Time period of a simple pendulum is $$T$$. The time taken to complete $$\frac{5}{8}$$ oscillations starting from mean position is $$\frac{\alpha}{12}T$$. The value of $$\alpha$$ is ______.

A simple pendulum starts from the mean position. We need to find the time taken to complete $$\frac{5}{8}$$ oscillations.

Starting from the mean position, the particle moves as: mean $$\to$$ positive extreme $$\to$$ mean $$\to$$ negative extreme $$\to$$ mean, completing one full oscillation in time $$T$$. In one complete oscillation, the total distance traversed is $$4A$$ (where $$A$$ is the amplitude).

Now, $$\frac{5}{8}$$ of an oscillation corresponds to $$\frac{5}{8} \times 4A = \frac{5A}{2}$$ of total path distance from the starting point.

Breaking this down: the first $$2A$$ of path (mean $$\to$$ +A $$\to$$ mean) takes time $$\frac{T}{2}$$. The remaining path distance is $$\frac{5A}{2} - 2A = \frac{A}{2}$$, which means the particle travels from the mean position toward the negative extreme and covers a displacement of magnitude $$\frac{A}{2}$$.

For motion from the mean position, $$y = A\sin(\omega t')$$, where $$t'$$ is the time measured from when the particle was at the mean position. Setting $$|y| = \frac{A}{2}$$, we get $$\sin(\omega t') = \frac{1}{2}$$, so $$\omega t' = \frac{\pi}{6}$$.

This gives $$t' = \frac{\pi}{6\omega} = \frac{\pi}{6 \cdot \frac{2\pi}{T}} = \frac{T}{12}$$.

The total time is $$\frac{T}{2} + \frac{T}{12} = \frac{6T + T}{12} = \frac{7T}{12} = \frac{\alpha T}{12}$$.

Therefore, $$\alpha = 7$$.