A sound source is releasing a signal of constant frequency 330 Hz. f& λare frequency and wavelength observed by an observer. The velocity of sound in air is 330 ms−1.The observer and source are stationary, wind isblowing towards source from observer.The source is stationary, wind is blowing from source to observer and observer is moving away from the source.The source is stationary and observer is moving towards source.

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P Q R S 3 2 4 1

P Q R S 3 1 4 3

P Q R S 3 2 4 2

P Q R S 3 none 4 1

answer is A.

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Detailed Solution

λ=vf=330330=1mP : As source and observer are moving towards each other f′=f0u+u0u-usFrequency increases and wavelength decreases.Q : As source and observer are at rest vO=0 and vS=0 wind is blowing from source to observer moving away from source= f′=f0u+uw+u0u+uw-us=f0Frequency remain same and wavelength also remain same.R : Observer is moving towards stationary source= f′=f0u+uw-u0u+uwFrequency decreases and wavelength remain same.S : f′=f0u+u0uFrequency increases and wavelength remain same.

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