JEE Advanced 2024 · Paper 1, Physics question 3

Equilibrium. The free bead sits diametrically opposite the fixed one. Let the free bead's angular position be $\pi + x$ measured from the fixed bead; the chord between them is $2R\cos\frac{x}{2}$.

Potential energy:
$$U = \frac{q^2}{4\pi\varepsilon_0\cdot 2R\cos(x/2)} \approx \frac{q^2}{8\pi\varepsilon_0 R}\left(1 + \frac{x^2}{8}\right).$$

Kinetic energy: $\frac12 mR^2\dot x^2$.

Comparing with SHM, $mR^2\omega^2 = U''(0) = \dfrac{q^2}{32\pi\varepsilon_0 R}$, so
$$\omega^2 = \frac{q^2}{32\pi\varepsilon_0 R^3 m}.$$