Interference — Question 6

Question 6

Three acoustic contributions can cancel without an opposite pair

Three loudspeakers produce equal pressure amplitudes $P$ at a microphone, with received phases $0$, $2\pi/3$ and $4\pi/3$. All have the same frequency. Treat source amplitudes as maintained by their drivers.

1. Find the total pressure phasor.

2. If the third speaker is switched off, find the new amplitude and phase. Why does removing a source increase the measured pressure amplitude?

Question 6 - Solution

Three acoustic contributions can cancel without an opposite pair

(a) Method and calculation

With the real-part convention,$$\widetilde p=P(1+e^{2\pi i/3}+e^{4\pi i/3}).$$$$\operatorname{Re}\widetilde p=P(1-\tfrac12-\tfrac12)=0,\quad\operatorname{Im}\widetilde p=P(\tfrac{\sqrt3}2-\tfrac{\sqrt3}2)=0.$$The three vectors form a closed triangle.

(b) Interpretation and further calculation

After removing the third contribution,$$\widetilde p=P(1+e^{2\pi i/3})=P(\tfrac12+i\tfrac{\sqrt3}2)=Pe^{i\pi/3}.$$The resultant amplitude is $P$ with phase $\pi/3$. Removing a canceling contribution destroys the null; local acoustic pressure is a coherent sum, not a sum of nonnegative amplitudes.

Physical check

A local pressure null does not mean the sources emit no sound or that pressure vanishes everywhere.