A) 6
B) 3
C) 0
D) 12
Correct Answer: A
Solution :
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 \[=\frac{2nv{{v}_{S}}}{{{v}^{2}}-v_{S}^{2}}=\frac{2nv{{v}_{S}}}{(v-{{v}_{S}})(v+{{v}_{S}})}=\frac{2\times 240\times 320\times 4}{(320-4)(320+4)}=6\]You need to login to perform this action.
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