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.…

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

$$mg = kx_0$$mg=kx0​

$k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000$k=x0​mg​=2×10−30.2×10​=1000 N/m

The frequency of oscillation is given by:

$$f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$$f=2π1​mk​​

$f = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi}$f=2π1​0.21000​​=2π1​5000​=2π1050​​=π550​​ Hz

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

Maximum extension $x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3}$xmax​=x0​+A=2 mm+2 mm=4 mm=4×10−3 m

Maximum energy in the spring $U_{max} = \dfrac{1}{2} k x_{max}^2$Umax​=21​kxmax2​

$$U_{max} = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2$$Umax​=21​×1000×(4×10−3)2

$U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3}$Umax​=500×16×10−6=8000×10−6=8×10−3 J

Answer: $\dfrac{5\sqrt{50}}{\pi}$π550​​ and $8 \times 10^{-3}$8×10−3