question_answer

A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5m/s. The speed of sound is 330 m/s. If the observer is between the wall and the source, then beats per second heard will be            [UPSEAT 2002]

A)            7.8 Hz                                      

B)            7.7 Hz

C)            3.9 Hz                                      

D)            Zero

Correct Answer: A

Solution :

           The observer will hear two sound, one directly from source and other from reflected image of sound Hence number of beats heard per second                        =\[\left( \frac{v}{v-{{v}_{S}}} \right)\,n-\left( \frac{v}{v+{{v}_{S}}} \right)\,n\]                        = \[\frac{2nv{{v}_{S}}}{{{v}^{2}}-v_{S}^{2}}=\frac{2\times 256\times 330\times 5}{335\times 325}\]= 7.8\[Hz\]