b
$y = 2\,\sin \,\left( {\frac{{\pi t}}{2} + \phi } \right)$
velocity of particle $\frac{{dy}}{{dt}} = 2 \times \frac{\pi }{2}\,\cos \,\left( {\frac{{\pi t}}{2} + \phi } \right)$
acceleration $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}}=-\frac{\pi^{2}}{2} \sin \left(\frac{\pi \mathrm{t}}{2}+\phi\right)$
Thus $a_{\max }=\frac{\pi^{2}}{2}$