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:

The graph in the figure represents:

Two particles $P$ and $Q$ start from the origin and execute simple harmonic motion along the $X$-axis with the same amplitude but with periods $3 \ s$ and $6 \ s$, respectively. The ratio of the velocities of $P$ and $Q$ when they meet is

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

$A$ body is performing simple harmonic motion with amplitude $A$ and time period $T$. The variation of its acceleration $(f)$ with time $(t)$ is shown in the figure. If at time $t$,the velocity of the body is $v$,which of the following graphs correctly represents the variation of velocity $(v)$ with time $(t)$?