Standing Waves — Question 9

Question 9

A stationary node can have nonzero elastic energy density

For $y=B\sin(kx)\cos\omega t$, inspect the node at $x=0$. Use $u_K=\mu y_t^2/2$ and $u_U=Fy_x^2/2$.

1. Find the two energy densities at the node as functions of time.

2. At the adjacent antinode $x=\pi/(2k)$, find the corresponding densities. Explain why zero displacement does not imply zero elastic energy.

Question 9 - Solution

A stationary node can have nonzero elastic energy density

Method and mathematical working

Evaluate the derivatives at each special position. Energy density depends on velocity and slope rather than displacement alone.$$\begin{aligned}u_K&=\tfrac12\mu B^2\omega^2\sin^2kx\sin^2\omega t,\\u_U&=\tfrac12FB^2k^2\cos^2kx\cos^2\omega t.\\[4pt]x=0&:\quad u_K=0,\quad u_U=\tfrac12FB^2k^2\cos^2\omega t,\\x=\pi/(2k)&:\quad u_U=0,\quad u_K=\tfrac12\mu B^2\omega^2\sin^2\omega t.\end{aligned}$$

(a) Calculation and conclusion

At the node $y_t=0$, so $u_K=0$. But $y_x=Bk\cos\omega t$, giving $u_U=FB^2k^2\cos^2\omega t/2$, generally nonzero.

(b) Interpretation and limits

At the antinode $y_x=0$, so $u_U=0$, while $u_K=\mu B^2\omega^2\sin^2\omega t/2$. Elastic energy reflects local stretching, controlled by slope. A material point can remain stationary while the segments adjacent to it change inclination and store elastic energy.

Physical check

At a node the kinetic density vanishes at all times, while elastic density generally does not. At an antinode the roles reverse. A stationary location can store elastic energy.