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 [E0 is the permittivity of free space.]

 Alright, let's tackle this physics problem step by step. I'm going to break it down to understand the situation and find the square of the angular frequency of the small oscillations for the movable bead.

Understanding the Problem

We have two beads on a circular hoop:

1. Bead 1: Glued to the hoop at a fixed point. It has charge $q$ and mass $m$.

2. Bead 2: Free to move along the hoop. It also has charge $q$ and mass $m$.

The hoop is:

• Horizontal (so gravity acts downward, but since it's horizontal, gravity doesn't affect motion along the hoop).

• Frictionless.

• Non-conducting (so charges don't redistribute).

Bead 2 is performing small oscillations about its equilibrium position. We need to find the square of the angular frequency (${\omega }^2$) of these small oscillations.

Visualizing the Setup

Imagine a circle (the hoop) lying flat on a table. One bead is stuck at, say, the "top" of the circle (let's call this position angle $\theta =0$). The other bead can move along the circumference. At equilibrium, the movable bead will be at the "bottom" of the circle (angle $\theta =\pi$), directly opposite the fixed bead, because like charges repel, and this is the farthest it can get from the fixed bead.

When the movable bead is displaced slightly from this equilibrium position, it oscillates back and forth. We're to find ${\omega }^2$ for these small oscillations.

Forces Acting on the Movable Bead

The only force acting on the movable bead along the hoop is the electrostatic repulsion from the fixed bead. Gravity acts downward, but since the hoop is horizontal, the normal force cancels gravity, and there's no component of gravity along the hoop's tangent.

The electrostatic force between two charges $q$ separated by a distance $d$ is:

$$F_e=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{d^2}$$

Here, the distance $d$ between the beads changes as the movable bead moves. If the movable bead is at an angle $\theta$ from the fixed bead (with $\theta =\pi$ at equilibrium), the distance $d$ is the chord length between them.

Calculating the Chord Length

For a circle of radius $R$, the chord length $d$ between two points separated by angle $\theta$ is:

$$d=2R\sin \left(\frac{\theta }{2}\right)$$

So the electrostatic force magnitude is:

$$F_e=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{{(2R\sin (\theta \mathrm{/}2))}^2}=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{4R^2{\sin }^2(\theta \mathrm{/}2)}=\frac{q^2}{16\pi {ϵ}_0{R}^2{\sin }^2(\theta \mathrm{/}2)}$$

Direction of the Force

The force is repulsive, so it acts along the line connecting the two beads. We're interested in the component of this force that is tangential to the hoop, which causes the bead to move along the circumference.

At angle $\theta$, the tangential component $F_t$ of $F_e$ is:

$$F_t=F_e\sin (\varphi )$$

where $\varphi$ is the angle between the line connecting the beads and the tangent direction.

From geometry, $\varphi =\frac{\theta }{2}$ (since the tangent is perpendicular to the radius, and the line connecting the beads makes an angle $\theta \mathrm{/}2$ with the diameter).

So:

$$F_t=F_e\sin \left(\frac{\theta }{2}\right)=\frac{q^2}{16\pi {ϵ}_0{R}^2{\sin }^2(\theta \mathrm{/}2)}\cdot \sin \left(\frac{\theta }{2}\right)=\frac{q^2}{16\pi {ϵ}_0{R}^2\sin (\theta \mathrm{/}2)}$$

Equilibrium Position

At equilibrium, the movable bead is at $\theta =\pi$. Let's define the displacement from equilibrium as $\alpha$, where:

$$\theta =\pi +\alpha$$

For small oscillations, $\alpha$ is small.

Now, $\theta \mathrm{/}2=\pi \mathrm{/}2+\alpha \mathrm{/}2$, so:

$$\sin (\theta \mathrm{/}2)=\sin (\frac{\pi }{2}+\frac{\alpha }{2})=\cos \left(\frac{\alpha }{2}\right)\approx 1$$

for small $\alpha$ (since $\cos (x)\approx 1$ for small $x$).

But this seems off because at $\alpha =0$, $\theta =\pi$, $\sin (\pi \mathrm{/}2)=1$, and as $\alpha$ increases, $\sin (\theta \mathrm{/}2)$ decreases from 1. Maybe a better approach is to consider the force near equilibrium.

Small Angle Approximation Near Equilibrium

Let’s set $\theta =\pi +\alpha$, where $\alpha$ is small.

Then:

$$\sin \left(\frac{\theta }{2}\right)=\sin (\frac{\pi }{2}+\frac{\alpha }{2})=\cos \left(\frac{\alpha }{2}\right)\approx 1-\frac{(\alpha \mathrm{/}2{)}^2}{2}\approx 1$$

for very small $\alpha$.

This suggests that to first order, $\sin (\theta \mathrm{/}2)$ doesn't change with $\alpha$, which would imply no restoring force, which can't be right. Maybe I need to consider the next term.

Alternatively, perhaps it's better to express the force in terms of $\alpha$ and then approximate.

Rewriting the Tangential Force

Given:

$$F_t=\frac{q^2}{16\pi {ϵ}_0{R}^2\sin (\theta \mathrm{/}2)}$$

and $\theta =\pi +\alpha$, so:

$$\sin (\theta \mathrm{/}2)=\cos (\alpha \mathrm{/}2)\approx 1-\frac{{\alpha }^2}{8}$$

But this still seems to suggest that $F_t$ doesn't change linearly with $\alpha$, which is needed for simple harmonic motion.

Perhaps I made a mistake in the direction of the tangential force. Let me reconsider the geometry.

Correcting the Tangential Force Direction

The electrostatic force is along the line connecting the two beads. The tangential component is the component of this force along the circumference.

The angle between the line connecting the beads and the tangent at the movable bead is $\frac{\pi -\theta }{2}$ (since the tangent is perpendicular to the radius, and the connecting line makes angle $\theta$ with the fixed radius).

Wait, perhaps a better way is to note that the tangential direction is perpendicular to the radius at the movable bead, and the angle between the connecting line and the radius at the movable bead is $\theta \mathrm{/}2$, so the angle between the connecting line and the tangent is $\pi \mathrm{/}2-\theta \mathrm{/}2$.

Thus:

$$F_t=F_e\cos \left(\frac{\theta }{2}\right)$$

Now, at $\theta =\pi$, $\cos (\pi \mathrm{/}2)=0$, which makes sense because at equilibrium, the force is purely radial (no tangential component).

For $\theta =\pi +\alpha$:

$$\cos \left(\frac{\theta }{2}\right)=\cos (\frac{\pi }{2}+\frac{\alpha }{2})=-\sin \left(\frac{\alpha }{2}\right)\approx -\frac{\alpha }{2}$$

So:

$$F_t\approx \frac{q^2}{16\pi {ϵ}_0{R}^2\sin (\theta \mathrm{/}2)}\cdot (-\frac{\alpha }{2})$$

And $\sin (\theta \mathrm{/}2)=\cos (\alpha \mathrm{/}2)\approx 1$, so:

$$F_t\approx -\frac{q^2}{32\pi {ϵ}_0{R}^2}\alpha$$

This is a restoring force proportional to $\alpha$, which is what we need for simple harmonic motion.

Equation of Motion

The tangential force causes angular acceleration:

$$F_t=m{a}_t=mR\frac{d^2\alpha }{d{t}^2}$$

So:

$$mR\frac{d^2\alpha }{d{t}^2}=-\frac{q^2}{32\pi {ϵ}_0{R}^2}\alpha$$$$\frac{d^2\alpha }{d{t}^2}=-\frac{q^2}{32\pi {ϵ}_{0}m{R}^3}\alpha$$

This is of the form:

$$\frac{d^2\alpha }{d{t}^2}=-{\omega }^2\alpha$$

where:

$${\omega }^2=\frac{q^2}{32\pi {ϵ}_{0}m{R}^3}$$

Final Answer

The square of the angular frequency of the small oscillations is:

$${\omega }^2=\frac{q^2}{32\pi {ϵ}_{0}m{R}^3}$$

Expressed in terms of $E_0$ (which is another notation for ${ϵ}_0$):

$${\omega }^2=\frac{q^2}{32\pi E_{0}m{R}^3}$$

However, looking back at the electrostatic force, the constant is $\frac{1}{4\pi {ϵ}_0}$, so perhaps there's a factor to adjust. But the calculation seems correct.

Wait, the distance between charges is $2R\sin (\theta \mathrm{/}2)$, and at equilibrium $\theta =\pi$, distance is $2R$. For small oscillations, the change in distance is small, but the force's tangential component is correctly approximated as linear in $\alpha$.

Thus, the final answer is:

$${\omega }^2=\frac{q^2}{32\pi {ϵ}_{0}m{R}^3}$$

Or, if $E_0={ϵ}_0$:

$${\omega }^2=\frac{q^2}{32\pi E_{0}m{R}^3}$$