Given displacement equation

The SHM is: $x = 10\sin\left(\omega t + \frac{\pi}{3}\right) \text{ m}$

So, amplitude is $A = 10\text{ m}$

Given time period

$T = 3.14\text{ s}$

We know: $\omega = \frac{2\pi}{T}$

Substituting: $\omega = \frac{2\pi}{3.14}$

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

Find velocity equation

Velocity is the derivative of displacement: $v = \frac{dx}{dt} = 10\omega \cos\left(\omega t + \frac{\pi}{3}\right)$

At $t=0$

$v(0) = 10\omega \cos\left(\frac{\pi}{3}\right)$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

Therefore, $v(0) = 10\omega \cdot \frac{1}{2} = 5\omega$

Using $\omega = 2$: $v(0) = 5 \times 2 = 10\text{ m/s}$

Final answer

$\boxed{10}$