Question

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air = 320 m s⁻¹.

Answer 1

Given:
Speed of sound in air v = 320 ms⁻¹
The path difference of the sound waves coming from the loudspeaker and reaching the person is given by:
Δx = 6.4 m − 6.0 m = 0.4 m
If $\left(f\right)$ is the frequency of either wave, then the wavelength of either wave will be:
$\mathrm{\lambda }=\frac{v}{f}=\frac{320}{f}$
For destructive interference, the path difference of the two sound waves reaching the listener should be an odd integral multiple of half of the wavelength.
$\therefore ∆x=(2n+1)\frac{\lambda }{2}$ , where n is an integer.
On substituting the respective values, we get:
$$\begin{gathered} 0.4\mathrm{m}=\left(2\mathrm{n}+1\right)\times \frac{320}{2f} \\ \Rightarrow f=\left(2n+1\right)\frac{320}{2\times 0.4} \\ \Rightarrow f=(2n+1)400\mathrm{Hz} \end{gathered}$$

Thus, on applying the different values of n, we find that the frequencies within the specified range that caused destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.