Standing Waves — Question 3

Question 3

Adjacent loops oscillate in opposite phase

A standing wave is $y=B\sin(kx)\cos\omega t$, with $B>0$. Consider points $x_1=\lambda/8$ and $x_2=3\lambda/8$, then $x_3=5\lambda/8$.

1. Compare the local amplitudes and phases at the three points.

2. What happens to the local phase at a node? Why is a smooth traveling-wave phase interpretation misleading here?

Question 3 - Solution

Adjacent loops oscillate in opposite phase

Method and mathematical working

Evaluate the signed spatial factor at each location. Its absolute value gives local amplitude, while its sign determines whether the motion is in phase or opposite in phase.$$\begin{aligned}\sin(2\pi/8)&=\sin(6\pi/8)=1/\sqrt2,\\\sin(10\pi/8)&=-1/\sqrt2.\\[4pt]y_1(t)=y_2(t)&=\frac{B}{\sqrt2}\cos\omega t,\qquad y_3(t)=-\frac{B}{\sqrt2}\cos\omega t.\end{aligned}$$

(a) Calculation and conclusion

The sine factors are $\sqrt2/2,\sqrt2/2,-\sqrt2/2$. All three amplitudes are $B/\sqrt2$. Points 1 and 2 move in phase; point 3 moves oppositely, with temporal phase difference $\pi$.

(b) Interpretation and limits

A node has zero oscillation amplitude, so its temporal oscillation phase is undefined. Within a loop all points share one temporal phase, and neighboring loops differ by $\pi$. This differs from a traveling wave whose phase progresses continuously with position.

Physical check

All three magnitudes agree, but the third point oscillates oppositely to the first two. At a permanent node the amplitude is zero, so assigning a local oscillation phase has no observable meaning.