Oscillations Part 1

Given that, a bob of mass m suspended by a thread of length l undergoes SHM with time period, $T_{1}=T=2\pi\sqrt{\frac{l}{g}}$ ...(i) In liquid, effective gravity, $g_{\text{eff}}=g\left(\frac{\sigma-\rho}{\sigma}\right)$ where, ρ = density of liquid and σ = density of body (bob). Given, $\rho=\frac{\sigma}{4}$ $\Rightarrow\sigma=4\rho$ Substituting this value, we get $g_{\text{eff}}=g\left(\frac{4\rho-\rho}{4\rho}\right)=\frac{3g}{4}$ Now, the new time period of pendulum in liquid is $T'=2\pi\sqrt{\frac{l'}{g_{\text{eff}}}}$ ...(ii) Here,$l'=l+\frac{l}{3}=\frac{4I}{3}$ Substituting the values in Eq. (ii), we get $T'=2\pi\sqrt{\frac{4l}{3}\times\frac{4}{3g}}$ $=\left(2\pi\sqrt{\frac{l}{g}}\right)\frac{4}{3}$ $\Rightarrow T'=\frac{4T}{3}$ [from Eq. (i)]