Oscillations of a Charged Arc

Consider that the charged arc is displaced by a small length r.Using the law of cosines the distance of the point (0,r) from the centre as origin to the small piece of charge is $s=\sqrt{R^{2}+r^{2}-2rRcos\theta}$ at some angle $\theta$ with the axis

Consider a small piece of the ring with charge $dq=\lambda Rd\theta$ the potential energy of the charge q as a fuction of r is therefore

$U(r)=2\int_{0}^{\theta}\frac{kQ\lambda(d\theta)}{s}=2kQ\lambda\int_{0}^{\theta}\frac{d\theta}{\sqrt{1+((\frac{r}{R})^{2}-\frac{2rcos\theta}{R})}}=I$ using approximation Taylors expansion of $(1-x)^{-0.5}=1-\frac{x}{2}+\frac{3x^{2}}{8}$ neglecting higher order terms obove 2nd order

we get

$U(r)=\frac{Q\lambda}{2\pi\varepsilon_{0}}\int_{0}^{\theta}(1-\frac{1}{2}((\frac{r}{R})^{2}-\frac{2rcos\theta}{R})+\frac{3}{8}(-\frac{2rcos\theta}{R})^{2})d\theta=\frac{Q\lambda}{2\pi\varepsilon_{0}}\int_{0}^{\theta}(1+\frac{r^{2}}{2R^{2}}(3cos^{2}\theta-1))d\theta$

$\frac{dU}{dr}=-F=\frac{Q\lambda}{2\pi\varepsilon_{0}}(1+\frac{r}{R^{2}}(\frac{3sin(2\theta)}{4}+\frac{\theta}{2}))=Ma$

If we take derivative of $\sqrt{\omega}$

$\frac{d\sqrt{\omega}}{d\theta}=0$ we get $\theta=\arccos(\frac{1}{\sqrt{3}})$ for maximum \omega

we can verify that it is maximum by second derivative test

$a=-\omega^{2}x$

$T=2\pi,\omega=1$

$M=\frac{Q\lambda}{2\pi\varepsilon_{0}R^{2}}(\frac{\arccos(\frac{1}{\sqrt{3}})}{2}+\sqrt{2})$