Solution: Resonance of a Coupled Electromechanical System

Let $Q(t)$ be the charge on the capacitor plate, $I(t) = \dot{Q}$ be the current, and $x(t)$ be the displacement of the particle from the center position.

1. Equations of Motion

Mechanical Equation (Newton’s Second Law):
The electric field between the plates is $E = Q/Cd$. The force on the particle is $F = qE$. $$ m\ddot{x} = \frac{qQ}{Cd} \quad \dots(1) $$

Circuit Equation (Conservation of Energy):
The total energy $U$ of the system is conserved. This includes magnetic energy, electrostatic potential energy, mechanical kinetic energy, and the interaction potential energy of the charge in the field ($U_{int} = -qEx$). $$ U = \frac{1}{2}L\dot{Q}^2 + \frac{Q^2}{2C} + \frac{1}{2}m\dot{x}^2 – \frac{qQx}{Cd} = \text{constant} $$ Differentiating with respect to time ($dU/dt = 0$): $$ \dot{Q} \left( L\ddot{Q} + \frac{Q}{C} – \frac{qx}{Cd} \right) + \dot{x} \left( m\ddot{x} – \frac{qQ}{Cd} \right) = 0 $$ The second term is zero by Newton’s law (Eq. 1). Therefore, the first term must also be zero: $$ L\ddot{Q} + \frac{Q}{C} = \frac{qx}{Cd} \quad \dots(2) $$

2. Decoupling the Equations

To eliminate $x$, we differentiate Eq. (2) twice with respect to time: $$ L \frac{d^4Q}{dt^4} + \frac{1}{C}\ddot{Q} = \frac{q}{Cd}\ddot{x} $$ Substituting $\ddot{x}$ from Eq. (1): $$ L \frac{d^4Q}{dt^4} + \frac{1}{C}\ddot{Q} – \frac{q^2}{mC^2d^2}Q = 0 $$

3. Solving for Frequency

For a resonating system, we assume a harmonic solution $Q(t) = Q_0 \cos(\omega t)$. Substituting $\frac{d^4Q}{dt^4} = \omega^4 Q$ and $\ddot{Q} = -\omega^2 Q$: $$ L\omega^4 – \frac{\omega^2}{C} – \frac{q^2}{mC^2d^2} = 0 $$ Let $\omega_0^2 = \frac{1}{LC}$. The equation becomes: $$ \omega^4 – \omega_0^2 \omega^2 – \frac{q^2 \omega_0^2}{mCd^2} = 0 $$ Solving this quadratic equation for $\omega^2$: $$ \omega^2 = \frac{\omega_0^2 + \sqrt{\omega_0^4 + \frac{4q^2 \omega_0^2}{mCd^2}}}{2} = \frac{\omega_0^2}{2} + \sqrt{\frac{\omega_0^4}{4} + \frac{q^2 \omega_0^2}{mCd^2}} $$