Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circula

Q: Why is the equilibrium position at the top?

Because the charges are identical, they repel each other. The free bead will move as far away from the fixed bead as possible. On a circle, the maximum distance is the diameter, which places the free bead exactly at the top.

Q: How did we get the distance $2R \cos(\theta/2)$?

If you draw a triangle with the two charges and the center of the hoop, you can use trigonometry. Alternatively, the angle subtended by the arc at the circumference is half the angle at the center. Using right triangles, the chord length is $2R \sin(\text{half-angle})$. Here, the angle from the bottom is $180^\circ - \theta$, so the distance is $2R \sin(90^\circ - \theta/2) = 2R \cos(\theta/2)$.

Q: Why do we only consider the tangential force?

The bead is constrained to move *on* the hoop. Any force perpendicular to the hoop (normal force) is canceled out by the structural reaction of the hoop itself. Only the force along the tangent can cause the bead to speed up or slow down.

Q: Why does $\cos(\theta/2)$ become 1 in the approximation?

For very small angles, the cosine curve is flat and close to its peak value of 1. Mathematically, the Taylor expansion is $\cos x \approx 1 - x^2/2$. When $x$ is tiny, $x^2$ is negligible, so we just use 1.

Q: What is the difference between $\theta$ and $x$ in the final equation?

$\theta$ is the angular displacement from the top, measured in radians. $x$ is the actual physical distance the bead has traveled along the arc of the hoop. They are related by the simple arc length formula $x = R\theta$.

Question

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.]

Options:

(A) $q^2 / (4\pi\varepsilon_0 R^3 m)$

(B) $q^2 / (32\pi\varepsilon_0 R^3 m)$

(C) $q^2 / (8\pi\varepsilon_0 R^3 m)$

(D) $q^2 / (16\pi\varepsilon_0 R^3 m)$

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine a frictionless, non-conducting circular hoop lying horizontally. We have a charge $+q$ fixed at the bottom of this hoop, and another identical charge $+q$ that is free to slide along it. Because both charges are positive, they repel each other with an electrostatic force.

### Finding the Equilibrium

Before we can talk about oscillations, we must find the equilibrium position. Since the charges repel, the free bead will naturally want to move as far away from the fixed bead as possible to minimize its potential energy. On a circular hoop, the maximum possible distance between any two points is the diameter. Therefore, the free bead will settle exactly at the top of the hoop. This top position is our stable equilibrium.

### The Slight Nudge

Now, let's disturb the peace. We displace the free bead by a small angle $	heta$ from its equilibrium position at the top. The electric force $\vec{F}$ acts along the straight line connecting the two charges.

To find the magnitude of this force, we need the distance between the charges. If you draw a triangle connecting the two charges and the center of the hoop, you can use trigonometry to find the chord length. The angle subtended by the arc at the circumference is half the angle at the center. Using right triangles, the distance $d$ is given by $2R \cos\left(\frac{	heta}{2}ight)$.

### The Restoring Force

The electric force pushes the bead outward, but the bead is constrained to move *on* the hoop. Any force perpendicular to the hoop (the normal component) is canceled out by the structural reaction of the hoop itself. Only the force along the tangent can cause the bead to accelerate.

By analyzing the geometry, the angle between the force vector and the tangent line is $90^\circ - \frac{	heta}{2}$. Therefore, the tangential component of the force is:

$$F_t = F \cos\left(90^\circ - \frac{	heta}{2}ight) = F \sin\left(\frac{	heta}{2}ight)$$

Substituting Coulomb's Law for $F$, we get:

$$F_t = \frac{q^2 \sin\left(\frac{	heta}{2}ight)}{16\pi\varepsilon_0 R^2 \cos^2\left(\frac{	heta}{2}ight)}$$

### The Magic of Small Angles

Here is where the physics gets elegant. Because the oscillations are very small, the angle $	heta$ approaches zero. We can apply the small-angle approximations derived from Taylor series expansions: $\sin\left(\frac{	heta}{2}ight) \approx \frac{	heta}{2}$ and $\cos\left(\frac{	heta}{2}ight) \approx 1$.

Plugging these approximations into our force equation, the complex trigonometric expression simplifies beautifully into a linear restoring force:

$$F_t \approx \frac{q^2 	heta}{32\pi\varepsilon_0 R^2}$$

### The Grand Finale

We are now in the familiar territory of mechanics. According to Newton's Second Law, this tangential force equals mass times tangential acceleration ($F_t = m a_t$).

Furthermore, the angular displacement $	heta$ is related to the physical arc length $x$ by the formula $x = R	heta$, which means $	heta = \frac{x}{R}$. Substituting this into our acceleration equation yields:

$$a_t = \left( \frac{q^2}{32\pi\varepsilon_0 m R^3} ight) x$$

This is the classic signature of Simple Harmonic Motion, where acceleration is directly proportional to displacement ($a = \omega^2 x$). The constant of proportionality is the square of the angular frequency, giving us our final answer:

$$\omega^2 = \frac{q^2}{32\pi\varepsilon_0 m R^3}$$

Select Answer:

Visualized Solution

\text{Equilibrium Position}

\text{Displacement and Restoring Force}

\text{Electric Force Magnitude}

\text{Tangential Component}

\text{Small Angle Approximation}

\text{Equation of Motion}

\text{Angular Frequency of SHM}

\text{The Way Forward}

The Sigma Insight: Simple Harmonic Motion (SHM)

Setting the Stage

Finding the Equilibrium

The Slight Nudge

The Restoring Force

The Magic of Small Angles

The Grand Finale

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