A particle executes simple harmonic motion with a time period of 16s. At time t = 2s, the particle crosses the mean position while at t = 4s, its velocity is $4 {\mathrm{ms}}^{-1}$. The amplitude of motion in meter is

$\begin{array}{l}\text{ At t}=2\text{sec},\text{ the particle crosses mean position. }\\ \text{ At t}=4\text{sec , its velocity is }4 {\text{ms}}^{-1}\\ \text{ For simple harmonic motion, y}=\text{a}\sin \left(\omega \text{t+}\varphi \right)\\ \therefore y=a\sin \left(\left(\frac{2\pi }{T}\right)t\text{+}\varphi \right)\\ \Rightarrow 0=a\sin [\left(\frac{2\pi }{16}\right)\times 2\text{+}\varphi ]\\ \Rightarrow \frac{\pi }{4}+\varphi =0\\ \Rightarrow \varphi =-\frac{\pi }{4}\\ y\text{ coordinate at t=4s is}\\ y=a\sin (\left(\frac{2\pi }{16}\right)\text{4}-\frac{\pi }{4})=\frac{a}{\sqrt{2}}\\ Now,v=\omega \sqrt{a^2-y^2}\\ \Rightarrow 4=\frac{2\pi }{16}\sqrt{a^2-\frac{a^2}{2}}\\ \Rightarrow a=\frac{32\sqrt{2}}{\pi }\\ \end{array}$