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.
Let us draw the figure according to question,
A gentle push on q along the axis of the ring gives rise to the situation shown in the figure below.
Taking line elements of charge at $A$ and $B$, having unit length, then charge on each elements.
$$d F=2\left(-\frac{Q}{2 \pi R}\right) q \times \frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cos \theta$$
Total force on the charge $q$, due to entire ring
$$\begin{aligned} & F=-\frac{Q q}{\pi R}(\pi R) \cdot \frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cdot \frac{2}{r} \\ \text{Here}, Z<< R, \quad & F=-\frac{Q q z}{4 \pi \varepsilon_0\left(Z^2+R^2\right)^{3 / 2}} \end{aligned}$$
where
$$\begin{gathered} \frac{Q q}{4 \pi \varepsilon_0 R^3}=\text { constant } \\ \Rightarrow\quad F \propto-Z \end{gathered}$$
Clearly, force on $q$ is proportional to negative of its displacement. Therefore, motion of $q$ is simple harmonic.
$$\begin{aligned} & \omega=\sqrt{\frac{K}{m}} \text { and } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \\ & T=2 \pi \sqrt{\frac{m 4 \pi \varepsilon_0 R^3}{Q q}} \\ \Rightarrow\quad & T=2 \pi \sqrt{\frac{4 \pi \varepsilon_0 m R^3}{Q q}} \end{aligned}$$