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BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    A person is observing two trains one coming towards him and other leaving with the same velocity 4 m/s. If their whistling frequencies are 240 Hz each, then the number of beats per second heard by the person will be (if velocity of sound is 320 m/s):

    A)  3                                            

    B)  6

    C)  9                                            

    D)  zero

    Correct Answer: B

    Solution :

                     Key Idea: The number of beats is the difference of apparent frequencies of two trains. From Doppler?s effect of sound when source is moving towards observer perceived frequency. \[n'=n=\frac{v}{v-{{v}_{s}}}\] where, v is velocity of sound,\[{{v}_{s}}\]is velocity of source.                 \[n'=\frac{320}{320-4}\times 240\]                 \[=\frac{320}{316}\times 240\] \[=243\text{ }Hz\] when source is moving away from observer, perceived frequency                 \[n''=n\frac{v}{v+{{v}_{s}}}\]                 \[=240\times \frac{320}{320+4}\] \[=237\text{ }Hz\] Number of beats per second (beat frequency) \[=n'-n''=\]difference of the frequency of sound source\[\Rightarrow \]\[=243-237=6\]


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