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(B) At equilibrium,the gravitational force is balanced by the spring force: $mg = Kx_0$,where $x_0 = 0.2 \ m$ is the extension.From this,we get the ra

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$\frac{2}{\sqrt{5}} \ s$

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(B) At equilibrium,the gravitational force is balanced by the spring force: $mg = Kx_0$,where $x_0 = 0.2 \ m$ is the extension.From this,we get the ratio $\frac{m}{K} = \frac{x_0}{g}$.Given $g = \pi^2$,we substitute the values: $\frac{m}{K} = \frac{0.2}{\pi^2}$.The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K}}$.Substituting the ratio: $T = 2\pi \sqrt{\frac{0.2}{\pi^2}}$.$T = 2\pi \cdot \frac{\sqrt{0.2}}{\pi} = 2\sqrt{0.2}$.$T = 2\sqrt{\frac{2}{10}} = 2\sqrt{\frac{1}{5}} = \frac{2}{\sqrt{5}} \ s$.

When a mass $m$ is attached to a spring,it extends by $0.2 \ m$. The mass $m$ is given a slight additional extension and released. Find the time period (take $g = \pi^2 \ m/s^2$).

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