Standing Waves — Question 4

Question 4

A straight snapshot does not turn every point into a node

A standing wave has $y=B\sin(kx)\cos\omega t$. At $t_* =\pi/(2\omega)$ the whole string lies along its equilibrium line.

1. Find the velocity and acceleration fields at that instant.

2. Distinguish the true nodes from the other zero-displacement points in this snapshot.

Question 4 - Solution

A straight snapshot does not turn every point into a node

Method and mathematical working

Evaluate the time derivatives at the specified zero of the displacement time factor. Distinguish instantaneous zero displacement from zero displacement at every time.$$\begin{aligned}y_t&=-B\omega\sin kx\sin\omega t,\\y_{tt}&=-\omega^2 y,\\t=\frac\pi{2\omega}&:\quad y=0,\quad y_t=-B\omega\sin kx,\quad y_{tt}=0.\\[4pt]\cos\omega t=0&\Rightarrow y=0\ \text{at one instant},\\\sin kx=0&\Rightarrow y=y_t=0\ \text{for every }t.\end{aligned}$$

(a) Calculation and conclusion

$y_t=-B\omega\sin(kx)\sin\omega t$, so at $t_*$ it is $-B\omega\sin(kx)$. Acceleration is $y_{tt}=-\omega^2y=0$ everywhere at that instant. Most points have maximum-magnitude velocity despite zero displacement.

(b) Interpretation and limits

True nodes have $\sin(kx)=0$ and remain at zero for all time, with zero velocity. The other zeros occur only because the common time factor vanishes. A single image cannot distinguish permanent nodes from momentary equilibrium crossings without motion information.

Physical check

At this instant the string passes through equilibrium with maximum local speed away from permanent nodes. This is the distributed analogue of an oscillator crossing its equilibrium position.