First, the initial charge of the capacitor is Q0, and the initial displacement of the inductor is 0. The magnetic field is perpendicular to the paper, and the inductance is on the right side of the capacitor. Then specify the symbol, the symbol of the charge on the capacitor plate is the symbol of the capacitor charge, the inductance displacement is positive to the right, and the current direction is positive counterclockwise.

The voltage across the capacitor at any moment is equal to the voltage across the inductor (including the self-induced electromotive force of the inductor and the induced electromotive force generated by cutting).

Bldx/dt-LdI/dt=Q/C ( 1)

Ampere force provides the acceleration of the sensor.

-bli = ma = MD 2x/(dt) 2 (2) (note that the current direction is counterclockwise! )

In ...

I=dQ/dt (3)

Initial condition

When t=0

x=0 (4)

v=dx/dt=0 (5)

Q=Q0 (6)

I=0 (7)

From the initial conditions of (1) ~ (3) and (4) ~ (6), we can get the displacement.

x(t)=At+Bsin(ωt)

In ...

A=Q0Bl/(B^2*l^2*C+m)

b=q0blmcl/((b^2*l^2*c+m)(b^2*l^2*c^2*ml+clm^2)^( 1/2))

ω=((b^2*l^2*c+m)/(cml))^( 1/2)

Therefore, the motion of the sensor is the superposition of uniform linear motion and simple harmonic motion.

Note: I sorted out the solutions of the above differential equations by myself and solved them with the mathematical software Marlab. You can do it by hand, but it's more troublesome.