  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  B
3.4 m/s  C
1.7 m/s  D
2.1 m/s  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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