Oscillations Part 1

Given, radius of earth $=R$ Distance of chord from centre of earth $=\frac{R}{2}$ Let $x_{1}$ be the radius of inner circle and $M$ be the mass of earth. $\therefore m' \left( \;\text{ effective mass of earth }\; \right) = \frac{M}{\frac{4}{3} \pi R^{3}} \cdot \frac{4}{3} \pi x_{1}^{3}$ $\Rightarrow m' = \frac{M}{R^{3}} x_{1}^{3}$ If $F$ is the gravitational force exerted by earth on particle at position $x$ and $\omega$ be the angular velocity in time period $T$, then $F = \frac{G m' m}{x_{1}^{2}} = \frac{G m}{x_{1}^{2}} \cdot \frac{M}{R^{3}} x_{1}^{3}$ $\Rightarrow m \omega^{2} x_{1} = \frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \omega = \sqrt{\frac{G M}{R^{3}}}$ $F = \frac{G m' m}{x_{1}^{2}} = \frac{G m}{x_{1}^{2}} \cdot R^{3} x_{1}^{3}$ $\Rightarrow m \omega^{2} x_{1} = \frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \omega = \sqrt{\frac{G M}{R^{3}}}$ Since, $\omega = \frac{2 \pi}{T}$ and $G M = g R^{2}$ Substituting the above values in Eq. (i), we get 1$\Rightarrow \frac{2 \pi}{T} = \sqrt{\frac{g R^{2}}{R^{3}}} \Rightarrow T = 2 \pi \sqrt{\frac{R}{g}}$