Energy in SHM question from MHT CET 2026 (14 May, Shift 2)

Answer: (c)

1. Starting from equilibrium: $x = A\sin\omega t$.
2. At $t = T/12$: $\omega t = \frac{2\pi}{T}\cdot\frac{T}{12} = \frac{\pi}{6}$, so $x = \frac{A}{2}$.
3. $\frac{KE}{PE} = \frac{A^2 - x^2}{x^2} = \frac{3A^2/4}{A^2/4} = 3$
4. So the answer is (c) 3 : 1.