We have in the circuit at a certain instant of time $$(t)$$, from Faraday's law of electromagnetic induction: $$L dfrac{di}{dt}= Bl dfrac{dx}{dt}$$ or $$L di=B l dx$$ As at $$t=0, xc=0,$$ so $$L i=B l x $$ or $$i=dfrac{B l}{L}x (1)$$ For the rod from the second law of motion $$F_x=m w_x$$ $$- i l B= m ddot x$$ Using Eqn. $$(1)$$, we get :$$ddot x=- left( dfrac{l^2 B62}{mL}\right)x=- -omega^2_0 x ...(2)$$ Where The solution of the above differential equation is of the from $$x= sin (omega_0 t+ alpha)$$ From the initial cosition art $$t=0, x=0$$ so $$alpha=0$$ Hence, $$x=a sin omega_0 t    (3)$$ Differentiating w.r.t. time , $$dot x= a omega_0 cosomega_0 t$$ But from the initial condition of the problem at $$t=0, dot x=v_0$$ Thus $$v_0-a omega_0$$ or $$a=v_0/ omega_0 (4)$$ Putting the value of $$a$$ from Eqsn $$(4)$$ into Eqn $$(3)$$, we obtained $$x=dfrac{v_0}{omega_0} sin omega_) t left( where omega_0=dfrac{lB}{sqrt{mL}}\right)$$