$S_1$ and $S_2$ are two identical sound sources of frequency $656 \ Hz$. The source $S_1$ is located at $O$ and $S_2$ moves anti$-$clockwise with a uniform speed $4 \sqrt{2} \ ms ^{-1}$ on a circular path around $O$, as shown in the figure. There are three points $P, Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$. The source $S_1$ can move along direction $O P.[$Given: The speed of sound in air is $324 \ ms ^{-1} ]$
$(1)$ When only $S_2$ is emitting sound and it is $Q$, the frequency of sound measured by the detector in $Hz$ is $. . . . .$
$(2)$ Consider both sources emitting sound. When $S_2$ is at $R$ and $S_1$ approaches the detector with a speed $4 \ ms ^{-1}$, the beat frequency measured by the detector is $Hz$.
$(1)$ When only $S_2$ is emitting sound and it is $Q$, the frequency of sound measured by the detector in $Hz$ is $. . . . .$
$(2)$ Consider both sources emitting sound. When $S_2$ is at $R$ and $S_1$ approaches the detector with a speed $4 \ ms ^{-1}$, the beat frequency measured by the detector is $Hz$.
IIT 2023, Advanced
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$(1) f ^{\prime}=\frac{ C }{ C + v \cos 45^{\circ}} f$
$=\frac{324}{324+4 \sqrt{2} \times \frac{1}{\sqrt{2}}} \times 656$
$=648 \ Hz $
$(2) f _{ \ce{Pfiom} s _2}=656 \ Hz$
$f _{\text {Pfrom } s_1}=\frac{ C }{ C - V } f $
$=\frac{656 \times 324}{324-4}=664.2$
$\Delta f =664.2-656$
$=8.2 \ Hz $
$=\frac{324}{324+4 \sqrt{2} \times \frac{1}{\sqrt{2}}} \times 656$
$=648 \ Hz $
$(2) f _{ \ce{Pfiom} s _2}=656 \ Hz$
$f _{\text {Pfrom } s_1}=\frac{ C }{ C - V } f $
$=\frac{656 \times 324}{324-4}=664.2$
$\Delta f =664.2-656$
$=8.2 \ Hz $
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