A) 1077 Hz
B) 1167 Hz
C) 985 Hz
D) 954 Hz
Correct Answer: A
Solution :
Since, train (source) is moving towards pedestrain (observer), the perceived frequency will be higher than the original. \[f=f\left( \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right)\] Here,\[{{v}_{o}}=0\](as observer is stationary) \[{{v}_{s}}=25\,\,m/s\](velocity of source) \[v=350\,\,m/s\](velocity of sound) and \[f=1kHz\](original frequency) Hence,\[f=1000\left( \frac{350+0}{350-25} \right)\] \[=1000\times \frac{350}{325}\] \[=1077\,\,Hz\]You need to login to perform this action.
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