A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field $$\vec{E}_1 = E_0\hat{x}$$. If another electric field $$\vec{E}_2 = 2E_0(\hat{y} + \hat{z})$$ is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

The restoring torque on a rigid electric dipole of moment magnitude $$p$$ kept in a uniform field $$\vec E$$ is $$\tau = -pE\sin\theta$$, where $$\theta$$ is the small angular displacement from the equilibrium direction.

For small oscillations $$\sin\theta \approx \theta$$, so the equation of rotational SHM is

$$I\frac{d^{2}\theta}{dt^{2}} = -pE\,\theta$$

Here $$I$$ is the moment of inertia of the dipole about its axis of rotation. Comparing with $$\ddot\theta + \omega^{2}\theta = 0$$, the angular frequency is

$$\omega = \sqrt{\frac{pE}{I}}$$

Case 1: Only the original field $$\vec{E}_1 = E_0\hat x$$ is present.
Magnitude $$E_1 = E_0$$

$$\omega_1 = \sqrt{\frac{pE_0}{I}}$$

Case 2: The additional field $$\vec{E}_2 = 2E_0(\hat y + \hat z)$$ is switched on.

The net field is the vector sum

$$\vec E_{\text{net}} = E_0\hat x + 2E_0\hat y + 2E_0\hat z$$

Magnitude

$$E_{\text{net}} = E_0\sqrt{1^{2} + 2^{2} + 2^{2}} = E_0\sqrt{1 + 4 + 4} = 3E_0$$

The new angular frequency is therefore

$$\omega_2 = \sqrt{\frac{pE_{\text{net}}}{I}} = \sqrt{\frac{p(3E_0)}{I}} = \sqrt{3}\,\omega_1$$

The ordinary (cyclic) frequency is proportional to angular frequency, so the same factor applies:

$$f_2 = \sqrt{3}\,f_1$$

Percentage change in frequency:

$$\frac{f_2 - f_1}{f_1}\times 100\% = (\sqrt{3}-1)\times 100\% \approx (1.732 - 1)\times 100\% \approx 0.732 \times 100\% \approx 73\%$$

Hence, the frequency increases by approximately $$73\%$$.

Option A which is: $$73\%$$