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(C) The displacement of the particle is given by $x = A \cos(\omega t)$,where $A = 3 	imes 10^{-3} 	ext{ m}$ and $\omega = 2 \pi 	ext{ rad/s}$.The

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$1/4 	ext{ s}$

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(C) The displacement of the particle is given by $x = A \cos(\omega t)$,where $A = 3 	imes 10^{-3} 	ext{ m}$ and $\omega = 2 \pi 	ext{ rad/s}$.The velocity $v$ of the particle is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -A \omega \sin(\omega t)$.The speed is $|v| = A \omega |\sin(\omega t)|$.Maximum speed occurs when $|\sin(\omega t)| = 1$,which means $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$.For the first time,$\omega t = \frac{\pi}{2}$.Substituting $\omega = 2 \pi$,we get $2 \pi t = \frac{\pi}{2}$.Solving for $t$,we get $t = \frac{1}{4} 	ext{ s}$.

$A$ mass attached to a spring performs $S.H.M.$ whose displacement is $x = 3 \times 10^{-3} \cos(2 \pi t) \text{ m}$. The time taken to obtain maximum speed for the first time is

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