From energy conservation (there are no dissipative forces)-

$E=\frac{1}{2}m{v}^2+\frac{1}{2}L{I}^2$

$\Rightarrow \frac{m}{2}\times 2v\frac{dv}{dt}+\frac{L}{2}\times 2I\frac{dI}{dt}=0$

$mv\frac{dv}{dt}=-IL\frac{dI}{dt}$

The emf induced in an inductor is

$E=L\frac{dI}{dt}=Blv$

$$\begin{gathered} \Rightarrow L\frac{dI}{dt}=Bl\frac{dx}{dt} \\ \Rightarrow LdI=Bldx \\ \Rightarrow I=\frac{Bl}{L}x \end{gathered}$$

The force acting on a current-carrying conductor is

$m\frac{dv}{dt}=-IBl$

$\Rightarrow \frac{dv}{dt}=-\frac{IBl}{m}$

Substituting the value of $I$ we get

$$\begin{gathered} \frac{dv}{dt}=-\left(\frac{{\displaystyle Bl}}{{\displaystyle L}}x\right)\frac{Bl}{m} \\ \Rightarrow a=-\left(\frac{{\displaystyle B^2{l}^2}}{{\displaystyle mL}}\right)x \end{gathered}$$

We got acceleration in the form of

$a=-{\omega }^{2}x$ 

this implies that the motion is SHM, with angular frequency

$\omega =\sqrt{\frac{B^2{l}^2}{mL}}$

The time period of motion is
$\Rightarrow T=2\pi \frac{\sqrt{mL}}{B\mathcal{l}}$