A particle executes simple harmonic motion with an amplitude of $$4$$ cm. At the mean position, velocity of the particle is $$10 \text{ cm s}^{-1}$$. The distance of the particle from the mean position when its speed becomes $$5 \text{ cm s}^{-1}$$ is $$\sqrt{\alpha}$$ cm, where $$\alpha =$$ ______.

In SHM, the velocity at displacement $$x$$ from mean position is:

$$v = \omega\sqrt{A^2 - x^2}$$

At mean position ($$x = 0$$): $$v_{max} = \omega A = 10$$ cm/s, with $$A = 4$$ cm.

So $$\omega = \frac{10}{4} = 2.5$$ rad/s.

When $$v = 5$$ cm/s:

$$5 = 2.5\sqrt{16 - x^2}$$

$$2 = \sqrt{16 - x^2}$$

$$4 = 16 - x^2$$

$$x^2 = 12$$

$$x = \sqrt{12}$$ cm

So $$\alpha = 12$$.

The answer is $$\boxed{12}$$.