$A$ particle is performing $S.H.M.$ with maximum velocity $v$. If the amplitude is tripled and periodic time is doubled,then the new maximum velocity will be:

The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

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

$A$ particle executes $S.H.M.$ and its position varies with time as $x = A \sin \omega t$. Its average speed during its motion from mean position to mid-point of mean and extreme position is

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

The displacement of a particle in $S.H.M.$ is given by $x = A \cos(\omega t + \pi/6)$. At what time will its speed be maximum?