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

The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

$A$ particle executes simple harmonic motion with an amplitude of $4 \ cm$. At the mean position,the velocity of the particle is $10 \ cm/s$. The distance of the particle from the mean position when its speed becomes $5 \ cm/s$ is $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

$A$ particle executes linear simple harmonic motion with an amplitude of $3\,cm$. When the particle is at $2\,cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds 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 . . . . . .

$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.