1. For a particle in simple harmonic motion, the maximum speed is

$v_{\max}=\omega A$

where:

• $A = 0.06\,\text{m}$
• $T = 3.14\,\text{s}$
• $\omega = \dfrac{2\pi}{T}$

2. Compute angular frequency:

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

3. Now calculate maximum velocity:

$v_{\max} = \omega A = 2 \times 0.06 = 0.12\,\text{m/s}$

4. Convert to cm/s:

$0.12\,\text{m/s} = 12\,\text{cm/s}$

Therefore, the maximum velocity is

$\boxed{12\,\text{cm/s}}$