22 Jun 2025

LC Circuit and Damped Oscillations

Electrical SHM in an ideal LC circuit and the underdamped, critically damped, and overdamped solutions.

oscillations lc-circuit damping

Ideal LC circuit

Let $q(t)$ be the signed charge coordinate of a capacitor $C$ connected to an ideal inductor $L$, and choose the loop orientation so that the current is

\[i=\frac{dq}{dt}.\]

With no resistance or source, Kirchhoff’s loop rule is

\[L\frac{di}{dt}+\frac qC=0.\]

Since $di/dt=d^2q/dt^2$,

\[L\ddot q+\frac qC=0.\]

Division by $L$ gives

\[\boxed{\ddot q+\omega_0^2q=0}, \qquad \boxed{\omega_0=\frac1{\sqrt{LC}}}.\]

Thus charge executes SHM:

\[q=Q_0\cos(\omega_0t+\phi).\]

The current is

\[i=\dot q=-\omega_0Q_0\sin(\omega_0t+\phi).\]

The capacitor’s electric energy and the inductor’s magnetic energy are

\[U_E=\frac{q^2}{2C}, \qquad U_B=\frac12Li^2.\]

Substitution of $q$ and $i$, together with $L\omega_0^2=1/C$, gives

\[U_E=\frac{Q_0^2}{2C}\cos^2(\omega_0t+\phi),\] \[U_B=\frac{Q_0^2}{2C}\sin^2(\omega_0t+\phi).\]

Therefore

\[\boxed{U_E+U_B=\frac{Q_0^2}{2C}=\text{constant}}.\]

The conservation can also be checked directly:

\[\frac{d}{dt}(U_E+U_B) =\frac qC\dot q+Li\dot i =i\left(\frac qC+L\dot i\right)=0\]

by Kirchhoff’s equation. The mechanical correspondences are $q\leftrightarrow x$, $i\leftrightarrow\dot x$, $L\leftrightarrow m$, and $1/C\leftrightarrow k$.

Free damped oscillator

For a mechanical oscillator with viscous resistance $-b\dot x$, where $b>0$, Newton’s equation is

\[m\ddot x=-kx-b\dot x.\]

Hence

\[m\ddot x+b\dot x+kx=0.\]

Define

\[2\beta=\frac bm, \qquad \omega_0^2=\frac km.\]

The equation becomes

\[\boxed{\ddot x+2\beta\dot x+\omega_0^2x=0}.\]

With $x=e^{\lambda t}$,

\[(\lambda^2+2\beta\lambda+\omega_0^2)e^{\lambda t}=0,\]

so the characteristic roots are

\[\boxed{\lambda=-\beta\pm\sqrt{\beta^2-\omega_0^2}}.\]

Underdamping: $0<\beta<\omega_0$

Now $\beta^2-\omega_0^2<0$. Define

\[\omega_d=\sqrt{\omega_0^2-\beta^2}.\]

Then $\lambda=-\beta\pm i\omega_d$, and the real solution is

\[\boxed{x=Ae^{-\beta t}\cos(\omega_dt+\phi)}.\]

The system oscillates at the damped angular frequency $\omega_d<\omega_0$ inside the exact amplitude envelopes $\pm Ae^{-\beta t}$.

Critical damping: $\beta=\omega_0$

The characteristic equation has the repeated root $\lambda=-\beta$. Two independent solutions are $e^{-\beta t}$ and $te^{-\beta t}$, so

\[\boxed{x=(C_1+C_2t)e^{-\beta t}}.\]

The displacement returns to equilibrium without oscillating.

Overdamping: $\beta>\omega_0$

Both characteristic roots are real and negative. Therefore

\[\boxed{x=C_1e^{(-\beta+\sqrt{\beta^2-\omega_0^2})t} +C_2e^{(-\beta-\sqrt{\beta^2-\omega_0^2})t}}.\]

This motion also returns to equilibrium without oscillating.

For the mechanical energy

\[E=\frac12m\dot x^2+\frac12kx^2,\]

the exact rate of change is

\[\frac{dE}{dt} =\dot x(m\ddot x+kx) =-b\dot x^2\le0.\]

Thus damping removes energy monotonically. The displacement envelope decays exactly as $e^{-\beta t}$ in the underdamped case. Only in the weak-damping limit, $\beta\ll\omega_0$, may the energy averaged over a cycle be written as $\langle E\rangle\propto e^{-2\beta t}$.

The LC equation and energy, the three damped solutions, and the mechanical-energy loss are verified in the Maxima worksheet; every printed residual is zero.