Answer
$T=2\pi\sqrt{\frac{l}{2g}}$
Work Step by Step
The natural period of oscillation can be determined as follows:
$\Sigma M=I\theta$
$\implies I\theta=-3Tsin(\phi)R$eq(1)
We know that
$\Sigma F=0$
$\implies 3T-mg=0$
$\implies T=\frac{mg}{3}$
We plug in the know values in eq(1) to obtain:
$\frac{1}{2}mR^2\theta+3(\frac{mg}{3})\times (\frac{R}{l}\theta)\times R=0$
$\implies \frac{1}{2}mR^2\theta+\frac{R^2}{l}mg\theta=0$
Now $\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$
$\implies \omega_n=\sqrt{\frac{mgR^2/l}{mR^2/2}}$
$\implies \omega_n=\sqrt{\frac{2g}{l}}$
Now $T=\frac{2\pi}{\omega}$
$\implies T=2\pi\sqrt{\frac{l}{2g}}$
