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Question

A source of sound attached to the bob of a simple pendulum executes $$S.H.M$$. The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the S.H.M. motion is $$2\%$$ of the natural frequency of the source. The velocity of the source at the mean position is:(velocity of sound in the air is $$340\ m/s$$)
[Assume velocity of sound source $$<<$$ velocity of sound in air]


A
1.4 m/s
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B
3.4 m/s
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C
1.7 m/s
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D
2.1 m/s
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Solution

The correct option is B $$3.4\ m/s$$
From doppler effect formula we know 
apparent frequency of sound as received by an observer during its approach $$f_{max}=f\dfrac{V}{V-V_S}$$   
apparent frequency of sound as received by an observer during its recession $$f_{min}=f\dfrac{V}{V+V_S}$$ 
where $$V$$ is velocity of sound and $$V_S$$ is velocity of source at mean position.
According to question

$$\dfrac{f_{max}-f_{min}}{f}=2\%=1/50$$

$$\dfrac{f\dfrac{V}{V-V_S}-f\dfrac{V}{V+V_S}}{f}=\dfrac{1}{50}$$

$$\dfrac{2V_S}{V}=\dfrac{1}{50}$$

$$V_S=\dfrac{V}{100}$$

$$V_S=\dfrac{340}{100}\, m/s$$

$$V_S=3.4\, m/s$$

Physics

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