Damping, resonance, and the quality factor

The puzzle: the note that breaks the glass¶

A singer holds one sustained note and a wine glass across the room shatters. The same singer, just as loud, a tone higher or lower — nothing. Not quieter destruction: no destruction. The glass is indifferent to almost every sound in the world and lethally sensitive to one.

Module 01 gave you an oscillator that ignores how hard you push it. This one is exquisitely sensitive to when you push it, and it is the same oscillator. Nothing has been added but a loss term and a hand on the other end.

The second half of the puzzle is where the physics is. If the response at the right frequency were truly unbounded, every bridge would fall and every radio would receive every station at once. Something finite is at work, and it turns out to be the same something that makes a struck glass eventually go quiet.

Predict before you calculate¶

Commit to an answer for each of these before running anything. Write them down.

You push a mass on a spring back and forth very slowly — far below its natural frequency. Does the mass move in step with your hand, opposite to it, or a quarter cycle behind? What about very fast pushing?
As you sweep the drive frequency, the response peaks somewhere. Is the peak below, at, or above the natural frequency $\wnat$ ?
You drive at exactly $\wnat$ and never stop. Does the amplitude grow without limit?
You double the damping. What happens to the number of visible cycles after a single plucked release, and what happens to the width of the resonance peak? Are those two answers related?

Explore the model¶

Two animations, both generated from the same wavelab.oscillators code the laboratory runs.

The same oscillator released from rest, over and over, as the damping is swept continuously from very light to very heavy. Left: the motion, inside its decaying envelope. Right: the same history in phase space, module 01’s closed ellipse unwinding into a spiral and finally into a curve that never completes a turn. The green mark on the top bar is critical damping — note that nothing discontinuous happens as the sweep crosses it. The system does not change kind; it simply runs out of cycles.

A drive slowly ramped through resonance. Left: the steady-state motion (blue) against the drive (grey) — watch the lag grow from nothing, through a quarter cycle, to a half cycle, so that far above resonance the mass moves against the hand pushing it. Right: the amplitude and lag curves being filled in as the ramp passes over them. The dotted line is $\wnat$ ; the dashed line is where the amplitude actually peaks. They are not the same line.

Model specification

System: a point mass $m$ on a massless linear spring of stiffness $k$ , with linear velocity damping $b$ and a sinusoidal driving force, in one dimension; the observables are position, velocity, the energies and the cycle-averaged power.
Dynamics: $m\ddot{x} = -kx - b\dot{x} + F_0\cos\omega t$ , solved in closed form where one exists and integrated by velocity Verlet elsewhere.
Boundary: none spatially; the environment appears only as a featureless sink that accepts energy and is never asked what it does with it.
Ensemble: a single deterministic trajectory per choice of parameters and initial conditions; measurement noise, where added, is Gaussian, independent and seeded.
Ignored: the microscopic mechanism behind $b$ , the drive’s back-reaction on whatever supplies it, the spring’s nonlinearity, and any heating of the system by its own dissipation.
Valid when: displacements stay in the linear regime of the real spring, the damping is genuinely viscous — proportional to velocity — and any claim about the steady state is made after the transient has died.
Failure modes: dry friction (a constant-magnitude force, not proportional to $v$ ), resonant amplitudes large enough to leave the linear regime, transient data read as steady state, and any question about where the dissipated energy goes.

Run the laboratory now, then come back. Three things are worth doing before you read on:

Sweep the damping slider slowly through the critical value and watch for the moment the motion stops being oscillatory. Try to catch it happening. You cannot — and that is the observation, not a failure of the slider.
Put the response explorer at a high $Q$ and find the peak by eye. Then lower $Q$ and watch the peak slide downhill in frequency until it falls off the edge and disappears.
Start from rest driving exactly at $\wnat$ , and count how many cycles pass before the amplitude settles. Compare that count with $Q$ .

Derive the result¶

System. Module 01’s mass and spring, plus two additions: a resistive force $-b\dot{x}$ and an applied force $F(t)$ . Newton’s second law reads

m\ddot{x} + b\dot{x} + kx = F(t), \qquad \wnat \equiv \sqrt{\frac{k}{m}}, \qquad \gamma \equiv \frac{b}{m} .

(1)

Note the course’s convention: $\gamma \equiv b/m$ , so amplitudes decay as $e^{-\gamma t/2}$ and energies as $e^{-\gamma t}$ . Several textbooks (French among them) define $\gamma = b/2m$ instead. Both are in print; mixing them within one calculation is the only real mistake.

Free decay: loss as an imaginary frequency. With $F = 0$ , try the module-00 move — assume the answer rotates — but allow the frequency to be complex, $x = \Real\!\left[C\,e^{-\ii\Omega t}\right]$ . Each derivative brings down a factor $-\ii\Omega$ , and the whole equation collapses to a quadratic in $\Omega$ :

\Omega^2 + \ii\gamma\Omega - \wnat^2 = 0 \qquad\Longrightarrow\qquad \Omega = \pm\,\omega_d - \frac{\ii\gamma}{2}, \qquad \omega_d = \sqrt{\wnat^2 - \frac{\gamma^2}{4}} .

(2)

Read what that says. The real part of $\Omega$ is a frequency of oscillation; the imaginary part is a decay rate, because $e^{-\ii(-\ii\gamma/2)t} = e^{-\gamma t/2}$ . Loss is the imaginary part of a frequency. Hold onto that sentence: it returns as the complex refractive index $n + \ii\kappa$ of an absorbing medium, and again as the linewidth of an optical cavity.

Three cases follow from the square root, and they are three cases of one formula:

Underdamped ( $\gamma < 2\wnat$ ): $\omega_d$ is real and the motion is $e^{-\gamma t/2}$ times a sinusoid at $\omega_d$ — ringing, inside a decaying envelope. Note $\omega_d < \wnat$ : damping slows the ringing as well as killing it.
Critical ( $\gamma = 2\wnat$ ): the two roots coincide, $\omega_d = 0$ , and the solution picks up a factor of $t$ : $(C + Dt)e^{-\wnat t}$ . This is the fastest possible return to equilibrium without overshoot, which is why it is what a door closer or a car suspension aims at.
Overdamped ( $\gamma > 2\wnat$ ): both roots are purely imaginary, the motion is a sum of two decaying exponentials, and it crosses zero at most once. Heavier damping now makes the return slower, not faster — the mass creeps home through treacle.

Quality factor. In the underdamped case the energy, which goes as amplitude squared, decays as $e^{-\gamma t}$ . Ask how much is lost per radian of oscillation and one dimensionless number falls out:

Q \equiv \frac{\wnat}{\gamma} = 2\pi \times \frac{\text{energy stored}}{\text{energy lost per cycle}} = \frac{\sqrt{mk}}{b} .

(3)

A struck oscillator rings for about $Q$ radians, which is $Q/2\pi$ cycles, before its energy falls by $1/e$ ; its amplitude survives about $Q/\pi$ cycles. A wine glass has $Q \sim 10^3$ , a quartz watch crystal 10⁵, the mirrors of a gravitational-wave detector 10⁷. Suspension dampers are built at $Q \approx 1$ on purpose.

Driven: the complex-amplitude method. Now switch the drive on, $F = F_0\cos\omega t$ , and wait for the transient to die. Whatever is left must repeat at the drive frequency — nothing else in the problem has a memory — so write both drive and response as rotating arrows, $F = \Real[F_0 e^{-\ii\omega t}]$ and $x = \Real\!\left[X e^{-\ii\omega t}\right]$ , with $X$ complex. Every derivative becomes a multiplication by $-\ii\omega$ , the exponential divides out of every term, and a differential equation becomes one line of arithmetic:

\left(-\omega^2 - \ii\gamma\omega + \wnat^2\right) X = \frac{F_0}{m} \qquad\Longrightarrow\qquad X(\omega) = \frac{F_0/m}{\wnat^2 - \omega^2 - \ii\gamma\omega} .

(4)

This is module 00 collecting its debt. The arrow’s length is the amplitude and its angle is the phase lag:

|X| = \frac{F_0/m}{\sqrt{\left(\wnat^2 - \omega^2\right)^2 + \gamma^2\omega^2}}, \qquad \tan\varphi_{\text{lag}} = \frac{\gamma\omega}{\wnat^2 - \omega^2} .

(5)

Reading the two curves. Three landmarks answer the first prediction question:

$\omega \ll \wnat$ : the denominator is $\wnat^2$ , so $|X| \to F_0/k$ with $\varphi_{\text{lag}} \to 0$ . Push slowly and the mass simply sits where the force puts it. The spring is in charge.
$\omega = \wnat$ : the real part of the denominator vanishes, leaving $|X| = Q\,F_0/k$ and $\varphi_{\text{lag}} = \pi/2$ exactly. Damping alone limits the response, and the mass runs a quarter cycle behind the hand — which is precisely the phase at which the drive does the most work per cycle.
$\omega \gg \wnat$ : $|X| \to F_0/m\omega^2$ with $\varphi_{\text{lag}} \to \pi$ . Inertia is in charge, and the mass moves against the force.

Where the peak really is. Maximising $|X|$ means minimising $\left(\wnat^2 - \omega^2\right)^2 + \gamma^2\omega^2$ . Differentiate with respect to $\omega^2$ and the answer is immediate:

\omega_{\text{peak}} = \sqrt{\wnat^2 - \frac{\gamma^2}{2}} = \wnat\sqrt{1 - \frac{1}{2Q^2}} .

(6)

Q, three ways. The same number can be read off three completely different experiments, and this is the module’s central claim:

Q = \frac{\wnat}{\gamma} \qquad\text{(ringdown: } E \propto e^{-\gamma t}\text{)},

(7)

Q = \frac{\wnat}{\Delta\omega} \qquad\text{(bandwidth: half-power full width } \Delta\omega = \gamma\text{)},

(8)

Q = \frac{\wnat}{2}\left.\frac{d\varphi_{\text{lag}}}{d\omega}\right|_{\wnat} \qquad\text{(phase slope: } d\varphi_{\text{lag}}/d\omega = 2/\gamma\text{)}.

(9)

The first is measured in the time domain by watching something die away; the second and third in the frequency domain by sweeping a drive and recording amplitude, or phase. Nothing guarantees in advance that three such different measurements return one number. That they do is the content of the theory, and the laboratory checks it on data.

Power and the Lorentzian. In the steady state, every joule the drive supplies is dissipated — the amplitude is not growing, so nothing is being stored on average. The cycle-averaged power is

\langle P \rangle = \tfrac12\,\gamma\, m\, \omega^2 |X|^2 = \frac{F_0^2}{2m}\,\frac{\gamma\omega^2}{\left(\wnat^2 - \omega^2\right)^2 + \gamma^2\omega^2} .

(10)

Divide numerator and denominator by $\omega^2$ and the frequency dependence becomes $\left[\left(\wnat^2 - \omega^2\right)/\omega\right]^2 + \gamma^2$ , which is smallest at $\omega = \wnat$ exactly, whatever the damping. Displacement resonance and power resonance are different frequencies. Near a sharp resonance the expression simplifies further:

The Lorentzian

For $Q \gg 1$ and $\omega$ near $\wnat$ , write $\wnat^2 - \omega^2 \approx 2\wnat(\wnat - \omega)$ and $\omega \approx \wnat$ everywhere else. Then

\langle P \rangle \approx \frac{F_0^2}{2m\gamma}\, \frac{\left(\gamma/2\right)^2}{\left(\omega - \wnat\right)^2 + \left(\gamma/2\right)^2},

(11)

a Lorentzian centred on $\wnat$ with full width at half maximum $\gamma$ — hence $Q = \wnat/\Delta\omega$ . The approximation is good to a few percent for $Q \gtrsim 10$ and worthless below $Q \sim 2$ , where the “peak” is wider than its own centre frequency. This curve is not a special fact about springs: it is the universal shape of a weakly damped resonance, and you will meet it again as an atom’s absorption line, a Fabry–Pérot transmission peak, and a laser cavity mode.

Transient plus steady state. The full solution of a linear equation with a drive is any particular solution plus the general solution of the undriven equation. Here that reads

x(t) = \underbrace{|X|\cos\left(\omega t - \varphi_{\text{lag}}\right)}_{\text{steady state}} \; + \; \underbrace{e^{-\gamma t/2}\left(A\cos\omega_d t + B\sin\omega_d t\right)}_{\text{transient}},

(12)

with $A$ and $B$ fixed by the initial conditions — and by nothing else, which is why the steady state is the same however you start. The transient lives for about $2/\gamma$ , or $Q/\pi$ periods. Starting from rest at exact resonance, the two pieces combine into a growing envelope

x(t) \approx Q\,\frac{F_0}{k}\left(1 - e^{-\gamma t/2}\right)\sin\wnat t .

(13)

There is the answer to the third prediction. The amplitude does not grow without limit: it climbs toward $Q$ times the static response and stops, taking about $Q/\pi$ cycles to get there. Resonance is $Q$ -fold amplification, paid for in patience. A high- $Q$ system is selective and loud and slow to respond — one number, three consequences, and the reason no receiver can be arbitrarily sharp and arbitrarily fast at once.

Verify computationally¶

Each check below is also a test in the project’s suite, so these claims cannot silently rot.

1. The closed forms, in every regime. Velocity-Verlet integration against damped_position for light, critical and heavy damping, including from both sides of the critical boundary — the three branches are one function, and agree where they meet to better than one part in 10⁶.

2. The peak position, swept. The falsifying experiment. Amplitude-response curves are computed at a range of dampings, the maximum of each is located numerically, and the results are plotted as $\omega_{\text{peak}}/\wnat$ against $1/Q^2$ .

3. Three ways to Q, on one dataset. q_from_ringdown, q_from_bandwidth and q_from_phase_slope are pointed at simulated measurements of the same oscillator — a decay record, an amplitude sweep, a phase sweep — with detector noise added and eight independent seeds. All three agree with $\sqrt{mk}/b$ , and with each other, to better than $1\%$ . The residual is not scatter: it is systematic, and each estimator’s is different. The bandwidth route reads low because the half-power width is only asymptotically $\gamma$ ; the ringdown route reads high because a noisy record’s mean square is its signal’s plus its noise’s. Telling the two kinds of error apart is what a measurement culture is.

4. Power balance. In the steady state, the cycle average of $F\dot{x}$ computed from an integration equals the cycle average of $b\dot{x}^2$ , and both equal power_absorbed, to two parts in 10³ at 800 steps per period. The drive is not storing energy anywhere; it is paying the damper, exactly.

Transfer the idea¶

A periodic but non-sinusoidal drive — a square wave, a shove once per cycle — is a comb of harmonics, and this response curve weights each one. That is the whole content of module 03: the resonator is a filter, and the Fourier series is the decomposition it filters.
The atom as an oscillator. Bind an electron with a spring and drive it with a light wave and you have the Lorentz model of a medium. Its absorption line is the $\langle P\rangle$ curve above, its refractive index the real part of the same response. This module is module 16-light-in-matter, with charge attached.
Cavities. A Fabry–Pérot resonator’s finesse and a laser cavity’s photon lifetime are $Q$ in other costumes; the transmission peak is this Lorentzian; the linewidth is $\Delta\omega = \wnat/Q$ .
The series RLC circuit. $L\ddot{q} + R\dot{q} + q/C = V(t)$ is this equation with $m \to L$ , $b \to R$ , $k \to 1/C$ . Tuning a radio is choosing $\wnat$ ; the station you do not hear is $\Delta\omega$ doing its job.
The tuned mass damper. The 660-tonne sphere near the top of Taipei 101 is a pendulum deliberately built to resonate with the tower and dissipate, moving out of phase with the building and taking the energy the wind puts in.

Check your understanding¶

A mass of 2.0 kg sits on a spring of stiffness 8.0 N/m with a damping coefficient of 10.0 kg/s. Which regime is this, and what does the motion look like after the mass is displaced and released?
A. Overdamped — it creeps back to equilibrium without oscillating, crossing zero at most once. B. Underdamped — it rings at a frequency slightly below sqrt(k/m) inside a decaying envelope. C. Critically damped — b is large enough that the motion just fails to oscillate. D. It depends on the release amplitude — heavy damping only matters for large displacements.
Check your answer
Correct answer: A — Overdamped — it creeps back to equilibrium without oscillating, crossing zero at most once.
- A. Right. Critical damping is b = 2 sqrt(mk) = 2 sqrt(16) = 8.0 kg/s, and 10.0 is above it, so gamma = 5.0 exceeds 2 omega0 = 4.0. Both roots are real and negative: two decaying exponentials, no ringing. Note that increasing b further makes the return slower, not faster.
- B. Check the boundary before choosing a regime: ringing requires b < 2 sqrt(mk) = 8.0 kg/s, and this system has 10.0. Comparing b with sqrt(mk) = 4.0 rather than with 2 sqrt(mk) is the usual way to land here.
- C. Critical is one exact value, b = 2 sqrt(mk) = 8.0 kg/s, not a range above some threshold. At 10.0 the system is past it and genuinely overdamped, which returns to equilibrium more slowly than the critical case, not faster.
- D. The equation is linear, so every amplitude scales the same solution: the regime is fixed by m, k and b alone. Amplitude-independence survived from module 01 intact; it is the timing of the drive, not the size of the motion, that this module makes decisive.
You drive an oscillator by hand. Compare the displacement with the driving force at three frequencies: far below omega0, exactly at omega0, and far above it. What is the phase lag in each case?
A. 0, then pi/2, then pi — in step, then a quarter cycle behind, then opposed. B. 0 throughout — the mass always follows the force, only the amplitude changes. C. pi/2 throughout — the response always lags a quarter cycle. D. pi, then pi/2, then 0 — the lag decreases as the drive gets faster.
Check your answer
Correct answer: A — 0, then pi/2, then pi — in step, then a quarter cycle behind, then opposed.
- A. Right, and each limit has a physical owner. Slowly, the spring is in charge and the mass sits where the force puts it. At omega0 the real part of the denominator vanishes and only the damping term survives, giving exactly pi/2 whatever the damping. Fast, inertia is in charge and the mass moves against the hand.
- B. Then the drive could never do net work over a cycle in the steady state, and the damping would have nothing to pay it with. The lag is not a detail: it is what lets energy flow in, and it is largest in effect at resonance where the force is in phase with the velocity.
- C. A quarter cycle is the value at omega0 exactly, and it is exact there for any damping — which is what makes it a good way to find omega0. Away from resonance the lag runs from 0 to pi, and how fast it turns is a third way of measuring Q.
- D. This is the sequence backwards. Test it against the slow limit: push a mass very gradually and it goes where you push it, which is a lag of zero, not of half a cycle. Lag grows with frequency because inertia progressively wins the argument.
A damped oscillator is driven at a range of frequencies and its steady-state amplitude is recorded. At which frequency is the amplitude largest?
A. Slightly below omega0, at omega0 sqrt(1 - 1/(2 Q^2)) — and for Q below 1/sqrt(2) there is no peak at all. B. Exactly at omega0 = sqrt(k/m), regardless of the damping. C. Slightly above omega0, because damping has to be overcome before the mass responds. D. At the damped free frequency omega_d = sqrt(omega0^2 - gamma^2/4).
Check your answer
Correct answer: A — Slightly below omega0, at omega0 sqrt(1 - 1/(2 Q^2)) — and for Q below 1/sqrt(2) there is no peak at all.
- A. Right. Maximising |X| means minimising (omega0^2 - omega^2)^2 + gamma^2 omega^2, which gives omega^2 = omega0^2 - gamma^2/2. The shift is 0.06% at Q = 20 — invisible, which is why the wrong answer survives — and 6.5% at Q = 2. Below Q = 1/sqrt(2) the square root has no real value and the response just falls away from F0/k.
- B. Exactly at omega0 is where the phase lag is pi/2 and where the absorbed power peaks — both of those are exact for any damping. The displacement amplitude is the one that shifts, because its denominator carries the extra factor of omega that the power expression cancels.
- C. Damping shifts the peak down, not up: it enters the denominator as -gamma^2/2 under the square root. The frequency that does sit above omega0 is the acceleration peak, at omega0 / sqrt(1 - 1/(2 Q^2)) — the mirror image of this one.
- D. A reasonable guess, and wrong by a factor of two in the correction: the ringing frequency of a free decay is omega_d, but the driven amplitude peaks at sqrt(omega0^2 - gamma^2/2), which is lower still. Three nearby frequencies — omega0, omega_d, omega_peak — that coincide only when the damping vanishes.
Starting from rest, you drive a damped oscillator at exactly its natural frequency and never stop. What happens to the amplitude?
A. It climbs toward Q times the static response F0/k and settles there, taking about Q/pi periods. B. It grows without limit — that is what resonance means. C. It grows for a while and then collapses, because the oscillator falls out of step with the drive. D. It reaches F0/k, the same amplitude a slowly applied force would produce.
Check your answer
Correct answer: A — It climbs toward Q times the static response F0/k and settles there, taking about Q/pi periods.
- A. Right. The envelope grows as Q (F0/k) (1 - e^(-gamma t / 2)): resonance is Q-fold amplification, not divergence. The build-up time 2/gamma is the same 2/gamma that sets how long a struck oscillator rings, which is why a high-Q system is loud and selective and slow.
- B. That is true only of the idealised undamped oscillator, which has no steady state at all. With any damping the energy lost per cycle grows as the amplitude squared while the energy supplied grows only linearly, so the two must balance — and they do, at Q F0/k.
- C. Nothing makes it fall out of step: the steady-state solution repeats at the drive frequency forever, locked at a quarter-cycle lag. Beating and collapse happen when the drive is detuned from omega0, not when it sits on it.
- D. F0/k is the answer far below resonance, where the spring alone opposes the force. At omega0 the spring and inertia terms cancel exactly and only the damping term survives in the denominator, leaving Q times as much.
A swept measurement of a resonator gives a power-response peak at 2400 Hz with a half-power full width of 30 Hz. What is its quality factor Q?
Check your answer
Correct answer: 80.0 (± 2.0)
Q = f0 / Delta f = 2400 / 30 = 80. The ratio is the same whether it is taken in hertz or in radians per second, because Q is dimensionless — that is exactly what lets one number describe the resonator regardless of the units on the axis.
A bell with Q = 30 is struck once. After roughly how many complete oscillations has its amplitude fallen to 1/e of its initial value?
Check your answer
Correct answer: 9.55 (± 1.0)
The amplitude decays as e^(-gamma t / 2), so it reaches 1/e at t = 2/gamma = 2Q/omega0. Divided by the period 2 pi/omega0 that is Q/pi = 30/pi = 9.5 oscillations. Note the factor of two: the energy falls to 1/e in half that time, Q/(2 pi) cycles, and quoting one number for the other is the commonest slip here.
You are handed three measurements of the same unlabelled resonator: a decay record after a single kick, a swept amplitude curve, and a swept phase curve. Describe how you would extract Q from each, and say what it would mean if the three answers disagreed by 30%.
Check your answer
From the decay record: fit the envelope’s logarithm to get gamma, count zero crossings to get omega_d, and report Q = omega0/gamma. From the amplitude sweep: find the peak and the two frequencies where the amplitude has fallen to its maximum over sqrt(2), then Q = omega_peak / (their separation). From the phase sweep: locate where the lag passes pi/2 and measure how fast it is turning there, since Q = (omega0/2) dphi/domega. The three are equivalent for a linear resonator with viscous damping, so a 30% disagreement is evidence against the model rather than against the apparatus — the likeliest culprits are a nonlinear spring at large amplitude, dry friction instead of viscous damping (its decay is linear in time, not exponential), or a second nearby resonance contaminating the sweep. A 1% disagreement means something different: that is the systematic error each estimator carries, and the module’s laboratory measures all three.

Exam-style problems for this module — the ones worth doing with a pen — are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each. No number will save you.

Explain to someone on a swing why pushing at one particular rhythm works and pushing harder at the wrong rhythm does not. Where in the cycle does a good push happen, and why is that the same as “a quarter cycle behind”?
Give three one-sentence definitions of $Q$ — one about time, one about frequency, one about phase — and then say what single physical fact makes them the same number.
Why does the amplitude peak sit below $\wnat$ while the absorbed power peaks exactly on it? Both are properties of the same motion.
Where does the dissipated energy go? Say what this model can tell you, and then say precisely which of its seven specification bullets makes the question unanswerable here.

Advanced: three resonances, and the world without one¶

Displacement, velocity, power. “The resonant frequency” is three different frequencies wearing one name. The displacement amplitude $|X|$ peaks at $\wnat\sqrt{1 - 1/(2Q^2)}$ , below $\wnat$ . The velocity amplitude $\omega|X|$ peaks at $\wnat$ exactly — and so, therefore, does the absorbed power, which goes as the square of the velocity. The acceleration amplitude $\omega^2|X|$ peaks above $\wnat$ , at $\wnat/\sqrt{1 - 1/(2Q^2)}$ , the mirror image of the displacement peak. At $Q = 20$ the three sit within a quarter of a percent of each other and the distinction is pedantry; at $Q = 2$ they are visibly different curves and the pedantry is the answer to an exam question.

That velocity resonance is exact is not an accident. Write the equation in terms of velocity and the oscillator becomes an impedance, $Z = b + \ii(m\omega - k/\omega)$ , whose magnitude is least when the reactive part cancels — at $\wnat$ , for any $b$ . That is the same mechanical impedance that will decide how much of a wave reflects at a boundary in module 10, and the same cancellation of two opposing reactances.

The world below $Q = 1/\sqrt{2}$ . Everything above assumed a peak exists. Below $Q = 1/\sqrt{2}$ — damping heavier than $\sqrt{2}\,\wnat$ — the response curve simply slides downhill from $F_0/k$ , and the system is a second-order low-pass filter with no resonance at all. Engineers reach for this deliberately: $Q = 1/\sqrt{2}$ is the flattest possible response with no peak, the Butterworth condition, and it is what a loudspeaker enclosure or an anti-aliasing filter is usually tuned to. Critical damping, $Q = 1/2$ , is flatter still and settles fastest without overshoot. Between them lies most of engineering; above them lies most of physics.

A harder question the model cannot answer. The damping term removes energy from the mass and the model stops there. In reality that energy goes into the molecular motion of the surrounding fluid, which then jostles the mass back — the same coupling that dissipates must also fluctuate. That is not a loose end but a theorem, the fluctuation–dissipation theorem, and it says that a resonator’s damping and its thermal noise are two faces of one coefficient. It is the reason a gravitational-wave detector’s mirror suspensions are made of fused silica with $Q \sim 10^7$ : not to ring longer, but to be quiet.