Beats — Question 4

Question 4

Initial phase shifts beat times without changing their spacing

Two nearby tones have frequency difference $\Delta f=5.0\,\mathrm{Hz}$ and relative phase $\delta(t)=2\pi\Delta f\,t+\pi/3$. Their received pressure amplitudes are equal.

1. Find the first nonnegative time of maximum envelope amplitude and the first time of zero envelope amplitude.

2. Find the intervals between successive maxima and between successive minima. Explain the role of initial phase.

Question 4 - Solution

Initial phase shifts beat times without changing their spacing

(a) Method and calculation

Constructive alignment requires $\delta=2\pi n$. The smallest nonnegative solution is$$t_{\max}=\frac{2\pi-\pi/3}{2\pi(5)}=\frac16\,\mathrm s.$$Cancellation requires $\delta=(2n+1)\pi$, giving$$t_{\min}=\frac{\pi-\pi/3}{10\pi}=\frac1{15}\,\mathrm s.$$

(b) Interpretation and further calculation

Adding $2\pi$ to either phase condition adds$$T_b=\frac{2\pi}{2\pi\Delta f}=0.200\,\mathrm s.$$Both sequences therefore repeat at $5.0\,\mathrm{Hz}$. Initial phase changes their time origin but not their spacing; the first observed maximum need not occur at emission or at $t=0$.

Physical check

The first minimum precedes the first maximum here by half a beat period, $0.100\,\mathrm s$.