Simple Harmonic Motion

Time period of oscillation of a pendulum in air is $T$.Density of the bob, $\rho=\;\frac{5000}{3}\;\mathrm{kg/m}^3$.Density of water, $\sigma=1000\;\mathrm{kg/m}^3$.We need to find the ratio $\;\frac{T}{t}$, where $t$ is the time period of the pendulum's oscillation in water.The time period of oscillation of a pendulum in water can be expressed as:$t=\;\frac{T}{\sqrt{1-\;\frac{\sigma}{\rho}}}$Substitute the given values:$t=\;\frac{T}{\sqrt{1-\;\frac{1000 \times 3}{5000}}}$This simplifies to:$t=\;\frac{T}{\sqrt{1-\;\frac{3000}{5000}}}$Simplifying further:$\;t=\;\frac{T}{\sqrt{1-0.6}}$$\;t=\;\frac{T}{\sqrt{0.4}}$The ratio $\;\frac{T}{t}$ is:$\;\frac{T}{t}\;=\sqrt{\;\frac{1}{0.4}}$$\;\frac{T}{t}\;=\sqrt{\;\frac{5}{2}}$Thus, the value of $\;\frac{T}{t}$ is $\sqrt{\;\frac{5}{2}}$.