Velocity & Acceleration in SHM question from MHT CET 2019 (2 May, Shift 2)

Answer: (a)

1. At $x = \frac{A}{2}$: $v = \omega\sqrt{A^2 - \frac{A^2}{4}} = \frac{\sqrt{3}}{2}\omega A$ and $a = \omega^2\frac{A}{2}$.
2. $\frac{\sqrt{3}}{2}\omega A = \frac{\omega^2 A}{2} \Rightarrow \omega = \sqrt{3}$ rad/s.
3. $T = \frac{2\pi}{\sqrt{3}}$ s.
4. So the answer is (a) $\frac{2\pi}{\sqrt{3}}$ s.