A) 436 to 586 Hz
B) 426 to 574 Hz
C) 436 to 574 Hz
D) 426 to 586 Hz
Correct Answer: A
Solution :
: Velocity of source\[{{v}_{s}}=R\omega \] \[=R\times 2\pi n\] \[=1.2\times 2\times 3.14\times \frac{400}{600}\] = 50 approximately \[\therefore \]Maximum apparent frequency\[=\frac{vn}{(v-{{v}_{s}})}\] \[=\frac{340\times 500}{(340-50)}=\frac{170000}{290}=586\,Hz\] Minimum apparent frequency\[=\frac{vn}{v+{{v}_{s}}}\] \[=\frac{340\times 500}{(340-50)}=\frac{17000}{390}=436\,Hz\] \[\therefore \]Range is 436 to 586 Hz.You need to login to perform this action.
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