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Manipal Medical Manipal Medical Solved Paper-2006

  • question_answer
    Velocity of sound waves in air is 330 m/s. For a particular sound in air, a path difference of 40 cm is equivalent to a phase difference of 1.6 tt. The frequency or the wave is:

    A)  165 Hz        

    B)  150 Hz

    C)  660 Hz        

    D)  330 Hz

    Correct Answer: C

    Solution :

     Phase difference of\[1.6\pi \]corresponds to path difference of 40 cm. Hence, phase difference of\[2\pi \]will correspond to a path difference of 50 cm, i.e.,\[K=50\]cm or 0.5 m. \[\therefore \] \[n=\frac{v}{\lambda }=\frac{330}{0.5}\] \[=660\text{ }Hz\]


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