If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

$A$ particle is executing the motion $x = A \cos (\omega t - \theta)$. The maximum velocity of the particle is

The force ($F$ in newton) acting on a particle of mass $90 \text{ g}$ executing simple harmonic motion is given by $F + 0.04 \pi^2 y = 0$, where $y$ is the displacement of the particle in meters. If the amplitude of the particle is $\frac{6}{\pi} \text{ m}$, then the maximum velocity of the particle is: (in $\text{ m/s}$)

The instantaneous displacement of a particle in $S.H.M.$ is $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. The time at which the velocity is maximum for the first time is

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

For a simple harmonic motion with amplitude $A$ and time period $T$,what is the velocity at $x = \frac{A}{2}$?