Here, m = 50 kg, Maximum extension, y = 20 – 0 = 20 cm = 0.2 m Maximum force, F = mg = 50 × 9.8 = 490 N ∴ Spring constant, $k = \frac{F}{y} = \frac{490}{0.2}$ $= \frac{490 \times 10}{2} = 2450 \mathrm{Nm}^{-1}$ When a body of mass M is suspended from the spring balance, it oscillates with a period of 0.6 s.∵ Time period, $T = 2\pi \sqrt{\frac{M}{k}}$ or $T^{2} = 4\pi^{2} \frac{M}{k}$$\therefore M = \frac{T^{2} k}{4\pi^{2}} = \frac{(0.6)^{2} \times 2450}{4 \times (3.14)^{2}}$M = 22.36 kg∴ Weight of the body,W = Mg = 22.36 × 9.8 = 219.1 N = 22.36 kgf