Question

Two identical coherent sound sources $R$ and $S$ with frequency $f$are $5\, m$ apart. An observer standing equidistant from the source and at a perpendicular distance of $12\, m$ from the line $RS$ hears maximum sound intensity.
When he moves parallel to $RS$, the sound intensity varies and is a minimum when he comes directly in front of one of the two sources. Then, a possible value of $f$ is close to (the speed of sound is $330 \,m/s$)

• A. 495 Hz
• B. 275 Hz
• C. 660 Hz
• D. 330 Hz

Answer 1

Answer: (A)

For a minima at $P$ path difference of sounds reaching $P$ must be an odd multiple of half wavelength
$ SP-RP=\left(2n+1\right)\frac{\lambda}{2}\ldots\left(i\right)$
where, $n =0,1,2,3\ldots$
from above figure,
$ SP=\sqrt{\left(RP\right)^{2}+\left(RS\right)^{2}}$
$SP=\sqrt{12^{2}+5^{2}=13}$
So,path differenece,
$ SP-RP=13-12=1 \,m$
hence, from Eq.$ \left(i\right)$ $1=\frac{\left(2n+1\right)v}{2f} $
or frequency of sound, $f=\left(\frac{2n+1}{2}\right)v$
Possible values of frequency of sound are for =$o,$
$f_{1} =\frac{v}{2}=\frac{330}{2}=165\, HZ$
For $n=1$
$f_{2}=\frac{3v}{2}=495 \,HZ,$.......,etc.
Hence, option $'a' $ matches with $f_2.$