Change in Frequency of Charged Ball's!

When the balls are uncharged, frequency $f_0 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

When the balls are identically charged,

$\large \frac{q^2}{4 \pi \epsilon_0 (\eta l)^2} = k(\eta l - l) .........(i)$

(where $l$ is the natural length of the spring and $\eta l$ is the new length of the spring after extension.)

or $\large l^3 = \frac{q^2}{4 \pi \epsilon_0 \eta^2 (\eta-1)k} ............(ii)$

When the ball 1 is displaced by a small distance $x$ from the equilibrium position to the right, the unbalanced force to the right is given by

$\large F_r = \frac{q^2}{4 \pi \epsilon_0 (\eta l + x)^2} - k(\eta l +x -l)$

Using Newton's law,

$\large m \frac{d^2 x}{dt^2} = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{\eta^2 l^2} (1+ \dfrac x {\eta l})^{-2} - kl(\eta-1) -kx$

Now expanding binomially and using equations $(i)$ & $(ii)$ , we get

$\large \frac{d^2 x}{dt^2} = -(\frac{3\eta-2}{\eta}) \frac{k}{m} x$

Now use the definition of Simple Harmonic Motion

$\large \frac{d^2 x}{dt^2} = - \omega^2 x$ to find $\omega$ and thus, the new frequency $f'$