In the arrangements shown in Fig. there is a uniform magnetic filed $B_0$ normal to the plane of paper. The connector is smooth and conducting and is has a mass $m$ and length $l$.

The connector is pushed against the spring so that the spring has compression $x_0$. The connector is released at $t = 0$. Find the time it will take to come to its original position again. The spring is non-conducting. The resistance of the rails is zero and neglect its self-inductance.

Let us say at any time $t$ the rod is at a distance $x$ from its original position.

$V=\frac{dx}{dt}$

$e\left(\text{induced emf}\right)=Blv=Bl\frac{dx}{dt}$

Charge on the capacitor $q=Ce=BlC\frac{dx}{dt}$

Current $i=\frac{dq}{dt}=BlC\frac{d^{2}x}{d{t}^2}$

Force on the conductor due to this current 

$F=ilB=B^2{l}^{2}C\frac{d^{2}x}{d{t}^2}$

Net force on the connector to the right, 

$m\frac{d^{2}x}{d{t}^2}=-B^2{l}^{2}C\frac{d^{2}x}{d{t}^2}+kx$

$\Rightarrow \frac{d^{2}x}{d{t}^2}=\frac{k}{m+B^2{l}^{2}C}x$

The acceleration is proportional to $x$ and opposite to it and hence we can say the motion will be SHM.

With $T=2\pi \sqrt{\frac{m+B^2{l}^{2}C}{k}}$

And the time taken will be $\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{m+B^2{l}^{2}C}{k}}.$