial x} = 0\, ,$$

that is, a Neumann boundary condition.

To solve this problem you do something similar to the case that you link to:

• Solve the differential equation for each segment. You would end up with different solutions because of the change in density.
• Apply the boundary condition on the left.
• Impose continuity in the interface between the two segments (in displacement and slope).
• Apply the boundary condition on the right to find the eigenvalues.
• Yeah, I thought about doing this - but is it ok if we only consider the time-independent equation describing our standing wave? Getting rid of the temporal part? – jambajuice May 14 at 11:30
• Yes, you could compute the Fourier transform or use the substitution $u(x, t) = w(x)e^{i\omega t}$. – nicoguaro May 14 at 15:03