A rigid dipole undergoes a simple harmonic motion about its centre...

The frequency of oscillation of a rigid dipole in an electric field is given by $f = \dfrac{1}{2\pi} \sqrt{\dfrac{pE}{I}}$, where $p is the dipole moment,$E$is the magnitude of the electric field, and$I$is the moment of inertia.<br><br>This implies$f \propto \sqrt{E}$.<br><br>The initial electric field is$\vec{E}_1 = E_0 \hat{x}$, so its magnitude is$E_1 = E_0$.<br><br>The initial frequency is$f_1 \propto \sqrt{E_0}$.<br><br>When the second electric field$\vec{E}_2 = 2E_0(\hat{y} + \hat{z})$is introduced, the net electric field becomes$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = E_0 \hat{x} + 2E_0 \hat{y} + 2E_0 \hat{z}$.<br><br>The magnitude of the net electric field is$E_{net} = \sqrt{E_0^2 + (2E_0)^2 + (2E_0)^2} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = 3E_0$.<br><br>The new frequency is$f_2 \propto \sqrt{3E_0}$.<br><br>The percentage change in the frequency is given by$\dfrac{f_2 - f_1}{f_1} \times 100\%$.<br><br>Substituting the values, we get \left( \dfrac{\sqrt{3E_0}}{\sqrt{E_0}} - 1 \right) \times 100\% = (\sqrt{3} - 1) \times 100\%$.

Using $\sqrt{3} \approx 1.732$, the percentage change is $(1.732 - 1) \times 100\% = 73.2\%$.

The approximate percentage change is $73\%$.

Answer: $73\%$