Two sources of sound $S_1$ અને $S_2$ are moving towards and away from a stationary observer with the same speed respectively. Observer detects $3$ beats per second. Find speed of source (approximately). (in $m/s$)
Given, $F 1= F 2=500\, Hz$. speed of air $=330\, m / s$
AIIMS 2019, Diffcult
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The expression for beats is given by,
$\left(F_{1}-F_{2}\right)=\left(\frac{330}{330-V}\right) \times 500-\left(\frac{330}{330+V}\right) \times 500$
Substitute $3$ for $\left(F_{1}-F_{2}\right)$
$3=\left(\frac{330}{330-V}\right) \times 500-\left(\frac{330}{330+V}\right) \times 500$
$=\frac{500}{330^{2}-V^{2}} \times 2 V \times 330$
Here $V^{2}<<330^{2}$ so,
$30=\frac{500 \times 330 \times 2 V}{330^{2}}$
$V=\frac{3}{2}\left(\frac{330}{500}\right)$
$\approx 1 m / s$
Further simplify the above
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(Speed of sound in air is $330 \,\, m s^{-1}$)