A grounded metallic sphere oscillating!

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Let us displace the sphere by an angle $\displaystyle \theta$

Note that when the displacement is small, the distance of the center of the sphere from the point charge $\displaystyle q$ is almost $\displaystyle L$

Now,
The trick here is to use Method of Images to find the induced charge on the grounded sphere.
It is pretty straightforward to prove that the induced charge has a magnitude of $\displaystyle Q = -\frac{R}{L}q$ and is located at a distance $\displaystyle d = \frac{R^2}{L}$ from the center of the sphere.

Thus,
The Force of attraction between the grounded sphere and $\displaystyle q$ is along the line joining the center and $\displaystyle q$ .

$F = \frac{kqQ}{\left(L-d\right)^2}$

$F = \frac{kq\left(\frac{R}{L}q\right)}{\left(L-\frac{R^2}{L}\right)^2}$

Considering the torque about the suspension point, (also use $\displaystyle \sin\theta \approx\theta$ )

$\tau_{net} = F(2\theta)L+mg(\theta)L$

Also, the moment of Inertia about the suspension point is,

$I_{net} = mL^2+ \frac{2}{5}mR^2$

Hence, using $\displaystyle \tau = I\alpha$ , we get,

$\left(\frac{2kq^2RL}{(L^2-R^2)^2}+mg\right)\theta = \left(mL^2+\frac{2}{5}mR^2\right)\alpha$

Substituting the values, we get,

$\alpha = (29.4545)\theta$

Since $\displaystyle\alpha = w^2\theta$ , we get,

$w = \sqrt{29.4545}$

Therefore, the Time Period of small oscillations is,

$T = \frac{2\pi}{w} = \frac{2\pi}{\sqrt{29.4545}} \approx \boxed{1.158sec}$