The displacement of a particle executing SHM is given by $$x = 10 \sin\left(\omega t + \frac{\pi}{3}\right)$$ m. The time period of motion is $$3.14$$ s. The velocity of the particle at $$t = 0$$ is _____ m/s.

Given $$x = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$ m with $$T = 3.14$$ s, we want to find the velocity at $$t = 0$$. First, we find the angular frequency $$\omega$$ by using $$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} = \frac{2 \times 3.14}{3.14} = 2 \text{ rad/s}$$.

The velocity is the time derivative of the displacement, so we have $$v = \frac{dx}{dt} = 10\omega\cos\left(\omega t + \frac{\pi}{3}\right)$$.

Evaluating this expression at $$t = 0$$ gives $$v(0) = 10 \times 2 \times \cos\frac{\pi}{3} = 20 \times \frac{1}{2} = 10 \text{ m/s}$$.

The correct answer is 10 m/s.