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EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    In a transverse progressive waves of amplitude A, the maximum particle velocity is four times wave velocity. Then the wave-length of wave is:

    A)  \[\pi \frac{A}{4}\]                                          

    B)  \[\pi \frac{A}{2}\]

    C)  \[\pi A\]                             

    D)  \[2\pi A\]

    Correct Answer: B

    Solution :

                     The maximum velocity of a wave is given by \[{{v}_{\max }}=A\omega \]       [Given: \[{{v}_{\max }}=4v\] ] \[\therefore \]  \[4v=A\omega \]                 \[4n\lambda =A.2\pi n\]                 when \[n=\]frequency                 \[\therefore \]  \[\lambda =\frac{A\pi }{2}\]


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