SHM in Electrostatics?

Magnitude of the force on the point charge due to an element $dq$ is $\large dF=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \frac { dqQ }{ { r }^{ 2 } }$ .

For every differential element of the ring, there is a diametrically opposite element. The components of forces along the axis will add up, while those perpendicular to it will cancel each other.

So, $\displaystyle F=\int { dF\cos { \theta =\cos { \theta \int { dF=\frac { x }{ r } \int { \frac { 1 }{ 4\pi { \varepsilon }_{ o } } \left[ -\frac { Qdq }{ { r }^{ 2 } } \right] } } } } }=-\frac { 1 }{ 4\pi { { \varepsilon }_{ o } } } \frac { Qqx }{ { \left( { a }^{ 2 }+{ x }^{ 2 } \right) }^{ 3/2 } }$ , where $r$ is the distance between the charge and the differential element of the ring, as ${ r }^{ 2 }=\left( { a }^{ 2 }+{ x }^{ 2 } \right)$ .

As the restoring force is not linear, the motion will be oscillatory. Also, ${ x }^{ 2 }<<{ a }^{ 2 }$ .

$\large F=-\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \frac { Qq }{ { a }^{ 3 } } =-kx$ and $\large k=\frac { Qq }{ 4\pi { \varepsilon }_{ o }{ a }^{ 3 } }$ .

Time period of simple harmonic motion is given by $\huge T=2\pi \sqrt { \frac { m }{ k } } =2\pi \sqrt { \frac { 4\pi { \varepsilon }_{ o }m{ a }^{ 3 } }{ qQ } }$ .