A mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period T . If the mass is increased by $\mathrm{m}_0$, the periodic time becomes $\frac{5 \mathrm{~T}}{4}$. The ratio $\frac{\mathrm{m}_0}{\mathrm{M}}$ is

The period of oscillation for a mass-spring system is given by $T = 2\pi\sqrt{\frac{m}{k}}$.

When the mass increases to $m + m_0$, the new period becomes $T' = \frac{5T}{4}$.

Taking the ratio of periods:

$\frac{T'}{T} = \sqrt{\frac{m + m_0}{m}}$

Substituting $T' = \frac{5T}{4}$ yields:

$\frac{5}{4} = \sqrt{\frac{m + m_0}{m}}$

Squaring both sides:

$\left(\frac{5}{4}\right)^2 = \frac{m + m_0}{m}$

Which simplifies to:

$\frac{25}{16} = 1 + \frac{m_0}{m}$

Solving for the ratio:

$\frac{m_0}{m} = \frac{25}{16} - 1 = \frac{9}{16}$

The ratio of additional mass to original mass is $\frac{9}{16}$.