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 = $320\frac{m}{s}$.

• A. 1.2, 2, 2.8, 3.6, 4.4 KHz
• B. 1.2, 3.6 KHz
• C. 1.2, 2.4, 3.6, 4.8 KHz
• D. None of these

Answer 1

The correct option is A

1.2, 2, 2.8, 3.6, 4.4 KHz

The path difference of the two sound waves is given by
$△=6.4-6.0=0.4m$

The wavelength of either wave $=\lambda =\frac{v}{f}=\frac{320}{f}\left(\frac{m}{s}\right)$

For destructive interference
$\Delta =\left(\frac{2n+1}{2}\right)\lambda =\left(\frac{2n+1}{2}\right)\frac{320}{f}$
$\Rightarrow f=\left(\frac{2n+1}{2}\right)\frac{320}{0.4}$
where n is an integer
For $n=0,f=400Hz$
The frequency doesn't fall under the frequency range of source.
Thus for $n=1,2,3,4and5$
Frequency $f=1200Hz,2000Hz,2800Hz,3600Hz,4400Hz$Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz.