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 $R S$ hears maximum sound intensity.When he moves parallel to $R S$, 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 ............ $Hz$ (the speed of sound is $330 \,m / s$ )
KVPY 2019, Advanced
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(a)

For a minima at $P$, path difference of sounds reaching $P$ must be an odd multiple of half wavelength.

So, $S P-R P=(2 n+1) \frac{\lambda}{2}$

where, $n=0,1,2,3 \ldots$

From above figure,

$S P=\sqrt{(R P)^{2}+(R S)^{2}}$

$S P=\sqrt{12^{2}+5^{2}}=13$

So, path difference,

$S P-R P=13-12=1 \,m$

$Hence, from Eq. (i),$

$1=\frac{(2 n+1) v}{2 f}$

Possible values of frequency of sound are for $n=0$,

$f_{1}=\frac{v}{2}=\frac{330}{2}=165 \,Hz$

For $n=1$

$f_{2}=\frac{3 v}{2}=495 Hz , \ldots \ldots, \text { etc. }$

Hence, option $'a'$ matches with $f_{2}$.

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