Physics - SHM Question with Solution | TestHub

From the figure, it can be concluded that

$\begin{array}{rcl}A& =& 1\\ \omega & =& \frac{2\pi }{8}\\ & =& \frac{\pi }{4}\end{array}$

Hence, the equation of motion of the particle is given by

$\begin{array}{rcl}x& =& A\sin \omega t\\ & =& \sin \frac{\pi }{4}t ...\left(1\right)\end{array}$

The velocity of the particle is, then, given by

$\begin{array}{rcl}v& =& \frac{dx}{dt}\\ & =& \frac{d}{dt}\left(\sin \frac{\pi }{4}t\right)\\ & =& \frac{\pi }{4}\cos \frac{\pi }{4}t ...\left(2\right)\end{array}$

Hence, the acceleration of the particle can be written as

$\begin{array}{rcl}a& =& \frac{dv}{dt}\\ & =& \frac{d}{dt}\left(\frac{\pi }{4}\cos \frac{\pi }{4}t\right)\\ & =& -\frac{{\pi }^2}{16}\sin \frac{\pi }{4}t ...\left(3\right)\end{array}$

Substitute $2$ for $t$ into equation (3) to calculate the required value of the acceleration.

$\begin{array}{rcl}a& =& -\frac{{\pi }^2}{16}\sin \left(\frac{\pi }{4}\times 2\right)\\ & =& -\frac{{\pi }^2}{16} {\mathrm{ms}}^{-2}\end{array}$