\subsection{LC Oscillations}

This subsection introduces the LC circuit -- a capacitor and inductor connected in a closed loop with no resistance -- derives the second-order differential equation governing charge evolution, identifies the sinusoidal oscillation of charge and current, defines the natural angular frequency $\omega = 1/\sqrt{LC}$, and analyses the continuous energy exchange between the capacitor and inductor.

\dfn{LC circuit}{An \emph{LC circuit} consists of an inductor of inductance $L$ and a capacitor of capacitance $C$ connected in a single closed loop. No resistor is present (ideal components). If the capacitor is initially charged and then connected to the inductor, the charge on the capacitor and the current through the inductor oscillate sinusoidally in time.}

\nt{Think of an LC circuit as the electrical analogue of a frictionless spring-mass system. The capacitor stores energy in its electric field (just as a spring stores potential energy), and the inductor stores energy in its magnetic field (just as a moving mass has kinetic energy). The charge oscillates between the two plates of the capacitor while the current alternates direction through the inductor, and the total energy remains constant.}

\thm{Oscillation of charge and current}{In a series LC circuit, let $L$ be the inductance and $C$ the capacitance. If at $t = 0$ the capacitor carries charge $q(0) = Q_{\max}$ and the current is $I(0) = 0$, then the charge on the capacitor at time $t$ is
\[
q(t) = Q_{\max}\,\cos(\omega t),
\]
and the current through the inductor is
\[
I(t) = \frac{dq}{dt} = -\,\omega\,Q_{\max}\,\sin(\omega t).
\]
Here $\omega = \dfrac{1}{\sqrt{LC}}$ is the \emph{angular frequency} of oscillation (in rad/s), $Q_{\max}$ is the maximum charge on the capacitor, and the phase constant is $\phi = 0$ for these initial conditions. The period of oscillation is
\[
T = \frac{2\pi}{\omega} = 2\pi\sqrt{LC}.
\]
The maximum current (current amplitude) is
\[
I_{\max} = \omega\,Q_{\max} = \frac{Q_{\max}}{\sqrt{LC}}.
\]}

\pf{Derivation of LC oscillations}{Apply Kirchhoff's voltage law around the loop. Going around, the voltage drop across the capacitor is $q/C$ and the voltage drop across the inductor is $L\,(dI/dt)$. Since the sum of voltage drops around a closed loop is zero:
\[
\frac{q}{C} + L\,\frac{dI}{dt} = 0.
\]
The current is the rate of change of charge on the capacitor. When the capacitor discharges, charge leaves the plate, so $I = -\,dq/dt$. (This sign convention is consistent: as $q$ decreases, $dq/dt < 0$ and $I > 0$, meaning positive current flows off the positively charged plate.) Substituting:
\[
\frac{q}{C} + L\,\frac{d}{dt}\!\left(-\,\frac{dq}{dt}\right) = 0,
\]
which simplifies to
\[
\frac{d^2q}{dt^2} + \frac{1}{LC}\,q = 0.
\]
This is the \emph{simple harmonic oscillator equation} (second-order linear ODE with constant coefficients). Comparing with the mechanical oscillator equation $\ddot{x} + (\kappa/m)\,x = 0$, the angular frequency is
\[
\omega = \frac{1}{\sqrt{LC}}.
\]
The general solution is
\[
q(t) = A\,\cos(\omega t) + B\,\sin(\omega t) = Q_{\max}\,\cos(\omega t + \phi),
\]
where $Q_{\max} = \sqrt{A^2 + B^2}$ is the amplitude and $\phi = \tan^{-1}(-B/A)$ is the phase constant. With $q(0) = Q_{\max}$ and $I(0) = 0$:
\[
q(0) = A = Q_{\max}, \qquad
I(0) = -\omega B = 0 \;\Rightarrow\; B = 0.
\]
Thus $\phi = 0$ and
\[
q(t) = Q_{\max}\,\cos(\omega t).
\]
The current is
\[
I(t) = \frac{dq}{dt} = -\,\omega\,Q_{\max}\,\sin(\omega t).
\]
The maximum current occurs when $|\sin(\omega t)| = 1$:
\[
I_{\max} = \omega\,Q_{\max} = \frac{Q_{\max}}{\sqrt{LC}}.
\]}

\nt{The initial conditions determine the amplitude $Q_{\max}$ and phase $\phi$. If the capacitor is initially charged to $q(0) = q_0$ with $I(0) = 0$, then $Q_{\max} = q_0$ and $\phi = 0$. If the current has an initial value $I(0) = I_0$ while $q(0) = 0$, then $Q_{\max} = I_0/\omega$ and $\phi = \pi/2$.}

\thm{Energy conservation in an LC circuit}{The energy stored in the capacitor at time $t$ is
\[
U_C(t) = \frac{q(t)^2}{2C} = \frac{Q_{\max}^2}{2C}\,\cos^2(\omega t),
\]
and the energy stored in the inductor is
\[
U_L(t) = \frac{1}{2}\,L\,I(t)^2 = \frac{1}{2}\,L\,\omega^2\,Q_{\max}^2\,\sin^2(\omega t).
\]
Using $\omega^2 = 1/(LC)$, the inductor energy becomes
\[
U_L(t) = \frac{Q_{\max}^2}{2C}\,\sin^2(\omega t).
\]
The total energy is therefore
\[
U_{\text{total}} = U_C(t) + U_L(t) = \frac{Q_{\max}^2}{2C}\,\bigl(\cos^2(\omega t) + \sin^2(\omega t)\bigr) = \frac{Q_{\max}^2}{2C}.
\]
The total energy is \emph{constant} and does not depend on time.}

\pf{Energy conservation}{The energy stored in a capacitor with charge $q$ is $U_C = q^2/(2C)$, and the energy stored in an inductor with current $I$ is $U_L = \tfrac{1}{2}LI^2$. Substituting the oscillation formulas:
\[
U_C(t) = \frac{Q_{\max}^2}{2C}\,\cos^2(\omega t),
\qquad
U_L(t) = \frac{1}{2}\,L\,\bigl(-\,\omega\,Q_{\max}\,\sin(\omega t)\bigr)^{\!2}
= \frac{1}{2}\,L\,\omega^2\,Q_{\max}^2\,\sin^2(\omega t).
\]
Since $\omega^2 = 1/(LC)$:
\[
U_L(t) = \frac{1}{2}\,L\cdot\frac{1}{LC}\cdot Q_{\max}^2\,\sin^2(\omega t)
= \frac{Q_{\max}^2}{2C}\,\sin^2(\omega t).
\]
Therefore,
\[
U_{\text{total}} = U_C + U_L = \frac{Q_{\max}^2}{2C}\,\bigl(\cos^2(\omega t) + \sin^2(\omega t)\bigr) = \frac{Q_{\max}^2}{2C}.
\]
This is time-independent, confirming energy conservation.}

\mprop{Energy exchange in an LC circuit}{The total energy is
\[
U = \frac{Q_{\max}^2}{2C} = \frac{1}{2}\,L\,I_{\max}^{\,2},
\]
where we used $I_{\max} = \omega Q_{\max} = Q_{\max}/\sqrt{LC}$, so $\tfrac{1}{2}LI_{\max}^{\,2} = \tfrac{1}{2}L\,(Q_{\max}^2/LC) = Q_{\max}^2/(2C)$.

The energy is purely capacitive at times $t$ when $\cos(\omega t) = \pm 1$ (i.e., $t = 0,\; T/2,\; T,\; \ldots$):
\[
U_C = \frac{Q_{\max}^2}{2C} = U_{\text{total}},
\qquad U_L = 0.
\]
The energy is purely inductive at times $t$ when $\sin(\omega t) = \pm 1$ (i.e., $t = T/4,\; 3T/4,\; \ldots$):
\[
U_L = \frac{Q_{\max}^2}{2C} = U_{\text{total}},
\qquad U_C = 0.
\]
At all other times, the energy is shared between the two elements.}

\cor{Analogy to spring-mass simple harmonic motion}{The LC circuit is directly analogous to a frictionless spring-mass oscillator:

\begin{center}
\begin{tabular}{c|c|c}
 & Spring--mass system & LC circuit \\ \hline
Displacement & $x(t) = A\,\cos(\omega t + \phi)$ & Charge $q(t) = Q_{\max}\,\cos(\omega t + \phi)$ \\
Velocity & $v(t) = -\,\omega A\,\sin(\omega t + \phi)$ & Current $I(t) = -\,\omega Q_{\max}\,\sin(\omega t + \phi)$ \\
Mass & $m$ & Inductance $L$ \\
Spring constant & $k$ & Inverse capacitance $1/C$ \\
Angular frequency & $\omega = \sqrt{k/m}$ & $\omega = 1/\sqrt{LC}$ \\
Potential energy & $\tfrac{1}{2}kx^2$ & $\tfrac{1}{2}q^2/C$ \\
Kinetic energy & $\tfrac{1}{2}mv^2$ & $\tfrac{1}{2}LI^2$ \\
Total energy & $\tfrac{1}{2}kA^2$ & $\dfrac{Q_{\max}^2}{2C} = \tfrac{1}{2}LI_{\max}^{\,2}$ \\
\end{tabular}
\end{center}
This analogy is useful for building physical intuition about LC circuits.}

\ex{Illustrative example}{An LC circuit has $L = 40\,\mathrm{mH}$ and $C = 2.0\,\mathrm{\mu F}$, initially charged to $Q_{\max} = 5.0\,\mathrm{\mu C}$. The angular frequency is $\omega = 1/\sqrt{LC} = 1/\sqrt{(40\times 10^{-3})(2.0\times 10^{-6})} = 1/\sqrt{8.0\times 10^{-8}} = 1/(8.94\times 10^{-4}) = 112\,\mathrm{rad/s}$. The total energy is $U = Q_{\max}^2/(2C) = (5.0\times 10^{-6})^2/(2\cdot 2.0\times 10^{-6}) = 6.25\times 10^{-6}\,\mathrm{J} = 6.25\,\mathrm{\mu J}$. The maximum current is $I_{\max} = \omega Q_{\max} = 112 \times 5.0\times 10^{-6} = 5.6\times 10^{-4}\,\mathrm{A} = 0.56\,\mathrm{mA}$.}

\qs{Worked example}{An LC circuit consists of an ideal inductor with inductance
\[
L = 25.0\,\mathrm{mH} = 25.0 \times 10^{-3}\,\mathrm{H},
\]
and an ideal capacitor with capacitance
\[
C = 5.00\,\mathrm{\mu F} = 5.00 \times 10^{-6}\,\mathrm{F}.
\]
At $t = 0$, the capacitor carries its maximum charge
\[
Q_{\max} = 10.0\,\mathrm{\mu C} = 1.00 \times 10^{-5}\,\mathrm{C},
\]
and the current is $I(0) = 0$. The phase constant is therefore $\phi = 0$, and
\[
q(t) = Q_{\max}\,\cos(\omega t),
\qquad
I(t) = -\,\omega\,Q_{\max}\,\sin(\omega t),
\]
with $\omega = 1/\sqrt{LC}$.

Find:
\begin{enumerate}[label=(\alph*)]
\item the oscillation frequency $f$ of the circuit,
\item the maximum current $I_{\max}$,
\item the energy $U_L$ stored in the inductor at time $t = T/4$ (one-quarter of the oscillation period), and
\item the energy $U_C$ stored in the capacitor at time $t_{1/2}$, the first instant at which the energy stored in the capacitor equals one-half of the total energy.
\end{enumerate}}

\sol \textbf{Part (a).} The angular frequency of oscillation is
\[
\omega = \frac{1}{\sqrt{LC}}.
\]
Substitute $L = 25.0\times 10^{-3}\,\mathrm{H}$ and $C = 5.00\times 10^{-6}\,\mathrm{F}$:
\[
LC = (25.0\times 10^{-3})(5.00\times 10^{-6}) = 1.25\times 10^{-7}\,\mathrm{H\cdot F}.
\]
Thus
\[
\omega = \frac{1}{\sqrt{1.25\times 10^{-7}}}
= \frac{1}{3.536\times 10^{-4}}
= 2828\,\mathrm{rad/s}.
\]
The oscillation frequency is
\[
f = \frac{\omega}{2\pi}
= \frac{2828}{2\pi}\,\mathrm{Hz}
= 450\,\mathrm{Hz}.
\]
Rounded to three significant figures:
\[
f = 450\,\mathrm{Hz}.
\]
(Equivalently, $4.50\times 10^2\,\mathrm{Hz}$.)

\textbf{Part (b).} The maximum current is
\[
I_{\max} = \omega\,Q_{\max}.
\]
Substitute $\omega = 2828\,\mathrm{rad/s}$ and $Q_{\max} = 1.00\times 10^{-5}\,\mathrm{C}$:
\[
I_{\max} = (2828)\,(1.00\times 10^{-5})\,\mathrm{A}
= 2.83\times 10^{-2}\,\mathrm{A}
= 28.3\,\mathrm{mA}.
\]

\textbf{Part (c).} At $t = T/4$, the angular argument is $\omega t = (2\pi/T)(T/4) = \pi/2$. Therefore:
\[
\cos\!\left(\frac{\pi}{2}\right) = 0,
\qquad
\sin\!\left(\frac{\pi}{2}\right) = 1.
\]
The charge and current at this instant are:
\[
q\!\left(\frac{T}{4}\right) = Q_{\max}\,\cos\!\left(\frac{\pi}{2}\right) = 0,
\qquad
I\!\left(\frac{T}{4}\right) = -\,\omega\,Q_{\max}\,\sin\!\left(\frac{\pi}{2}\right)
= -\,\omega\,Q_{\max}
= -\,I_{\max}.
\]
The energy stored in the capacitor is $U_C = q^2/(2C) = 0$, so \emph{all} the energy is in the inductor:
\[
U_L\!\left(\frac{T}{4}\right) = U_{\text{total}} = \frac{Q_{\max}^2}{2C}.
\]
Substitute the values:
\[
U_L\!\left(\frac{T}{4}\right)
= \frac{(1.00\times 10^{-5}\,\mathrm{C})^2}{2\,(5.00\times 10^{-6}\,\mathrm{F})}
= \frac{1.00\times 10^{-10}}{1.00\times 10^{-5}}\,\mathrm{J}
= 1.00\times 10^{-5}\,\mathrm{J}.
\]

\textbf{Part (d).} We seek the first time $t_{1/2}$ such that the capacitor energy is half the total:
\[
U_C = \frac{q^2}{2C} = \frac{1}{2}\,U_{\text{total}}
= \frac{1}{2}\cdot\frac{Q_{\max}^2}{2C}
= \frac{Q_{\max}^2}{4C}.
\]
Multiply both sides by $2C$:
\[
q^2 = \frac{Q_{\max}^2}{2},
\qquad
q = \frac{Q_{\max}}{\sqrt{2}}.
\]
Now use $q(t) = Q_{\max}\,\cos(\omega t)$:
\[
Q_{\max}\,\cos(\omega t_{1/2}) = \frac{Q_{\max}}{\sqrt{2}},
\qquad
\cos(\omega t_{1/2}) = \frac{1}{\sqrt{2}}.
\]
The first solution (smallest positive angle) is $\omega t_{1/2} = \pi/4$. We will not need the numerical value of $t_{1/2}$ to find the energy. The capacitor energy at this instant is
\[
U_C = \frac{1}{2}\,U_{\text{total}}
= \frac{1}{2}\cdot\frac{Q_{\max}^2}{2C}
= \frac{Q_{\max}^2}{4C}.
\]
Substitute the values:
\[
U_C = \frac{(1.00\times 10^{-5}\,\mathrm{C})^2}{4\,(5.00\times 10^{-6}\,\mathrm{F})}
= \frac{1.00\times 10^{-10}}{2.00\times 10^{-5}}\,\mathrm{J}
= 5.00\times 10^{-6}\,\mathrm{J}.
\]

\textbf{Check.} At $t = T/4$, the charge is zero and the energy is entirely in the inductor: $U_L = 1.00\times 10^{-5}\,\mathrm{J} = U_{\text{total}}$, which checks out. At $t = t_{1/2}$, the capacitor holds half the total energy, so $U_C = 5.00\times 10^{-6}\,\mathrm{J}$, and the inductor holds the other half: $U_L = U_{\text{total}} - U_C = 5.00\times 10^{-6}\,\mathrm{J}$, as expected.

Therefore,
\[
f = 450\,\mathrm{Hz},
\qquad
I_{\max} = 28.3\,\mathrm{mA},
\qquad
U_L\!\left(\frac{T}{4}\right) = 1.00\times 10^{-5}\,\mathrm{J},
\qquad
U_C(t_{1/2}) = 5.00\times 10^{-6}\,\mathrm{J}.
\]
