A mass of 0 .2 k g is attached to the lower end of a massless spring of force constant 200 Nm^-1 , the upper end of which is fixed to a rigid support. Study the following statements. 

If in equilibrium position, elongation of the spring is equal to ${\mathrm{x}}_0,$ then ${\mathrm{Kx}}_0=\mathrm{mg}\text{ or }{\mathrm{x}}_0=\frac{\mathrm{mg}}{\mathrm{K}}=1\mathrm{cm}$ 

If the block is raised till spring becomes unstretched and then released, then
during subsequent motion maximum elongation of the spring ( y ) will be
calculated by energy conservation law.

Loss of PE of block $\left(\mathrm{mgy}\right)=\mathrm{strain}\mathrm{energ}\mathrm{y}\left(\frac{1}{2}{\mathrm{Ky}}^2\right)$

$\mathrm{y}=\frac{2\mathrm{mg}}{\mathrm{K}}=2\mathrm{cm}$  Hence option ( B ) is correct.

Frequency of oscillations will be, $\mathrm{f}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}=5\mathrm{Hz}$

Hence option ( C ) is correct. Since frequency, f , does not depend upon
gravitational acceleration, therefore frequency will remain unchanged, even if the
system is taken to moon. Hence option ( D ) is also correct.