Coupled LC circuit with a moving core and capacitor plate

A series LC circuit is fed by a voltage source. The coil's iron core and one capacitor plate are both attached to a mass on a spring and damper, so both $L$ and $C$ depend on the position $x$. The paper derives the coupled electromechanical equations carefully, and it rightly questions the sign of the given $L(x)$. Its simulations are reproduced well. Its most interesting observation is also right: the electrical oscillation dies out although the circuit has no resistance. The energy leaks through the coupling into the mechanical damper. However, the paper's explanations of how the parameters affect this decay are mostly wrong. A stronger damper, for example, makes the decay slower. The interactive simulator shows the energy flow and a decay-time formula that explains all the trends.

Problem statement

The capacitance $C$ and inductance $L$ of the system in the figure depend on the position $x$:

$$C(x) = \frac{C_0d_0}{d_0 + x},\qquad L(x) = L_0\Big(1 - \frac{x}{x_0}\Big)$$

where $C_0$, $L_0$, $d_0$ and $x_0$ are constants and $x$ is the displacement of the core. Determine the equations of motion and draw the dynamic structure, assuming the spring is relaxed at $x = 0$.

The mass $m$ carries the right plate of the capacitor and the iron core of the coil. Positive $x$ widens the capacitor gap and pushes the core further into the coil.
The mass is held by a spring $k$ and a damper $B$.
The capacitor and coil form a series circuit with a voltage source $U$ (a step, or a sine $U_0\sin\omega t$). The circuit has no resistor.

The generalized coordinates are $x$ and the charge $q$ (one loop, so one charge coordinate).

Figure 1. Capacitor plate and coil core attached to the mass m (spring k, damper B). The series circuit C(x)–L(x) is fed by U. Positive x widens the capacitor gap and pushes the core into the coil.

The sign of L(x)

The paper first checks the given dependences against the geometry. For the capacitor, $C = \varepsilon_0S/(d_0+x)$ leads exactly to the given $C(x) = C_0d_0/(d_0+x)$.

For the coil the paper applies Ampère's law. A coil of length $l$ has an iron-filled part (permeability $\mu_0\mu_j$) and an air part of length $x_0 - x$. If the iron's permeability is much larger, only the air part counts:

$$L = \mu_0\frac{N^2S}{x_0 - x} = L_0\frac{x_0}{x_0-x} \approx L_0\Big(1 + \frac{x}{x_0}\Big)\quad(x\ll x_0)$$

Pushing the core further in ($x > 0$) must increase the inductance. The given $L_0(1 - x/x_0)$ has the wrong sign for the drawn geometry. The paper notices this and uses $L_0(1+x/x_0)$. That is a sound engineering decision, well argued, and it should be the first thing a reader of this problem notices. In this system the choice hardly changes the results, because the magnetic force is small: the current is almost zero in the steady state. The simulator lets you switch between the given, linearized and exact forms.

Equations of motion

TerminologyAs in the course: repulsive coenergy $\mathcal{R}^* = \tfrac12m\dot x^2$, attractive energy $\mathcal{A} = \tfrac12kx^2$, magnetic coenergy $W_m^* = \tfrac12L(x)\dot q^2$, electric energy $W_e = q^2/(2C(x))$, with $\mathcal{L} = \mathcal{R}^* + W_m^* - \mathcal{A} - W_e$.

$$\mathcal{L} = \tfrac12m\dot x^2 + \tfrac12L_0\Big(1+\frac x{x_0}\Big)\dot q^2 - \tfrac12kx^2 - \frac{q^2(d_0+x)}{2C_0d_0},\qquad \mathcal{D} = \tfrac12B\dot x^2$$

The voltage $U$ is the generalized force of $q$. Lagrange's equations give:

(1)$$m\ddot x + B\dot x + kx = \frac{L_0}{2x_0}\dot q^2 - \frac{q^2}{2C_0d_0}$$

(2)$$L_0\Big(1+\frac x{x_0}\Big)\ddot q + \frac{L_0}{x_0}\dot x\dot q + \frac{q\,(d_0+x)}{C_0d_0} = U$$

Mechanical equation (1). The right-hand side holds the two electrical forces. The magnetic one pulls the core into the coil ($+x$). The electric one pulls the plate towards the fixed plate ($-x$).
Electrical equation (2). It contains the motional EMF $\dot L\dot q$ and the capacitor voltage $q/C(x)$.

The derivation and the physical interpretation of each term are correct.

Steady state for a DC voltage

With all derivatives zero, $\dot q = 0$ and therefore the magnetic force vanishes:

$$q^* = \frac{UC_0d_0}{d_0+x^*},\qquad x^* = -\frac{q^{*2}}{2C_0d_0k}$$

For the base parameters ($U=5$ V, $C_0=5\cdot10^{-5}$ F, $d_0=5$ mm, $k=300$ N/m) this gives $x^* = -0.52$ mm and $q^* = 2.8\cdot10^{-4}$ C. $L$, $m$ and $B$ do not appear, as the paper correctly notes.

Time scales

With $L_0 = 0.1$ H the electrical circuit oscillates at $\omega_e = 1/\sqrt{L_0C_0} = 447$ rad/s (period 14 ms). The mechanical part has $\omega_n = \sqrt{k/m} = 55$ rad/s and $\zeta = B/(2\sqrt{km}) = 27$, so it is heavily overdamped. Its relaxation time is about $B/k = 1$ s.

Simulink model and results

The model solves (1) and (2) for $\ddot x$ and $\ddot q$, using seven constants $C_1\ldots C_7$ and four subsystems for the nonlinear terms. Reference parameters: $L_0 = 0.1$ H, $x_0 = 0.1$ m, $m = 0.1$ kg, $k = 300$ N/m, $C_0 = 5\cdot10^{-5}$ F, $d_0 = 5$ mm, $B = 300$ N·s/m, a step $U = 5$ V, and for the sine $U_0 = 1$ V, $\omega = 1$ rad/s. Each figure varies one parameter around these values.

Varied	Paper's observation (settling of q)	Simulation of (1)–(2)
Reference	q oscillates between 0 and 5·10⁻⁴ C and settles in ≈ 25 s; x dips to −0.64 mm and settles at −0.52 mm	Reproduced
L₀ ↑	Slower settling, "because the natural frequency increases"	Almost no effect (slightly faster); a larger L lowers ω_e
m ↑	"Drastically" faster settling	Slightly slower
k ↑	No effect on settling; smaller \|x\| and q	Smaller \|x*\| ✓; settling slightly slower
B ↑	Faster settling: "B is the only dissipative element"	Slower settling: B = 350 is slower, and with B = 30 the oscillation left after 15 s is about 60× smaller
U ↑	Larger q, \|x\|, faster settling: "the source has more power"	Faster ✓, but for a different reason (section 5)
Sine U₀ sin ωt	q follows the voltage; x oscillates at about 2ω around a shifted mean; the x amplitude falls with ω	Reproduced

The conclusion stresses that the electrical oscillation settles although the circuit has no resistor. The energy is drained through the coupling into the mechanical damper. This observation is correct and is the most interesting result of the paper.

Why the oscillation decays, and how fast

After the voltage step the charge oscillates about $q^*$: $q \approx q^* + a\cos\omega_et$. The electric force contains $q^2 \approx q^{*2} + 2q^*a\cos\omega_et + \ldots$, so the electrical oscillation shakes the mass at $\omega_e$ with a force amplitude

$$F_1 = \frac{q^*a}{C_0d_0}$$

The mass responds with velocity $F_1/|Z|$, where $Z = B + i\,(m\omega_e - k/\omega_e)$ is the mechanical impedance. The damper dissipates $\tfrac12B|v|^2$. Dividing by the electrical energy $\tfrac12a^2/C_0$ gives the decay rate of the oscillation amplitude:

$$\sigma \approx \frac{B\,q^{*2}}{2C_0d_0^2\,\big[B^2 + (m\omega_e - k/\omega_e)^2\big]}$$

For the reference values $\sigma \approx 0.1\ \text{s}^{-1}$, i.e. a time constant of about 10 s, which matches the simulations. The formula explains all the trends:

Damper $B$: $\sigma$ is largest for $B = |m\omega_e - k/\omega_e| \approx 44$ N·s/m. Above that, a larger $B$ reduces the decay rate, since $\sigma\propto1/B$ for large $B$. A very strong damper barely lets the mass move and therefore dissipates almost nothing, just as in problem 16.
Voltage $U$ and capacitance $C_0$: they act through $q^{*2}$. A larger static charge couples the circuit more strongly to the mass.
Mass $m$: a larger mass makes the mass harder to shake at $\omega_e$, so the decay is slightly slower.
Inductance $L_0$: it only enters through $\omega_e$ in the small reactive term, so its effect is very small.

Even a small resistance changes the picture completely. A series resistor $R$ adds $\sigma_R = R/(2L_0)$, which is 10 s⁻¹ for $R = 2\ \Omega$, a hundred times faster than the mechanical leak (preset "R = 2 Ω"). Any real coil has some resistance.

Critical review

The damping explanation contradicts the physics

The paper states that a larger $B$ shortens the settling time because $B$ is the only dissipative element. With $B = 300$ N·s/m the system is far on the "too strong" side of the optimum (≈ 44 N·s/m). Our simulation of the paper's own equations gives slower settling for $B = 350$ than for $B = 300$, and with $B = 30$ the oscillation remaining after 15 s is about 60 times smaller. The paper's plot for this case is either mislabelled or misread.

Mass, inductance and voltage: wrong reasons

Mass: a heavier mass does not make the conversion of energy "faster". It reduces the coupling, and the simulated effect is small and in the opposite direction.
Inductance: increasing $L_0$ lowers, not raises, the natural frequency $1/\sqrt{LC}$, and it hardly affects the settling.
Voltage: a higher voltage speeds up the settling because the static charge, and with it the coupling force, grows, not because the source "has more power".
Frequency: under a sine input the $x$ amplitude falls with frequency because the overdamped mechanics act as a low-pass filter with corner frequency $k/B = 1$ rad/s, not because of shorter half-periods.

A circuit without resistance

The model has no resistor. The paper admits that its parameters are "not quite realistic". This choice makes the slow mechanical drain the only loss and hides the fact that a real coil's resistance would end the oscillation within a fraction of a second. A short remark with an estimate ($\tau = 2L/R$) would put the result in perspective.

Smaller points

The steady-state section does not mention that, with a DC step, the steady state is reached only after the electrical oscillation has decayed, which takes tens of seconds here. The plots of $q$ are dense bands of oscillations at 447 rad/s, and a plot of their envelope or of the energy would show the decay much better. The plot colours of several figures are hard to tell apart.

What is goodThe derivation of $L(x)$ from Ampère's law, and the resulting correction of the sign in the given formula, is excellent and well argued. The Lagrangian, the equations of motion and the steady state are correct. The Simulink model is reproduced, and the key observation about energy draining through the coupling is correct and interesting.

Interactive simulator

The simulator integrates (1)–(2) with 4th-order Runge–Kutta, using a time step of at most 0.1 ms, which resolves the 14 ms electrical period. It also tracks the electric, mechanical and dissipated energies.

Cases

Voltage

Inductance

Parameters

L₀ [H]

x₀ [m]

C₀ [F]

d₀ [m]

m [kg]

k [N/m]

B [N·s/m]

R [Ω] (extension)

U [V] (step)

U₀ [V] (sine)

ω [rad/s] (sine)

Duration [s]

Charge q(t)

Displacement x(t)

Current i(t)

Energy flow

The electrical energy is gradually transferred to the damper.

Animation

Results table

Review questions

Why must the inductance increase when the core moves into the coil?

Iron has a far higher permeability than air, so the magnetic reluctance is mainly in the air-filled part of the coil. Pushing the core in shortens that part, the reluctance drops and $L = N^2/\mathcal{R}_m$ rises.

Which way do the two electrical forces push the mass?

The magnetic force $\tfrac12i^2\,dL/dx$ pulls the core into the coil ($+x$). The electric force $-q^2/(2C_0d_0)$ pulls the plate towards the fixed plate ($-x$). In the DC steady state only the electric force remains.

How can an LC circuit without a resistor settle?

Through the coupling. The oscillating charge produces an oscillating electric force on the mass, the mass moves a little, and the damper dissipates that energy. The electrical oscillation loses energy to the mechanics.

Why does a larger damper constant make that decay slower here?

The mass is driven at 447 rad/s. With a huge $B$ it barely moves, so the dissipated power $B\dot x^2$ is small. The power is largest when $B$ matches the reactive part $|m\omega_e - k/\omega_e|$.

Why does the steady state not depend on $L$, $m$ and $B$?

In the steady state there is no current and no velocity, so the energies stored in the inertial elements are zero and nothing is dissipated. Only the capacitor, the spring and the source determine the equilibrium.