Percentage Change in Frequency of Dipole Oscillations in Electric Fields

To determine the percentage change in the frequency of oscillation of the dipole, we analyze the equation of motion for angular oscillations in a uniform electric field.

1. Dynamics of Dipole Oscillation in an Electric Field

The torque $\vec{\tau}$ acting on a dipole with dipole moment $\vec{p}$ in a uniform electric field $\vec{E}$ is given by: $\vec{\tau} = \vec{p} \times \vec{E}$

For a small angular displacement $\theta$ from its equilibrium position (where $\vec{p}$ aligns with $\vec{E}$), the magnitude of the restoring torque is: $\tau \approx -p E \theta$

Using Newton's second law for rotation, $I \alpha = \tau$, where $I$ is the moment of inertia of the dipole about its center and $\alpha = \frac{d^2\theta}{dt^2}$ is the angular acceleration: $I \frac{d^2\theta}{dt^2} = -p E \theta \implies \frac{d^2\theta}{dt^2} + \left(\frac{p E}{I}\right)\theta = 0$

This is the standard differential equation for Simple Harmonic Motion (SHM). The angular frequency of oscillation $\omega$ and linear frequency $f$ are given by: $\omega = \sqrt{\frac{p E}{I}}$ $f = \frac{1}{2\pi}\sqrt{\frac{p E}{I}}$

Thus, the frequency of oscillation is directly proportional to the square root of the magnitude of the net electric field: $f \propto \sqrt{E}$

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2. Initial State

Initially, the electric field present is: $\vec{E}_1 = E_0 \hat{x}$

The magnitude of the initial electric field is: $E_i = |\vec{E}_1| = E_0$

The initial frequency of oscillation $f_1$ is: $f_1 = \frac{1}{2\pi}\sqrt{\frac{p E_0}{I}}$

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3. Final State

When another electric field $\vec{E}_2 = 2E_0 (\hat{y} + \hat{z})$ is introduced, the total net electric field becomes: $\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = E_0 \hat{x} + 2E_0 \hat{y} + 2E_0 \hat{z}$

The magnitude of the new net electric field $E_f$ is: $E_f = |\vec{E}_{\text{net}}| = \sqrt{E_0^2 + (2E_0)^2 + (2E_0)^2} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = \sqrt{9E_0^2} = 3E_0$

The new frequency of oscillation $f_2$ is: $f_2 = \frac{1}{2\pi}\sqrt{\frac{p (3E_0)}{I}} = \sqrt{3} f_1$

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4. Percentage Change in Frequency

The percentage change in the frequency of oscillation is calculated as: $\text{Percentage Change} = \frac{f_2 - f_1}{f_1} \times 100\% = \frac{\sqrt{3}f_1 - f_1}{f_1} \times 100\% = (\sqrt{3} - 1) \times 100\%$

Taking the approximate value $\sqrt{3} \approx 1.732$: $\text{Percentage Change} \approx (1.732 - 1) \times 100\% = 73.2\% \approx 73\%$

Thus, the correct option is A.