As shown in figure, the two parallel conducting rails, in a...

From energy conservation (there are no dissipative forces)-

$E = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}LI^{2}$

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

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

The emf induced in an inductor is

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

$⇒ L\dfrac{dI}{dt} = Bl\dfrac{dx}{dt} ⇒ LdI = Bldx ⇒ I = \dfrac{Bl}{L}x$

The force acting on a current-carrying conductor is

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

$⇒ \dfrac{dv}{dt} = - \dfrac{IBl}{m}$

Substituting the value of $I$ we get

$\dfrac{dv}{dt} = - \left(\dfrac{Bl}{L}x\right)\dfrac{Bl}{m} ⇒ a = - \left(\dfrac{B^{2}l^{2}}{mL}\right)x$

We got acceleration in the form of

$a = - {ω}^{2}x$

this implies that the motion is& SHM, with angular frequency

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

The time period of motion is
$⇒ T = 2\pi\dfrac{\sqrt{mL}}{Bl}$