What is the maximum acceleration of the particle doing the SHM $\mathrm{y}=2\sin \left[\frac{\mathrm{\pi t}}{2}+\mathrm{\phi }\right]$ where 2 is in cm

Comparing given equation with standard equation, 
 $\mathrm{y}=A\sin (\mathrm{\omega t}+\mathrm{\phi }),$ we get, $\mathrm{A}=2\mathrm{cm},$ $\mathrm{\omega }=\frac{\mathrm{\pi }}{2}$

$\therefore {\mathrm{a}}_{\max }={\mathrm{\omega }}^2\mathrm{A}$ $={\left(\frac{\mathrm{\pi }}{2}\right)}^2\times 2$ $=\frac{{\mathrm{\pi }}^2}{2}\mathrm{cm}/{\mathrm{s}}^2$