$A$ simple pendulum performs $S.H.M$ about $x = 0$ with an amplitude $A$ and time-period $T$. The speed of the pendulum at $x = \frac{A}{2}$ will be

$A$ 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:

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \,m \,s^{-1}$ and $10 \,m \,s^{-2}$, then the time period of oscillation of the particle is

The position,velocity,and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \ m$,$2 \ ms^{-1}$,and $16 \ ms^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x} \ m$,where $x$ is . . . . . .

The kinetic energy of a simple harmonic oscillator is oscillating with an angular frequency of $176 \ rad/s$. The frequency of this simple harmonic oscillator is . . . . . . $Hz$. $\left[\text{take } \pi=\frac{22}{7}\right]$

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