A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through…

Maximum kinetic energy in SHM relates to amplitude and angular frequency through the equation $KE_{\text{max}} = \frac{1}{2} m (A\omega)^2$.

Substituting known values $m = 0.2\ \text{kg}$, $A = 0.2\ \text{m}$, and $KE_{\text{max}} = 16 \times 10^{-3}\ \text{J}$:

$16 \times 10^{-3} = \frac{1}{2} (0.2) (0.2\omega)^2$

Solving for $\omega^2$:

$\omega^2 = \frac{16 \times 10^{-3}}{0.1 \times 0.04} = 4$

Thus $\omega = 2\ \text{rad/s}$.

The displacement equation becomes $Y = 0.2 \sin(2t)$, which corresponds to option D.