Awesome SHM!

Relevant wiki: Simple Harmonic Motion - Problem Solving

Here is my solution which is compact enough.

Lets displace the ring through an angle $d \theta$ , which implies that the string moved a distance of $2 \times a \times d \theta$ [As is rotates about the end so distance is $2a$ ]

Which means that the spring is compressed by $2 a d \theta$ .

So our first equation can be $T=k \times 2a d \theta$ .

Now lets write the equation for rotational motion.

For the ring rotating about the end in vertical plane.. we can find it by this method.

By perpendicular axis theorem..

$2I_{x=y} = I_{cm}$

$I_{x=y}= \dfrac{ma^2}{2}$

$\implies I_{\text{about end and moving in vertical plane}} = ma^2 + \dfrac{1}{2} ma^2$

$\implies I_{needed}=\dfrac{3}{2} mr^2$

So we can write the equation of rotation as

$T \times 2a = I_{needed} \alpha$

$k \times 2a \times d \theta \times 2a = \dfrac{3}{2} ma^2 \alpha$

$\implies \dfrac{\alpha}{\theta} = \dfrac{8k}{3m}$

$\implies \boxed{\boxed{T= 2 \pi \sqrt{\dfrac{3m}{8k}}}}$

First let's find the initial extension of the spring $x_{0}$

Balancing torques about the hinge we get :

$x_{0}=\dfrac{mg}{2k}$

Consider the situation when the ring makes an angle $\theta$ with the horizontal. Since there is no dissipation of energy , we have :

$mga\sin\theta+\dfrac{1}{2}k(x_{0}-2a\sin\theta)^2+\dfrac{3}{4}ma^2\omega^2=C$

Where C is a constant.

Using the small angle approximation $\sin\theta\approx\theta$ and differentiating the above equation we get:

$\alpha=-\dfrac{8k}{3m}\theta$

Comparing the above equation with :

$\alpha=-\omega^2\theta$

$\omega=\sqrt{\dfrac{8k}{3m}}$

$\boxed{T=2π\sqrt{\dfrac{3m}{8k}}}$

First diagram is in equilibrium that is possible if torque on ring is zero

2a×T(tension in string) = mg ×a (torque due to mass of ring)

And tension = spring force kx0( x0 = initial extension)

Now let the ring is displaced with a small angle theta and it's point at which string is connected is at a distance x below initial position. Length of string is same so extension in spring is x + x0

Now torque on ring about an axis passing through at hinge's position and parallel to the axis parallel to the plane of ring passing through its centre.

Mgacosϴ -T×2acosϴ =I×alpha

T=mg/2 also T=kx0

Kx0= mg/2 -------(1)

Mgacosϴ -2kaxcosϴ-2kx0×acosϴ= I alpha From (1)

-2kaxcosϴ =3/2 ma² alpha

-4ka²sinϴcosϴ=3/2ma²× alpha

( Since sinϴ =x/2a )

For small angle sinϴ~ϴ and cosϴ~1

Alpha = -8/3 k/m ϴ

W =√8k/3m

Time period = 2π √3m/8k