Simple Pendulum question from MHT CET 2021 (23 Sep, Shift 2)

Answer: (d)

1. In SHM, $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ with $\omega = \frac{2\pi}{T}$.
2. At $x = \frac{A}{2}$: $A^2 - x^2 = \frac{3A^2}{4}$.
3. $KE = \frac{1}{2}m\cdot\frac{4\pi^2}{T^2}\cdot\frac{3A^2}{4} = \frac{3m\pi^2A^2}{2T^2}$
4. So the answer is (d) $\frac{3m\pi^2A^2}{2T^2}$.