Frequency and Maximum Energy of Loaded Vertical Spring System

To find the spring constant $k$, we apply the equilibrium condition $m g = k x_0$ with $m = 0.2\text{ kg}$, $g = 10\text{ m/s}^2$, and $x_0 = 2 \times 10^{-3}\text{ m}$: $k = \frac{m g}{x_0} = \frac{0.2 \times 10}{2 \times 10^{-3}} = 1000\text{ N/m}$

The frequency of the oscillating system is: $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{1000}{0.2}} = \frac{5\sqrt{50}}{\pi}\text{ Hz}$

The maximum displacement of the spring from its natural length occurs at the lowest position, $x_{\max} = x_0 + A = 2\text{ mm} + 2\text{ mm} = 4 \times 10^{-3}\text{ m}$. The maximum energy stored in the spring is: $E_{\max} = \frac{1}{2} k x_{\max}^2 = \frac{1}{2}(1000)(4 \times 10^{-3})^2 = 8 \times 10^{-3}\text{ J}$

Thus, the frequency and maximum energy are $\frac{5\sqrt{50}}{\pi}\text{ Hz}$ and $8 \times 10^{-3}\text{ J}$, respectively, which corresponds to Option A.