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

[𝜀₀ is the permittivity of free space.]

The correct option is:
$q^2/\left(32\pi {\epsilon }_0{R}^{3}m\right)$

Let us derive this result step-by-step.
Consider a circular, non-conducting hoop of radius $R$ placed horizontally in a frictionless environment.
One bead with charge $q$ is glued at a fixed point on the hoop. Let this point be the origin of angular coordinates, $\theta =0$.
The second bead, also with charge $q$ and mass $m$, is free to move along the hoop.

Since both beads have charges of the same sign, they repel each other. The equilibrium position of the movable bead must be at the diametrically opposite point, which is $\theta =\pi$ relative to the fixed bead.

Let the angular position of the movable bead be $\theta$. The straight-line distance $d$ between the fixed bead at $\theta =0$ and the movable bead at $\theta$ can be determined geometrically using the chord length formula:
$d=2R\sin \left(\frac{\theta }{2}\right)$

The electrostatic potential energy $U\left(\theta \right)$ of the system of the two charges is given by Coulomb's law:
$U\left(\theta \right)=\frac{q^2}{4\pi {\epsilon }_{0}d}=\frac{q^2}{8\pi {\epsilon }_{0}R\sin (\theta /2)}$

To analyze small oscillations about the equilibrium position at ${\theta }_0=\pi$, we define the angular displacement $x$ such that:
$\theta =\pi +x$
where $x$ is very small ($x\ll 1$).

Substituting this into the expression for potential energy:
$U\left(x\right)=\frac{q^2}{8\pi {\epsilon }_{0}R\sin \left(\frac{\pi +x}{2}\right)}=\frac{q^2}{8\pi {\epsilon }_{0}R\cos (x/2)}$

Using the Taylor series expansion for $\cos (x/2)$ for small $x$:
$\cos \left(\frac{x}{2}\right)\approx 1-\frac{x^2}{8}$
Therefore, the potential energy is:
$U\left(x\right)\approx \frac{q^2}{8\pi {\epsilon }_{0}R}{(1-\frac{x^2}{8})}^{-1}\approx \frac{q^2}{8\pi {\epsilon }_{0}R}(1+\frac{x^2}{8})=U_0+\frac{1}{2}k{x}^2$

Comparing this to the standard form of potential energy for angular oscillations $U\left(x\right)=U_0+\frac{1}{2}\kappa x^2$, we find the torsional/angular spring constant $\kappa$ to be:
$\kappa =\frac{q^2}{32\pi {\epsilon }_{0}R}$

The moment of inertia of the movable bead of mass $m$ rotating about the center of the hoop of radius $R$ is:
$I=m{R}^2$

The square of the angular frequency ${\omega }^2$ of the small oscillations is given by:
${\omega }^2=\frac{\kappa }{I}=\frac{q^2}{32\pi {\epsilon }_{0}R\cdot \left(m{R}^2\right)}=\frac{q^2}{32\pi {\epsilon }_0{R}^{3}m}$