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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Waves

  • question_answer
    A source of sound attached to the bob of a simple pendulum execute SHM. The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the SHM 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

    Correct Answer: B

    Solution :

    [b] \[{{f}_{1}}={{f}_{0}}\left( \frac{{{V}_{0}}}{{{V}_{0}}-V} \right){{f}_{2}}={{f}_{0}}\left( \frac{{{V}_{0}}}{{{V}_{0}}+V} \right)\] \[{{f}_{1}}-{{f}_{2}}={{f}_{0}}{{V}_{0}}\left( \frac{1}{{{V}_{0}}-V}-\frac{1}{{{V}_{0}}+V} \right)\] \[={{f}_{0}}{{V}_{0}}\left( \frac{{{V}_{0}}+V-{{V}_{0}}+V}{V_{0}^{2}-{{V}^{2}}} \right)\] \[={{f}_{0}}{{V}_{0}}\times \frac{2V}{V_{0}^{2}}={{f}_{0}}\frac{2V}{{{V}_{0}}}\] Given \[\frac{2V{{f}_{0}}}{{{V}_{0}}}=0.02\times {{f}_{0}}\Rightarrow V=0.01\,{{V}_{0}}=3.4\,m/s\]


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