A particle performs S.H.M. of amplitude A and | vedclass

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(C) For a wave represented by $y = A \sin(\omega t - kx)$,the wave velocity is given by $V = \frac{\omega}{k}$.We know that the wave number $k = \frac

Vedclass · Accepted Answer

$
u = \frac{2 \pi A}{\lambda} V$

Vedclass · Answer

(C) For a wave represented by $y = A \sin(\omega t - kx)$,the wave velocity is given by $V = \frac{\omega}{k}$.We know that the wave number $k = \frac{2 \pi}{\lambda}$.Thus,$V = \frac{\omega}{2 \pi / \lambda} = \frac{\omega \lambda}{2 \pi}$.The maximum particle velocity in $S.H.M.$ is given by $
u = A \omega$.From this,we have $\omega = \frac{
u}{A}$.Substituting the value of $\omega$ in the expression for $V$:$V = \frac{(
u / A) \lambda}{2 \pi} = \frac{\lambda 
u}{2 \pi A}$.Rearranging for $
u$:$
u = \frac{2 \pi A}{\lambda} V$.Therefore,the correct relation is $
u = \frac{2 \pi A}{\lambda} V$.

$A$ particle performs $S.H.M.$ of amplitude $A$ and the wave has a wavelength $\lambda$. If $V$ is the wave velocity and $\nu$ is the maximum particle velocity,then they are related as:

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