Standing Waves — Question 5

Question 5

A pure standing mode trades total kinetic and elastic energy

A fixed-end string of length $L$ supports $y=B\sin(n\pi x/L)\cos\omega t$, where $\omega=v n\pi/L$ and $v^2=F/\mu$.

1. Integrate the kinetic and elastic energies over the whole string.

2. Find their sum and explain why the traveling-wave pointwise equality of the two energy densities does not apply.

Question 5 - Solution

A pure standing mode trades total kinetic and elastic energy

Method and mathematical working

Integrate the local kinetic and elastic densities over the string. For an allowed nonzero mode, the integrals of both squared spatial factors equal $L/2$.$$\begin{aligned}\int_0^L\sin^2(kx)\,dx&=\int_0^L\cos^2(kx)\,dx=L/2,\\K&=\frac{\mu LB^2\omega^2}{4}\sin^2\omega t,\\U&=\frac{FB^2k^2L}{4}\cos^2\omega t.\\[4pt]E=K+U&=\frac{\mu LB^2\omega^2}{4}(\sin^2\omega t+\cos^2\omega t)=\frac{\mu LB^2\omega^2}{4}.\end{aligned}$$

(a) Calculation and conclusion

Using $\int_0^L\sin^2(n\pi x/L)dx=\int_0^L\cos^2(n\pi x/L)dx=L/2$, $K=\mu L B^2\omega^2\sin^2\omega t/4$ and $U=\mu L B^2\omega^2\cos^2\omega t/4$.

(b) Interpretation and limits

The constant total is $E=\mu L B^2\omega^2/4$. Kinetic and elastic energies exchange as the mode evolves. Locally their spatial and temporal factors differ, unlike a single traveling profile where $y_t=\mp vy_x$ enforces equality. A standing wave is a sum of opposite directions.

Physical check

The dispersion relation makes the two prefactors equal. Their sum is time independent because $\sin^2\omega t+\cos^2\omega t=1$, even though each contribution oscillates.