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(D) The maximum velocity of a particle in $S.H.M.$ is given by $V = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.We know

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$6V$

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(D) The maximum velocity of a particle in $S.H.M.$ is given by $V = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.We know that $\omega = \frac{2\pi}{T}$,where $T$ is the periodic time.Given that the new periodic time $T' = \frac{1}{3}T$ and the new amplitude $A' = 2A$.The new angular frequency is $\omega' = \frac{2\pi}{T'} = \frac{2\pi}{\frac{1}{3}T} = 3 \left(\frac{2\pi}{T}ight) = 3\omega$.The new maximum velocity $V'$ is given by $V' = A'\omega'$.Substituting the values,$V' = (2A) 	imes (3\omega) = 6(A\omega) = 6V$.

The maximum velocity of a particle performing $S.H.M.$ is $V$. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ of its original value and the amplitude is doubled,then the new maximum velocity of the particle will be:

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