Total charge $-\,Q$ is uniformly spread along length of a ring of radius $R$. A small test charge $+q$ of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.

$(a) $ Show that the particle executes a simple harmonic oscillation.

$(b)$ Obtain its time period.

The force acting on a point charge $q$ at point $\mathrm{P}$ at distance $x<<<<\mathrm{R}$ from the centre of ring of radius $\mathrm{R}$ on its axis by $(-d \mathrm{Q})$ charge at $\mathrm{A}$ on ring is,

$d \mathrm{~F}=k \frac{(-d \mathrm{Q}) q}{\left(\sqrt{\mathrm{R}^{2}+x^{2}}\right)^{2}}=-k \frac{(d \mathrm{Q}) q}{\left(\mathrm{R}^{2}+x^{2}\right)}$

As shown in figure, $d \mathrm{Fsin} \theta$ are same in magnitude but opposite in direction. Hence, they cancelled the effect of each other. Hence, $F$ will be the sum of only $d \mathrm{~F} \cos \theta $ components which are parallel to axis and towards centre $\mathrm{O}$.

$\mathrm{F}=\oint d \mathrm{~F} \cos \theta$

$=\oint-k \frac{(d \mathrm{Q}) q}{\left(\mathrm{R}^{2}+x^{2}\right)} \times \frac{x}{\sqrt{\mathrm{R}^{2}+x^{2}}}$

$=k q \frac{x}{\left(\mathrm{R}^{2}+x^{2}\right)^{3 / 2}} \oint(-d \mathrm{Q})$

$=k q \frac{x}{\left(\mathrm{R}^{2}+x^{2}\right)^{3 / 2}}(-\mathrm{Q})$