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(C) The time period of a mass $m$ attached to a spring of spring constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For the initial mass $m$,$T =

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$\frac{9}{16}$

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(C) The time period of a mass $m$ attached to a spring of spring constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For the initial mass $m$,$T = 2\pi \sqrt{\frac{m}{k}}$.When the mass is increased by $m_0$,the new mass is $m' = m + m_0$. The new time period $T'$ is given as $\frac{5T}{4}$.So,$T' = 2\pi \sqrt{\frac{m + m_0}{k}} = \frac{5}{4} T$.Substituting the expression for $T$: $2\pi \sqrt{\frac{m + m_0}{k}} = \frac{5}{4} 	imes 2\pi \sqrt{\frac{m}{k}}$.Squaring both sides: $\frac{m + m_0}{k} = \frac{25}{16} \frac{m}{k}$.Canceling $k$ from both sides: $m + m_0 = \frac{25}{16} m$.Rearranging for $m_0$: $m_0 = \frac{25}{16} m - m = \frac{9}{16} m$.Therefore,the ratio $\frac{m_0}{m} = \frac{9}{16}$.

$A$ mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released,so that the mass executes $S.H.M.$ with a time period $T$. If the mass is increased by $m_0$,the time period becomes $\frac{5T}{4}$. The ratio $\frac{m_0}{m}$ is

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