The variation of the acceleration $a$ of a particle executing $S.H.M.$ with displacement $x$ is as shown in the figure. Which of the following graphs correctly represents this relationship?

$A$ 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:

Obtain instantaneous acceleration of a $SHM$ particle with the help of reference circle.

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,

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

$A$ 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: