High School Physics · Grades 10–12

31Free Electrical Oscillations: RLC

Turn the heavy dial of an attic radio and the stations file past the needle: a voice, violins, static, a voice again. Behind the panel there is no motor and no computer — a coil, a capacitor whose interleaved plates the dial rotates, and a length of aerial wire. This chapter connects last chapter’s two energy tanks (Chapter 30) head to head and finds the fastest pendulum ever built: charge swinging back and forth, up to millions of times per second, at a tempo the dial chooses.

31.1 An electrical pendulum: the LC circuit

Charge a capacitor $C$ to a voltage $U_0$ — its plates then hold $\pm Q_0$ with $Q_0 = C U_0$ — and at $t = 0$ switch it onto an ideal coil $L$ , of zero resistance. No source anywhere in the loop: whatever happens next, the circuit does on its own.

The LC circuit: a capacitor charged to Q_0, a switch, an ideal coil. Closing K lets the charge spill through the coil — and the coil, hating change, will not let it stop. — The LC circuit: a capacitor charged to $Q_0$ , a switch, an ideal coil. Closing $K$ lets the charge spill through the coil — and the coil, hating change, will not let it stop.

Proposition 31.1 (The LC equation)

In the ideal LC loop the charge $q(t)$ of the upper plate obeys

L\,\frac{d^2q}{dt^2} + \frac{q}{C} = 0, \qquad\text{i.e.}\qquad \frac{d^2q}{dt^2} = -\frac{1}{LC}\,q .

Proof. Around the loop the voltages sum to zero (Proposition 12.4): $u_L + u_C = 0$ . The capacitor gives $u_C = q/C$ and the coil $u_L = L\,di/dt$ (Definition 30.10), with $i = dq/dt$ counting the charge arriving on the upper plate (Proposition 30.3); substitute. ∎

Theorem 31.2 (Free oscillations of the LC circuit)

Released at $t = 0$ with $q(0) = Q_0$ and $i(0) = 0$ , the circuit oscillates:

q(t) = Q_0 \cos\!\left(\frac{2\pi t}{T_0}\right), \qquad i(t) = -I_0 \sin\!\left(\frac{2\pi t}{T_0}\right), \qquad T_0 = 2\pi\sqrt{LC},

with peak current $I_0 = 2\pi Q_0/T_0 = Q_0/\sqrt{LC}$ .

Proof. Verify by substitution. Differentiating, $i = dq/dt = -\frac{2\pi}{T_0} Q_0 \sin(2\pi t/T_0)$ — the stated $i(t)$ — and again $\frac{d^2q}{dt^2} = -(\frac{2\pi}{T_0})^2 q$ , which equals $-q/(LC)$ exactly when $(2\pi/T_0)^2 = 1/(LC)$ , i.e. $T_0 = 2\pi\sqrt{LC}$ . And $q(0) = Q_0$ , $i(0) = 0$ . That no other function fits the equation and the start is admitted, exactly as for the spring (Remark 28.9). ∎

Definition 31.3 (Natural period and frequency)

The natural period of an LC circuit is $T_0 = 2\pi\sqrt{LC}$ ; its natural frequency is

f_0 = \frac{1}{T_0} = \frac{1}{2\pi\sqrt{LC}} .

Dimensional check: $1\,\mathrm{H} = 1\,\mathrm{V}\,\mathrm{s}/\mathrm{A}$ and $1\,\mathrm{F} = 1\,\mathrm{C}/\mathrm{V}$ , so $LC$ carries $\mathrm{s} \times \mathrm{C}/\mathrm{A} = \mathrm{s}^{2}$ — $\sqrt{LC}$ is a time, as it must be.

Example 31.4 (First numbers)

$L = 10\,\mathrm{mH}$ with $C = 1.0\,\text{µ}\mathrm{F}$ : $\sqrt{LC} = 1.0 \times 10^{-4}\,\mathrm{s}$ , so $T_0 = 0.63\,\mathrm{ms}$ and $f_0 \approx 1.6\,\mathrm{kHz}$ — an audible tone. $L = 1.0\,\mathrm{mH}$ with $C = 100\,\mathrm{pF}$ : $T_0 = 2.0\,\text{µ}\mathrm{s}$ , $f_0 \approx 0.50\,\mathrm{MHz}$ — a radio frequency. Only the product $LC$ sets the tempo.

Remark 31.5 (A quarter period out of step)

The current is a quarter period out of phase with the charge: $i = 0$ when $q = \pm Q_0$ (plates full, flow reversing) and $i = \pm I_0$ when $q = 0$ (plates empty, flow at full tilt) — exactly the pendulum, motionless at the extremes and fastest at the bottom.

Charge and current of the free LC circuit: the current peaks each time the charge crosses zero, a quarter period out of step.

31.2 The energy waltz

Proposition 31.6 (Conservation of the stored energy)

In the ideal LC circuit the total stored energy (Proposition 30.15)

E \;=\; E_C + E_L \;=\; \frac{q^2}{2C} + \tfrac12\,L i^2 \;=\; \frac{Q_0^2}{2C} \;=\; \tfrac12\,L I_0^2

is constant: entirely in the capacitor when $q = \pm Q_0$ , entirely in the coil when $i = \pm I_0$ .

Proof. Differentiate: $\frac{dE}{dt} = \frac{q}{C}\frac{dq}{dt} + Li\frac{di}{dt} = i\left(\frac{q}{C} + L\frac{d^2q}{dt^2}\right) = 0$ by Proposition 31.1. Its value is read at $t = 0$ (all capacitor), and again at the all-current instant $q = 0$ — whence $\tfrac12 L I_0^2 = Q_0^2/2C$ , recovering $I_0 = Q_0/\sqrt{LC}$ by energy alone. ∎

One period of the waltz: energy sloshes from plates to coil and back twice per period, the sum pinned at Q_02/2C — the very figure of the spring–mass oscillator, relabeled. — One period of the waltz: energy sloshes from plates to coil and back twice per period, the sum pinned at $Q_0^2/2C$ — the very figure of the spring–mass oscillator, relabeled.

Example 31.7 (The waltz, measured)

$C = 1.0\,\text{µ}\mathrm{F}$ charged to $U_0 = 10\,\mathrm{V}$ , switched onto $L = 10\,\mathrm{mH}$ : energy $E = \tfrac12 C U_0^2 = 5.0 \times 10^{-5}\,\mathrm{J}$ , period $T_0 = 0.63\,\mathrm{ms}$ , and a peak current $I_0 = U_0\sqrt{C/L} = 0.10\,\mathrm{A}$ reached a quarter period ( $0.16\,\mathrm{ms}$ ) after the switch closes.

31.3 Real circuits: damped oscillations

Every real coil is wound from resistive wire: the honest loop is a series RLC circuit, and the resistance taxes the waltz.

Definition 31.8 (Pseudo-periodic and aperiodic regimes)

In a series RLC circuit the resistor dissipates energy by Joule heating (Chapter 12): the free oscillations are damped. Two regimes, read off an oscilloscope:

pseudo-periodic (small $R$ ): the circuit still oscillates inside a shrinking envelope, with a repeat time — the pseudo-period — close to $T_0 = 2\pi\sqrt{LC}$ when the damping is light;
aperiodic (large $R$ ): the charge creeps back to zero without ever overshooting — no oscillation at all.

The exact mirror of the damped mechanical oscillator (Definition 28.12).

The two fates of a real RLC circuit: oscillations dying inside an exponential envelope (dashed), or a creep to zero with no oscillation. Lightly damped, the pseudo-period stays T_0. — The two fates of a real RLC circuit: oscillations dying inside an exponential envelope (dashed), or a creep to zero with no oscillation. Lightly damped, the pseudo-period stays $\approx T_0$ .

Remark 31.9 (Maintaining oscillations)

To keep the amplitude alive, feed the circuit exactly the energy the resistance burns, once per cycle and in step with the swing: an amplifier does for the RLC loop what the escapement’s push does for the pendulum (Chapter 28). Every quartz watch and every radio transmitter is such a maintained oscillator, running at its natural frequency $f_0$ .

31.4 The mechanical twin

Proposition 31.10 (The electrical–mechanical dictionary)

The LC equation is the spring–mass equation (Proposition 28.8) word for word:

LC circuit		spring–mass glider
[2pt] charge $q$	$\longleftrightarrow$	position $x$
current $i = dq/dt$	$\longleftrightarrow$	velocity $v = dx/dt$
inductance $L$	$\longleftrightarrow$	mass $m$
inverse capacitance $1/C$	$\longleftrightarrow$	stiffness $k$
resistance $R$	$\longleftrightarrow$	friction
$L\,\frac{d^2q}{dt^2} + \frac{q}{C} = 0$	$\longleftrightarrow$	$m\,\frac{d^2x}{dt^2} + kx = 0$
[2pt] $T_0 = 2\pi\sqrt{LC}$	$\longleftrightarrow$	$T_0 = 2\pi\sqrt{m/k}$
[2pt] $E_C = q^2/2C$	$\longleftrightarrow$	$E_p = \tfrac12 kx^2$
$E_L = \tfrac12 Li^2$	$\longleftrightarrow$	$E_k = \tfrac12 mv^2$

Same equation, same solutions: every result about either oscillator translates instantly into the other language.

Proof. Compare Proposition 31.1 with $m\,d^2x/dt^2 = -kx$ : substituting $q \to x$ , $L \to m$ , $1/C \to k$ turns one into the other, and $LC \to m/k$ turns $2\pi\sqrt{LC}$ into $2\pi\sqrt{m/k}$ . ∎

Example 31.11 (Tuning a radio)

An aerial feeds a whisper of every station into an LC loop, but only the one broadcasting near $f_0$ pushes in step and builds up — resonance (Definition 28.13). The dial rotates the plates of a variable capacitor: changing $C$ moves $f_0 = 1/(2\pi\sqrt{LC})$ , and the needle hands the earphone one station at a time. On the transmitting side, a maintained oscillator (Remark 31.9) sets the broadcast frequency; in a watch, a quartz crystal — a mechanical twin of breathtaking stiffness — plays the same role.

31.5 Exercises

Exercise 31.1 ★

Compute $T_0$ and $f_0$ for (a) $L = 10\,\mathrm{mH}$ , $C = 1.0\,\text{µ}\mathrm{F}$ ; (b) $L = 1.0\,\mathrm{mH}$ , $C = 100\,\mathrm{pF}$ . By what factor do the frequencies differ?

Solution

Solution of Exercise 31.1.

(a) $\sqrt{LC} = 1.0 \times 10^{-4}\,\mathrm{s}$ : $T_0 = 0.63\,\mathrm{ms}$ , $f_0 \approx 1.6\,\mathrm{kHz}$ . (b) $\sqrt{LC} = 3.16 \times 10^{-7}\,\mathrm{s}$ : $T_0 = 2.0\,\text{µ}\mathrm{s}$ , $f_0 \approx 0.50\,\mathrm{MHz}$ — a factor $\sqrt{10^5} \approx 316$ .

Exercise 31.2 ★

Using $1\,\mathrm{H} = 1\,\mathrm{V}\,\mathrm{s}/\mathrm{A}$ and $1\,\mathrm{F} = 1\,\mathrm{C}/\mathrm{V}$ , show unit by unit that $\sqrt{LC}$ is a time. Why does the $2\pi$ change nothing?

Solution

Solution of Exercise 31.2.

$LC$ : $(\mathrm{V}\,\mathrm{s}/\mathrm{A})(\mathrm{C}/\mathrm{V}) = \mathrm{s} \times \mathrm{C}/\mathrm{A} = \mathrm{s} \times \mathrm{s} = \mathrm{s}^{2}$ , so $\sqrt{LC}$ is in seconds. $2\pi$ is a pure number — no dimension to change.

Exercise 31.3 ★

A $470\,\mathrm{nF}$ capacitor is charged to $6.0\,\mathrm{V}$ . Compute $Q_0$ and the stored energy; describe, without computation, what follows when it is switched onto an ideal coil.

Solution

Solution of Exercise 31.3.

$Q_0 = CU_0 = 2.8\,\text{µ}\mathrm{C}$ ; $E = \tfrac12 CU_0^2 = \tfrac12 \times 4.7 \times 10^{-7} \times 36 \approx 8.5\,\text{µ}\mathrm{J}$ . The charge oscillates at $f_0 = 1/(2\pi\sqrt{LC})$ , current a quarter period out of phase, energy sloshing between plates and coil, total constant.

Exercise 31.4 ★

An LC circuit follows $q(t) = 2.0\cos(2\pi t/T_0)$ in microcoulombs, with $T_0 = 4.0\,\mathrm{ms}$ . Read off $Q_0$ ; compute $f_0$ , $q(T_0/2)$ , $i(0)$ , and the peak current $I_0 = 2\pi Q_0/T_0$ .

Solution

Solution of Exercise 31.4.

$Q_0 = 2.0\,\text{µ}\mathrm{C}$ ; $f_0 = 1/T_0 = 250\,\mathrm{Hz}$ ; $q(T_0/2) = -2.0\,\text{µ}\mathrm{C}$ ; $i(0) = 0$ (cosine flat at $t=0$ ); $I_0 = 2\pi \times 2.0 \times 10^{-6}/4.0 \times 10^{-3} \approx 3.1\,\mathrm{mA}$ .

Exercise 31.5 ★

At which instants of the cycle is the energy entirely in the capacitor? Entirely in the coil? What is the mechanical counterpart of each moment for a pendulum?

Solution

Solution of Exercise 31.5.

All in the capacitor when $q = \pm Q_0$ , $i = 0$ ( $t = 0$ , $T_0/2$ , $T_0$ , …): the pendulum’s extremes. All in the coil when $q = 0$ , $i = \pm I_0$ ( $t = T_0/4$ , $3T_0/4$ , …): the pendulum’s lowest, fastest point.

Exercise 31.6 ★★

Verify by substitution that $q(t) = Q_0\cos(2\pi t/T_0)$ satisfies $L\,d^2q/dt^2 + q/C = 0$ exactly when $T_0 = 2\pi\sqrt{LC}$ .

Solution

Solution of Exercise 31.6.

$d^2q/dt^2 = -(2\pi/T_0)^2\,Q_0\cos(2\pi t/T_0) = -(2\pi/T_0)^2 q$ , so $L\,d^2q/dt^2 + q/C = q\,[\,1/C - L(2\pi/T_0)^2\,] = 0$ for all $t$ exactly when $(2\pi/T_0)^2 = 1/(LC)$ , i.e. $T_0 = 2\pi\sqrt{LC}$ .

Exercise 31.7 ★★

A long-wave station broadcasts at $198\,\mathrm{kHz}$ . With $C = 220\,\mathrm{pF}$ , what inductance tunes to it?

Solution

Solution of Exercise 31.7.

$L = \dfrac{1}{4\pi^2 f_0^2 C} = \dfrac{1}{39.5 \times (1.98 \times 10^{5})^2 \times 2.2 \times 10^{-10}} \approx 2.9\,\mathrm{mH}$ .

Exercise 31.8 ★★

$C = 1.0\,\text{µ}\mathrm{F}$ charged to $10\,\mathrm{V}$ discharges into $L = 10\,\mathrm{mH}$ . Compute the stored energy, the peak current (by energy conservation), and the instant it is first reached.

Solution

Solution of Exercise 31.8.

$E = \tfrac12 CU_0^2 = 5.0 \times 10^{-5}\,\mathrm{J}$ ; $\tfrac12 LI_0^2 = E$ gives $I_0 = \sqrt{2E/L} = \sqrt{1.0 \times 10^{-2}} = 0.10\,\mathrm{A}$ , first reached at $T_0/4 = \tfrac14 \times 0.63\,\mathrm{ms} \approx 0.16\,\mathrm{ms}$ .

Exercise 31.9 ★★

Explain why $E_C$ and $E_L$ oscillate at frequency $2f_0$ , not $f_0$ . Show that at $t = T_0/8$ (started from $q = Q_0$ ) the energy is split exactly in half.

Solution

Solution of Exercise 31.9.

$E_C \propto \cos^2(2\pi t/T_0)$ , and $\cos^2$ repeats every half turn: two fillings of the capacitor per period, hence $2f_0$ . At $t = T_0/8$ the phase is $\pi/4$ : $\cos^2(\pi/4) = \tfrac12$ , so $E_C = E_L = E/2$ .

Exercise 31.10 ★★

On an oscillogram of a lightly damped RLC circuit, successive maxima read $4.0\,\mathrm{V}$ then $3.0\,\mathrm{V}$ . Assuming the ratio stays constant, predict the sixth maximum after the $4.0\,\mathrm{V}$ one. What fraction of the energy survives each pseudo-period? What can you say of the pseudo-period itself?

Solution

Solution of Exercise 31.10.

Ratio $3.0/4.0 = 0.75$ per pseudo-period: sixth maximum $4.0 \times 0.75^6 \approx 0.71\,\mathrm{V}$ . Energy $\propto$ amplitude squared: $0.75^2 \approx 56\%$ survives each period. The pseudo-period stays $\approx T_0 = 2\pi\sqrt{LC}$ : light damping shrinks the amplitude, hardly the tempo.

Exercise 31.11 ★★

An LC circuit has $L = 0.50\,\mathrm{H}$ and $C = 20\,\text{µ}\mathrm{F}$ . Compute $T_0$ ; then, using the dictionary (Proposition 31.10), find the stiffness $k$ giving a $0.50\,\mathrm{kg}$ glider the same period, and check it.

Solution

Solution of Exercise 31.11.

$T_0 = 2\pi\sqrt{0.50 \times 2.0 \times 10^{-5}} = 2\pi \times 3.16 \times 10^{-3} \approx 20\,\mathrm{ms}$ . Dictionary: $k = m/(LC) = 0.50/1.0 \times 10^{-5} = 5.0 \times 10^{4}\,\mathrm{N}/\mathrm{m}$ ; check $2\pi\sqrt{m/k} = 2\pi\sqrt{1.0 \times 10^{-5}}$ — the same $20\,\mathrm{ms}$ .

Exercise 31.12 ★★★

A tuner uses $L = 0.20\,\mathrm{mH}$ and a variable capacitor sweeping $50\text{ to }500\,\mathrm{pF}$ . Compute the two extreme frequencies. Show that $f_0 \propto 1/\sqrt{C}$ , so tenfold in $C$ gives only $\sqrt{10} \approx 3.2$ in frequency.

Solution

Solution of Exercise 31.12.

$C = 500\,\mathrm{pF}$ : $f_0 = 1/(2\pi\sqrt{1.0 \times 10^{-13}}) \approx 0.50\,\mathrm{MHz}$ ; $C = 50\,\mathrm{pF}$ : $f_0 \approx 1.6\,\mathrm{MHz}$ . At fixed $L$ , $f_0 = (2\pi\sqrt{L})^{-1} C^{-1/2} \propto 1/\sqrt{C}$ : tenfold $C$ , only $\sqrt{10} \approx 3.2$ in $f_0$ — as found.

Exercise 31.13 ★★★

A real coil makes the circuit lose $5.0\%$ of its energy each pseudo-period. After how many periods has the energy halved? What is the amplitude then, as a fraction of the start? At $f_0 = 1.0\,\mathrm{MHz}$ , how long is that in microseconds?

Solution

Solution of Exercise 31.13.

$0.95^n = \tfrac12$ : $n = \ln 2/\lvert\ln 0.95\rvert \approx 13.5$ , so about $14$ periods. Amplitude $\propto \sqrt{E}$ : down to $1/\sqrt{2} \approx 0.71$ of the start. At $1.0\,\mathrm{MHz}$ , $T_0 = 1.0\,\text{µ}\mathrm{s}$ : about $14\,\text{µ}\mathrm{s}$ .

Exercise 31.14 ★★★

A watch oscillator stores $E = 1.0 \times 10^{-9}\,\mathrm{J}$ and loses $2.0\%$ of it per cycle at $f_0 = 32\,768\,\mathrm{Hz}$ . What average power must the maintaining circuit supply? Why does the push have to arrive in step with the oscillation?

Solution

Solution of Exercise 31.14.

Loss per cycle $0.020 \times 1.0 \times 10^{-9}\,\mathrm{J} = 2.0 \times 10^{-11}\,\mathrm{J}$ , at $32\,768$ cycles per second: $P = 32\,768 \times 2.0 \times 10^{-11} \approx 6.6 \times 10^{-7}\,\mathrm{W}$ — under a microwatt, years on a button cell. In step, because only a push at the right phase adds energy (resonance); out of step it would brake the swing.

Exercise 31.15 ★★★

After each edge of a square signal, an experimenter’s wiring “rings” at $25\,\mathrm{MHz}$ ; the stray inductance is about $1.0\,\text{µ}\mathrm{H}$ . Estimate the stray capacitance responsible. Which regime should a small added series resistor push the circuit into, and why does that cure the ringing?

Solution

Solution of Exercise 31.15.

$C = \dfrac{1}{4\pi^2 f_0^2 L} = \dfrac{1}{39.5 \times (2.5 \times 10^{7})^2 \times 1.0 \times 10^{-6}} \approx 41\,\mathrm{pF}$ . The resistor should push the loop into the aperiodic regime: the charge then settles without overshooting — no oscillation, no ringing.

31.6 Problem: Building a crystal radio

Problem 31.1

Weekend problem — building a crystal radio: one coil, one variable capacitor, a diode and an earphone — no battery, no amplifier, and the evening news arrives on the energy of the wave itself

A kit from a grandparent’s attic: a coil $L = 0.20\,\mathrm{mH}$ wound on a cardboard tube, a variable capacitor whose plates rotate from $20\,\mathrm{pF}$ to $480\,\mathrm{pF}$ , a crystal diode, an earphone, and $20\,\mathrm{m}$ of aerial wire. The stations to catch broadcast between $530\,\mathrm{kHz}$ and $1700\,\mathrm{kHz}$ .

Part I — The tuning loop.

The capacitor, charged, is left to discharge into the coil: write the loop law and turn it into the differential equation for $q(t)$ .
Verify by substitution that $q(t) = Q_0\cos(2\pi t/T_0)$ solves it, provided $T_0 = 2\pi\sqrt{LC}$ .
Check by dimensional analysis that $\sqrt{LC}$ is a time.
Dial set to $C = 330\,\mathrm{pF}$ : compute $T_0$ and $f_0$ . Is the loop tuned inside the band?
Find the capacitance that tunes to exactly $1.00\,\mathrm{MHz}$ .

Part II — The variable capacitor.

Compute the capacitance needed for $530\,\mathrm{kHz}$ .
Compute the capacitance needed for $1700\,\mathrm{kHz}$ .
Compute the two frequencies the kit’s $20\text{ to }480\,\mathrm{pF}$ capacitor actually reaches. Does it cover the whole band?
Show that $f_0 \propto 1/\sqrt{C}$ at fixed $L$ ; by what factor must $C$ change to double $f_0$ ?
The plates open linearly, so $C$ falls steadily as the dial turns. Explain why the stations then crowd together at one end of the dial — and which end.

Part III — Energy and damping. Tuned to $1.00\,\mathrm{MHz}$ ( $C = 127\,\mathrm{pF}$ ), the aerial momentarily charges the capacitor to $U_0 = 20\,\mathrm{mV}$ .

Compute the charge $Q_0$ and the stored energy $E$ .
Compute the peak current $I_0$ , using energy conservation.
The coil’s resistance dissipates $4.0\%$ of the stored energy per cycle: after how many cycles has the energy halved?
How long is that in microseconds? Why can this “ringing” alone not play the evening news?
Explain why the arriving wave keeps the loop swinging only when tuned to that station — and what maintains the oscillation at the transmitter (Remark 31.9).

Part IV — The mechanical twin.

Using the dictionary (Proposition 31.10) with a glider of mass $m = 0.20\,\mathrm{kg}$ , compute the stiffness $k$ of the spring matching the tuned circuit.
A stiff car spring has $k \approx 5 \times 10^{4}\,\mathrm{N}/\mathrm{m}$ . Comment: can a bench-top spring–mass system oscillate at megahertz?
Keep a reasonable $k = 100\,\mathrm{N}/\mathrm{m}$ instead: what mass oscillates at $1.00\,\mathrm{MHz}$ ?
What real object is such a featherweight, ultra-stiff mechanical oscillator, and where did you meet it earlier this year (Chapter 28)?
Spec sheet, one sentence: the band your radio sweeps, the capacitance that picks $1.00\,\mathrm{MHz}$ , how long a lone ring lasts, and what plays the news anyway.

Solution

Solution of Problem 31.1.

1. Loop law: $u_L + u_C = 0$ , with $u_C = q/C$ , $u_L = L\,di/dt$ , $i = dq/dt$ : $L\,d^2q/dt^2 + q/C = 0$ .

2. $d^2q/dt^2 = -(2\pi/T_0)^2 q$ , so the equation reads $q\,[\,1/C - L(2\pi/T_0)^2\,] = 0$ , true for all $t$ exactly when $T_0 = 2\pi\sqrt{LC}$ .

3. $LC$ : $(\mathrm{V}\,\mathrm{s}/\mathrm{A})(\mathrm{C}/\mathrm{V}) = \mathrm{s} \times \mathrm{C}/\mathrm{A} = \mathrm{s}^{2}$ — $\sqrt{LC}$ is a time.

4. $LC = 6.6 \times 10^{-14}\,\mathrm{s}^{2}$ : $T_0 = 2\pi \times 2.57 \times 10^{-7} \approx 1.6\,\text{µ}\mathrm{s}$ , $f_0 \approx 620\,\mathrm{kHz}$ — inside $530\text{ to }1700\,\mathrm{kHz}$ .

5. $C = \dfrac{1}{4\pi^2 f_0^2 L} = \dfrac{1}{39.5 \times 1.0 \times 10^{12} \times 2.0 \times 10^{-4}} \approx 127\,\mathrm{pF}$ .

6. Same formula at $530\,\mathrm{kHz}$ : $C \approx 451\,\mathrm{pF}$ .

7. At $1700\,\mathrm{kHz}$ : $C \approx 44\,\mathrm{pF}$ .

8. $C = 480\,\mathrm{pF}$ : $f_0 \approx 514\,\mathrm{kHz}$ ; $C = 20\,\mathrm{pF}$ : $f_0 \approx 2.5\,\mathrm{MHz}$ — the whole band, with margin at both ends.

9. $f_0 = (2\pi\sqrt{L})^{-1}C^{-1/2} \propto 1/\sqrt{C}$ ; doubling $f_0$ requires dividing $C$ by $4$ .

10. Since $f_0 \propto 1/\sqrt{C}$ , equal steps of $C$ move $f_0$ slowly at large $C$ and fast at small $C$ : a given spacing between stations then takes little dial travel — they crowd at the high-frequency (low- $C$ ) end.

11. $Q_0 = CU_0 = 1.27 \times 10^{-10} \times 0.020 \approx 2.5 \times 10^{-12}\,\mathrm{C}$ ; $E = \tfrac12 CU_0^2 \approx 2.5 \times 10^{-14}\,\mathrm{J}$ .

12. $\tfrac12 LI_0^2 = E$ : $I_0 = U_0\sqrt{C/L} = 0.020\sqrt{6.35 \times 10^{-7}} \approx 16\,\text{µ}\mathrm{A}$ .

13. $0.96^n = \tfrac12$ : $n = \ln 2/\lvert\ln 0.96\rvert \approx 17$ cycles.

14. At $1.00\,\mathrm{MHz}$ , $T_0 = 1.0\,\text{µ}\mathrm{s}$ : the ring halves in about $17\,\text{µ}\mathrm{s}$ . A lone ring dies in microseconds; music needs the loop fed continuously.

15. The wave pushes millions of times per second; only when its frequency matches $f_0$ do the pushes arrive in step and build the swing — resonance; off-tune stations push as often against as with it. At the transmitter, an amplifier returns the dissipated energy each cycle: a maintained oscillator.

16. $LC = 2.54 \times 10^{-14}\,\mathrm{s}^{2}$ : $k = m/(LC) = 0.20/2.54 \times 10^{-14} \approx 7.9 \times 10^{12}\,\mathrm{N}/\mathrm{m}$ .

17. About $10^8$ times the car spring: no bench-top spring–mass system reaches megahertz — everyday stiffnesses and masses are hopelessly slow.

18. $m = \dfrac{k}{(2\pi f_0)^2} = \dfrac{100} {(6.28 \times 10^{6})^2} \approx 2.5 \times 10^{-12}\,\mathrm{kg}$ — a few nanograms, a speck of dust.

19. A quartz crystal: a featherweight, ultra-stiff sliver singing at its natural frequency — the timekeeper met earlier this year.

20. With $L = 0.20\,\mathrm{mH}$ and $20\text{ to }480\,\mathrm{pF}$ the loop sweeps about $0.51\text{ to }2.5\,\mathrm{MHz}$ , covering the band; $C \approx 127\,\mathrm{pF}$ picks $1.00\,\mathrm{MHz}$ ; left alone the ring halves in $\approx 17\,\text{µ}\mathrm{s}$ , but the station’s own wave, resonating in step, re-feeds the loop — the news plays on the energy of the wave itself.