$A$ 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:
- A$\frac{V}{2}$
- B$\frac{V}{3}$
- C$6V$
- D$\frac{2V}{3}$
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