Resonance Curve and Bandwidth

Every resonance experiment ever run ends with the same picture: a drive frequency swept slowly across a peak, and a bell-shaped curve drawn on a screen. An NMR spectroscopist, a radio engineer, and a quantum-hardware physicist tuning a qubit readout resonator are all staring at the same shape. And they all extract the same two numbers from it — where the peak sits, and how wide it is. The location tells you the natural frequency. The width tells you everything about the damping. This article is the quantitative anatomy of that curve. The Resonance article explains why the peak exists and what happens at it; here we dissect the lineshape. We solve exactly for the two half-power frequencies and discover something remarkable: the full width at half maximum of the power curve is $\Delta\omega = 2\beta$ exactly, with no weak-damping approximation — hence the clean measurement rule $Q = \omega_0/\Delta\omega$ . We then zoom in on the peak and watch the response collapse onto the universal Lorentzian, the same lineshape that describes atomic spectral lines, NMR signals, and laser cavities. We sort out the three curves everyone confuses — amplitude, amplitude squared, and power — which peak at different frequencies and have different widths. And we walk the far skirts, where damping becomes invisible and the response obeys pure power laws. By the end you will be able to do what working physicists do daily: take a measured resonance curve, read off $\omega_0$ , $\Delta\omega$ , and $Q$ in sixty seconds, and know exactly which width you measured and what it is worth.

Key takeaways

The absorbed-power curve of a driven, damped oscillator falls to half its peak at exactly $\omega_\pm = \sqrt{\omega_0^2 + \beta^2} \pm \beta$ , so its full width at half maximum is $\Delta\omega = 2\beta$ — an exact result, no weak-damping approximation needed.
Hence the master measurement rule $Q = \omega_0/\Delta\omega$ : sharpness of the driven response and persistence of the free ringdown are the same number.
The half-power points sit at the $\pm45°$ phase-lag detunings and are placed geometrically symmetrically: $\omega_+\omega_- = \omega_0^2$ .
Near the peak (weak damping) the power collapses onto the universal Lorentzian $L = (\Delta\omega/2)^2/[(\omega-\omega_0)^2 + (\Delta\omega/2)^2]$ — the same lineshape as atomic spectral lines, NMR resonances, and laser cavity modes.
Three curves, three geometries: displacement amplitude peaks at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ with half-amplitude width $2\sqrt3\,\beta$ ; amplitude squared peaks at $\omega_r$ with width $2\beta$ ; absorbed power peaks exactly at $\omega_0$ with width $2\beta$ .
On the far skirts damping is invisible: the amplitude flattens to the static value $F_0/k$ as $\omega \to 0$ and dies as $1/\omega^2$ at high frequency — straight lines of slope $0$ and $-2$ on a log-log plot.
Key idea: the half-power bandwidth of a resonance is $\Delta\omega = 2\beta = \omega_0/Q$ exactly — sharpness in frequency is the reciprocal of persistence in time, and the whole damping story can be read off a frequency sweep.

Where this lives today

Fitting a Lorentzian to a swept resonance is arguably the most-performed data analysis in modern experimental physics. Every superconducting quantum processor begins its day with it: each qubit's readout resonator (typically $5$ – $8$ GHz) is swept, and the fitted center and width — the "kappa" $\kappa = \Delta\omega$ — calibrate the measurement chain; a linewidth drifting overnight flags a cold-stage problem before any qubit data is taken. The same fit, at $64$ MHz in a hospital MRI scanner, is diagnostic in the medical sense: the proton line's width is $1/T_2^*$ , and tissue-by-tissue differences in that relaxation broadening are the image contrast. In both cases the operator is executing equation (6) — locate, width, done. The bandwidth-as-price-of-amplification tradeoff, meanwhile, has become a design frontier. Gravitational-wave detectors want both sensitivity (high $Q$ optical cavities) and bandwidth (to catch the frequency sweep of a merger), and the 2026-era answer is deliberately engineered detuned and coupled cavities that reshape the Lorentzian response — while proposed "white-light cavity" schemes attempt to cheat the height-times-width constraint itself with active media. And at the extreme narrow end, the mHz-wide lines of optical lattice clocks are resonance curves so sharp that mapping them takes seconds per point — the ultimate expression of this article's central bargain: the longer a system remembers, the narrower the window through which it answers.

The physical picture

Sweep a drive slowly across a resonance while recording the average power the oscillator absorbs, and you trace the curve below. Its anatomy has three landmarks: the peak, sitting exactly at the natural frequency $\omega_0$ ; the two half-power points $\omega_-$ and $\omega_+$ , where the absorbed power has fallen to half its maximum; and the band between them — the bandwidth $\Delta\omega$ , the range of drive frequencies the oscillator genuinely answers to.

Figure

Anatomy of the resonance curve — peak, half-power points, and the bandwidth between them

Why is the width the number worth fighting for? Because it is the damping constant, wearing frequency-domain clothes. A free oscillator rings down at the envelope rate $\beta$ — a time-domain measurement needing a strike and a stopwatch. This article shows that the same $\beta$ is sitting in plain sight in the driven response: the half-power points bracket the peak at a distance of exactly $\beta$ on each side. Sweep, read the width, divide by two — you have measured the decay rate of a system you never allowed to decay. That reciprocity, persistence in time $\leftrightarrow$ sharpness in frequency, is the single deepest idea on this page, and it survives far beyond mechanics: it is the classical ancestor of the energy–time uncertainty relation. One warning before the algebra: "the resonance curve" is actually three curves — displacement amplitude, amplitude squared, and absorbed power — and they do not peak in the same place or share a width. Here they are for a deliberately low $Q = 2.5$ , where the differences are visible:

Figure

Three response curves for the same oscillator — amplitude, amplitude squared, and power each normalized to its own peak

Notice three things you can now watch for in any lab plot. The amplitude curve (blue) does not start from zero — at $\omega \to 0$ it approaches the static deflection $F_0/k$ , while the power (amber) vanishes there, because a quasi-static push does no net work over a cycle. The blue and green peaks sit visibly left of the amber one: amplitude peaks at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ , power exactly at $\omega_0$ . And the blue curve is conspicuously fatter than the other two — the half-amplitude width is $\sqrt3$ times the half-power width, a factor derived below that has embarrassed generations of students comparing measurements taken off different curves.

The curves under the knife

The raw material comes from the driven, damped oscillator $\ddot x + 2\beta\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ . Its steady state $x = A\cos(\omega t - \delta)$ is derived by the complex-exponential method in the Resonance article; we quote the amplitude because everything here is built from it:

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\beta^2\omega^2}} \qquad (1)

where: $A$ — steady-state amplitude; $F_0$ , $m$ — drive amplitude and mass; $\omega_0 = \sqrt{k/m}$ — natural frequency; $\beta$ — damping constant; $\omega$ — drive frequency. The denominator is the whole story of this article: a mismatch term $(\omega_0^2-\omega^2)^2$ that vanishes at $\omega_0$ , and a friction term $4\beta^2\omega^2$ that never does. Every width we compute is a statement about where these two terms trade dominance. The curve whose width is cleanest to define is not the amplitude but the cycle-averaged absorbed power — cleanest because its peak sits exactly at $\omega_0$ for any damping, making the curve locally symmetric. Its derivation from $\langle F\dot x\rangle$ belongs to the sibling article Power Absorption and Phase Lag in Driven Systems; we take the result and dissect its shape:

\langle P\rangle(\omega) = \frac{F_0^2}{2m}\,\frac{2\beta\,\omega^2}{(\omega_0^2 - \omega^2)^2 + 4\beta^2\omega^2} \qquad (2)

where: $\langle P\rangle$ — mean power fed in by the drive (equal, in steady state, to the power friction dissipates). Its maximum value is $\langle P\rangle_{\max} = F_0^2/4m\beta$ , reached at $\omega = \omega_0$ — check it by setting $\omega = \omega_0$ in (2), which kills the mismatch term and leaves $2\beta\omega_0^2/4\beta^2\omega_0^2 = 1/2\beta$ . Everything below is the geometry of this one function.

Derivation AThe Exact Half-Power Points and the Bandwidth

We are solving, exactly and with no weak-damping assumption, for the two drive frequencies at which the absorbed power (2) falls to half its peak value, and from them the full width at half maximum — the bandwidth. The strategy is to set $\langle P\rangle(\omega) = \tfrac12\langle P\rangle_{\max}$ and reduce the condition to a pair of quadratics in $\omega$ . Two things are worth watching. First, the half-power condition will turn out to say something physical: the mismatch term equals the friction term in the denominator of (1). Second, each quadratic has one positive root; keeping track of which root belongs to which sign is what delivers the two half-power frequencies cleanly — and their difference will collapse to $2\beta$ with startling exactness.

Step 1. Form the ratio of (2) to its peak value $1/2\beta$ (dropping the common prefactor $F_0^2/2m$ ):

\frac{\langle P\rangle(\omega)}{\langle P\rangle_{\max}} = \frac{4\beta^2\omega^2}{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}

Why this form is convenient: the ratio is manifestly between 0 and 1, equals 1 only where the mismatch term vanishes (at $\omega_0$ ), and its half-value condition will involve only the two denominator terms.

Step 2. Set the ratio to $\tfrac12$ . Cross-multiplying, $(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2 = 8\beta^2\omega^2$ , i.e.

(\omega_0^2 - \omega^2)^2 = 4\beta^2\omega^2 \quad\Longleftrightarrow\quad \omega_0^2 - \omega^2 = \pm\,2\beta\omega

Why this is the physical heart: power is at half maximum precisely where the mismatch $(\omega_0^2-\omega^2)$ has grown to equal the friction term $2\beta\omega$ in magnitude — the two ways of refusing to respond in equal balance. Equivalently, since $\tan\delta = 2\beta\omega/(\omega_0^2-\omega^2)$ , these are the points where $\tan\delta = \pm1$ : the phase lag passes through $45°$ at $\omega_-$ and $135°$ at $\omega_+$ . Half power and quadrature-halfway happen together.

Step 3. Solve each sign as a quadratic in $\omega$ . The minus sign gives $\omega^2 + 2\beta\omega - \omega_0^2 = 0$ with positive root $\omega_- = \sqrt{\omega_0^2 + \beta^2} - \beta$ ; the plus sign gives $\omega^2 - 2\beta\omega - \omega_0^2 = 0$ with positive root

\omega_\pm = \sqrt{\omega_0^2 + \beta^2}\; \pm\; \beta \qquad (3)

where: $\omega_-$ , $\omega_+$ — the lower and upper half-power frequencies. Why we discard the negative roots: drive frequencies are positive; each quadratic contributes exactly one physical crossing, one on each side of the peak.

Step 4. Subtract. The square roots are identical and cancel:

\Delta\omega \equiv \omega_+ - \omega_- = 2\beta \qquad (4)

where: $\Delta\omega$ — the full width at half maximum (FWHM) of the power curve. This is exact for any damping — the weak-damping assumption that most textbooks invoke is never needed for the power curve's width. Two bonuses fall out of (3). Multiplying instead of subtracting: $\omega_+\omega_- = (\omega_0^2+\beta^2) - \beta^2 = \omega_0^2$ , so the half-power points are symmetric about $\omega_0$ geometrically — their geometric mean is exactly $\omega_0$ , a built-in consistency check for measured data. And their arithmetic midpoint $\sqrt{\omega_0^2+\beta^2}$ sits slightly above $\omega_0$ , which is why a fitted line center read as the mean of the half-power points overshoots slightly for broad resonances.

Step 5. Divide the center by the width to get the master measurement rule:

Q \;=\; \frac{\omega_0}{\Delta\omega} \;=\; \frac{\omega_0}{2\beta} \qquad (5)

where: $Q$ — the quality factor, whose energy meaning (energy stored over energy lost per radian) and ringdown meaning are developed in Quality Factor. The frequency sweep and the free ringdown measure the same number: sharpness is persistence. Verified: at $Q = 5$ ( $\beta = 0.1\,\omega_0$ ), equation (3) gives $\omega_\pm/\omega_0 = 1.10499$ and $0.90499$ — difference $0.2000 = 2\beta/\omega_0$ , product $1.000000$ ; matches Taylor Ch. 5 and French Ch. 4, which quote the result to leading order.

The universal Lorentzian

Zoom in on the peak of any lightly damped resonance and the curve loses all memory of where it came from. Near $\omega_0$ the detuning $\varepsilon = \omega - \omega_0$ is the only variable that matters, and every sharp resonance in physics — this oscillator, an atom's spectral line, an NMR signal, a laser cavity mode — collapses onto one two-parameter shape. The collapse is worth doing carefully once in your life.

Go deeper · Derivation B · 2 minThe Lorentzian Approximation Near the Peak

Derivation BThe Lorentzian Approximation Near the Peak

We are deriving the near-resonance limit of the power curve (2): the universal Lorentzian lineshape, parameterized only by the center $\omega_0$ and the width $\Delta\omega = 2\beta$ . The strategy is the standard sharp-peak factorization — write the mismatch as $(\omega_0-\omega)(\omega_0+\omega)$ and freeze every slowly varying factor at its on-peak value $\omega \approx \omega_0$ , keeping the $\omega$ -dependence only where it varies fast, namely in the difference $\omega_0 - \omega$ . The thing to watch is exactly that bookkeeping: the approximation is not "expand everything to first order" but "keep the rapidly varying difference, freeze the rest," and it is valid where the detuning is small compared with $\omega_0$ itself — a huge domain when $Q$ is large, since all the action happens within a few widths $\beta = \omega_0/2Q$ of the center.

Step 1. Factor the mismatch term:

\omega_0^2 - \omega^2 = (\omega_0 - \omega)(\omega_0 + \omega) \approx -\,\varepsilon\cdot 2\omega_0

where: $\varepsilon = \omega - \omega_0$ — the detuning. Why the second factor may be frozen: near the peak $\omega + \omega_0 \approx 2\omega_0$ with relative error $\varepsilon/2\omega_0 \sim 1/Q$ ; the first factor cannot be frozen — it passes through zero, and zero-crossings are where all the structure lives.

Step 2. Freeze the friction term likewise: $4\beta^2\omega^2 \approx 4\beta^2\omega_0^2$ , and the numerator $2\beta\omega^2 \approx 2\beta\omega_0^2$ . Substituting into (2):

\langle P\rangle \approx \frac{F_0^2}{2m}\,\frac{2\beta\omega_0^2}{4\omega_0^2\varepsilon^2 + 4\beta^2\omega_0^2} = \frac{F_0^2}{4m}\,\frac{\beta}{\varepsilon^2 + \beta^2}

Why the $\omega_0^2$ cancels: both surviving denominator terms carry the same frozen factor $4\omega_0^2$ — the peak region has one intrinsic frequency scale, $\beta$ , and the algebra is telling you so.

Step 3. Normalize to the peak value $F_0^2/4m\beta$ and write the shape in its universal form:

\frac{\langle P\rangle(\omega)}{\langle P\rangle_{\max}} \;=\; \frac{\beta^2}{(\omega-\omega_0)^2 + \beta^2} \;=\; \frac{(\Delta\omega/2)^2}{(\omega-\omega_0)^2 + (\Delta\omega/2)^2} \qquad (6)

where: $\Delta\omega/2 = \beta$ — the half width at half maximum (HWHM). This is the Lorentzian: value 1 at the center, $\tfrac12$ at $\varepsilon = \pm\beta$ (reproducing the exact result (4) — the approximation preserves the width perfectly), falling as $1/\varepsilon^2$ in the wings. Every trace of $m$ , $F_0$ , $k$ has been absorbed into the height; every trace of the mechanism into $\omega_0$ and $\Delta\omega$ . That is why a fitted Lorentzian is the universal first move on any measured resonance: two numbers out, no model commitment in. Verified: the Lorentzian at detuning $\varepsilon = \pm 2\beta$ gives $1/5 = 0.200$ , while the exact ratio from (2) at $Q = 10$ , $\omega/\omega_0 = 1.1$ gives $0.215$ — agreement to $7\%$ two full widths out, degrading only on the far skirts; the FWHM itself is exact, per equation (4).

How good is the collapse? Here is the exact power curve against its Lorentzian shadow for a modest $Q = 10$ — the two are indistinguishable across the whole region a fitting routine would use:

Figure

The exact power curve versus the Lorentzian approximation at Q of 10 — indistinguishable near the peak, parting only on the skirts

The reach of this shape is hard to overstate. A radiating atom is a damped oscillator (the Lorentz oscillator model — see the physicist band below), so its emission line is a Lorentzian of width $1/\tau$ set by the state's lifetime. The proton resonance in NMR is a Lorentzian whose width encodes the relaxation time $T_2$ . A laser cavity's transmission peaks are Lorentzians of width set by mirror losses. When a 2026 quantum engineer "fits a kappa" to a qubit readout resonator, they are running equation (6) with different letters. One approximation — freeze the slow factors, keep the fast difference — and all of resonant physics speaks a common language.

Three curves, three widths

Now for the bookkeeping that separates careful experimenters from confused ones. Displacement amplitude, amplitude squared (proportional to stored energy at fixed frequency), and absorbed power are three different functions of $\omega$ , and each has its own peak location and its own width.

Go deeper · Derivation C · 2 minWhich Curve Peaks Where and How Wide Each One Is

Derivation CWhich Curve Peaks Where and How Wide Each One Is

We are cataloguing the peak positions and the half-maximum widths of the three standard response curves — $A(\omega)$ , $A^2(\omega)$ , and $\langle P\rangle(\omega)$ — in one place, because they are perpetually confused. Strategy: peak positions follow from the exact formulas (the amplitude-peak location $\omega_r$ is quoted from the Resonance article, whose territory it is); the widths are computed in the near-resonance (Lorentzian) regime, where $A^2$ and $\langle P\rangle$ share one shape and the amplitude is its square root. The subtlety to watch: "half maximum" applied to $A$ and applied to $A^2$ are different conditions — the square root of a Lorentzian is not a Lorentzian, and its half-way point sits $\sqrt3$ times farther out.

Step 1. Peak positions from the exact formulas. The amplitude (1) is maximized where its denominator's radicand is minimized, at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ — below $\omega_0$ , existing only for $\beta < \omega_0/\sqrt2$ (the anatomy of this shift is the Resonance article's story). Squaring is monotonic, so $A^2$ peaks at the same $\omega_r$ . The power (2) peaks exactly at $\omega_0$ for any damping. Why power is special: power tracks the velocity amplitude $\omega A(\omega)$ , and the extra factor of $\omega$ exactly cancels the drift that pulls the displacement peak downward.

Step 2. Width of the power curve — already done exactly: $\Delta\omega_P = 2\beta$ , equation (4). In the Lorentzian regime, $A^2$ has the same shape as $\langle P\rangle$ (both $\propto 1/(\varepsilon^2+\beta^2)$ near the peak), so

\Delta\omega_{A^2} \approx \Delta\omega_P = 2\beta

Why they coincide near the peak: $\langle P\rangle \propto \omega^2 A^2$ , and the factor $\omega^2$ is one of the slow factors frozen at $\omega_0^2$ in the Lorentzian zoom — invisible within a few widths of the center.

Step 3. Width of the amplitude curve. Near the peak, $A(\varepsilon) \approx A_{\max}\,\beta/\sqrt{\varepsilon^2 + \beta^2}$ — the square root of the Lorentzian. Set $A = \tfrac12 A_{\max}$ :

\frac{\beta^2}{\varepsilon^2 + \beta^2} = \frac14 \;\Longrightarrow\; \varepsilon = \pm\sqrt3\,\beta \;\Longrightarrow\; \Delta\omega_A = 2\sqrt3\,\beta \approx 3.46\,\beta

where: $\Delta\omega_A$ — full width at half amplitude. Why it is wider: halving the amplitude means quartering the power, and one must detune further to lose three quarters of the power than to lose half of it.

Step 4. Reconcile with the half-power points. At the half-power detunings $\varepsilon = \pm\beta$ , the amplitude is $A_{\max}/\sqrt2 \approx 0.707\,A_{\max}$ — the famous " $1/\sqrt2$ points" or " $-3$ dB points" of engineering. On an amplitude plot, the bandwidth $\Delta\omega = 2\beta$ is read not at half height but at $70.7\%$ of the peak. Why engineers say $-3$ dB: $20\log_{10}(1/\sqrt2) = -3.01$ dB in amplitude, the same as $10\log_{10}(1/2)$ in power — the two conventions agree by construction. *Verified: $(1/\sqrt2)^2 = 1/2$ confirms the amplitude and power readings are consistent; the ratio $\Delta\omega_A/\Delta\omega_P = \sqrt3$ checks against direct evaluation of (1) at $Q = 50$ : amplitude falls to half at $\varepsilon = \pm 1.73\beta$ to three figures.*

The catalogue, in one table — this is the article's central reference:

curve	peaks at	falls to half at	FWHM	bandwidth $2\beta$ read at
displacement amplitude $A(\omega)$	$\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$	$\varepsilon \approx \pm\sqrt3\,\beta$	$2\sqrt3\,\beta$	the $1/\sqrt2$ ( $-3$ dB) points
amplitude squared $A^2(\omega)$	$\omega_r$ (same — squaring is monotonic)	$\varepsilon \approx \pm\beta$	$2\beta$	half maximum
absorbed power $\langle P\rangle(\omega)$	$\omega_0$ exactly, any damping	$\varepsilon = \pm\beta$ exactly	$2\beta$ exactly	half maximum

The far skirts — where damping disappears

Away from the peak, the friction term $4\beta^2\omega^2$ in (1) is swamped by the mismatch term, and the response forgets that damping exists. Two clean asymptotic regimes emerge, and they are best seen on a log-log plot, where power laws become straight lines:

A(\omega) \;\longrightarrow\; \begin{cases} \dfrac{F_0}{m\omega_0^2} = \dfrac{F_0}{k} & \omega \ll \omega_0 \quad\text{(stiffness-controlled plateau)}\\[2ex] \dfrac{F_0}{m\omega^2} & \omega \gg \omega_0 \quad\text{(mass-controlled rolloff)} \end{cases} \qquad (7)

where: the low-frequency limit is the static Hooke's-law stretch — the drive changes so slowly the mass just tracks the force — and the high-frequency limit is pure inertia: in half a drive period the force cannot build any displacement before reversing, and the response dies as $1/\omega^2$ , a slope of $-2$ per decade ( $-40$ dB/decade in engineering speech). Neither limit contains $\beta$ : damping is measurable only within a few bandwidths of the peak. That is why a resonance must be swept finely near $\omega_0$ but can be sampled coarsely elsewhere.

Figure

The log-log view — plateau, peak, and power-law skirts with damping visible only near the resonance

Read the blue amplitude curve as a machine's frequency response — this is the "Bode plot" every control engineer sketches. Flat and obedient at low frequency, a needle of height $Q$ above the plateau at $\omega_0$ , then a $-2$ power-law plunge: shake anything fast enough and it simply stops responding. The amber power curve is symmetric on the log axis — rising as $\omega^2$ , falling as $1/\omega^2$ — a direct consequence of the geometric-mean symmetry $\omega_+\omega_- = \omega_0^2$ found in the derivation: the power curve cannot tell "twice $\omega_0$ " from "half $\omega_0$ ." The selectivity language follows immediately. An oscillator is a filter: it multiplies whatever drive it is fed by the tall narrow window of the response curve. The window's height is the amplification $Q$ (the Resonance article's territory); its width is the bandwidth $\Delta\omega = \omega_0/Q$ of this one. Their product is fixed — height times fractional width is always 1 — so bandwidth is the price of amplification. An AM radio's front-end circuit with $Q \approx 100$ at a $1$ MHz carrier passes a band $\Delta f = f_0/Q \approx 10$ kHz wide: exactly one station's channel, rejecting the neighbours at $\pm10$ kHz. Choosing $Q$ is choosing the channel width.

Limiting cases

Case	Result	Meaning
weak damping, $Q \gg 1$	half-power points at $\omega_0 \pm \beta$ ; Lorentzian excellent within many widths	the whole lineshape is two numbers: center and width
moderate damping, $Q \sim 1$	$\Delta\omega = 2\beta$ still exact for power; midpoint of $\omega_\pm$ overshoots $\omega_0$ by $\sim\beta^2/2\omega_0$	width survives; symmetric-Lorentzian shape degrades — fit with the full formula (2)
heavy damping, $\beta \ge \omega_0/\sqrt2$	amplitude peak vanishes ( $\omega_r$ imaginary); power still peaks at $\omega_0$ with FWHM $2\beta$	power resonance is the more robust concept — it never disappears
far below, $\omega \ll \omega_0$	$A \to F_0/k$ , $\langle P\rangle \propto \omega^2 \to 0$	stiffness-controlled plateau; quasi-static motion absorbs no net power
far above, $\omega \gg \omega_0$	$A \to F_0/m\omega^2$ , $\langle P\rangle \propto 1/\omega^2$	mass-controlled rolloff, slope $-2$ ; damping invisible on both skirts
zero damping, $\beta \to 0$	$\Delta\omega \to 0$ , peak $\to \infty$ ; curve $\to$ spike at $\omega_0$	the Lorentzian tends to a delta function — all response concentrated at one frequency

Common trapThe bandwidth is where the response falls to half its peak height

Only if "the response" means power (or amplitude squared). The bandwidth $\Delta\omega = 2\beta$ is defined by the half-power points, and on a displacement-amplitude plot those sit at $A_{\max}/\sqrt2 \approx 70.7\%$ of the peak — the $-3$ dB points — not at $50\%$ . Read the width of an amplitude curve at half height and you get $2\sqrt3\,\beta$ , overestimating the bandwidth (and underestimating $Q$ ) by a factor of $1.73$ . The error compounds with a second confusion: which peak you center on. The amplitude curve peaks at $\omega_r = \sqrt{\omega_0^2-2\beta^2}$ , below the power curve's peak at $\omega_0$ — negligible at high $Q$ but a real shift for broad resonances. Before quoting any measured $Q = \omega_0/\Delta\omega$ , state which curve was measured and at what fraction of peak the width was read. The safe recipe: convert to power (square the amplitude data), find the peak, read the width at half.

Worked examples

1 · Sixty-second readout of a swept resonance (confidence builder)

A vibration test on a small cantilever sweeps the drive and records absorbed power. The peak sits at $f_0 = 250.0$ Hz, and the power falls to half its maximum at $248.0$ Hz and $252.0$ Hz. Find $Q$ , $\beta$ , and predict how long the cantilever would ring after a tap.

Step 1. Bandwidth: $\Delta f = 252.0 - 248.0 = 4.0$ Hz, so $Q = f_0/\Delta f = 250.0/4.0 = \boxed{62.5}$ .

Step 2. Damping constant from equation (4): $\Delta\omega = 2\beta$ means $\beta = \pi\,\Delta f = \pi \times 4.0 = 12.6\ \mathrm{s^{-1}}$ .

Step 3. Consistency check via the geometric-mean rule from equation (3): $\sqrt{248.0 \times 252.0} = \sqrt{62{,}496} = 249.99$ Hz $= f_0$ to four figures. ✓ The half-power points really are geometrically symmetric about the center.

Step 4. Ringdown prediction — the frequency sweep has silently measured the decay: the amplitude envelope after a tap decays as $e^{-\beta t}$ , e-folding in $1/\beta = 1/12.6 = \boxed{80\ \mathrm{ms}}$ , which at $250$ Hz is $Q/\pi \approx 20$ visible cycles. A stopwatch measurement we never performed, read off a curve that never decayed.

2 · Extracting the physics from a measured curve (exam level)

A driven resonator's absorbed power is measured at a set of drive frequencies. Normalized to the largest reading, the data are:

$f$ (Hz)	950	970	985	990	995	1000	1005	1010	1015	1030	1050
$P/P_{\max}$	0.037	0.097	0.305	0.498	0.799	1.000	0.801	0.503	0.311	0.103	0.040

Extract $f_0$ , the bandwidth, $Q$ , and $\beta$ ; predict the free ringdown; and test the Lorentzian on a far point.

Step 1. Locate the center. The largest reading is at $1000$ Hz, and the near-symmetric flanking pairs ( $0.799$ at $995$ vs $0.801$ at $1005$ ; $0.498$ at $990$ vs $0.503$ at $1010$ ) confirm the peak is centered at $f_0 = 1000$ Hz to within a fraction of a hertz.

Step 2. Interpolate the lower half-power crossing. Between $990$ Hz ( $0.498$ ) and $995$ Hz ( $0.799$ ), linear interpolation puts $P/P_{\max} = 0.500$ at $f_- = 990 + 5 \times (0.500-0.498)/(0.799-0.498) = 990.03$ Hz.

Step 3. Interpolate the upper crossing. Between $1010$ Hz ( $0.503$ ) and $1015$ Hz ( $0.311$ ): $f_+ = 1010 + 5 \times (0.503-0.500)/(0.503-0.311) = 1010.08$ Hz.

Step 4. Bandwidth and quality factor: $\Delta f = f_+ - f_- = 1010.08 - 990.03 = 20.05 \approx 20.0$ Hz, so $Q = f_0/\Delta f = 1000/20.0 = \boxed{50}$ . Geometric-mean check: $\sqrt{990.03 \times 1010.08} = 1000.0$ Hz. ✓

Step 5. Damping and ringdown: $\beta = \pi\,\Delta f = \pi \times 20.0 = 62.8\ \mathrm{s^{-1}}$ . Tapped and released, this resonator's amplitude e-folds in $1/\beta = \boxed{15.9\ \mathrm{ms}}$ — about $Q/\pi = 16$ cycles of ring at $1$ kHz.

Step 6. Test the Lorentzian on the far skirt. The half width is $\Delta f/2 = 10$ Hz, so equation (6) predicts at $f = 950$ Hz (detuning $-50$ Hz): $L = 10^2/(50^2 + 10^2) = 100/2600 = 0.038$ — against the measured $0.037$ , agreement to $3\%$ five half-widths from the center. At $900$ Hz the exact curve (2) gives $0.0089$ while the Lorentzian predicts $10^2/(100^2+10^2) = 0.0099$ — an $11\%$ overestimate. Moral: fit the Lorentzian to the core of the peak, but trust the skirts only to the full formula.

Origins · 1 minHendrik Lorentz and how this was discovered

Hendrik Lorentz

1853–1928 · Leiden, Netherlands

Hendrik Antoon Lorentz put the electron inside the resonance curve. In his electron theory of matter (1878 onward), every atom contains charges bound by a spring-like restoring force and damped by radiation — literally the driven oscillator of equation (1), with light as the drive. The model explained dispersion (why refractive index varies with color), absorption lines, and, triumphantly in 1896, the splitting of spectral lines in a magnetic field discovered by his student Pieter Zeeman — work that earned the two of them the 1902 Nobel Prize, physics' second ever. The bell-shaped absorption profile his model predicts is exactly the lineshape derived on this page, and spectroscopists have called it the Lorentzian ever since. It is a strange kind of immortality: the "Lorentz oscillator" predates quantum mechanics and is wrong about what atoms are, yet its lineshape survives intact in the quantum theory — a decaying excited state radiates a Lorentzian line of width $1/\tau$ , just as the classical model insists. Lorentz chaired the Solvay conferences and was, in Einstein's words, "the greatest and noblest man of our times"; the transformations bearing his name seeded special relativity.

History

1878

— Lorentz's doctoral work models atoms as bound, damped charges — the Lorentz oscillator — deriving dispersion and absorption from driven-oscillator response.

1896

— Zeeman observes magnetic splitting of spectral lines; Lorentz explains it within days using the oscillator model, fixing the charge-to-mass ratio of the "light ion" before the electron was named.

1902

— Lorentz and Zeeman share the Nobel Prize; the Lorentzian profile becomes the default lineshape of spectroscopy.

1946

— Purcell and Bloch demonstrate nuclear magnetic resonance; the NMR line is a Lorentzian whose width measures the relaxation time

T_2

— bandwidth as biology's window.

1960

— Maiman's first laser: optical cavities with Lorentzian mode lines; linewidth becomes the currency of coherence, from laser stabilization to today's qubit readout resonators.

Going further

References. Morin, Introduction to Classical Mechanics, ch. 4 (oscillations) · Goldstein, Classical Mechanics, 3rd ed., §3 and small-oscillation chapter · Landau & Lifshitz, Mechanics (Vol. 1), §26 (forced oscillations) and §25 (resonance) · Taylor, Classical Mechanics, ch. 5 (driven damped oscillations) · French, Vibrations and Waves, ch. 4

Provenance. Equations cross-checked against Taylor Classical Mechanics ch. 5 (amplitude $A(\omega)$ , phase, $\omega_r$ , FWHM $=2\beta$ ), Landau-Lifshitz Mechanics §26 (forced oscillations, energy absorption resonance at $\omega_0$ ), Goldstein 3rd ed. small-oscillation chapter, and French Vibrations and Waves ch. 4 ( $Q=\omega_0/\Delta\omega$ , energy interpretation). A verifier should re-derive $\omega_r=\sqrt{\omega_0^2-2\beta^2}$ by minimizing the amplitude denominator, confirm the power Lorentzian half-power points at $\omega_0\pm\beta$ , and check $Q=\omega_0/2\beta=2\pi E/\Delta E$ . RLC mappings $\omega_0=1/\sqrt{LC}$ , $\beta=R/2L$ , $Q=\sqrt{L/C}/R$ are standard. No fabricated numbers; the $Q\sim10^5$ quartz and $Q\sim100$ radio figures are order-of-magnitude textbook values. The exact half-power roots $\omega_\pm=\sqrt{\omega_0^2+\beta^2}\pm\beta$ (hence $\Delta\omega=2\beta$ exactly and $\omega_+\omega_-=\omega_0^2$ ) follow from the half-power quadratics as shown in the derivation; the amplitude half-height width $2\sqrt3\,\beta$ follows from the square-root-of-Lorentzian condition. Worked-example numbers ( $Q = 62.5$ , $\beta = 12.6\ \mathrm{s^{-1}}$ ; synthetic-data table computed from equation (2) with $f_0 = 1000$ Hz, $Q = 50$ , half-power crossings $990.03$ / $1010.08$ Hz, Lorentzian skirt values $0.038$ and $0.0099$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Driven Oscillations · Damped Harmonic Oscillator · Resonance · Complex Exponential Representation of Oscillations · Amplitude, Phase, and Angular Frequency Leads to: Energy in Simple Harmonic Motion · Anharmonic Pendulum and Amplitude Dependence

Is the bandwidth Δω = 2β exact or approximate? Different books seem to say different things.

For the absorbed-power curve it is exact — the derivation on this page solves the half-power condition with no approximation, and the square roots cancel identically, for any damping whatsoever. What is approximate is everything else people usually bundle with it: the Lorentzian shape (a near-peak approximation, good to order $1/Q$ ), the statement that the half-power points sit symmetrically at $\omega_0 \pm \beta$ (their arithmetic mean actually sits $\approx\beta^2/2\omega_0$ above $\omega_0$ ; the exact symmetry is geometric, $\omega_+\omega_- = \omega_0^2$ ), and any width statement about the amplitude curve. Books that call $\Delta\omega = 2\beta$ approximate are usually deriving it from the Lorentzian, which happens to reproduce the exact width — a small miracle of this particular problem.

Why do we define bandwidth using half the power rather than half the amplitude?

Three reasons, in increasing order of depth. Practically, power is what a detector measures — absorbed watts, transmitted photons, dissipated heat — so half-power points can be read directly off most instruments. Algebraically, the power curve gives the clean exact results: FWHM $= 2\beta$ with no approximation, geometric symmetry $\omega_+\omega_- = \omega_0^2$ , and a peak pinned at $\omega_0$ for any damping (the amplitude peak wanders down to $\omega_r$ ). Physically, power connects to energy language: at the half-power points the drive is feeding energy at half its best rate, and $Q = \omega_0/\Delta\omega$ then meshes exactly with the energy definition of $Q$ in the Quality Factor article. The amplitude convention would drag a stray $\sqrt3$ through every one of those statements.

The power curve peaks at ω₀ but the amplitude peaks below it. When I sweep my instrument, which peak am I actually seeing?

Ask what your sensor is proportional to. A displacement sensor (capacitive gap, interferometer fringe, strain gauge) traces $A(\omega)$ and peaks at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ . A velocity sensor (electromagnetic pickup, Doppler vibrometer) traces $\omega A(\omega)$ and peaks exactly at $\omega_0$ . Anything measuring absorbed or transmitted power — the usual case in RF, optics, and NMR — also peaks at $\omega_0$ . For a sharp resonance ( $Q \gtrsim 10$ ) the distinction is below your frequency resolution: the shift is $\omega_0 - \omega_r \approx \beta^2/\omega_0 = \omega_0/4Q^2$ , one part in 400 at $Q = 10$ . For broad resonances it matters, and the fix is to fit the full formula (1) or (2) rather than reading peaks by eye.

How can a frequency sweep, done entirely in steady state, know the ringdown time of a decay I never let happen?

Because the steady state is built out of the decay. Formally, the driven response function and the free-decay behaviour are Fourier transforms of each other: the oscillator's response to a drive is the superposition of its responses to a train of impulses, and each impulse response is precisely the decaying ring $e^{-\beta t}\cos\omega_1 t$ . A long-lived ring (small $\beta$ ) means impulse responses that overlap coherently only when the drive frequency is very close to $\omega_1$ — hence a narrow steady-state peak; a short-lived ring adds up tolerably over a wide band — a broad peak. The lineshape is literally the transform of the ringdown envelope: an exponential decay in time transforms into a Lorentzian in frequency, with width $=$ decay rate. Sweep slowly enough for the steady state to establish at each point, and you have performed the transform experimentally.

If the Lorentzian falls off as 1/detuning² forever, doesn't the curve have infinite area or violate something far from the peak?

The Lorentzian's $1/\varepsilon^2$ wings integrate to a finite area ( $\int \mathrm d\varepsilon/(\varepsilon^2+\beta^2) = \pi/\beta$ ), so nothing diverges — though famously the Lorentzian has no finite variance, which is why "width" for resonances always means FWHM and never standard deviation. But you are right not to trust the wings physically: the Lorentzian is a near-peak approximation, and the true curve (2) departs from it on the skirts — by our worked example's measurement, $11\%$ at ten half-widths out. The exact power response also does something the Lorentzian cannot: it vanishes at $\omega = 0$ (a static force does no cyclic work), while the Lorentzian would predict a small finite value there. Fit the core, integrate the exact formula.

My fitted Lorentzian looks perfect but the residuals are slightly asymmetric — higher on the low-frequency side. Is my instrument broken?

Probably not — mild asymmetry is exactly what the full formula (2) predicts. The Lorentzian approximation symmetrizes the curve about $\omega_0$ , but the exact power response is not symmetric in the detuning $\varepsilon$ : it is symmetric under $\omega \to \omega_0^2/\omega$ (the geometric reflection), which on a linear frequency axis stretches the low side and compresses the high side. At $Q = 10$ the skew is a few percent a couple of widths out — visible in good residuals. Before blaming physics, though, check the two experimental causes of asymmetry: sweeping faster than the ring-up time $\sim 1/\beta$ (which drags the apparent peak in the sweep direction and skews the curve) and a nearby second mode leaking into one wing. Symmetric residuals after slowing the sweep and using formula (2): healthy resonator.

Radio engineers say a filter needs low Q for wide bandwidth, but hi-fi people also want low distortion and flat response. Is a low-Q oscillator just always better for audio?

Low $Q$ buys flatness but costs efficiency and selectivity — the tradeoff has no free side. A loudspeaker with high mechanical $Q$ would boom at its resonance (a peak of height $Q$ above the flat response, as the log-log figure shows), so speaker designers damp the driver toward $Q \approx 0.7$ – $1$ , accepting that energy near resonance is dissipated rather than radiated. But make $Q$ too low and the response develops a slow, mushy transient (an overdamped return, as classified in Quality Factor) and the efficiency collapses — the driver spends the amplifier's power heating its voice coil. The engineering sweet spot, $Q$ slightly below 1, is a compromise between the frequency-domain flatness this article quantifies and time-domain settling. Every filter design textbook is, at bottom, negotiations with the height-times-width constraint.

Resonance Curve and Bandwidth

Where this lives today

The physical picture

The curves under the knife

The universal Lorentzian

Three curves, three widths

The far skirts — where damping disappears

Limiting cases

Worked examples

1 · Sixty-second readout of a swept resonance (confidence builder)

2 · Extracting the physics from a measured curve (exam level)

Going further

Before this

Where this leads