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 .... $rad\, s^{-1}$.

$A$ $0.10\, kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10\,cm)\cos [(10\,rad/s)\,t + \pi /2\,rad]$. What is the maximum acceleration experienced by the block?

$A$ 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

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = 4/3 \,s$ is

$A$ simple harmonic oscillator has an amplitude of $0.5 \ m$ and a time period of $2 \ s$. What is the magnitude of acceleration when it is displaced from the mean position by $0.25 \ m$?

$A$ particle of mass $0.3 \, kg$ is subjected to a force $F = -kx$ with $k = 15 \, N/m$. What will be its initial acceleration if it is released from a point $20 \, cm$ away from the origin? (in $m/s^2$)