A) 500 Hz
B) 506 Hz
C) 512 Hz
D) 494 Hz
Correct Answer: C
Solution :
\[f=500Hz\] Case 1: When source is moving towards stationary listener apparent frequency \[\eta '=\eta \left( \frac{v}{v-{{v}_{s}}} \right)\] \[=500\left( \frac{340}{336} \right)=506Hz\] Case 2: When source is moving away from the stationary listener \[\eta ''=\eta \left( \frac{v}{v+{{v}_{s}}} \right)=500\left( \frac{340}{344} \right)=494\,Hz\] In case 1 number of beats heard is 6 and in case 2 number of beats heard is 18 therefore frequency of the source at B = 512 HzYou need to login to perform this action.
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