Q 11-13-021NEETJEE MainMedium
A block hangs from a light spring, which stretches it by $4$ cm at equilibrium. If the block is pulled down slightly and released, its period of oscillation is ($g = \pi^2\ \text{m/s}^2$)
Answer: (C) $0.4$ s
At equilibrium $kx_0 = mg$, so $\dfrac{m}{k} = \dfrac{x_0}{g}$.
$$T = 2\pi\sqrt{\frac{x_0}{g}} = 2\pi\sqrt{\frac{0.04}{\pi^2}} = 2 \times 0.2 = 0.4\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics