To solve this problem, we need to use the equations of motion for simple harmonic motion (SHM).$\\$Given:$\\$• Amplitude $A = 3$ cm$\\$• Displacement from mean position $x = 2$ cm$\\$In SHM, the velocity $v$ and acceleration $a$ are given by:$\\ v = \omega \sqrt{A^2 - x^2} \\ a = \omega^2 x \\$According to the problem, the magnitude of velocity is equal to the magnitude of acceleration.$\\$Thus, we have:$\\ \omega \sqrt{A^2 - x^2} = \omega^2 x \\$Substitute the given values:$\\ \omega \sqrt{3^2 - 2^2} = \omega^2 \cdot 2 \\$Simplify:$\\ \omega \sqrt{9 - 4} = 2 \omega^2 \\ \omega \sqrt{5} = 2 \omega^2 \\$Divide both sides by $\omega: \\ \sqrt{5} = 2 \omega \\$Solve for $\omega: \\ \omega = \frac{\sqrt{5}}{2} \\$The angular frequency $\omega$ is related to the time period $T$ by the equation:$\\ \omega = \frac{2\pi}{T} \\$Substitute the value of $\omega: \\ \frac{\sqrt{5}}{2} = \frac{2\pi}{T} \\$Rearrange to solve for $T: \\ T = \frac{4\pi}{\sqrt{5}} \\$Therefore, the time period of the particle is $\frac{4\pi}{\sqrt{5}}$ seconds. This corresponds to Option 2.