The correct option of this question will be (C).
Solution —
If $A$ is the amplitude of simple harmonic motion, then
$v=A \omega$
But $\omega=\frac{2 \pi}{T}$ where $T$ is the time period of oscillation.
$\therefore \quad v=A \frac{2 \pi}{T}$
When the amplitude is doubled and the time period decreased to $\frac{T}{3}$, the maximum velocity becomes
$v^{\prime}=2 A\left(\frac{2 \pi}{T / 3}\right)=6\left(A \frac{2 \pi}{T}\right)=6 v$
(using (i))