Creepy Oscillations (Part 1)

Nice problem!

At any general time $t$ , the rod $OA$ makes an angle $\theta$ with the horizontal. The coordinates of the point $A$ are $(\cos{\theta},\sin{\theta})$ .

Calculating the potential energy of the system of charges:

$\mathcal{V} = \frac{Kq^2}{2} + \frac{Kq^2}{\sqrt{\cos^2{\theta}+ (1-\sin{\theta})^2}} + \frac{Kq^2}{\sqrt{\cos^2{\theta}+ (1+\sin{\theta})^2}}$ $\implies \mathcal{V} = \frac{Kq^2}{2} +\frac{Kq^2}{\sqrt{2(1-\sin{\theta})}} + \frac{Kq^2}{\sqrt{2(1+\sin{\theta})}}$

The velocity components of the charge at $A$ is $(-\dot{\theta}\sin{\theta},\dot{\theta}\cos{\theta})$ . Therefore, the kinetic energy is:

$\mathcal{T} = \frac{m}{2} \dot{\theta}^2$

Now, since there are no dissipative forces acting on the system, its energy is conserved. This means:

$E = \mathcal{V} + \mathcal{T}$

$E$ is a constant. Now, differentiating both sides with respect to time gives:

$0 = \frac{d \mathcal{T}}{dt} + \frac{d \mathcal{V}}{dt}$

$\implies \dot{\theta} \left(m \ddot{\theta} + \frac{Kq^2}{2\sqrt{2}}\cos{\theta} \left((1-\sin{\theta})^{-3/2} - (1+\sin{\theta})^{-3/2}\right) \right)=0$

Now $\dot{\theta} \ne 0$ otherwise the system would not be in motion. Therefore:

$m \ddot{\theta} + \frac{Kq^2}{2\sqrt{2}}\cos{\theta} \left((1-\sin{\theta})^{-3/2} - (1+\sin{\theta})^{-3/2}\right) =0$

We are asked to analyse small oscillations. In this event, we assume $\sin{\theta} \approx \theta$ and $\cos{\theta} \approx 1$ . This transforms the equation of motion to:

$m \ddot{\theta} + \frac{Kq^2}{2\sqrt{2}} \left((1-\theta)^{-3/2} - (1+\theta)^{-3/2}\right) =0$

By doing a binomial expansion and neglecting terms of higher powers since $\theta$ is small leads to:

$m \ddot{\theta} + \frac{Kq^2}{2\sqrt{2}} \left(1 + \frac{3 \theta}{2} - 1 + \frac{3 \theta}{2}\right) =0$ $m \ddot{\theta} + \frac{3Kq^2}{2\sqrt{2}}\theta = 0$

This equation represents a simple harmonic oscillation of natural angular frequency:

$\omega^2 = \frac{3Kq^2}{2m\sqrt{2}}$ $\implies \omega = \sqrt{\frac{3Kq^2}{2m\sqrt{2}}}$ $\implies 2 \pi f = \sqrt{\frac{3Kq^2}{2m\sqrt{2}}}$ $\implies\boxed{ f = \frac{1}{2 \pi} \sqrt{\frac{3Kq^2}{2m\sqrt{2}}}}$