$A$ particle executes harmonic motion with an angular velocity and maximum acceleration of $3.5\, rad/s$ and $7.5\, m/s^2$ respectively. The amplitude of oscillation is .... $m$

$A$ particle executes $SHM$ on a straight line path. The amplitude of oscillation is $2 \, cm$. When the displacement of the particle from the mean position is $1 \, cm$,the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of velocity. The frequency of $SHM$ (in $s^{-1}$) is:

The displacement of a particle varies with time as $x = 12 \sin \omega t - 16 \sin^3 \omega t$ (in $cm$). If its motion is $S.H.M.$,then its maximum acceleration is

For a particle in $SHM$,if the amplitude of the displacement is $a$ and the amplitude of velocity is $v$,the amplitude of acceleration is

$A$ point mass oscillates along the $x$-axis according to $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$. If the acceleration of the point mass is written as $a = A \sin (\omega t + \delta)$,then:

In simple harmonic motion,the ratio of acceleration of the particle to its displacement at any time is a measure of