Two beads, each with charge $$q$$ and mass $$m$$, are on a horizontal, frictionless, non-conducting, circular hoop of radius $$R$$. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by

[$$\varepsilon_0$$ is the permittivity of free space.]

Let the fixed bead be at point $$A$$ on the hoop, whose centre is $$O$$. Measure the angular position of the movable bead by angle $$\theta$$ made by $$OA$$ and the radius $$OP$$ that joins the centre to the movable bead. Equilibrium occurs when the tangential component of the electrostatic force on the movable bead is zero.

Because the only force is the repulsion from the fixed bead, the force line is along the straight chord joining the two beads. At $$\theta = \pi$$ (diametrically opposite to the fixed bead) this chord passes through the centre, so its direction is purely radial and has no tangential component. Hence the equilibrium position is $$\theta_0 = \pi$$.

Introduce a small angular displacement $$\phi$$ about this equilibrium: $$\theta = \pi + \phi \qquad(|\phi|\ll 1).$$ We shall find the potential energy as a function of $$\phi$$ upto the term in $$\phi^2$$, and compare it with the quadratic form $$\tfrac12 I\omega^2\phi^2$$ of simple harmonic motion.

Magnitude of separation between the beads (straight line, not arc): The angle subtended at the centre between the two position vectors is $$\delta=\theta = \pi+\phi$$, hence $$r = 2R\sin\frac{\delta}{2}=2R\sin\!\left(\frac{\pi+\phi}{2}\right) = 2R\cos\frac{\phi}{2}.$$

Electrostatic potential energy of the two-charge system: $$U(\phi)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r} =\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2R\cos(\phi/2)}.$$

Expand for small $$\phi$$ using $$\cos(\phi/2)\approx 1-\frac{\phi^2}{8}$$ and for any small $$x$$, $$\frac1{1-x}\approx 1+x$$: $$\frac{1}{\cos(\phi/2)}\approx 1+\frac{\phi^2}{8}.$$ Therefore $$U(\phi)\approx\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{2R}\Bigl(1+\frac{\phi^2}{8}\Bigr) =U_0+\frac{q^2}{4\pi\varepsilon_0}\,\frac{\phi^2}{16R},$$ where $$U_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{2R}$$ is the minimum (constant) part.

The change in potential energy near equilibrium is thus $$\Delta U=\frac{q^2}{4\pi\varepsilon_0}\frac{\phi^2}{16R}.$$

The movable bead of mass $$m$$ moves along a circle of radius $$R$$, so its moment of inertia about the centre is $$I=mR^2.$$ For small angular oscillations, $$\Delta U=\tfrac12 I\omega^2\phi^2$$. Equating the coefficients of $$\phi^2$$:

$$\frac{1}{2}mR^2\omega^2=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16R}$$ $$\Rightarrow\;\omega^2=\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{8mR^3} =\frac{q^2}{32\pi\varepsilon_0 R^3 m}.$$

Hence the square of the angular frequency is $$\boxed{\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}}.$$

Option B which is: $$\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}$$