1. For SHM, the displacement is given by $x = A\sin \omega t$

2. The potential energy of the oscillator is $U = \frac{1}{2}kx^2$ Using $k = m\omega^2$ and $x = A\sin\omega t$, $U = \frac{1}{2}m\omega^2A^2\sin^2\omega t$

3. We need the slope of the $U$-$t$ curve, i.e. $\frac{dU}{dt}$

   Differentiate: $\frac{dU}{dt} = \frac{1}{2}m\omega^2A^2 \cdot \frac{d}{dt}(\sin^2\omega t)$

   $\frac{d}{dt}(\sin^2\omega t) = 2\sin\omega t\cos\omega t\cdot \omega$

   Hence, $\frac{dU}{dt} = m\omega^3A^2\sin\omega t\cos\omega t$

   Using $2\sin\theta\cos\theta = \sin 2\theta$, $\frac{dU}{dt} = \frac{1}{2}m\omega^3A^2\sin 2\omega t$

4. The slope is maximum when $\sin 2\omega t = 1$ So, $2\omega t = \frac{\pi}{2} + 2n\pi$

   For the first positive time, $2\omega t = \frac{\pi}{2}$ $\omega t = \frac{\pi}{4}$ $t = \frac{\pi}{4\omega}$

5. Since the time period is $T = \frac{2\pi}{\omega}$ we get $t = \frac{\pi}{4\omega} = \frac{1}{8}\cdot \frac{2\pi}{\omega} = \frac{T}{8}$

6. Comparing with $t = \frac{T}{\beta}$ we get $\beta = 8$

Therefore, the required integer is $8$.