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

If a simple harmonic oscillator has a displacement of $0.02\, m$ and an acceleration equal to $2.0\, m\, s^{-2}$ at any time,the angular frequency of the oscillator is equal to .... $rad\, s^{-1}$.

The displacement of an oscillating particle varies with time $t$ (in seconds) according to the equation $y (cm) = \sin \left[ \frac{\pi}{2} \left( \frac{t}{2} + \frac{1}{3} \right) \right]$. The maximum acceleration of the particle is approximately ..... $cm/s^2$.

$A$ point mass oscillates along the $X$-axis according to the law $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$. If the acceleration of the particle is written as $a=A \cos (\omega t-\delta)$, then

For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

The average acceleration of a particle performing $SHM$ over one complete oscillation is: