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AMU Medical AMU Solved Paper-1999

  • question_answer
    A whistle of frequency 500 Hz tied to the end of a string of length 1.2 m revolves at 400 revolutions per minute. If the velocity of sound is 340 m/sec, a listener standing some distance away in the plane of rotation of the whistle hears the frequencies in the range

    A)  436 to 586 Hz   

    B)  426 to 574 Hz

    C)  436 to 574 Hz   

    D)  426 to 586 Hz

    Correct Answer: A

    Solution :

    : Velocity of source\[{{v}_{s}}=R\omega \] \[=R\times 2\pi n\] \[=1.2\times 2\times 3.14\times \frac{400}{600}\] = 50 approximately \[\therefore \]Maximum apparent frequency\[=\frac{vn}{(v-{{v}_{s}})}\] \[=\frac{340\times 500}{(340-50)}=\frac{170000}{290}=586\,Hz\] Minimum apparent frequency\[=\frac{vn}{v+{{v}_{s}}}\] \[=\frac{340\times 500}{(340-50)}=\frac{17000}{390}=436\,Hz\] \[\therefore \]Range is 436 to 586 Hz.


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