A spring stretches by \(2\) mm when it is loaded with a mass of \(200\) g. From equilibrium position the mass is further pulled down by \(2\) mm and released. The frequency associated with the system and maximum energy in the spring are __________ Hz and __________ J, respectively. (Take g \(= 10\) m/s\(^2\))

At equilibrium, the restoring force of the spring balances the weight of the mass:

\(mg = kx_0\)

\(k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000\) N/m

The frequency of oscillation is given by:

\(f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}\)

\(f = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi}\) Hz

The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.

Maximum extension \(x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3}\) m

Maximum energy in the spring \(U_{max} = \dfrac{1}{2} k x_{max}^2\)

\(U_{max} = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2\)

\(U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3}\) J

Answer: \(\dfrac{5\sqrt{50}}{\pi}\) and \(8 \times 10^{-3}\)