A ring of radius R carries a uniformly distributed charge +Q. A point charge $-q$ is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle executes a ismple harmonic motion along the axis of the ring.

Solution: KEY CONCEPT: Force on charge $\left(-q\right)$ due to small charge $dq$ istuated at length $dl$ is given by

$dF=k\frac{qdq}{5R^2}$
Resolving this force into two parts $dF\cos \theta$ and $dFisn\theta$ as shown in figure.
If we take another diametrically oppoiste lenth $dl$, the charge on it being $dq$. Then the force on charge $\left(-q\right)$ by this small charge $dq$ will be
$dF=k\frac{qdq}{5R^2}$
Again resolving this force, we find $dFisn\theta$ components of the two forces cancel out and $dF\cos \theta$ component adds up.
$\therefore$ The total force
$F={\int }_0^{2\pi R}dF\cos \theta ={\int }_0^{2\pi R}\frac{kqdq}{5R^2}\times \frac{2R}{\sqrt{5}R}$
charge on length $2\pi R=Q$
$\therefore$ Charge on lenth $dl=\frac{Qdl}{2\pi R}=dq$
$\therefore$ $F={\int }_0^{2\pi R}\frac{2kq}{5\sqrt{5}{R}^2}\times \frac{Qdl}{2\pi R}$
$=\frac{2kQq}{5\sqrt{5}\times 2\pi R^3}\times 2\pi R=\frac{2kQq}{5\sqrt{5}{R}^2}$
This is not an equation of ismple harmonic motion.