The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:

$$x = A \sin(\omega t + \phi)$$

Where:

$$A$$ is the amplitude of the SHM,

$$\omega$$ is the angular frequency,

$$t$$ is the time,

$$\phi$$ is the phase constant (phase angle at $$t = 0$$).

In the given equation, $$x = 10 \sin(\omega t + \frac{\pi}{3})$$ m, the amplitude $$A = 10$$ m and the phase constant $$\phi = \frac{\pi}{3}$$. The time period $$T = 3.14$$ s is given, from which we can find the angular frequency $$\omega$$ using the relationship:

$$\omega = \frac{2\pi}{T}$$

Substituting the given $$T = 3.14$$ s:

$$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$$

To find the velocity of the particle, we differentiate the displacement $$x$$ with respect to time $$t$$. The derivative of the displacement gives the velocity:

$$v = \frac{dx}{dt}$$

So, for $$x = 10 \sin(\omega t + \frac{\pi}{3})$$:

$$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$$

Applying differentiation, we get:

$$v = 10\omega \cos(\omega t + \frac{\pi}{3})$$

Plug in the value of $$\omega = 2$$ rad/s and evaluate it at $$t = 0$$ to find the initial velocity:

$$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$$

$$v = 20 \cos(\frac{\pi}{3})$$

$$\cos(\frac{\pi}{3}) = \frac{1}{2}$$, therefore:

$$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$$

Thus, the velocity of the particle at $$t = 0$$ is $$10$$ m/s.