Standing Waves — Question 6

Question 6

Zero mean power permits instantaneous energy exchange within a loop

For $y=B\sin(kx)\cos\omega t$, use $P=-F y_xy_t$ with rightward flux positive.

1. Derive the instantaneous power and its time average.

2. Locate positions where power vanishes for all time. Does zero mean flux imply that every interior point exchanges no energy with its neighbors?

Question 6 - Solution

Zero mean power permits instantaneous energy exchange within a loop

Method and mathematical working

Multiply the spatial and temporal derivatives in the signed power formula. Use double-angle identities to expose the separate spatial and temporal zero factors.$$\begin{aligned}P&=-F(Bk\cos kx\cos\omega t)(-B\omega\sin kx\sin\omega t)\\&=\frac{FB^2k\omega}{4}\sin(2kx)\sin(2\omega t).\\[4pt]\langle P\rangle&=\frac1T\int_0^T P(x,t)\,dt=0,\qquad\sin2kx=0\Rightarrow P(x,t)=0.\end{aligned}$$

(a) Calculation and conclusion

$P=FB^2k\omega\sin(kx)\cos(kx)\sin\omega t\cos\omega t=(FB^2k\omega/4)\sin(2kx)\sin(2\omega t)$. Its average over a period is zero.

(b) Interpretation and limits

Permanent zero-flux positions have $\sin(2kx)=0$, including both displacement nodes and antinodes. Between them the flux reverses sign periodically, redistributing energy locally. Zero cycle-averaged transport is compatible with nonzero instantaneous transfer.

Physical check

The time average of $\sin(2\omega t)$ is zero. At nodes and antinodes the spatial factor also vanishes, so power there is zero instantaneously, not merely on average.