$A$ 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.

An object is executing simple harmonic motion with an angular frequency $\omega$. If the maximum velocity is $v_{\max}$,then the maximum acceleration of the object is:

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 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

In case of a simple harmonic motion,if the velocity is plotted along the $X$-axis and the displacement (from the equilibrium position) is plotted along the $Y$-axis,the resultant curve is an ellipse with the ratio:
Major axis (along $X$) $= 20 \pi \times$ Minor axis (along $Y$).
What is the frequency of the simple harmonic motion?

Maximum speed of a particle in simple harmonic motion is $v_{max}.$ Then average speed of a particle in one complete oscillation is equal to