The acceleration-displacement graph of a particle executing $SHM$ is shown in the figure. The time period of the simple harmonic motion 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 .... $rad\, s^{-1}$.

$A$ particle is executing simple harmonic motion with an amplitude of $0.1 \, m$. At a certain instant when its displacement is $0.02 \, m$,its acceleration is $0.5 \, m/s^2$. The maximum velocity of the particle is (in $m/s$):

$A$ 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:

Obtain the formula for the acceleration of a particle from the formula of displacement of $SHM$.

$A$ point mass oscillates along the $x$-axis according to the law $x = x_0 \cos(\omega t + \pi/4)$. If the acceleration of the particle is written as $a = A \cos(\omega t + \delta)$,then: