Physics - SHM Question with Solution | TestHub

The general equation of SHM is

$x=A\sin \left[\omega t+\varphi \right]$, here

$\omega =\frac{2\pi }{16}=\frac{\pi }{8} \mathrm{rad} {\mathrm{s}}^{-1}$

At $t=2 \mathrm{s}$, the particle is at the mean position. Hence,

$0=A\sin \left[\frac{\pi }{8}\times 2+\varphi \right]$

$\Rightarrow \varphi =-\frac{\pi }{4}$

So the equation of SHM will become

$x=A\sin \left[\frac{\pi }{8}t-\frac{\pi }{4}\right]$

$v=\frac{\pi A}{8}\cos \left[\frac{\pi }{8}t-\frac{\pi }{4}\right]$

At $t= 4 \mathrm{s}$

$4=\frac{\pi A}{8}\cos \left[\frac{\pi }{8}\times 4-\frac{\pi }{4}\right]$

$4=\frac{\pi A}{8}\cos \left[\frac{\pi }{4}\right]$

$\Rightarrow A=\frac{32\sqrt{2}}{\pi } \mathrm{m}$