Small Oscillations of Charged Bead on Circular Hoop

To find the square of the angular frequency of small oscillations, we analyze the forces acting on the moving bead along the circular hoop.

1. System Geometry and Equilibrium Position

Let the circular hoop lie in the $xy$-plane with its center at the origin $(0,0)$ and radius $R$.

• Let the glued bead be fixed at the point $A(-R, 0)$, corresponding to an angular position $\theta = 0$.
• The moving bead of mass $m$ and charge $q$ is located at point $B(R\cos\theta, R\sin\theta)$.

The distance $d$ between the two beads as a function of the angle $\theta$ is given by: $d = \sqrt{(R\cos\theta + R)^2 + (R\sin\theta)^2} = \sqrt{2R^2(1 + \cos\theta)} = 2R \cos\left(\frac{\phi}{2}\right)$ where $\phi$ is the angular displacement from the diametrically opposite point ($\theta = \pi$), so $\theta = \pi + \phi$.

The equilibrium position of the moving bead occurs at $\phi = 0$ ($\theta = \pi$), where the electrostatic repulsion from the fixed bead is directed purely radially outwards, resulting in zero tangential force along the hoop.

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2. Tangential Restoring Force

When the bead is displaced by a small angle $\phi$ from its equilibrium position:

• Position vector of the fixed bead: $\vec{r}_A = (-R, 0)$
• Position vector of the moving bead: $\vec{r}_B = (R\cos(\pi+\phi), R\sin(\pi+\phi)) = (-R\cos\phi, -R\sin\phi)$

The vector pointing from $A$ to $B$ is: $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = R(1 - \cos\phi)\hat{i} - R\sin\phi\,\hat{j}$

The magnitude of $\vec{r}_{AB}$ is: $|\vec{r}_{AB}| = 2R \sin\left(\frac{\pi+\phi}{2}\right) = 2R \cos\left(\frac{\phi}{2}\right)$

The unit tangent vector to the circle at point $B$ in the direction of increasing $\phi$ is: $\hat{t} = \sin\phi\,\hat{i} - \cos\phi\,\hat{j}$

The electrostatic force exerted by charge $A$ on charge $B$ is: $\vec{F}_e = \frac{q^2}{4\pi\varepsilon_0 |\vec{r}_{AB}|^3} \vec{r}_{AB}$

The tangential force component $F_t$ driving the motion along the hoop is: $F_t = \vec{F}_e \cdot \hat{t} = \frac{q^2}{4\pi\varepsilon_0 |\vec{r}_{AB}|^3} \left(\vec{r}_{AB} \cdot \hat{t}\right)$

Evaluating the dot product: $\vec{r}_{AB} \cdot \hat{t} = R(1 - \cos\phi)\sin\phi + R\sin\phi\cos\phi = R\sin\phi$

Substituting this into the tangential force expression gives: $F_t = \frac{q^2 R \sin\phi}{4\pi\varepsilon_0 \left(2R \cos\left(\frac{\phi}{2}\right)\right)^3} = \frac{q^2 \sin\phi}{32\pi\varepsilon_0 R^2 \cos^3\left(\frac{\phi}{2}\right)}$

Since the tangential force opposes an increase in $\phi$ when moving away from equilibrium, the restoring force is: $F_{\text{restoring}} = -F_t = -\frac{q^2 \sin\phi}{32\pi\varepsilon_0 R^2 \cos^3\left(\frac{\phi}{2}\right)}$

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3. Equation of Motion for Small Oscillations

For small angular displacements ($\phi \ll 1$), we use the approximations $\sin\phi \approx \phi$ and $\cos\left(\frac{\phi}{2}\right) \approx 1$: $F_{\text{restoring}} \approx -\frac{q^2}{32\pi\varepsilon_0 R^2} \phi$

Applying Newton's second law along the arc length $s = R\phi$: $m \frac{d^2 s}{dt^2} = F_{\text{restoring}}$ $m R \ddot{\phi} = -\frac{q^2}{32\pi\varepsilon_0 R^2} \phi$ $\ddot{\phi} + \left(\frac{q^2}{32\pi\varepsilon_0 R^3 m}\right) \phi = 0$

Comparing this with the standard Simple Harmonic Motion equation $\ddot{\phi} + \omega^2 \phi = 0$, the square of the angular frequency is: $\omega^2 = \frac{q^2}{32\pi\varepsilon_0 R^3 m}$

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Correct Answer:

(B) $\frac{q^2}{32\pi\varepsilon_0 R^3 m}$