A body of mass 2 kg hangs from a spring and oscillates with a time period of $\pi$ seconds. If the b

Question

A body of mass 2 kg hangs from a spring and oscillates with a time period of $\pi$ seconds. If the ball is removed, the length of the spring will decrease by

Professor PC Thomas Classes · Answer

(a) : Given Mass $m = 2 \mathrm{~kg}$ and time period $T = \pi \mathrm{~s}$. The formula for time period is $T = 2\pi\sqrt{\frac{m}{k}}$. Substituting the values: $\pi = 2\pi\sqrt{\frac{2}{k}} \Rightarrow \frac{1}{2} = \sqrt{\frac{2}{k}} \Rightarrow \frac{1}{4} = \frac{2}{k} \Rightarrow k = 8 \mathrm{~N/m}$. The force exerted by the mass is $F = mg = 2g$. By Hooke's Law, extension $l = \frac{F}{k} = \frac{2g}{8} = \frac{g}{4}$. Thus, the length will decrease by $g/4$ metres when the mass is removed.