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

In a simple harmonic motion,let $f$ be the acceleration and $T$ be the time period. If $x$ denotes the displacement,then the $|fT|$ vs. $x$ graph will look like,

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 variation of displacement with time of a simple harmonic motion $(SHM)$ for a particle of mass $m$ is represented by $y = 2 \sin \left(\frac{\pi t}{2} + \phi\right) \text{ cm}$. The maximum acceleration of the particle is

$A$ 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:

$A$ particle is performing $S.H.M.$ starting from the extreme position. The graphical representation shows that between displacement and acceleration,there is a phase difference of: