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JIPMER Jipmer Medical Solved Paper-2002

  • question_answer
    A sound wave of wavelength \[\lambda \] travels towards the right horizontally with a velocity \[\upsilon ,\]. It strikes and reflects from a vertical plane surface, travelling at a speed \[\upsilon \] towards the left. The number of positive crests striking in a time interval of these second on the wall is:

    A)  \[\frac{(\upsilon -\upsilon )}{3\lambda }\]                          

    B)  \[\frac{3(\upsilon +\upsilon )}{\lambda }\]         

    C)         \[\frac{(\upsilon +\upsilon )}{3\lambda }\]         

    D)         \[\frac{3(\upsilon -\upsilon )}{\lambda }\]

    Correct Answer: B

    Solution :

    We know that the relative velocity of sound waves with respect to the wall is given by \[\upsilon +\upsilon \] so, the apparent frequency of the waves striking. The surface of the wall is \[\frac{(\upsilon +\upsilon )}{\lambda }.\] Therefore, the number of positive crests striking per second is the same as frequency. So, in three seconds the number is \[\frac{(\upsilon +\upsilon )}{\lambda }.\] So, the number of positive crests striking per second is the same as frequency. In three seconds the number is\[\frac{3\,(\upsilon +\upsilon )}{\lambda }\].


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