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Haryana PMT Haryana PMT Solved Paper-2001

  • question_answer
    A body is walking away from a wall towards an observer at a speed of 1 m/s and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observer per second is  (velocity of sound in air = 340 m/s)

    A)  4                                            

    B)  8

    C)  2                                            

    D)  zero

    Correct Answer: A

    Solution :

                    When source and observer are moving close to each other, then apparent frequency \[n=\left( \frac{\upsilon +{{\upsilon }_{o}}}{\upsilon -{{\upsilon }_{s}}} \right)\times n\]              ...(1)    Here, \[{{\upsilon }_{s}}=1m/s,\,{{\upsilon }_{0}}=1m/s,n=680Hz\] So, from equation (1)                 \[n=\left( \frac{340+1}{340-1} \right)680=684\] Number of beats per second                 \[=684-680=4\text{ }beats/sec\]


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