Energy in SHM question from JEE Main 2023 (30 Jan, Shift 1)

Answer: 8

1. $U = \frac{1}{2}kA^2\sin^2\omega t$, so $\frac{dU}{dt} = \frac{1}{2}kA^2\omega\sin 2\omega t$.
2. The slope is maximum when $\sin 2\omega t = 1$: $2\omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4\omega} = \frac{T}{8}$.
3. So $\beta = 8$.