$A$ body of mass $5\, g$ is executing $S.H.M.$ about a point with amplitude $10\, cm$. Its maximum velocity is $100\, cm/s$. Its velocity will be $50\, cm/s$ at a distance of:

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

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $12 \ s$. After $2 \ s$,its velocity is $\pi \ m/s$. The amplitude of the oscillation is: $[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3}/2]$

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are $2 \ m/s$ and $4 \ m/s^2$,respectively. Then the angular velocity will be ..... $rad/s$.

$A$ particle is executing a linear simple harmonic motion. Let $V_1$ and $V_2$ be its speeds at distances $x_1$ and $x_2$ from the equilibrium position,respectively. The amplitude of oscillation is

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.