Q.A particle executes S.H.M. with amplitude $5\ \text{cm}$ and period $2\ \text{s}$. Find its maximum velocity and maximum acceleration.

Differentiating $x(t)=A\cos(\omega t+\phi)$ gives velocity $v(t)=-A\omega\sin(\omega t+\phi)$ and acceleration $a(t)=-A\omega^2\cos(\omega t+\phi)=-\omega^2x(t)$, restating the defining property of S.H.M. kinematically. Velocity is maximum ($v_{\max}=A\omega$) at the mean position and zero at the extremes; a useful relation is $v=\omega\sqrt{A^2-x^2}$. Acceleration is zero at the mean position and maximum ($a_{\max}=A\omega^2$) at the extremes, where the restoring force is strongest. Velocity leads displacement by a quarter cycle; acceleration is exactly out of phase with displacement.