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(C) Given:- The initial maximum velocity is $V = A\omega$.- The amplitude doubles: $A' = 2A$.- The period becomes one-third: $T' = \frac{T}{3}$.Step $

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$6V$

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(C) Given:- The initial maximum velocity is $V = A\omega$.- The amplitude doubles: $A' = 2A$.- The period becomes one-third: $T' = \frac{T}{3}$.Step $1$: Calculate the new angular frequency $\omega'$.Since $\omega = \frac{2\pi}{T}$,the new angular frequency is:$\omega' = \frac{2\pi}{T'} = \frac{2\pi}{T/3} = 3 \left(\frac{2\pi}{T}\right) = 3\omega$.Step $2$: Calculate the new maximum velocity $V'$.The formula for maximum velocity is $V_{\max} = A\omega$.$V' = A' \omega' = (2A)(3\omega) = 6(A\omega)$.Since $V = A\omega$,we get:$V' = 6V$.

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

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