A particle undergoes SHM governed by the equation $\mathrm{x}={\mathrm{x}}_0\cos \left(\omega \mathrm{t}-\frac{\pi }{4}\right)$. Also, if acceleration of the particle can be written as $\mathrm{a}=\mathrm{Acos}(\omega \mathrm{t}+\delta )$, then

Given. $\mathrm{x}={\mathrm{x}}_0\cos \left(\omega \mathrm{t}-\frac{\pi }{4}\right)$

Velocity, $\nu =\frac{\mathrm{dx}}{\mathrm{dt}}=-{\mathrm{x}}_0\omega \sin \left(\omega t-\frac{\pi }{4}\right)$

Acceleration,

$\begin{array}{l}a=\frac{\mathrm{dv}}{\mathrm{dt}}=-{\mathrm{x}}_0{\omega }^2\cos \left(\omega \mathrm{t}-\frac{\pi }{4}\right)\\ {\mathrm{x}}_0{\omega }^2\cos \left[\pi +\left(\omega \mathrm{t}-\frac{\pi }{4}\right)\right]\\ {\mathrm{x}}_0{\omega }^2\cos \left[\omega t+\frac{3\pi }{4}\right]\end{array}$

Comparing it with acceleration,

$\mathrm{a}=\mathrm{Acos}(\omega \mathrm{t}+\delta ),$ we get

$\mathrm{A}={\mathrm{x}}_0{\omega }^2,\delta =\left(\frac{3\pi }{4}\right)$