$A$ 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:

$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

$A$ particle executes $S.H.M.$ with a period of $6 \, s$ and amplitude of $3 \, cm$. Its maximum speed in $cm/s$ is

$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

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 executes simple harmonic motion with a time period $0.6 \,s$ and amplitude $10 \,cm$. Find the mean velocity of the particle over the time interval during which it travels a distance $5 \,cm$ starting from the equilibrium position.