Velocity & Acceleration in SHM question from JEE Main 2017 (8 Apr, Shift 1)

Answer: (b)

1. $\frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} = \omega = 10$ rad/s.
2. $x = A\sin(\omega t + \frac{\pi}{4})$; at $t = 0$: $5 = \frac{A}{\sqrt{2}} \Rightarrow A = 5\sqrt{2}$ m.
3. $a_{max} = \omega^2 A = 100\times 5\sqrt{2} = 500\sqrt{2}$ m/s$^2$.
4. So the answer is (b) $500\sqrt{2}$ m/s$^2$.