$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

The maximum velocity of a particle,executing $S$.$H$.$M$. with an amplitude $7 \ mm$ is $4.4 \ ms^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$. (in $s$)

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

For a simple harmonic motion with amplitude $A$ and time period $T$,at what position is the velocity half of the maximum velocity?

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is

$A$ particle executes $S.H.M.$ starting from the mean position. Its amplitude is '$a$' and its periodic time is '$T$'. At a certain instant,its speed '$u$' is half that of maximum speed $V_{\text{max}}$. The displacement of the particle at that instant is