Energy in SHM question from MHT CET 2026 (15 Apr, Shift 1)

Answer: (a)

1. $K = \frac{1}{2}m\omega^2(A^2 - x^2)$, so $K = 50\%$ of $E$ when $x^2 = \frac{A^2}{2}$, i.e. $x = \frac{A}{\sqrt{2}}$.
2. From mean position $x = A\sin\omega t \Rightarrow \sin\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$.
3. $t = \frac{T}{8} = \frac{6}{8} = 0.75$ s
4. So the answer is (a) $0.75$ s.