A) 12
B) Zero
C) 6
D) 3
Correct Answer: C
Solution :
Key Idea: The apparent frequency heard by the man depends on the relative motion between sound source and man. The apparent frequency heard by the man is given by \[f{{}_{app}}=\left( \frac{v+{{v}_{0}}}{v\pm {{v}_{s}}} \right)\] where v is velocity of sound, \[{{v}_{0}}\]is velocity of observer (man), and \[{{v}_{s}}\] is velocity of sound source (train) In our case, v^, = 0 \[\therefore \] \[{{f}_{app}}=f\left( \frac{v}{v\pm {{v}_{s}}} \right)\] Given, \[v=300\,m/s,\,{{v}_{s}}=4\,m/s,\,f=240\,Hz\] In first case train is approaching the man, so frequency heard \[{{f}_{1}}=f\left( \frac{v}{v-{{v}_{s}}} \right)\] \[=240\left( \frac{300}{300-4} \right)\] \[=\frac{240\times 300}{296}=243.24\,Hz\] In second case another train is going away from the man, so frequency heard \[{{f}_{2}}=f\left( \frac{v}{v+{{v}_{s}}} \right)\] \[=240\left( \frac{300}{300+4} \right)\] \[=\frac{240\times 300}{304}=236.84\,Hz\] Hence, number of beats heard by the man per second \[={{f}_{1}}-{{f}_{2}}\] \[=243.24-236.84\] = 6.4 \[\approx 6\]
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