Frequency of spring waves

Consider transverse waves on a stretched spring (just as in the demonstration given in lecture.) The same wave equation applies as for strings, namely

$$\tau \frac{\partial^2 y}{\partial x^2} = \mu \frac{\partial^2 y}{\partial t^2},$$

where $\mu$ is the mass per unit length and $\tau$ is the tension in the spring. Note that for springs the tension comes not from an external applied pull, as it does in our realization for strings, but from the stretch of the spring with proportionality constant $K$. Thus, $\tau = K (L - L_o)$, where $L$ is the length of the spring set by the distance between Prof. Arias and the wall, and $L_o$ is its equilibrium length. Note that for low amplitude waves, the change in the length of the spring created by the waves is relatively small and can be ignored for the purpose of this problem.

(a)

Taking the spring to have mass $M$ and considering only small amplitude waes, compute the angular frequency $\omega$ of the lowest frequency standing waves for boundary conditions with both ends fixed. Express your result in terms of only $M$, $L$, $L_o$ and $K$.

(b)

Show that when the spring is stretched very far (so that $L>>L_o$), the frequency $\omega$ from (a) becomes independent of the length $L$!