A particle performing $SHM$ is found at its equilibrium at $t=1sec$ and it is found to have a speed of $0.25m/s$ at $t=2s$. If the period of oscillation is $6sec$, calculate amplitude of oscillation.

Since a SHM can be represented by $x=A\sin \left(\omega t+\varphi \right)$

$\Rightarrow x=A\sin \left(\left(\frac{2\pi }{T}\right)t+\varphi \right)$ ; $T=6s$

so the equation becomes
$x=A\sin \left(\frac{\pi t}{3}+\varphi \right)$

We are given that at $t=1s,x=0$; so we have
$$\begin{gathered} 0=A\sin \left(\frac{\pi }{3}+\varphi \right) \\ \Rightarrow \left(\frac{\pi }{3}+\varphi \right)=0 \\ \Rightarrow \varphi =-\frac{\pi }{3} \\ \Rightarrow x=A\sin \left(\frac{\pi t}{3}-\frac{\pi }{3}\right) \\ \Rightarrow v=\frac{dx}{dt}=\frac{d}{dt}\left(A\sin \left(\frac{\pi t}{3}-\frac{\pi }{3}\right)\right)=\frac{\pi A}{3}\cos \left(\frac{\pi t}{3}-\frac{\pi }{3}\right) \end{gathered}$$
Now we are given that at $t=2s,\left|v\right|=0.25m/s$, so we have
$$\begin{gathered} 0.25=\left|\frac{\pi A}{3}\cos \left(\frac{2\pi }{3}-\frac{\pi }{3}\right)\right| \\ \Rightarrow 0.25=\left|\frac{\pi A}{3}\cos \left(\frac{\pi }{3}\right)\right| \\ \Rightarrow 0.25=\frac{\pi A}{6} \\ \Rightarrow A=\frac{3}{2\pi } \end{gathered}$$