LC circuit — oscillations (electrical analog of SHM): Visual walkthrough

Step 1 — The two boxes and what lives in each

WHAT: We name the two things that change with time: $q$ (how full the bucket is) and $i$ (how fast charge moves).

WHY: Everything that follows is a story about these two numbers trading places. If we know what pushes them, we know the whole motion.

PICTURE: In the figure, the top plate carries $+ q$ , the bottom $- q$ . The coil sits on the right. No current flows yet — the switch is open.

Figure — LC circuit — oscillations (electrical analog of SHM)

Step 2 — What each box pushes with (the two voltages)

WHAT: We write down how hard each box pushes.

WHY these two forms, and not others?

$V_{C} = q / C$ because a fuller capacitor pushes back harder — double the charge, double the push. The number $C$ (capacitance, see Capacitance and Energy in Capacitors) says how much charge you need to build one volt. Big $C$ = easy to fill = weak push per charge.
$V_{L} = L \frac{d i}{d t}$ because an inductor doesn't care about current, only about change in current (see Inductance and Self-Induction). The $\frac{d i}{d t}$ is "how fast the current is changing per second." The number $L$ (inductance) says how stubbornly it resists that change.

PICTURE: The capacitor's push (magenta arrow) drives current out of the top plate. The inductor's back-push (violet arrow) points against whatever the current is trying to do.

Step 3 — Walking the loop (Kirchhoff's rule)

WHAT: We add the two pushes around the loop and set the sum to zero.

WHY: With no battery feeding energy in, there is nothing else in the loop. The capacitor's outward push and the inductor's back-push are the only two voltages, and around any complete loop voltages must cancel — otherwise charge would gain energy from nowhere on each trip.

PICTURE: The green arrow traces the loop. As we pass the capacitor we count $+ q / C$ ; as we pass the inductor we count $+ L \frac{d i}{d t}$ . Arrive back where we started → net push $= 0$ .

Step 4 — Turning it into one equation for $q$ alone

WHAT: We eliminate $i$ so the equation talks about $q$ only.

WHY: The symbol $\frac{d ^{2} q}{d t ^{2}}$ (say "d-two-q-d-t-squared") means the rate of change of the rate of change — the acceleration of charge. One equation in one unknown is something we can actually solve.

Reading the boxed result term by term:

$\frac{d ^{2} q}{d t ^{2}}$ — how fast the current is changing (charge's "acceleration").
$- \frac{1}{L C}$ — a positive number with a minus sign in front. The minus is the whole story.
$q$ — the current charge.

PICTURE: The plot shows: whenever $q$ is positive (bucket has positive charge), the acceleration arrow points down toward zero. Whenever $q$ is negative, it points up. The push always aims back toward empty — that is what a restoring force does.

Step 5 — Recognising an old friend: this IS the SHM equation

WHAT: We compare our boxed equation to the SHM equation, symbol for symbol.

WHY: If two equations have the same shape, they have the same solution. We never have to solve a differential equation from scratch — we just read off the match. The thing multiplying $q$ (or $x$ ) with a minus sign is always $ω^{2}$ .

The match forces: $ω^{2} = \frac{1}{L C} ⟹ ω = \frac{1}{L C}$

Why the square root? Because the equation gives us $ω^{2}$ , not $ω$ . To get $ω$ we undo the squaring — that is exactly what does.

PICTURE: Two panels side by side — a mass bouncing on a spring, and our LC loop — with the dictionary $q \leftrightarrow x$ , $L \leftrightarrow m$ , $\frac{1}{C} \leftrightarrow k$ drawn as matching arrows.

Step 6 — The solution and the 90° dance of $q$ and $i$

Reading it:

$q_{0}$ — the starting charge (height of the cosine).
$cos (ω t)$ — starts at $1$ , so $q$ starts full. ✔ matches our setup.
$- q_{0} ω sin (ω t)$ — starts at $0$ , so $i$ starts at rest. ✔
$i_{0} = q_{0} ω$ — the peak current, reached when the cosine crosses zero.

WHY the 90° offset? Look at any cosine: at its peak its slope is flat (zero). At its zero crossing its slope is steepest. Since current is the slope of charge, current is zero when charge is maximum and maximum when charge is zero. That quarter-cycle gap is the $9 0^{\circ}$ phase difference.

PICTURE: Both curves on one axis. Vertical dashed lines mark the four instants: (a) $q$ max, $i = 0$ ; (b) $q = 0$ , $i$ most negative; (c) $q$ most negative, $i = 0$ ; (d) $q = 0$ , $i$ most positive.

Step 7 — Energy: the total never moves (the degenerate/limiting check)

WHAT: We add the two energies and watch the wobble cancel.

WHY: The identity $cos^{2} θ + sin^{2} θ = 1$ makes the time-dependence vanish. With no resistor there is nowhere for energy to leak, so the sum must be flat. This is the ultimate sanity check — if it weren't constant, we'd have made an error. (Add a resistor and you get Damped Oscillations / LCR Circuit instead.)

The limiting cases the picture must show:

At $t = 0$ : all energy is electric ( $U_{E}$ full, $U_{B} = 0$ ) — bucket full, flywheel still.
At $ω t = π /2$ : all energy is magnetic ( $U_{E} = 0$ , $U_{B}$ full) — bucket empty, flywheel spinning fastest.
Where they cross: $cos^{2} = sin^{2} \Rightarrow ω t = π /4 \Rightarrow t = T /8$ . Energy split 50/50.

PICTURE: $U_{E}$ (magenta) and $U_{B}$ (violet) each wobble, but the orange line $U_{E} + U_{B}$ is dead flat. Crossings marked at $T /8, 3 T /8, \dots$

The one-picture summary

Recall Feynman retelling of the whole walkthrough

Picture a bucket of water joined to a heavy paddle-wheel by a pipe, with no friction anywhere. Start with the bucket full and the wheel still (Step 1–2). The full bucket pushes water out — that push is $V_{C} = q / C$ (fuller = harder push). But the heavy wheel resists suddenly spinning up; its resistance to change is $V_{L} = L d i / d t$ (Step 2). Since it's a closed loop with nothing else, those two pushes must exactly cancel around the loop (Step 3, Kirchhoff). Rewrite "flow" as "how fast the bucket empties," and you get one clean law: the charge's acceleration is minus a constant times the charge itself (Step 4). That minus sign means the circuit always yanks the charge back toward empty — a spring in disguise. We've seen that exact equation before — it's a mass on a spring (Step 5). Matching them term by term, the frequency has to be $1/ L C$ : a heavier wheel ( $L$ ) or a bigger bucket ( $C$ ) makes a slower sway. Solve it and charge is a cosine, current its slope — a sine (Step 6). So the water rushes fastest exactly when the bucket is empty (steepest slope) and stops flowing exactly when the bucket is full (flat top). That's the $9 0^{\circ}$ dance. Finally, add the bucket's stored energy to the wheel's spinning energy (Step 7): thanks to $cos^{2} + sin^{2} = 1$ , the total never changes. With no friction the sway continues forever, energy forever pouring from bucket to wheel and back.

Recall Quick self-check

The boxed equation of motion is? ::: $\frac{d ^{2} q}{d t ^{2}} = - \frac{1}{L C} q$ Why must $ω$ carry a square root? ::: The matched equation gives $ω^{2} = 1/ L C$ ; taking undoes the square. When is current maximum? ::: When charge passes through zero (steepest slope of the cosine). What keeps the total energy constant? ::: The identity $cos^{2} + sin^{2} = 1$ and the absence of any resistor. At what time is energy shared 50/50? ::: $t = T /8$ (where $ω t = π /4$ ).

Connections

Simple Harmonic Motion — the equation we matched against in Step 5.
Capacitance and Energy in Capacitors — source of $V_{C} = q / C$ and $U_{E} = q^{2} /2 C$ .
Inductance and Self-Induction — origin of $V_{L} = L d i / d t$ .
Kirchhoff's Voltage Law — the loop rule of Step 3.
Damped Oscillations / LCR Circuit — what Step 7's flat line becomes once $R \neq = 0$ .
Resonance and AC Circuits — $ω = 1/ L C$ reappears as the resonant frequency.