Comparing the given equation with standard velocity-displacement equation of simple harmonic motion, i.e., $\text{v}=\omega \sqrt{{\text{A}}^2-{\text{x}}^2}$, we get

$\omega =\text{1 rad/s}$ and $\mathrm{A}=3 \mathrm{m}$

The magnitude of maximum acceleration of the particle in SHM is$={\omega }^2\text{A}$

$={\left(1\right)}^2\left(3\right) {\text{m/s}}^2=3 {\text{m/s}}^2$