A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $$\frac{1}{a}$$ s. The value of $$a$$ to the nearest integer is ________.

For a particle in simple harmonic motion with period $$T = 2 \text{ s}$$, the displacement from the mean position is given by $$x = A\sin(\omega t)$$, where $$\omega = \dfrac{2\pi}{T} = \pi \text{ rad/s}$$.

We need the time to reach displacement $$x = \dfrac{A}{2}$$ from the mean position. Setting $$\dfrac{A}{2} = A\sin(\pi t)$$, we get $$\sin(\pi t) = \dfrac{1}{2}$$, so $$\pi t = \dfrac{\pi}{6}$$, giving $$t = \dfrac{1}{6} \text{ s}$$.

Since the time is given as $$\dfrac{1}{a}$$ seconds, we have $$a = 6$$.

The answer is $$\boxed{6}$$.