A particle executes simple harmonic motion. Its amplitude is $$8$$ cm and time period is $$6$$ s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ______ s.

The displacement of a particle in SHM is given by: $$x = A\cos(\omega t)$$ where $$A = 8$$ cm is the amplitude and $$T = 6$$ s is the time period. Since $$\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$$ rad/s.

At $$t = 0$$, the particle is at maximum displacement: $$x_0 = A = 8$$ cm. We need to find the time when $$x = \frac{A}{2} = 4$$ cm.

Setting $$\frac{A}{2} = A\cos(\omega t)$$ gives $$\cos(\omega t) = \frac{1}{2}$$. Therefore, $$\omega t = \frac{\pi}{3}$$ and hence $$t = \frac{\pi}{3\omega} = \frac{\pi}{3 \times \frac{\pi}{3}} = \frac{\pi}{\pi} = 1$$ s.

The answer is $$\boxed{1}$$ s.