LC Oscillations

Every circuit in this chapter so far has been driven — an AC supply setting the rhythm. Now unplug the supply entirely. Take a capacitor charged to some voltage, connect it across a coil, and stand back. Common sense predicts an anticlimax: the capacitor dumps its charge through the coil, the current dies, the end. What actually happens is one of the prettiest phenomena in circuit physics: the charge rings. Current builds, overshoots, recharges the capacitor the other way round, reverses, and swings back — an electrical oscillation at a frequency the circuit chooses for itself, $\omega_0 = 1/\sqrt{LC}$ , with no source anywhere in sight. The Energy in an Inductor module left this as a teaser: what happens when a coil's energy store and a capacitor's energy store are given nothing to do but trade with each other? This module cashes it. The answer is a perfect electrical copy of the spring-block oscillator — charge playing position, current playing velocity, the inductor's inertia playing mass, the capacitor's stiffness playing the spring — a translation so exact that every SHM result this track owns carries across word for word. The LC oscillation is also the natural heartbeat the resonance module drove from outside: a circuit responds hugely to driving at $1/\sqrt{LC}$ precisely because that is the rhythm it rings at when left alone. Here we watch the unforced ring itself — the four-frame energy slosh, the frequency, the peak current — and meet the honest reason real rings fade.

Key takeaways

A charged capacitor across an inductor does not just discharge — it oscillates: charge and current swing sinusoidally, forever in the ideal case.
The energy sloshes between two stores: the capacitor's electric field ( $\tfrac{1}{2}q^2/C$ ) and the inductor's magnetic field ( $\tfrac{1}{2}Li^2$ ), with the total constant.
The frequency is $\omega_0 = \dfrac{1}{\sqrt{LC}}$ , $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ — the same expression that marks the resonance peak, now as the circuit's natural rhythm.
The mapping to the spring-block oscillator is exact: $q \leftrightarrow x$ , $i \leftrightarrow v$ , $L \leftrightarrow m$ , $1/C \leftrightarrow k$ — an electrical SHM, solved by translation rather than by new mathematics.
Peak current follows from energy conservation alone: $\tfrac{1}{2}q_0^2/C = \tfrac{1}{2}Li_{\max}^2 \Rightarrow i_{\max} = q_0/\sqrt{LC}$ .
Key idea: an LC circuit is a spring-block oscillator built from fields — charge sloshes at $\omega_0 = 1/\sqrt{LC}$ , with $q$ and $i$ a quarter cycle apart, until resistance (always present in reality) bleeds the ring away.

The core insight

Left to themselves, a capacitor and an inductor trade a fixed stock of energy back and forth — electric field to magnetic field and back — making charge and current oscillate at the natural frequency $\omega_0 = 1/\sqrt{LC}$ ; the system is term-for-term the spring-block oscillator ( $q\leftrightarrow x$ , $i\leftrightarrow v$ , $L\leftrightarrow m$ , $1/C\leftrightarrow k$ ), so every SHM result transfers by dictionary, including $i_{\max} = q_0/\sqrt{LC}$ from energy conservation.

Where this lives today

The LC oscillation is the metronome of the wireless world. Every radio transmitter needs a circuit that wants to oscillate at the carrier frequency — an LC pair with its losses continuously topped up — and every receiver needs the same pair, tuned, as its front door. The 13.56 MHz of an NFC card, the kilohertz hum inside a wireless charger, the local oscillator in every radio ever sold: LC rings, sustained. Where frequency stability matters more than tunability, a quartz crystal — a mechanical oscillator with extreme $Q$ , the dictionary run back into mechanics — takes the LC pair's seat: the 32,768 Hz heartbeat of a wristwatch is exactly such a resonator. At the exotic end, the superconducting LC circuits inside quantum computers ring with so little loss that their oscillations must be treated as quantum objects. In this chapter's own storyline, the machinery now turns from oscillation to the workhorse that built the modern grid: The Transformer.

The physical picture: the four-frame slosh

Charge a capacitor to $q_0$ and connect it across an ideal coil at $t = 0$ . Follow one cycle in four freeze-frames, a quarter period apart.

Frame 1 ( $t = 0$ ). All charge on the plates, no current yet. The energy — all of it — sits in the capacitor's electric field: $E = \tfrac{1}{2}q_0^2/C$ (the ledger of Energy Stored in a Capacitor). The capacitor begins to drive charge through the coil — but the coil, true to Self-Inductance & Inductors, fights change: the current builds only gradually, electrical inertia in action.

Frame 2 ( $t = T/4$ ). The capacitor has fully emptied: $q = 0$ , no voltage anywhere. But the current is now at its maximum — the step where intuition stumbles. Why doesn't everything stop? Because stopping would mean changing the current from $i_{\max}$ to zero, and the coil resists that change exactly as it resisted the build-up. The energy audit agrees: the capacitor's store is empty, yet the total must survive — it has moved, in full, into the coil's magnetic field, $\tfrac{1}{2}Li_{\max}^2$ (the store built in Energy in an Inductor & the Magnetic Field).

Frame 3 ( $t = T/2$ ). The coil's collapsing field has driven the current onward — piling charge onto the opposite plate. The current is back to zero and the capacitor fully charged at $-q_0$ : same size, reversed polarity, all the energy electric again — a mirror image of the start.

Frame 4 ( $t = 3T/4$ ). The reversed capacitor discharges back through the coil; the current peaks in the opposite direction; the energy is all magnetic once more. A quarter period later the system is exactly at Frame 1 — and, with no resistance to tax the traffic, repeats. Forever.

Figure

One cycle of the LC slosh — four freeze-frames of charge, current and energy

Notice the quadrature in the waveforms: charge at an extreme whenever current is zero, and vice versa. Position and velocity of an oscillating mass obey exactly this relationship — not an analogy but, as the next section shows, an identity.

The dictionary: LC oscillation is SHM in translation

Set the two systems side by side, quantity by quantity. This mapping — not a differential equation — is how the frequency is obtained at this level, and it is exact.

mechanical (spring-block)	electrical (LC)	shared role
position $x$	charge $q$	the displaced quantity that wants to return
velocity $v$	current $i$	the rate the displacement changes
mass $m$	inductance $L$	inertia: resists changes of $v$ / of $i$
spring constant $k$	$1/C$	stiffness: restoring push per unit displacement
$\tfrac{1}{2}mv^2$	$\tfrac{1}{2}Li^2$	kinetic-type store, largest at mid-swing
$\tfrac{1}{2}kx^2$	$\tfrac{1}{2}q^2/C$	potential-type store, largest at the extremes
$\omega = \sqrt{k/m}$	$\omega_0 = \sqrt{(1/C)/L} = 1/\sqrt{LC}$	natural frequency

Figure

The same oscillator in two costumes — spring-block and LC, matched part for part

Argue the two load-bearing rows honestly rather than by pattern-matching. Inertia: a mass resists having its velocity changed; an inductor resists having its current changed — its back-EMF fights every adjustment of $i$ (the whole message of Self-Inductance & Inductors). Same role, so $L$ sits where $m$ sits. Stiffness: a stretched spring pushes back with force $kx$ ; a charged capacitor pushes back with voltage $q/C$ — and the stiffer spring corresponds to the smaller capacitance, since a small $C$ builds a large opposing voltage from little charge. Same role, so $1/C$ sits where $k$ sits. With the dictionary certified, The Spring–Block Oscillator's $\omega = \sqrt{k/m}$ translates immediately:

\omega_0 = \sqrt{\frac{1/C}{L}} = \frac{1}{\sqrt{LC}}, \qquad f_0 = \frac{1}{2\pi\sqrt{LC}}, \qquad T = 2\pi\sqrt{LC}

where: $L$ is in henry and $C$ in farad — precisely the resonance frequency of the driven circuit, as it must be: the resonance module's peak sits where the driving matches this natural ring. Everything else transfers too: sinusoidal $q(t)$ with $i(t)$ a quarter cycle offset, energy trading at twice the oscillation frequency, amplitude set by initial conditions. (CBSE and ISC flag this topic "qualitative treatment only" — the four-frame story, the dictionary and $\omega_0 = 1/\sqrt{LC}$ are exactly the demanded scope. The energy numericals below are where JEE goes beyond it.)

The flagship algebra: peak current from the energy books

Derivation APeak current from conservation of the sloshing energy

The goal is $i_{\max} = q_0/\sqrt{LC}$ — the one quantitative result JEE extracts from this module — derived purely from energy conservation between two freeze-frames, with no equation of motion anywhere. The strategy: the total energy is fixed (ideal circuit, nothing dissipates), so equate the all-electric frame to the all-magnetic frame and solve. Watch how little machinery this needs.

Step 1. Frame 1 audit. At $t = 0$ : charge $q_0$ , current zero. Total energy $E = \dfrac{q_0^2}{2C}$ — entirely in the capacitor's electric field.

Step 2. Frame 2 audit. A quarter period later: charge zero, current at its peak $i_{\max}$ . Total energy $E = \tfrac{1}{2}Li_{\max}^2$ — entirely in the coil's magnetic field.

Step 3. No resistance exists to take a cut between the frames, so the two audits describe the same energy:

\frac{q_0^2}{2C} = \frac{1}{2}Li_{\max}^2 \;\Rightarrow\; i_{\max} = \frac{q_0}{\sqrt{LC}} = \omega_0\, q_0

Why the last rewriting matters: $i_{\max} = \omega_0 q_0$ is the dictionary's $v_{\max} = \omega A$ from SHM Kinematics, translated — a consistency check between the energy route and the oscillation picture, passed automatically.

Step 4. The same two-frame method answers every intermediate question. At the moment the charge has fallen to $q_0/2$ : electric store $= \dfrac{(q_0/2)^2}{2C} = \dfrac{E}{4}$ , so the magnetic store holds the remaining $\dfrac{3E}{4}$ , giving $\tfrac{1}{2}Li^2 = \tfrac{3}{4}\cdot\tfrac{1}{2}Li_{\max}^2$ , hence $i = \dfrac{\sqrt{3}}{2}\,i_{\max} \approx 0.87\,i_{\max}$ — half the charge gone, yet the current already 87% of its peak. Energy bookkeeping, nothing else. Verified: with $C = 25\ \mathrm{\mu F}$ charged to 300 V and $L = 0.1$ H — $q_0 = CV = 7.5$ mC, $E = \tfrac{1}{2}CV^2 = 1.125$ J; $i_{\max} = q_0/\sqrt{LC} = 7.5\times10^{-3}/1.581\times10^{-3} = 4.74$ A; cross-check $\tfrac{1}{2}Li_{\max}^2 = \tfrac{1}{2}(0.1)(4.74)^2 = 1.125$ J ✓; and $i$ at $q = q_0/2$ : $0.866 \times 4.74 = 4.11$ A ✓.

The honest footnote: why real rings fade

Every real coil has winding resistance, and the moment $R$ enters, the slosh pays a toll on each pass: the oscillation still runs at (very nearly) $1/\sqrt{LC}$ , but its amplitude decays, cycle by cycle, as $I^2R$ heat leaks out of the trading stock. Strike the circuit once and it rings like a struck bell — a sinusoid inside a dying envelope, the electrical twin of the damped oscillator from Damped & Forced Oscillations, Resonance. (The decay law and the critical-resistance story are UG fare; at this level, "rings at $\omega_0$ , fades exponentially, faster for larger $R$ " is the full honest statement.) Two consequences are worth carrying. First, this is exactly why the driven circuits of the resonance module need a source — to top up, every cycle, what $R$ bleeds; an LC circuit with its losses replenished at its own frequency is a radio's tuned circuit in operation, and the Tuning & Resonance Problems gym at the chapter's end trades on that. Second, a charge oscillating at megahertz rates does one more thing this chapter cannot yet price — an accelerating charge radiates. The spark-gap Leyden-jar rings of the 1890s flung their energy into space as electromagnetic waves; how, and why, is the EM-waves chapter's business.

Limiting cases

case	result	meaning
$t = 0, T/2, T, ...$	$q = \pm q_0$ , $i = 0$	all energy electric; turning points of the slosh
$t = T/4, 3T/4, ...$	$q = 0$ , $i = \pm i_{\max}$	all energy magnetic; mid-swing
$L$ increased	$\omega_0$ down, $i_{\max}$ down	heavier inertia: slower, gentler slosh
$C$ increased	$\omega_0$ down, $i_{\max}$ up ( $q_0 = CV$ fixed $V$ )	softer spring: slower ring, more charge to move
$R = 0$ (ideal)	constant amplitude forever	the assumed case of every formula above
real $R > 0$	ring decays exponentially	energy taxed each cycle; needs a source to sustain

Common trapWhen the capacitor is empty the oscillation is over

The empty-capacitor moment is the middle of the story, not its end — the point where the current, and the magnetic store, are at their absolute maximum. The error comes from casting the capacitor as the only actor: "charge gone, nothing left to push, so everything stops." But stopping is precisely what the inductor forbids — killing a current of $i_{\max}$ in an instant would take an infinite back-EMF; the coil instead keeps the charge moving, spending its stored $\tfrac{1}{2}Li^2$ to pile charge onto the far plate until the roles reverse. Three exam injuries follow. First: "find the current when $q = 0$ " — the answer is $i_{\max}$ , not zero. Second: "where is the energy when the capacitor is uncharged?" — in the coil's magnetic field, all of it. Third: energy-sharing numericals require trusting that the two stores genuinely trade — the two-frame audit — rather than assuming the capacitor's store drains away. The mechanical translation kills the confusion permanently: a mass at mid-swing has zero displacement and maximum speed; nobody expects a pendulum to halt at the bottom of its arc.

Worked examples

Board level — the slosh, narrated and quantified (CBSE/ISC pattern, 3 marks)

A 25 μF capacitor is charged to 300 V and then connected across a 0.1 H ideal inductor. (a) Describe the energy transformations over one full cycle. (b) Find the frequency of the oscillation. (c) Find the total energy of the circuit.

Step 1. Part (a), in three sentences: initially all energy is electric, $q_0^2/2C$ . At quarter-period the capacitor is empty and all energy is magnetic, $\tfrac{1}{2}Li_{\max}^2$ ; the coil's collapsing field then recharges the capacitor with opposite polarity (all electric again at half-period). The process reverses and repeats — electric → magnetic → electric → magnetic, twice per cycle, total constant.

Step 2. Part (b): $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.1 \times 25\times10^{-6}}} = \dfrac{1}{1.581\times10^{-3}} = 632.5$ rad/s, so $f_0 = \omega_0/2\pi = \boxed{100.7\ \mathrm{Hz}}$ .

Step 3. Part (c): $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 25\times10^{-6} \times (300)^2 = \boxed{1.125\ \mathrm{J}}$ — the fixed trading stock for the whole performance.

Competitive level — energy audits mid-slosh (JEE-Main pattern, ~3 minutes)

In the circuit above ( $C = 25\ \mathrm{\mu F}$ at 300 V, $L = 0.1$ H), find (a) the maximum current, (b) the current at the instant the capacitor's charge has fallen to half its initial value, (c) the earliest time at which the energy is entirely magnetic.

Step 1. Part (a): equate the two full stores. $q_0 = CV = 7.5\times10^{-3}$ C; $i_{\max} = \dfrac{q_0}{\sqrt{LC}} = \dfrac{7.5\times10^{-3}}{1.581\times10^{-3}} = \boxed{4.74\ \mathrm{A}}$ . (Audit: $\tfrac{1}{2}Li_{\max}^2 = \tfrac{1}{2}(0.1)(22.5) = 1.125$ J = the total ✓.)

Step 2. Part (b): at $q = q_0/2$ , the electric store is $(1/2)^2 = 1/4$ of the total, so the magnetic store holds $3/4$ of it: $i = \sqrt{3/4}\;i_{\max} = 0.866 \times 4.74 = \boxed{4.11\ \mathrm{A}}$ . Energy fractions go as the squares of $q$ and $i$ — never linearly.

Step 3. Part (c): all-magnetic first occurs a quarter period in: $T = 2\pi\sqrt{LC} = 2\pi \times 1.581\times10^{-3} = 9.93$ ms, so $t = T/4 = \boxed{2.48\ \mathrm{ms}}$ . Timing note: 3 minutes. The whole example is two energy audits and one period formula — resist any urge to write $q(t)$ expressions and chase phases. The square-law step in (b) is the mark-separator: half the charge means a quarter of the electric energy. An answer of $i_{\max}/2$ is the exact error the setter priced in.

Exam traps

$q = 0$ means $i = i_{\max}$ . The empty-capacitor instant is maximum current, maximum magnetic energy — mid-swing, not full stop. The single most tested fact in the module.

Energy goes as the square. Half the charge → quarter of the electric energy; half the current → quarter of the magnetic energy. Linear energy-sharing answers are always wrong.

Same formula, two meanings. $1/2\pi\sqrt{LC}$ is the natural frequency here and the resonance frequency in the driven module — one number wearing two hats. Exams exploit the connection: "an LC circuit rings at 1 MHz; at what frequency would a source drive maximum current?" Answer: 1 MHz, no computation.

The oscillation needs no source. "What drives the oscillation?" — the initial stored energy, traded between the two fields. Inserting a phantom EMF into an energy audit is a conceptual error examiners flag.

$T = 2\pi\sqrt{LC}$ arithmetic. Multiply $L \times C$ first, square-root, then scale — the same $\sqrt{LC}$ discipline as the resonance module, with the same inverse-square-root traps on " $C$ is quadrupled" questions ( $T$ doubles; $f$ halves).

Ideal versus real, one honest line. If the problem says the coil has resistance, say the oscillation decays as heat leaks each cycle — but the frequency stays (very nearly) $1/\sqrt{LC}$ . Claiming the ring stops instantly, or that $R$ shifts the frequency dramatically, both lose the mark.

Boards and depth. CBSE and ISC both state "LC oscillations (qualitative treatment only)": the required scope is the energy-slosh narrative, the SHM correspondence, and $f_0 = 1/2\pi\sqrt{LC}$ — this module's spine, with the dictionary standing in for calculus. JEE-Main and JEE-Advanced go quantitative exactly where the flagship goes: energy-conservation numericals ( $i_{\max}$ , mid-slosh audits, timing); the JEE-Advanced syllabus names "LC circuits with d.c. sources" explicitly. NEET stays near the CBSE line. AP Physics 2 has no AC or LC oscillations on its syllabus, and IB (2023 onward) dropped AC circuits entirely — this module is CBSE/ISC/JEE/NEET territory.

Origins · 1 minWilliam Thomson (Lord Kelvin) and how this was discovered

William Thomson (Lord Kelvin)

1824–1907 · University of Glasgow

In 1853, decades before anyone could watch it happen, the 29-year-old William Thomson worked out on paper that this module's slosh must exist. The puzzle of the day: what actually happens when a Leyden jar discharges through a wire? The assumption was a one-way rush, like a bucket emptying. Thomson, in "On Transient Electric Currents", ran the energy bookkeeping — the jar's electrostatic store against the "electro-dynamic" energy of the current in the coil — and showed the discharge should overshoot: the current's inertia recharges the jar in reverse, and the system rings at a frequency set by the jar's capacity and the coil's inductance, $1/\sqrt{LC}$ in modern dress. Deduced from conservation principles, not observed: Feddersen's rotating-mirror photographs confirmed it within five years, and the ringing discharge became the beating heart of early radio — Hertz's sparks and Marconi's tuning both stand on this paper. Thomson appears throughout this track wearing different hats — thermodynamics, the Atlantic cable, electrometers; this band is specifically the young theorist who heard a circuit ring in the mathematics before any instrument could hear it at all.

History

1853

— Thomson predicts, from energy bookkeeping alone, that a capacitor discharging through a coil must oscillate

1858

— Feddersen photographs the oscillatory spark in a rotating mirror: the ring is real

1887

— Hertz's spark-gap oscillators — driven LC rings — generate the first deliberate electromagnetic waves

1900

s — every radio transmitter and receiver is built around sustained or tuned LC oscillations

1920

s — quartz crystals begin replacing LC pairs where extreme frequency stability is needed

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.8, LC oscillations box) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, LC oscillations (qualitative) · Thomson, "On Transient Electric Currents" (1853), for the founding prediction.

Provenance. All results standard. A verifier should recompute: $q_0 = 7.5$ mC; $E = 1.125$ J; $\sqrt{LC} = 1.581$ ms; $\omega_0 = 632.5$ rad/s, $f_0 = 100.7$ Hz, $T = 9.93$ ms, $T/4 = 2.48$ ms; $i_{\max} = 4.74$ A with $\tfrac{1}{2}Li_{\max}^2 = 1.125$ J ✓; $i(q_0/2) = 0.866 \times 4.74 = 4.11$ A ✓.

Prerequisites. Energy in an Inductor & the Magnetic Field ( $\tfrac{1}{2}Li^2$ — and the teaser this module cashes) · Energy Stored in a Capacitor ( $\tfrac{1}{2}q^2/C$ ) · The Spring–Block Oscillator (the mechanical original). Leads to. The Transformer (next) · ARCHETYPE: Tuning & Resonance Problems (the ring put to work) · LC & RLC Oscillations (UG deep dive).

What physically keeps the current flowing when the capacitor voltage reaches zero?

The inductor's inertia. Current through a coil cannot change abruptly — any change is fought by the back-EMF in proportion to how fast it is attempted (Self-Inductance & Inductors). At the zero-voltage instant the current is large, and stopping would mean changing from $i_{\max}$ to zero — exactly what the coil resists hardest. So the coil, spending its stored $\tfrac{1}{2}Li^2$ , acts momentarily as the source, driving charge onto the far plate against the capacitor's growing opposition. The mechanical translation makes it obvious: at mid-swing the spring pulls nothing, yet the mass sails through at top speed. The capacitor is the spring; the coil is the mass; zero voltage is mid-swing, not the end.

Why does the frequency come out the same as the resonance frequency of the driven LCR circuit?

Because resonance is this ring, located from outside. Left alone, the LC pair trades energy at its own rhythm $1/\sqrt{LC}$ — this module. Driven, the response is largest exactly when the driving rhythm matches the natural one — the resonance module. The mechanical twins are the free pendulum and the pushed swing: one frequency, discovered two ways. The correspondence pays in the exam: any question giving a "natural ringing frequency" has silently given you the resonance frequency (and vice versa), and the tuning problems in the chapter's final gym switch between the two phrasings freely.

Where does the energy go in the long run if even a tiny resistance is present?

Into heat, a slice per cycle. Each round of the slosh pushes the full current through $R$ , and the $I^2R$ toll comes straight out of the trading stock: the amplitude decays smoothly, cycle by cycle, until the stock is exhausted — the struck-bell ring fading. The frequency barely shifts (for small $R$ the ring still runs at essentially $1/\sqrt{LC}$ ); it is the envelope that dies, faster for larger $R$ . Two footnotes live beyond this level: at high frequencies some energy also leaves as electromagnetic radiation — the fate that made spark-gap rings usable as transmitters — and if $R$ grows large enough the circuit stops ringing altogether and oozes to zero, a regime change the UG treatment quantifies. For every HS exam purpose: rings at $\omega_0$ , decays exponentially, heat in $R$ is where the joules went.

Is the LC oscillation really SHM, or just something that looks like it?

Really SHM, in the strict sense. SHM is defined by a restoring influence proportional to displacement acting on an inertia — that structure produces sinusoidal motion at $\sqrt{\text{stiffness}/\text{inertia}}$ , regardless of what the "displacement" physically is. Here the displaced quantity is charge, the restoring influence is the capacitor's voltage $q/C$ (proportional to $q$ — the Hooke's-law shape), and the inertia is the inductance. Every SHM theorem then applies verbatim: sinusoidal time-dependence, quadrature between $q$ and $i$ , energy trading at $2f_0$ , $i_{\max} = \omega_0 q_0$ . Only the costume changes — no positions, no metres — which is why the module never solved anything new: the spring-block oscillator's mathematics recycled wholesale through the dictionary.

Can I use the SHM dictionary in the exam, or do I need the calculus derivation?

Use the dictionary — with the correspondence argued, not just asserted. For CBSE and ISC ("qualitative treatment only"), the expected answer is exactly: energy sloshes between $\tfrac{1}{2}q^2/C$ and $\tfrac{1}{2}Li^2$ ; $L$ is the inertia (a coil resists current changes as mass resists velocity changes) and $1/C$ the stiffness (a capacitor opposes charge with voltage $q/C$ as a spring opposes displacement with $kx$ ); hence $f_0 = 1/2\pi\sqrt{LC}$ by substitution into $\omega = \sqrt{k/m}$ . That is a full-marks derivation at this level. If you know calculus, the loop equation gives the same result in three lines and no board objects — but nothing in these exams requires it, and the energy-audit method is faster than calculus for every numerical the module faces.

Did anyone actually see these oscillations before electronics existed?

Yes — through heroic ingenuity. Thomson's 1853 prediction was pure theory, but within five years Berend Feddersen aimed a rotating mirror at the spark of a discharging Leyden jar: successive instants of the spark smeared across a photographic plate, and the image showed it flickering — extinguishing and reigniting as the current reversed — resolving oscillations at hundreds of kilohertz. The frequency matched the capacity-and-coil prediction, closing Thomson's loop. Half a century later the same physics became technology: Hertz's sparks and the first spark-gap radio transmitters were violently struck LC circuits flinging their ring into space. A Leyden jar with a simple wire loop rings at a few megahertz — which is why early radio lived on those wavelengths.