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

Frequency of sound heard by the man from approaching trainna=n vv−vs=240 320320−4=243 Hz  Frequency of sound heard by the man from receding train nr=n vv+vs=240 320320+4=237HzHence, number of beats heard by man per sec   =na−nr=243−237=6Short trick : Number of beats heard per sec   =2nvvSv2−vS2=2nvvS(v−vS)(v+vS)=2×240×320×4(320−4)(320+4)=6