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))