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(C) The equilibrium condition for the spring-mass system is given by $Kx = mg$,where $K$ is the spring constant,$x$ is the extension,and $g$ is the ac

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$\frac{2\pi}{7}\, s$

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(C) The equilibrium condition for the spring-mass system is given by $Kx = mg$,where $K$ is the spring constant,$x$ is the extension,and $g$ is the acceleration due to gravity.From this,we get the ratio $\frac{m}{K} = \frac{x}{g}$.The time period $T$ of a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{K}}$.Substituting $\frac{m}{K} = \frac{x}{g}$ into the formula,we get $T = 2\pi \sqrt{\frac{x}{g}}$.Given $x = 0.2\, m$ and taking $g = 9.8\, m/s^2$,we have $T = 2\pi \sqrt{\frac{0.2}{9.8}}$.Simplifying the fraction,$\frac{0.2}{9.8} = \frac{2}{98} = \frac{1}{49}$.Therefore,$T = 2\pi \sqrt{\frac{1}{49}} = 2\pi 	imes \frac{1}{7} = \frac{2\pi}{7}\, s$.

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

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