Question

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place $6.0m$ one of the speakers and $6.4m$ from the other. If the second signal is continously 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 on air $=320m/s$

Answer 1

The path difference of the two sound waves is given by
$\Delta L=6.4-6.0=0.4m$
The wavelength of either wave $=\lambda =\frac{V}{f}=\frac{320}{f}\left(m/s\right)$

For destructive interference $\Delta L=\frac{(2n+1)\lambda }{2}$ where $n$ is an integers.

or $0.4m=\frac{2n+1}{2}\times \frac{320}{f}$

$\Rightarrow f=\frac{320}{0.4}\frac{2n+1}{2}Hz=800\frac{2n+1}{2}Hz=(2n+1)400Hz$

Thus the frequency with the specified range which cause destructive interference are $1200Hz,2000Hz,2800Hz,3600Hz$ and $4400Hz$.