A particle is executing the motion $x = A\cos (\omega \,t - \theta )$. The maximum velocity of the particle is
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Velocity of a particle executing SHM is given by,
$v =\omega \sqrt{ A ^2- x ^2}$
For maximum velocity we need $x=0$. So, by putting $x=0$ in equation $(1)$ we have,
$v_{\max }=A \omega$ which is our required answer.
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