When a mass $m$ is attached to a spring, it normally extends by $0.2\, m$. The mass $m$ is given a slight addition extension and released, then its time period will be
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(c) $Kx = mg$
$\Rightarrow \frac{m}{K} = \frac{x}{g}$
So $T = 2\pi \sqrt {\frac{m}{K}} = 2\pi \sqrt {\frac{x}{g}} = 2\pi \sqrt {\frac{{0.2}}{{9.8}}} = \frac{{2\pi }}{7}\sec $
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