In Simple Harmonic Motion (SHM), the magnitude of velocity ($v$) and acceleration ($a$) at a displacement $x$ from the mean position are given by: $v = \omega \sqrt{A^2 - x^2}$ $a = \omega^2 x$ where $A$ is the amplitude and $\omega$ is the angular frequency.

Given that the magnitude of velocity equals the magnitude of acceleration ($|v| = |a|$) at $x = 2 \text{ cm}$ and $A = 3 \text{ cm}$: $\omega \sqrt{A^2 - x^2} = \omega^2 x$

Dividing both sides by $\omega$ (since $\omega \neq 0$): $\sqrt{A^2 - x^2} = \omega x$

Squaring both sides: $A^2 - x^2 = \omega^2 x^2$

Substituting the values $A = 3$ and $x = 2$: $3^2 - 2^2 = \omega^2 (2^2)$ $9 - 4 = 4\omega^2$ $5 = 4\omega^2 \implies \omega^2 = \frac{5}{4} \implies \omega = \frac{\sqrt{5}}{2} \text{ rad/s}$

The time period $T$ is given by: $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{5}/2} = \frac{4\pi}{\sqrt{5}} \text{ s}$