A man is watching two trains, one leaving and the other coming in with equal speeds of $4\, m/sec$. If they sound their whistles, each of frequency $240 Hz$, the number of beats heard by the man (velocity of sound in air $= 320 m/sec$) will be equal to
Diffcult
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(a) Frequency of sound heard by the man from approaching train
${n_a} = n\,\left( {\frac{v}{{v - {v_s}}}} \right) = 240\,\left( {\frac{{320}}{{320 - 4}}} \right) = 243\,Hz$
Frequency of sound heard by the man from receding train
${n_r} = n\,\left( {\frac{v}{{v + {v_s}}}} \right) = 240\,\left( {\frac{{320}}{{320 + 4}}} \right) = 237Hz$
Hence, number of beats heard by man per sec
$ = {n_a} - {n_r} = 243 - 237 = 6$
Short trick : Number of beats heard per sec
$ = \frac{{2nv{v_S}}}{{{v^2} - v_S^2}} = \frac{{2nv{v_S}}}{{(v - {v_S})(v + {v_S})}} = \frac{{2 \times 240 \times 320 \times 4}}{{(320 - 4)(320 + 4)}} = 6$
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