The general displacement of a simple harmonic oscillator is $$x = A \sin \omega t$$. Let $$T$$ be its time period. The slope of its potential energy $$(U)$$ - time $$(t)$$ curve will be maximum when $$t = \frac{T}{\beta}$$. The value of $$\beta$$ is ______.

For a SHM oscillator with $$x = A\sin\omega t$$, the potential energy is given by $$U = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 \sin^2\omega t$$.

Differentiating this with respect to time yields $$\frac{dU}{dt} = \frac{1}{2}m\omega^2 A^2 \times 2\sin\omega t \times \omega\cos\omega t = \frac{1}{2}m\omega^3 A^2 \sin 2\omega t$$, which represents the slope of the U-t curve.

The slope $$\frac{dU}{dt}$$ is maximum when $$\sin 2\omega t = 1$$, i.e., when $$2\omega t = \frac{\pi}{2}$$, so that

$$t = \frac{\pi}{4\omega} = \frac{T}{8}$$ (since $$T = \frac{2\pi}{\omega}$$).

Therefore, $$\beta = \boxed{8}$$.