question_answer

A and B are two sources generating sound waves. A listener is situated at C. The frequency of the source at A is 500 Hz A, now, moves towards C with a speed 4 m/s. The number of beats heard at C is 6. When A moves away from C with speed 4 m/s, the number of beats heard at C is 18. The speed of sound is 340 m/s. The frequency of the source at B is:                     JEE Main  Online Paper (Held On 22 April 2013)

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 Hz