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(A) The apparent frequency $f'$ heard by a stationary observer from a moving source is given by $f' = f \left( \frac{v}{v \mp v_s} \right)$,where $v$

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(A) The apparent frequency $f'$ heard by a stationary observer from a moving source is given by $f' = f \left( \frac{v}{v \mp v_s} ight)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.For source $S_1$ moving towards the observer,the apparent frequency is $f_1' = f \left( \frac{v}{v - V} ight)$.For source $S_2$ moving away from the observer,the apparent frequency is $f_2' = f \left( \frac{v}{v + V} ight)$.The beat frequency is the difference between these two frequencies:$|f_1' - f_2'| = 3$$f \left( \frac{v}{v - V} - \frac{v}{v + V} ight) = 3$Substituting the given values $f = 500 \, Hz$ and $v = 330 \, m/s$:$500 \left( \frac{330(v + V) - 330(v - V)}{v^2 - V^2} ight) = 3$$500 \left( \frac{660V}{330^2 - V^2} ight) = 3$Since $V^2 \ll 330^2$,we can approximate $330^2 - V^2 \approx 330^2$:$500 \left( \frac{660V}{330^2} ight) = 3$$500 \left( \frac{2V}{330} ight) = 3$$V = \frac{3 	imes 330}{1000} = \frac{990}{1000} = 0.99 \, m/s \approx 1 \, m/s$.

Two sources of sound $S_1$ and $S_2$ are moving towards and away from a stationary observer with the same speed $V$ respectively. The observer detects $3$ beats per second. Find the speed of the source (approximately) in $m/s$. Given: $f_1 = f_2 = 500 \, Hz$,speed of sound in air $= 330 \, m/s$.

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