Question 3

Let the glued bead be at angular position $\theta = 0$. The equilibrium position of the free bead will be at $\theta = \pi$ (diametrically opposite) due to electrostatic repulsion. Let the free bead be displaced by a small angle $\delta\theta$ from its equilibrium, so its position is $\theta = \pi + \delta\theta$. The distance $d$ between the beads is $d = 2R \sin(\theta/2) = 2R \sin(\frac{\pi + \delta\theta}{2}) = 2R \cos(\frac{\delta\theta}{2})$. The electrostatic potential energy is: $U = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d} = \frac{q^2}{8\pi\varepsilon_0 R \cos(\frac{\delta\theta}{2})}$ For small $\delta\theta$, $\cos(\frac{\delta\theta}{2}) \approx 1 - \frac{(\delta\theta/2)^2}{2} = 1 - \frac{\delta\theta^2}{8}$. $U \approx \frac{q^2}{8\pi\varepsilon_0 R} (1 - \frac{\delta\theta^2}{8})^{-1} \approx \frac{q^2}{8\pi\varepsilon_0 R} (1 + \frac{\delta\theta^2}{8}) = \frac{q^2}{8\pi\varepsilon_0 R} + \frac{q^2}{64\pi\varepsilon_0 R} \delta\theta^2$ The kinetic energy of the bead is $K = \frac{1}{2} m v^2 = \frac{1}{2} m (R \dot{\delta\theta})^2 = \frac{1}{2} m R^2 \dot{\delta\theta}^2$. The total energy $E = K + U$ is constant: $E = \frac{1}{2} (m R^2) \dot{\delta\theta}^2 + \frac{1}{2} \left( \frac{q^2}{32\pi\varepsilon_0 R} \right) \delta\theta^2 + \text{const}$ Comparing with the standard SHM energy $\frac{1}{2} I \omega^2 + \frac{1}{2} k_\theta \theta^2$, we have the square of the angular frequency: $\omega_{osc}^2 = \frac{k_\theta}{I} = \frac{q^2 / (32\pi\varepsilon_0 R)}{m R^2} = \frac{q^2}{32\pi\varepsilon_0 R^3 m}$