Wavelength and Frequency — Question 5

Question 5

A displacement equality is weaker than an equal phase

Two points on $y=A\cos(kx-\omega t)$ have equal displacement at one instant. Their separation is unknown.

1. Give an example in which their phases differ but their displacements agree.

2. What additional local measurement, with known propagation direction and $\omega$, distinguishes the two phases except at degenerate endpoints?

Question 5 - Solution

A displacement equality is weaker than an equal phase

Method and mathematical working

Use both the sine and cosine of phase. Position alone determines cosine and therefore leaves a sign ambiguity in sine.$$\begin{aligned}\cos\Phi&=\frac yA,\qquad \sin\Phi=\frac{y_t}{A\omega},\\ \Phi&=\operatorname{atan2}\!\left(\frac{y_t}{A\omega},\frac yA\right).\\[4pt]\cos(\pi/3)&=\cos(-\pi/3)=1/2,\\y_t(\pi/3)&=+\frac{\sqrt3}{2}A\omega,\qquad y_t(-\pi/3)=-\frac{\sqrt3}{2}A\omega.\end{aligned}$$

(a) Calculation and conclusion

Phases $\pi/3$ and $-\pi/3$ both give $y=A/2$ but differ by $2\pi/3$, not a full cycle. Equality of a cosine value alone does not establish equal phase or an integer-wavelength spacing.

(b) Interpretation and limits

The local velocity is $\partial_t y=A\omega\sin\Phi$, which has opposite signs at those phases. Together displacement and velocity determine sine and cosine, hence phase modulo $2\pi$. At turning points the sign test degenerates, but the full state still fixes phase modulo a cycle.

Physical check

Two states with the same position but opposite velocities are physically different. Phase is still determined only modulo a full cycle, even when both state measurements are available.