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$$^2$$)

Mass, $$ m = 200 \text{ g} = 0.2 \text{ kg} $$

Initial stretch, $$ x_0 = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} $$

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

$$ k x_0 = mg $$

$$ k = \frac{mg}{x_0} $$

$$ k = \frac{0.2 \times 10}{2 \times 10^{-3}} = \frac{2}{2 \times 10^{-3}} = 1000 \text{ N/m} $$

The natural frequency of the spring-mass system is given by the formula:

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

$$ f = \frac{1}{2\pi} \sqrt{\frac{1000}{0.2}} $$

$$ f = \frac{1}{2\pi} \sqrt{5000} $$

$$ f = \frac{50\sqrt{2}}{2\pi} $$

$$ f = \frac{25\sqrt{2}}{\pi} \text{ Hz} $$

The maximum extension of the spring from its natural, unstretched length ($$x_{max}$$) occurs at the lowest point of the oscillation:

$$ x_{max} = x_0 + A $$

$$ x_{max} = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3} \text{ m} $$

The maximum elastic potential energy stored in the spring is:

$$ U_{max} = \frac{1}{2} k (x_{max})^2 $$

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

$$ U_{max} = 500 \times (16 \times 10^{-6}) $$

$$ U_{max} = 8000 \times 10^{-6} $$

$$ U_{max} = 8 \times 10^{-3} \text{ J} $$