Question
Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0m from one of the speakers and 6.4m from the other. If the sound signal is continuously varied from 500Hz to 5000Hz, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air = 320m/s.

Answer

The path difference of the two sound waves is given by$\triangle\text{L}=6.4-6.0=0.4\text{m}$
The wavelength of either wave = $\lambda=\frac{\text{V}}{\rho}=\frac{320}{\rho}(\text{m/s})$ For destructive interference $\triangle\text{L}=\frac{(2\text{n}+1)\lambda}{2}$ where n is an integers. Or $0.4\text{m}=\frac{2\text{n}+1}{2}\times\frac{320}{\rho}$$\Rightarrow\rho=\text{n}=\frac{320}{0.4}=800\frac{2\text{n}+1}{2}\text{Hz}=(2\text{n}+1)400\text{Hz}$
Thus the frequency within the specified range which cause destructive interference are 1200Hz, 2000Hz, 2800Hz, 3600Hz and 4400Hz.

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