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Question

Two tuning forks with natural frequencies 340 Hz each move relative to stationary observer. One fork moves away from the observer while the other moves towards him at the same speed. The observer hears beats of frequency 3 Hz. Find the speed of the tuning fork (velocity of sound in air is 340 m/s).

A
1.5 m/s
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B
3 m/s
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C
4.5 m/s
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D
6 m/s
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Solution

The correct option is A 1.5 m/s
Given,
Velocity of sound, v=340 m/s
Natural frequency of tuning fork, f0=340 Hz

Let,
f1= frequency observed by observer due to fork 1,
f2= frequency observed by observer due to fork 2,
vs= velocity of tuning fork,


As we know that,
f1=f0(vvvs) and

f2=f0(vv+vs)

According to the problem, the beat frequency,
f1f2=3

Substituting the values we get,
f0(vvvs)f0(vv+vs)=3

⎢ ⎢1(1vsv)11+vsv⎥ ⎥f0=3

[(1vsv)1(1+vsv)1]f0=3

By binomial expansion with neglecting higher terms,

[(1+vsv)(1vsv)]f0=3

2vsf0v=3

vs=3v2f0

Substituting the values, we get

vs=(3)(340)(2)(340)=1.5 m/s

Hence, correct option is (a).

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