Oscillation Frequency of Charged Mass Attached to Spring with Dipole Interaction

To find the frequency of oscillation of the particle, we analyze the forces acting on the particle at its equilibrium position and when it is slightly displaced from it.

1. Equilibrium Position

Let the origin $(0,0)$ be the position of the fixed point dipole $\vec{p} = p\hat{i}$ and the fixed end of the spring. The particle of mass $m$ and positive charge $q$ is at a distance $x$ along the positive x-axis.

The electric field at position $x$ along the axial line of the dipole is: $E(x) = \frac{1}{4\pi\varepsilon_0} \frac{2p}{x^3}$

The electrostatic force acting on the charge $q$ is directed away from the origin (along $+\hat{i}$): $F_e(x) = q E(x) = \frac{1}{4\pi\varepsilon_0} \frac{2pq}{x^3}$

Since the spring has negligible unstretched length, the spring force acts towards the origin (along $-\hat{i}$): $F_s(x) = kx$

At the equilibrium position $x = l$, the net force on the particle is zero: $F_e(l) - F_s(l) = 0$ $\frac{1}{4\pi\varepsilon_0} \frac{2pq}{l^3} - kl = 0$ $\implies \frac{1}{4\pi\varepsilon_0} \frac{2pq}{l^3} = kl \quad \text{--- (1)}$

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2. Perturbation and Equation of Motion

Let the particle be displaced slightly by a distance $\Delta x = \Delta l$ along the line of equilibrium such that its new position is $x = l + \Delta x$, where $\Delta x \ll l$.

The net restoring force acting on the particle at this new position is: $F_{\text{net}} = F_e(l + \Delta x) - F_s(l + \Delta x)$ $F_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \frac{2pq}{(l + \Delta x)^3} - k(l + \Delta x)$

Using binomial approximation $(1 + y)^{-3} \approx 1 - 3y$ for $y = \frac{\Delta x}{l} \ll 1$: $\frac{1}{(l + \Delta x)^3} = \frac{1}{l^3} \left(1 + \frac{\Delta x}{l}\right)^{-3} \approx \frac{1}{l^3} \left(1 - 3\frac{\Delta x}{l}\right)$

Substituting this back into the net force equation: $F_{\text{net}} \approx \left(\frac{1}{4\pi\varepsilon_0} \frac{2pq}{l^3}\right) \left(1 - 3\frac{\Delta x}{l}\right) - kl - k\Delta x$

Using relation $(1)$, we substitute $\frac{1}{4\pi\varepsilon_0} \frac{2pq}{l^3} = kl$: $F_{\text{net}} \approx kl \left(1 - 3\frac{\Delta x}{l}\right) - kl - k\Delta x$ $F_{\text{net}} \approx kl - 3k\Delta x - kl - k\Delta x$ $F_{\text{net}} \approx -4k\Delta x$

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3. Calculation of Oscillation Frequency

The net restoring force is proportional to the displacement $\Delta x$, demonstrating simple harmonic motion: $m \frac{d^2(\Delta x)}{dt^2} = -4k \Delta x$ $\frac{d^2(\Delta x)}{dt^2} + \left(\frac{4k}{m}\right)\Delta x = 0$

The angular frequency of oscillation is: $\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}$

The frequency of oscillation $f$ is given by: $f = \frac{\omega}{2\pi} = \frac{2\sqrt{\frac{k}{m}}}{2\pi} = \frac{1}{\pi}\sqrt{\frac{k}{m}}$

Comparing this with the given expression $f = \frac{1}{\delta}\sqrt{\frac{k}{m}}$, we get: $\delta = \pi \approx 3.14$