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(A) Let the speed of sound be $v = 320 \, m/s$,the speed of the source be $v_s = 4 \, m/s$,and the source frequency be $n = 240 \, Hz$.For the train m

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$6$

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(A) Let the speed of sound be $v = 320 \, m/s$,the speed of the source be $v_s = 4 \, m/s$,and the source frequency be $n = 240 \, Hz$.For the train moving towards the man,the apparent frequency is $n_1 = \frac{v}{v - v_s} \cdot n = \frac{320}{320 - 4} \cdot 240 = \frac{320}{316} \cdot 240 \approx 243.038 \, Hz$.For the train moving away from the man,the apparent frequency is $n_2 = \frac{v}{v + v_s} \cdot n = \frac{320}{320 + 4} \cdot 240 = \frac{320}{324} \cdot 240 \approx 237.037 \, Hz$.The number of beats heard per second is the difference between the apparent frequencies: $n_{beat} = n_1 - n_2$.$n_{beat} = 320 \cdot 240 \cdot \left( \frac{1}{316} - \frac{1}{324} \right) = 76800 \cdot \left( \frac{324 - 316}{316 \cdot 324} \right) = 76800 \cdot \frac{8}{102384} \approx 6 \, Hz$.

$A$ man is standing between two trains moving away from and towards him,both with a speed of $4 \, m/s$. If both trains produce a sound of frequency $240 \, Hz$,calculate the number of beats heard by the man. (Speed of sound in air = $320 \, m/s$)

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