Answer : (c) $\;2 \pi\;\sqrt{\large\frac{ 2 \in_{0}\;m\;R^2}{\lambda\;Q}}$

Explanation :

Electric potential on the axis of ring at a distance x is =$\; \large\frac{k\;q}{(x^2+R^2)^{\large\frac{1}{2}}}$

and Electric field on the axis of ring is $\;E_{x}=-\large\frac{\partial V}{\partial x } \hat{i}$

$E_{x}=-(-\large\frac{1}{2}\;\large\frac{k\;q\;2\;x}{(x^2R^2)^{\large\frac{3}{2}}})$

$E_{x}=\large\frac{k\;q\;x}{(x^2+R^2)^{\large\frac{3}{2}}}=\large\frac{\lambda R x}{2 \in_{0} (x^2+R^2)^{\large\frac{3}{2}} }$

Restoring force $F_{res}=-QE_{x}$

$=-\large\frac{Q \lambda R x }{2 \in_{0} (x^2+R^2)^{\large\frac{3}{2}} }$

Since x < < R

$F_{res} \approx -\large\frac{Q \lambda R x }{2 \in_{0} R^3}$

$F_{res}\;\propto \; -x\;.$ Thus the particle perform simple harmonic motion

$a=\large\frac{F_{res}}{m}=-\large\frac{Q \lambda R x}{m R^3 2 \in_{0}}=-w^2 x$

$w=\sqrt{\large\frac{Q \lambda R}{2 \in_{0} m R^3}}$

$T=\large\frac{2 \pi}{w}=2 \pi \sqrt{\large\frac{2 \in_{0} m R^2}{Q \lambda}}$