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

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

EAMCETEAMCET 1998

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} $