Electromechanical Oscillation

Consider the mass to be at a distance $x$ from the wall. Let the natural length of the spring be $L_o$ . At a general instant of time $t$ let the rod be moving rightward with a speed $\dot{x}$ . As the rod moves in the magnetic field, a voltage is induced across its ends. The magnitude of this voltage is:

$E = BH\dot{x}$ $E = \dot{x}$

Let the charge on the capacitor be $Q$ and the current through the inductor be $I$ . Using Kirchoff's laws gives:

$-\dot{x} + \dot{I} + Q=0 \ \dots (1)$ $I = \dot{Q}\ \dots (2)$

Now, as the rod moves rightward since a current flows through it, it experiences two forces which oppose its motion. One is the spring force while the other is the magnetic force $IHB$ . Applying Newton's second law gives: $m\ddot{x} = -K(x - L_o) - IHB$

$\ddot{x} = -x + L_o -I \ \dots (3)$

Double differentiating (3) gives:

$\frac{d^4x}{dt^4} + \frac{d^2x}{dt^2} = - \frac{d^2I}{dt^2} \ \dots (4)$

Differentiating (1) gives:

$\frac{d^2x}{dt^2} - I = \frac{d^2I}{dt^2}$

Substituting the above in (4) gives:

$\frac{d^4x}{dt^4} + \frac{d^2x}{dt^2} = -\frac{d^2x}{dt^2} + I$

Now substituting (3) above gives:

$\frac{d^4x}{dt^4} + 2\frac{d^2x}{dt^2} = -x + L_o -\frac{d^2x}{dt^2}$ $\implies \frac{d^4x}{dt^4} + 3\frac{d^2x}{dt^2} +x = L_o$

Finally, substituting $x-L_o = y$ in the above differential equation gives:

$\boxed{\frac{d^4y}{dt^4} + 3\frac{d^2y}{dt^2} +y=0}$

Its characteristic equation is:

$m^4 + 3m^2 + 1 = 0$

Finally, This is a standard linear fourth-order differential equation with a general solution:

$y = A\sin\left(\omega_1 t\right) + B\cos\left(\omega_1 t\right) + C\sin\left(\omega_2 t\right) + D\cos\left(\omega_2 t\right)$

Here: $A$ , $B$ , $C$ and $D$ are arbitrary constants and:

$\omega_1 = \sqrt{\frac{3+\sqrt{5}}{2}}$ $\omega_2 = \sqrt{\frac{3-\sqrt{5}}{2}}$

Finally, after simplifying:

$\boxed{(\omega_1+\omega_2)^2 = 5}$

solution will be posted later