Interference of Waves — Question 9

Question 9

A small frequency mismatch makes a prepared null temporary

At a fixed point, equal-amplitude contributions have phases $2\pi f_1t$ and $2\pi f_2t+\pi$, with $f_1=50.0\,\mathrm{Hz}$ and $f_2=50.2\,\mathrm{Hz}$. They are opposite in phase at $t=0$.

1. Find the relative phase as a function of time and the first time the two phases align.

2. Explain why a fixed path adjustment cannot keep the cancellation exact indefinitely when the two frequencies differ.

Question 9 - Solution

A small frequency mismatch makes a prepared null temporary

Method and mathematical working

Track relative phase as a function of time. Starting at opposition, the first alignment occurs after a half-cycle of relative phase accumulation.$$\begin{aligned}\Delta\omega&=2\pi(50.2-50)=0.4\pi\,\mathrm{rad/s},\\t_{\mathrm{align}}&=\frac\pi{|\Delta\omega|}=2.5\,\mathrm s.\\[4pt]\Delta\phi(t)&=\pi+2\pi(0.20)t,\qquad\Delta\phi(t)=2\pi\Rightarrow t=2.5\,\mathrm s.\end{aligned}$$

(a) Calculation and conclusion

The relative phase is $\Delta(t)=\pi+2\pi(.2)t$. The first alignment occurs when $\Delta=2\pi$, giving $t=2.50\,\mathrm{s}$. The initial opposite-phase relation is not stationary.

(b) Interpretation and limits

A fixed path adjustment supplies a constant phase offset for each steady frequency; it cannot remove a relative phase rate $2\pi(f_2-f_1)$. Sustained exact cancellation requires matching frequencies as well as amplitudes and arrival phases. An accidental zero instantaneous displacement is weaker than a persistent null.

Physical check

The full relative-phase repeat takes $1/0.2=5$ s. A fixed initial phase can choose the starting point of that cycle but cannot stop drift caused by unequal frequencies.