$A$ particle executes simple harmonic motion with a periodic time of $0.05 \, s$ and an amplitude of $4 \, cm$. What is its maximum velocity?

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

$A$ particle performing $SHM$ has a time period of $\frac{2 \pi}{\sqrt{3}} \,s$ and a path length of $4 \,cm$. The displacement from the mean position at which the magnitude of acceleration is equal to the magnitude of velocity is (in $\,cm$)

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?

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 . . . . . .

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