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Question

Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0 m/s so that he approaches one tuning fork while receding from the other (figure). Find the beat frequency observed by the listener. Speed of sound in air = 332 m/s.



A

5.2

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B

4.2

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C

4.6

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D

5.6

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Solution

The correct option is A

4.6


Here given velocity of the sources vs = 0

Velocity of the observer v0 = 3 m/s

So, the apparent frequency heard by the man from the tuning fork in front= f=(332+3332)×256= 258.3

the apparent frequency heard by the man from the tuning fork behind him=f=[(3323)332]×256=253.7Hz.

So, beat produced by them = 258.3 - 253.7 = 4.6 Hz


Physics

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