We know E on the axis of the ring is: E=\\dfrac {k q x}{(r^2+x^2)^{3/2}} For small amplitudes r>>x : \\Rightarrow E=\\dfrac {k q x}{r^3} We can see that force is always applied in the opposite direction of displacement i.e: F=\\dfrac {-k q x}{r^3}(1\ imes 10^{-6})C \\Rightarrow a=\\dfrac {-k q x (1\ imes 10^{-6})}{m r^3} We know for a particle in S. H. M: a=-\\omega^2 x \ herefore \\omega^2=\\dfrac {k q (1\ imes 10^{-6})}{m r^3}=\\dfrac {9\ imes 10^9\ imes 1\ imes 10^{-5}\ imes 1\ imes 10^{-6}}{0\\cdot 9\ imes 10^{-3}\ imes 1^3}=100 \\Rightarrow \\omega=10 Now, T=\\dfrac {2\\pi}{w}=\\dfrac {2\ imes 3\\cdot 14}{10}=0.63s