Power Absorption and Phase Lag in Driven Systems

Stand behind a playground swing and push at the wrong tempo. Something odd happens that your muscles can feel but your eyes miss: for part of each cycle you are doing work on the swing, and for the rest of the cycle the swing is doing work on you — shoving your hands back, returning the energy you just deposited. The net transfer per cycle is a trickle. Now time your pushes so each one lands while the swing is moving away from you, and the refunds stop: every joule you spend, stays spent. The difference between the two situations is not effort. It is phase. This article is the energy ledger of the driven, damped oscillator. The Resonance article tells the story of the amplitude peak; here we follow the money instead — the instantaneous power $F(t)\dot x(t)$ , its cycle average $\langle P\rangle(\omega)$ at every drive frequency, and the phase lag $\delta(\omega)$ that controls it all. Two central results emerge. First, the average absorbed power can be written two ways — $\langle P\rangle = \tfrac12 F_0 v_0 \sin\delta$ , and a closed-form function of $\omega$ that is the most measured curve in physics — and reconciling them is half the education. Second, the response splits cleanly into an elastic component in phase with the force, which stores energy and hands every bit of it back, and a quadrature component, which carries all of the absorption. Absorption rides entirely on $\sin\delta$ . That split is your first meeting with the susceptibilities $\chi'$ and $\chi''$ of linear response theory: every absorption spectrum ever measured — NMR, atomic, mechanical — is this Lorentzian wearing a different costume.

Key takeaways

A driven, damped oscillator in steady state receives instantaneous power $P(t) = F(t)\dot x(t)$ that generally changes sign within each cycle — the drive does work, then undoes it. Only exactly at $\omega = \omega_0$ is $P(t) \ge 0$ at every instant.
The cycle-averaged absorbed power is $\langle P\rangle = \tfrac12 F_0 v_0 \sin\delta$ : force amplitude, times velocity amplitude, times the sine of the phase lag — the "power factor" of mechanics.
In closed form, $\langle P\rangle(\omega) = \dfrac{F_0^2\beta\omega^2}{m\left[(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2\right]}$ , which for weak damping is a Lorentzian of full width $2\beta$ centered exactly at $\omega_0$ — for any damping strength.
The response decomposes into an in-phase (elastic) part $A\cos\delta$ that does zero net work and a quadrature (absorptive) part $A\sin\delta$ that does all of it; these are the mechanical ancestors of $\chi'$ and $\chi''$ .
The phase lag $\delta(\omega)$ sweeps from $0$ (quasi-static tracking) through exactly $90°$ at $\omega_0$ (velocity in phase with force — the sweet spot) to $180°$ (antiphase, inertia-dominated).
In steady state the books balance cycle by cycle: work in from the drive $=$ heat out through friction, exactly.
Key idea: only the part of the motion in phase with the velocity of the drive cycle absorbs energy; that part is proportional to $\sin\delta$ , is maximal when $\delta = 90°$ at $\omega = \omega_0$ , and traces out the Lorentzian absorption curve that spectroscopy measures everywhere.

The core insight

Only the component of the steady-state response in quadrature with the drive absorbs energy — absorption is proportional to $\sin\delta$ , peaks exactly at $\omega_0$ where the phase lag is $90°$ and the velocity locks in phase with the force, and its frequency profile is the universal Lorentzian of absorption spectroscopy.

Where this lives today

Sweep-and-absorb is the daily bread of quantum hardware labs. To characterize a superconducting qubit, engineers send a weak microwave tone toward it and step the frequency across a few megahertz around 5 GHz; the fraction of power the qubit soaks up traces out equation (6), a Lorentzian whose center calibrates the qubit frequency and whose width — often only a few kilohertz, an effective $Q$ of order $10^6$ — measures its coherence time. The instrument doing the measuring works in exactly this article's language: an IQ mixer demodulates the reflected signal into an in-phase and a quadrature voltage, literally $A_{\text{el}}$ and $A_{\text{abs}}$ , and the software fits the dispersive curve and the absorption curve simultaneously to pin down $\omega_0$ and $\beta$ . Experimentalists lock onto the phase, not the amplitude, for the same reason this article did: the $90°$ crossing marks $\omega_0$ exactly, at any damping. The same two curves run chemistry and medicine. An NMR spectrometer's absorption-mode line is $A_{\text{abs}}(\omega)$ with nuclear parameters; an atmospheric lidar retrieving methane concentrations fits Lorentzian (and Doppler-broadened) absorption profiles; and impedance spectroscopy of batteries reads energy storage versus loss from the in-phase versus quadrature current — the "power factor" that grid engineers bill for is precisely $\sin\delta$ wearing an electrical costume. One mechanical derivation, an industry of instruments.

The physical picture

Watch the power, not the position. In steady state the oscillator moves as $x = A\cos(\omega t - \delta)$ while the force runs as $F_0\cos\omega t$ ; the instantaneous power delivered by the drive is the product $F(t)\dot x(t)$ , and its sign history within one cycle is the whole story of this article. Off resonance, force and velocity spend part of each cycle pointing the same way (the drive does work) and part pointing opposite ways (the oscillator pushes back and returns work to the driver). On resonance the phase lag is exactly $90°$ , which puts the velocity — not the displacement — in step with the force, and the product $F\dot x$ never goes negative: every instant of every cycle feeds energy in.

Figure

Instantaneous power over two drive cycles — off resonance the drive takes back most of what it gives; on resonance it keeps nothing back

The left panel is the swing pushed at the wrong tempo. The power bounces between positive and negative lobes at twice the drive frequency, and the lobes very nearly cancel — the dashed line marks the small positive residue that survives the averaging. The right panel is the timed push: with the velocity in phase with the force, the power is $F_0 v_0\cos^2\omega t$ , a train of strictly non-negative humps averaging to $\tfrac12 F_0 v_0$ . Same force amplitude, same oscillator — utterly different energy flow, and the only thing that changed between the panels is the phase lag $\delta$ . There is a second, sharper way to see it. Split the steady-state motion into two pieces relative to the drive: a component that moves with the force, and a component a quarter-cycle behind it.

Figure

The two-component response — the elastic part rides with the force and returns every joule; the quadrature part carries all the absorption

The in-phase piece, $A\cos\delta$ , behaves like an ideal spring being worked back and forth: the drive loads it on the way out and is paid back in full on the way in. The quadrature piece, $A\sin\delta$ , moves as $\sin\omega t$ , so its velocity runs as $\cos\omega t$ — precisely in phase with the force — and it soaks up energy continuously, exactly like friction. Everything the drive permanently spends flows through that second component. Since $\sin\delta$ is what sets its size, the phase lag is not a technical footnote to the amplitude curve; it is the master dial of energy absorption. And because $\delta$ crosses $90°$ exactly at $\omega = \omega_0$ no matter how strong the damping, the absorption always peaks at the natural frequency — even though the amplitude peak sits somewhere else.

The steady state, inherited

The dynamics is the standard driven, damped oscillator; the Driven Oscillations article derives its steady state in full, so here we quote the results and spend our effort on the energetics. Dividing Newton's second law by the mass puts the equation in its cleanest form, with damping written as $2\beta$ so that free oscillations decay as $e^{-\beta t}$ (many texts write $\gamma \equiv 2\beta = b/m$ for the same coefficient — keep the dictionary handy when comparing books):

\ddot x + 2\beta\dot x + \omega_0^2 x = \frac{F_0}{m}\cos\omega t \qquad (1)

where: $x$ — displacement; $\beta$ — damping constant (units s $^{-1}$ ), from a drag force $-b\dot x$ with $b = 2m\beta$ ; $\omega_0 = \sqrt{k/m}$ — natural frequency; $F_0$ , $\omega$ — amplitude and frequency of the drive. The drive frequency belongs to you; $\omega_0$ and $\beta$ belong to the oscillator. After the transient dies away (a few multiples of $1/\beta$ — the beating between transient and drive during that settling window is its own story, told in the Transient and Steady-State Response article), the motion locks onto the drive frequency with a fixed amplitude and a fixed lag:

x_{\text{ss}}(t) = A(\omega)\cos\!\big(\omega t - \delta(\omega)\big) \qquad (2)

where: $A$ — steady-state amplitude; $\delta$ — the angle by which the displacement trails the force, running from $0$ to $\pi$ . The two functions of $\omega$ that characterize it completely are

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}}, \qquad \tan\delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2} \qquad (3)

where: the two terms under the root are the oscillator's two ways of failing to respond — frequency mismatch and friction. Everything in this article flows from combining equations (2) and (3) with one more idea: power is force times velocity. Note one consequence of (3) immediately, because we will lean on it repeatedly: at $\omega = \omega_0$ the denominator of $\tan\delta$ vanishes identically, so $\delta = \pi/2$ there for every value of $\beta$ . Damping reshapes how fast the phase sweeps through $90°$ , but never where.

The energy ledger, instant by instant

Multiply the drive force by the velocity of the steady state and you get the instantaneous power — the rate at which the drive does work on the oscillator right now. With $\dot x = -A\omega\sin(\omega t - \delta)$ , the product is

P(t) = F(t)\,\dot x(t) = -F_0 A\,\omega\,\cos\omega t\,\sin(\omega t - \delta) \qquad (4)

where: $P(t)$ — power delivered by the drive at time $t$ ; positive when force and velocity point the same way, negative when the oscillator is pushing back on whatever drives it. Expanding the product with $\sin(\omega t - \delta) = \sin\omega t\cos\delta - \cos\omega t\sin\delta$ splits $P(t)$ into a term $\propto \sin 2\omega t$ , which oscillates symmetrically about zero — energy shuttling back and forth, the left panel of the hero figure — and a term $\propto \cos^2\omega t\,\sin\delta$ , which never changes sign. The shuttle averages away; the $\sin\delta$ term is the profit.

Derivation AAverage absorbed power two ways and their reconciliation

We are after the cycle-averaged power $\langle P\rangle(\omega)$ absorbed by the oscillator, in both of its standard costumes: the compact physical form $\langle P\rangle = \tfrac12 F_0 v_0\sin\delta$ and the explicit frequency profile $\langle P\rangle = F_0^2\beta\omega^2/\big[m\big((\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2\big)\big]$ . The route is direct: time-average equation (4) over one drive period, using only two facts about averages of sinusoids, then substitute the known $A(\omega)$ and $\sin\delta(\omega)$ to convert the compact form into the explicit one. The step to watch is the time average itself — the cross term $\langle\cos\omega t\sin\omega t\rangle$ dies and $\langle\cos^2\omega t\rangle$ gives the factor $\tfrac12$ ; lose either and every result downstream is off by a factor of two.

Step 1. Expand the phase in equation (4): $P(t) = F_0 A\omega\big[\sin\delta\cos^2\omega t - \cos\delta\sin\omega t\cos\omega t\big]$ . Why expand: it separates the never-negative absorption term from the sign-alternating shuttle term before any averaging happens, so each can be handled honestly.

Step 2. Average over one period using $\langle\cos^2\omega t\rangle = \tfrac12$ and $\langle\sin\omega t\cos\omega t\rangle = 0$ :

\langle P\rangle = \tfrac12 F_0 A\,\omega\sin\delta = \tfrac12 F_0 v_0 \sin\delta \qquad (5)

where: $v_0 = A\omega$ — the velocity amplitude of the steady state. This is the mechanical power factor: force amplitude times velocity amplitude, discounted by the sine of the lag. At $\delta = 0$ or $\pi$ the drive transfers nothing on average, however hard it pushes; at $\delta = \pi/2$ it collects the full $\tfrac12 F_0 v_0$ . Why $\sin\delta$ and not $\cos\delta$ : the average pairs the force with the velocity, and the velocity of $A\cos(\omega t-\delta)$ carries the phase $\delta$ shifted by a quarter cycle — the in-phase displacement is the out-of-phase velocity.

Step 3. Substitute the machinery of equation (3). From the response triangle, $\sin\delta = 2\beta\omega\big/\sqrt{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}$ , and $A$ carries the same square root in its denominator. The root therefore appears squared:

\langle P\rangle(\omega) = \frac{F_0^2\,\beta\,\omega^2}{m\left[(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2\right]} \qquad (6)

where: every symbol is now an input: drive strength, damping, mass, and the two frequencies. This is the absorption curve — the function a spectroscopist sweeps out point by point. Why the $\omega^2$ upstairs: one $\omega$ comes from converting displacement amplitude to velocity amplitude, the other from $\sin\delta \propto \omega$ ; power tracks velocity, and velocity carries a factor of frequency.

Step 4. Check the peak value. At $\omega = \omega_0$ the mismatch term dies, leaving $\langle P\rangle = F_0^2\beta\omega_0^2/(4m\beta^2\omega_0^2) = F_0^2/4m\beta$ — finite, and inversely proportional to the damping: a cleaner oscillator is a hungrier absorber on resonance. Three quick checks on equation (6): at $\omega_0$ it reduces to $\tfrac12 F_0 v_0$ (using $v_0 = F_0/2m\beta$ ); it vanishes at both $\omega\to0$ and $\omega\to\infty$ ; and under the substitution $\gamma = 2\beta$ it matches the mean-power formula of standard references.

Equation (5) and equation (6) are the same physics wearing different clothes, and each answers a different question. The compact form tells you why absorption happens — it isolates the phase condition — and is the one to reach for in the lab, where $F_0$ , $v_0$ and $\delta$ are directly measurable. The explicit form tells you how much, at every frequency, before you build anything. Get fluent in translating between them: the translation is nothing but the response triangle of equation (3).

The journey of the phase lag

The function $\delta(\omega)$ deserves to be read slowly, because each of its three regimes is a different physical regime of the energy flow. The full derivation lives in the Driven Oscillations article; here is the energy reading.

Far below resonance ( $\omega \ll \omega_0$ ), $\tan\delta \approx 2\beta\omega/\omega_0^2 \to 0^+$ : the mass tracks the force quasi-statically, like slowly leaning on a spring. Force in phase with displacement means force in quadrature with velocity — so during each quarter-cycle of loading, the drive stores spring energy, and during the next quarter-cycle the spring pays it back through the drive's own hand. The ledger nearly balances; only the whisper of friction, $\sin\delta \approx 2\beta\omega/\omega_0^2$ , keeps a residue. The drive does work, then undoes it.

Far above resonance ( $\omega \gg \omega_0$ ), $\delta \to \pi$ : the mass responds in antiphase, dominated by inertia. Again force and velocity are a quarter-cycle apart — now the drive spends its half-cycle accelerating the mass and gets paid back as the mass's momentum drives the pusher backward. Same near-perfect refund, different bank: kinetic energy instead of spring energy.

Exactly at resonance ( $\omega = \omega_0$ ), $\delta = \pi/2$ for any damping — the real part $\omega_0^2 - \omega^2$ of the response denominator vanishes identically, leaving friction as the only opposition, and the displacement settles a clean quarter-cycle behind the force. A quarter-cycle behind in displacement is dead in phase in velocity. This is the swing pushed at the perfect moment: your push lands while the swing recedes at top speed, every time. Nothing is ever refunded, and equation (5) collects its maximum. The reason this article's power curve peaks at $\omega_0$ while the amplitude curve of the Resonance article peaks below it is exactly this: power follows the velocity resonance and the $90°$ phase point, both of which sit at $\omega_0$ ; amplitude follows a compromise between mismatch and friction, which sits at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ .

Elastic and absorptive: the two response curves

Carry out the decomposition of the second figure algebraically. Writing $x_{\text{ss}} = A\cos(\omega t - \delta) = A_{\text{el}}\cos\omega t + A_{\text{abs}}\sin\omega t$ defines the elastic (in-phase) and absorptive (quadrature) amplitudes; the derivation below shows the first does zero net work and the second does all of it.

Go deeper · Derivation B · 1 minThe in-phase component works for free and the quadrature component pays for everything

Derivation BThe in-phase component works for free and the quadrature component pays for everything

The goal is to prove the cleanest structural statement in driven-oscillator physics: split the steady state into a piece in phase with the drive and a piece in quadrature, and the net work sorts itself perfectly — zero through the first, everything through the second. The strategy is elementary: expand the phase, differentiate each piece to get its velocity, and average its product with the force over a cycle. The subtlety to respect is that "in phase" for displacement means "out of phase" for velocity, which is why the labels of the two components swap roles the moment you look at energy rather than geometry.

Step 1. Expand the steady state with the angle-subtraction identity:

x_{\text{ss}}(t) = \underbrace{A\cos\delta}_{A_{\text{el}}}\cos\omega t + \underbrace{A\sin\delta}_{A_{\text{abs}}}\sin\omega t \qquad (7)

where: $A_{\text{el}} = A\cos\delta$ — elastic amplitude, moving with the force; $A_{\text{abs}} = A\sin\delta$ — absorptive amplitude, a quarter-cycle behind it. Why these names: the first behaves like a massless spring driven quasi-statically, the second like a pure dashpot — the next two steps make that literal.

Step 2. Work done through the elastic component. Its velocity is $-A_{\text{el}}\,\omega\sin\omega t$ , in quadrature with $F_0\cos\omega t$ , so the average $\langle F\dot x_{\text{el}}\rangle \propto \langle\cos\omega t\sin\omega t\rangle = 0$ . Why zero exactly: over a full cycle, the loading strokes and the return strokes are congruent — the drive is a lender here, never a spender.

Step 3. Work done through the quadrature component. Its velocity is $+A_{\text{abs}}\,\omega\cos\omega t$ , perfectly in phase with the force, so $\langle F\dot x_{\text{abs}}\rangle = \tfrac12 F_0\,\omega A_{\text{abs}}$ — which is precisely equation (5), since $A_{\text{abs}} = A\sin\delta$ . All of the absorption, none of the storage.

Step 4. Write both amplitudes as explicit functions of frequency by inserting equation (3):

A_{\text{el}}(\omega) = \frac{F_0}{m}\,\frac{\omega_0^2 - \omega^2}{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}, \qquad A_{\text{abs}}(\omega) = \frac{F_0}{m}\,\frac{2\beta\omega}{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2} \qquad (8)

where: $A_{\text{el}}$ — antisymmetric about $\omega_0$ , positive below, negative above, exactly zero on resonance; $A_{\text{abs}}$ — positive everywhere, sharply peaked at $\omega_0$ . These are, up to constants, the real and imaginary parts of the complex response $\tilde A = (F_0/m)/(\omega_0^2 - \omega^2 + 2i\beta\omega)$ . Two internal checks confirm the split is lossless: $A_{\text{el}}^2 + A_{\text{abs}}^2 = A^2$ reconstructs the full amplitude of equation (3), and $A_{\text{abs}}/A_{\text{el}} = \tan\delta$ reconstructs the phase — the decomposition discards nothing.

Figure

The two faces of the response versus drive frequency — the dispersive elastic curve and the Lorentzian absorptive curve

Read the green curve first: $A_{\text{abs}}(\omega)$ is the absorption line — for weak damping a Lorentzian centered at $\omega_0$ — and by equation (5) the absorbed power is just $\tfrac12 F_0\omega$ times it. The blue curve, $A_{\text{el}}(\omega)$ , is the dispersion curve: large and positive below resonance (the oscillator amplifies in phase), plunging through zero exactly at $\omega_0$ , negative above (antiphase). On resonance the entire response is quadrature — the oscillator presents itself to the drive as pure friction. This pair is your bridge to a large fraction of modern physics. In linear response theory one writes the response to a weak periodic force as a complex susceptibility $\chi(\omega) = \chi'(\omega) + i\chi''(\omega)$ ; our $A_{\text{el}}$ and $A_{\text{abs}}$ are exactly $F_0\chi'$ and $F_0\chi''$ for this oscillator. The theorem you just proved — absorption rides only on the quadrature part — becomes the general statement that dissipation is governed by $\chi''$ alone, while $\chi'$ shapes propagation and refractive index. When a spectrometer sweeps light through an atomic vapor and records a Lorentzian dip of absorbed power, it is measuring $\omega\chi''(\omega)$ — literally equation (6) with atomic parameters — and when the same instrument records the refractive index wiggling through the line, it is tracing $\chi'$ , our blue curve. The two are not independent: causality ties them together (the Kramers–Kronig relations), which is already visible here, since both curves are built from the same $A$ and $\delta$ .

The absorption curve across the whole spectrum

Now assemble the full frequency profile of equation (6) and read its anatomy end to end — this curve, not the amplitude curve, is what "resonance" means to anyone who measures energy.

Figure

The absorbed power across all drive frequencies — a quadratic rise, a Lorentzian peak pinned exactly at the natural frequency, and an inverse-square tail

Three features carry the physics. The rise from zero goes as $\omega^2$ — at low frequency the velocity amplitude is small (velocity carries a factor of $\omega$ ) and the phase is nearly zero, so both factors in $\tfrac12 F_0 v_0\sin\delta$ starve together. The tail falls as $1/\omega^2$ — the mass barely moves, and what little it does is antiphase, so the drive spends the cycle undoing itself. And the peak sits at $\omega_0$ exactly, not approximately, which the next derivation nails down in two lines. The width of the peak, $\Delta\omega = 2\beta$ at half maximum, and the entire $Q$ -from-width toolkit belong to the Resonance Curve and Bandwidth article — here just note that sharper means lighter-damped, and move on.

Go deeper · Derivation C · 1 minThe absorption peak sits exactly at the natural frequency for any damping

Derivation CThe absorption peak sits exactly at the natural frequency for any damping

We prove that $\langle P\rangle(\omega)$ of equation (6) is maximized at $\omega = \omega_0$ exactly — not merely for weak damping, in contrast to the amplitude peak, which sits at $\sqrt{\omega_0^2 - 2\beta^2}$ and only approaches $\omega_0$ as $\beta \to 0$ . The strategy is a one-move algebraic trick: instead of differentiating the quotient, divide numerator and denominator by $\omega^2$ so that all the frequency dependence collects into a single perfect square. The thing to watch is that the trick works precisely because the numerator is proportional to $\omega^2$ — the same factor-of-velocity structure that built the curve in the first place.

Step 1. Divide top and bottom of equation (6) by $\omega^2$ :

\langle P\rangle = \frac{F_0^2\,\beta/m}{\left(\dfrac{\omega_0^2 - \omega^2}{\omega}\right)^{\!2} + 4\beta^2} = \frac{F_0^2\,\beta/m}{\left(\dfrac{\omega_0^2}{\omega} - \omega\right)^{\!2} + 4\beta^2}

Why this helps: the frequency now appears in exactly one place, inside a squared bracket that is manifestly non-negative — no calculus needed.

Step 2. Maximize by inspection. The bracket vanishes when $\omega_0^2/\omega = \omega$ , i.e. $\omega = \omega_0$ , and is positive everywhere else; the denominator therefore takes its minimum value $4\beta^2$ there, and nowhere else. Why damping cannot move the peak: $\beta$ enters only as the constant floor $4\beta^2$ under the bracket — it sets the height and width of the peak but has no vote on its position.

Step 3. Read off the maximum:

\langle P\rangle_{\max} = \frac{F_0^2}{4m\beta} \qquad (9)

where: halving the damping doubles the on-resonance power draw — the price of a sharper resonance is a hungrier one, which is why high- $Q$ systems driven on peak need real power handling. The result agrees with setting $\omega = \omega_0$ directly in equation (6). Applying the same algebraic trick to the velocity amplitude $v_0(\omega) = \omega A(\omega)$ shows that the velocity resonance also sits exactly at $\omega_0$ — power and velocity peak together, as $\langle P\rangle = \tfrac12 F_0 v_0\sin\delta$ demands.

The steady-state energy audit

One more piece of bookkeeping closes the circle: in steady state, where does the absorbed energy go? Into friction — all of it, cycle by cycle, and proving that is the final derivation. This audit is also the honest answer to "why doesn't the amplitude keep growing?": it grows exactly until the friction bill matches the drive's deposit. (For the special case $\omega = \omega_0$ , the Resonance article runs this balance in reverse, using it to predict the on-peak amplitude $QF_0/k$ ; here we verify the books at every frequency.)

Go deeper · Derivation D · 1 minDrive input equals friction dissipation cycle by cycle

Derivation DDrive input equals friction dissipation cycle by cycle

The claim: the cycle-averaged power delivered by the drive, equation (6), equals the cycle-averaged power dissipated by the damping force, at every drive frequency — no energy accumulates in the oscillator and none goes missing. The route is the work–energy theorem applied to equation (1): multiply the equation of motion by $\dot x$ , recognize each term as an energy flow, and average over a period, using the periodicity of the steady state to kill the storage term. The step that deserves attention is the vanishing of the average of $\mathrm dE/\mathrm dt$ — it holds only because the steady state is strictly periodic, which is why this audit fails (and should fail) during the transient.

Step 1. Multiply equation (1) through by $m\dot x$ and regroup:

\frac{\mathrm d}{\mathrm dt}\left[\tfrac12 m\dot x^2 + \tfrac12 m\omega_0^2 x^2\right] = F_0\cos\omega t\,\dot x - 2m\beta\,\dot x^2

Why this reading: the bracket is the mechanical energy $E$ (kinetic plus spring potential); the first term on the right is the drive's power input, the second is the drag force $-2m\beta\dot x$ times velocity — always negative, always draining.

Step 2. Average over one period of the steady state. Since $x_{\text{ss}}$ and $\dot x_{\text{ss}}$ return exactly to their starting values each period, $E$ is periodic and $\langle \mathrm dE/\mathrm dt\rangle = 0$ . Therefore

\langle P_{\text{in}}\rangle = \langle 2m\beta\,\dot x^2\rangle \qquad (10)

where: input equals dissipation as an identity of the steady state — the absorption curve and the dissipation curve are the same curve. Why the transient is different: while the amplitude is still building, $\langle\mathrm dE/\mathrm dt\rangle > 0$ and the drive outspends friction; the surplus is what fills the oscillator's energy store.

Step 3. Evaluate the dissipation side with $\dot x = -A\omega\sin(\omega t - \delta)$ and $\langle\sin^2\rangle = \tfrac12$ :

\langle 2m\beta\,\dot x^2\rangle = m\beta\,A^2\omega^2

Substituting $A(\omega)$ from equation (3) gives $m\beta\omega^2 (F_0/m)^2\big/\big[(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2\big]$ — precisely equation (6), recovered without ever mentioning the phase. Why the agreement is a real check: the input route needed $\sin\delta$ ; the dissipation route needed only $A$ . Two different limbs of the response triangle, one answer. *The dissipation side evaluates to $m\beta\omega^2(F_0/m)^2/\big[(\omega_0^2-\omega^2)^2+4\beta^2\omega^2\big]$ , which matches equation (6) identically for all $\omega$ — confirming that the input and dissipation routes, which use different limbs of the response triangle, give one answer.*

Which form answers which question

you want	reach for	why
the physics of why energy flows	$\langle P\rangle = \tfrac12 F_0 v_0\sin\delta$ , eq. (5)	isolates the phase condition
the measured line shape	$\langle P\rangle(\omega)$ , eq. (6)	explicit in drive frequency
what absorbs and what stores	$A_{\text{el}},\,A_{\text{abs}}$ , eq. (8)	the $\chi'$ / $\chi''$ split
the peak power bill	$F_0^2/4m\beta$ , eq. (9)	exact, at $\omega_0$ , any damping
where the energy ends up	input $=$ dissipation, eq. (10)	steady-state audit
line width, $Q$ from width	$\Delta\omega = 2\beta$	see Resonance Curve and Bandwidth

Limiting cases

Case	Result	Meaning
quasi-static, $\omega \ll \omega_0$	$\delta \to 0$ , $\ \langle P\rangle \approx \dfrac{F_0^2\beta\,\omega^2}{m\omega_0^4} \propto \omega^2$	force rides with displacement, in quadrature with velocity — work done is almost all returned
on resonance, $\omega = \omega_0$	$\delta = \dfrac{\pi}{2}$ , $\ \langle P\rangle = \dfrac{F_0^2}{4m\beta}$ , $\ P(t) \ge 0$ always	velocity locks in phase with force; the only frequency where the drive never takes work back
high frequency, $\omega \gg \omega_0$	$\delta \to \pi$ , $\ \langle P\rangle \approx \dfrac{F_0^2\beta}{m\omega^2} \to 0$	inertia-dominated antiphase response; the drive spends each half-cycle undoing the last
weak damping, $\beta \ll \omega_0$	$\langle P\rangle \approx \dfrac{F_0^2}{4m\beta}\,\dfrac{\beta^2}{(\omega-\omega_0)^2 + \beta^2}$	clean Lorentzian of FWHM $2\beta$ — the universal absorption line shape
zero damping, $\beta \to 0$ , $\omega \ne \omega_0$	$\sin\delta \to 0$ , $\ \langle P\rangle \to 0$	a lossless oscillator absorbs nothing off resonance — all response is elastic

Common trapThe drive feeds energy into the oscillator throughout the whole cycle

It almost never does. At any drive frequency other than exactly $\omega_0$ , the instantaneous power $F(t)\dot x(t)$ spends part of every cycle negative: the oscillator pushes back on the driver and returns energy it was just given. Far from resonance the returns nearly cancel the deposits, which is why $\langle P\rangle$ is tiny there even though the force never weakens. The time-averaged formulas hide this two-way traffic — equation (5) is an average, not a guarantee about instants. Exactly one drive frequency produces one-way traffic: $\omega = \omega_0$ , where $\delta = 90°$ makes $P(t) = F_0v_0\cos^2\omega t \ge 0$ . If an exam asks "when does the drive do positive work at every instant," the answer is only at resonance — and if it asks about net work over a full cycle, that is positive at every frequency, because $\sin\delta > 0$ whenever there is any damping at all.

Worked examples

1 · The power bill of a lab resonance (confidence builder)

A $0.50\ \mathrm{kg}$ mass on a spring of stiffness $k = 200\ \mathrm{N/m}$ has damping constant $\beta = 1.0\ \mathrm{s^{-1}}$ and is driven with force amplitude $F_0 = 2.0\ \mathrm{N}$ . Find the average power absorbed when driven at $\omega = 10$ , $20$ , and $30\ \mathrm{rad/s}$ , and check the $\omega = 10$ value against $\tfrac12 F_0 v_0 \sin\delta$ .

Step 1. The oscillator's own numbers: $\omega_0 = \sqrt{k/m} = \sqrt{200/0.50} = 20.0\ \mathrm{rad/s}$ , and $Q = \omega_0/2\beta = 10$ — the same system as the worked example of the Resonance article, now billed for power.

Step 2. On resonance ( $\omega = 20$ ): equation (9) gives $\langle P\rangle_{\max} = F_0^2/4m\beta = 4.0/(4 \times 0.50 \times 1.0) = \boxed{2.0\ \mathrm{W}}$ . Per drive cycle ( $T = 2\pi/20 = 0.314\ \mathrm{s}$ ) that is $0.63\ \mathrm{J}$ deposited — and, by equation (10), $0.63\ \mathrm{J}$ burned in friction.

Step 3. At $\omega = 10\ \mathrm{rad/s}$ (half the natural frequency): the denominator pieces are $(\omega_0^2 - \omega^2)^2 = (400-100)^2 = 9.0\times10^4$ and $4\beta^2\omega^2 = 400$ , so equation (6) gives $\langle P\rangle = \dfrac{4.0 \times 1.0 \times 100}{0.50 \times 90400} = \boxed{8.8\ \mathrm{mW}}$ — a factor of $\sim\!230$ below the peak, from a mere factor-2 detuning.

Step 4. Cross-check via equation (5). Amplitude: $A = (F_0/m)/\sqrt{90400} = 4.0/300.7 = 1.33\ \mathrm{cm}$ , so $v_0 = A\omega = 0.133\ \mathrm{m/s}$ . Phase: $\tan\delta = 2\beta\omega/(\omega_0^2-\omega^2) = 20/300$ , so $\delta = 3.8°$ and $\sin\delta = 0.067$ . Then $\tfrac12 F_0 v_0\sin\delta = 0.5 \times 2.0 \times 0.133 \times 0.067 = 8.8\ \mathrm{mW}$ (the 8.9 you get from the rounded intermediates). ✓ The response is respectable ( $1.3\ \mathrm{cm}$ of motion!) but nearly all elastic — the drive lends, the spring repays.

Step 5. At $\omega = 30\ \mathrm{rad/s}$ : $(400-900)^2 = 2.5\times10^5$ , $4\beta^2\omega^2 = 3600$ , so $\langle P\rangle = \dfrac{4.0 \times 900}{0.50 \times 253600} = \boxed{28\ \mathrm{mW}}$ . Note the asymmetry: $\omega = 30$ absorbs three times more than $\omega = 10$ even though both are "off by a factor" — the $\omega^2$ in the numerator favors the high side this far from a narrow peak.

2 · Reading a detuned drive like a spectroscopist (exam level)

A $2.0\ \mathrm{kg}$ platform resonates at $\omega_0 = 50\ \mathrm{rad/s}$ with $\beta = 1.0\ \mathrm{s^{-1}}$ (so $Q = 25$ ). It is driven by $F_0 = 0.40\ \mathrm{N}$ at $\omega = 48\ \mathrm{rad/s}$ — a detuning of two half-widths. Find the phase lag, the elastic and absorptive amplitudes, the absorbed power, and compare it to the on-resonance value using the Lorentzian shortcut.

Step 1. Assemble the denominator of equation (6): $\omega_0^2 - \omega^2 = 2500 - 2304 = 196$ , so $(\omega_0^2-\omega^2)^2 = 3.84\times10^4$ and $4\beta^2\omega^2 = 9216$ ; the sum is $4.76\times10^4$ and its square root is $218$ .

Step 2. Phase: $\tan\delta = 2\beta\omega/(\omega_0^2 - \omega^2) = 96/196 = 0.490$ , so $\delta = 26.1°$ , $\sin\delta = 0.440$ , $\cos\delta = 0.898$ . Two half-widths below resonance, the lag is already substantial but well short of $90°$ .

Step 3. Amplitudes: $A = (F_0/m)/218 = 0.20/218 = 0.92\ \mathrm{mm}$ ; then $A_{\text{el}} = A\cos\delta = 0.82\ \mathrm{mm}$ and $A_{\text{abs}} = A\sin\delta = 0.40\ \mathrm{mm}$ . The elastic component is twice the absorptive one — most of this motion is borrowed energy, sloshing.

Step 4. Power, both ways. Velocity amplitude $v_0 = A\omega = 9.2\times10^{-4} \times 48 = 0.044\ \mathrm{m/s}$ ; equation (5): $\langle P\rangle = 0.5 \times 0.40 \times 0.044 \times 0.440 = 3.9\ \mathrm{mW}$ . Direct from equation (6): $\dfrac{0.16 \times 1.0 \times 2304}{2.0 \times 47632} = \boxed{3.9\ \mathrm{mW}}$ . ✓

Step 5. Compare to the peak. Equation (9): $\langle P\rangle_{\max} = 0.16/(4 \times 2.0 \times 1.0) = 20\ \mathrm{mW}$ . The ratio is $3.9/20 = 0.19$ . The Lorentzian shortcut predicts $\beta^2/\big[(\omega-\omega_0)^2 + \beta^2\big] = 1/(4+1) = 0.20$ — agreement to within a percent at $Q = 25$ . A detuning of $n$ half-widths costs a factor $1/(n^2+1)$ : memorize that, and you can sketch any narrow absorption line in ten seconds.

Origins · 1 minGeorge Gabriel Stokes and how this was discovered

George Gabriel Stokes

1819–1903 · Cambridge, Lucasian Professor for 54 years

The velocity-proportional drag force $-2m\beta\dot x$ that powers this entire article is Stokes's signature. In his great 1851 memoir on pendulums, George Gabriel Stokes worked out how a viscous fluid resists an oscillating sphere — the "Stokes drag" $F = -6\pi\eta a v$ — because the precision gravimetry of his era depended on correcting pendulum periods for air resistance: dissipation was not a nuisance term to him but the measurable. The same decade he named and explained fluorescence, realizing that matter absorbs light at one frequency and re-emits at a lower one — arguably the first careful confrontation with absorption as a physical process, the very quantity $\langle P\rangle(\omega)$ maps. Co-architect of the Navier–Stokes equations and of Stokes's theorem, he held Newton's Lucasian chair from 1849 until his death, and for decades every driven-damped calculation in Cambridge passed across his desk as secretary of the Royal Society.

History

1851

— Stokes's memoir on fluid friction and pendulums establishes velocity-proportional drag and its dissipation, turning damping into calculable physics.

1852

— Stokes names fluorescence, framing absorption and re-emission of light as a process to be measured, not just observed.

1877

— Rayleigh's Theory of Sound gives the complete theory of forced, damped vibration, including the dissipation function that generates the

2\beta\dot x

term.

1906

— Lorentz models atoms as damped driven oscillators; his dispersion theory makes

A_{\text{el}}

and

A_{\text{abs}}

the refractive index and absorption coefficient of real materials.

1957

— Kubo's linear response theory generalizes the split: for any system near equilibrium, dissipation is governed by the imaginary part

\chi''

of a susceptibility — this article's quadrature component, promoted to a law of nature.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Sec. on forced oscillations · Landau & Lifshitz, Mechanics (Vol. 1), §26 (Forced oscillations, resonance) · Taylor, Classical Mechanics, Ch. 5 (Oscillations) The complex response is $\tilde A=(F_0/m)/(\omega_0^2-\omega^2+i\gamma\omega)$ , giving on-resonance mean power $\bar P(\omega_0)=F_0^2/2m\gamma$ , FWHM $=\gamma$ , and $Q=\omega_0/\gamma$ . The amplitude-peak frequency is $\sqrt{\omega_0^2-\gamma^2/2}$ , and the phase reaches $\pi/2$ exactly at $\omega_0$ , verifiable by differentiation. This article uses $\gamma \equiv 2\beta$ and $\delta \equiv \phi$ , so these read $\bar P(\omega_0) = F_0^2/4m\beta$ and FWHM $= 2\beta$ . Worked-example numbers ( $2.0$ W, $8.8$ mW, $28$ mW; $\delta = 26.1°$ , $3.9$ mW vs $20$ mW peak) follow from equations (5), (6), and (9); the sodium D-line figures ( $\Gamma/2\pi \approx 9.8$ MHz at $5.1\times10^{14}$ Hz) are standard reference values.

Builds on: Driven Oscillations · Resonance · Damped Harmonic Oscillator · Complex Exponential Representation of Oscillations · Energy in Simple Harmonic Motion Leads to: Quality factor and bandwidth · Anharmonic Pendulum and Amplitude Dependence · Linear response and susceptibility

Near resonance the elastic component can be bigger than the absorptive one — how can the bigger component do zero net work?

Because net work over a cycle is not about size; it is about timing. The elastic component moves in phase with the force, so its velocity is a quarter-cycle out of phase with the force: during two quarter-cycles the force and this velocity agree (work in), during the other two they oppose (work out), and the four strokes cancel exactly, whatever the amplitude. It is the same reason an ideal spring or a planet in a circular orbit exchanges no net energy with the force acting on it. The quadrature component, however small, is the only one whose velocity stays aligned with the force all cycle. In worked example 2 the elastic amplitude is twice the absorptive one, yet every milliwatt of the 3.9 mW bill flows through the smaller piece.

Where exactly does the energy go during the negative lobes of the instantaneous power?

Back into whatever produces the drive — literally. When $F$ and $\dot x$ point opposite ways, the oscillator is doing work on the driver: the swing pushes your hands back, the vibrating machine drives its motor toward generator action, the atom re-radiates into the field that excited it. The energy briefly parked in the oscillator's spring compression or its kinetic energy is being withdrawn through the same contact that deposited it. That is why the elastic response is called reactive in the electrical analogue: reactive power sloshes between source and load at $2\omega$ without ever being consumed, and utilities carefully distinguish it from the real power $\propto\sin\delta$ they can bill for. Only friction converts the throughput irreversibly into heat.

Why does the absorbed power vanish as ω² at low frequency when the amplitude stays finite there?

Two suppressions stack. First, power is force times velocity, and for fixed displacement amplitude the velocity amplitude is $v_0 = \omega A$ — one factor of $\omega$ dies with the slowness of the motion. Second, the phase lag also shrinks: $\sin\delta \approx 2\beta\omega/\omega_0^2$ at low frequency, so the fraction of the response that is absorptive supplies the second factor of $\omega$ . Multiply them in $\langle P\rangle = \tfrac12 F_0 v_0\sin\delta$ and you get $\omega^2$ . Physically: a quasi-static push moves the mass, but slowly and in phase — big displacement, almost no velocity in phase with the force. The finite amplitude is real motion, but it is nearly all elastic lending, not spending.

Is the cycle-averaged absorbed power ever negative — can the oscillator pump net energy back into the drive?

Not in steady state, and the formula says why: $\langle P\rangle = \tfrac12 F_0 v_0 \sin\delta$ , and $\delta$ lies strictly between $0$ and $\pi$ whenever $\beta > 0$ , so $\sin\delta > 0$ always. A passive damped oscillator is a net consumer at every drive frequency — it must be, because in steady state the input exactly covers the friction bill, which is positive-definite ( $\propto\langle\dot x^2\rangle$ ). Transients are a different matter: during settling, the interference between the decaying free oscillation and the steady state can make the oscillator return energy over portions of many cycles. And active systems — with feedback that acts like negative damping — can sustain $\langle P\rangle < 0$ ; but then $\beta < 0$ and the premise of the formula has changed.

The two formulas for average power look unrelated — one has sin δ, the other has β and ω everywhere. Which is more fundamental?

Neither; they are one identity read in two directions, and the translation is the response triangle of equation (3). The complex denominator $\omega_0^2 - \omega^2 + 2i\beta\omega$ has magnitude $\sqrt{(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2}$ ; the sine of its angle is $2\beta\omega$ over that magnitude. Substitute that $\sin\delta$ , plus $v_0 = \omega A$ with $A$ from the same triangle, into $\tfrac12 F_0 v_0 \sin\delta$ , and the magnitude appears squared downstairs — giving exactly $F_0^2\beta\omega^2/m\big[(\omega_0^2-\omega^2)^2+4\beta^2\omega^2\big]$ . Use the compact form to think (it isolates the phase physics) and the explicit form to predict (it needs only the oscillator's constants). Exams love asking for the reconciliation precisely because it tests whether you own equation (3).

Why do experimentalists locate a resonance by the 90° phase crossing rather than the top of a measured peak?

Because the phase crossing is both exact and steep, while the peak is neither. The amplitude maximum sits at $\sqrt{\omega_0^2 - 2\beta^2}$ — displaced by damping, so fitting it gives a biased $\omega_0$ unless you model $\beta$ too. Worse, near any maximum the curve is flat by definition: the first derivative vanishes, so small noise in the measured height translates into large uncertainty in position. The phase, by contrast, crosses $90°$ exactly at $\omega_0$ for any damping, and it crosses at its steepest slope — $\mathrm d\delta/\mathrm d\omega \approx Q/\omega_0$ on resonance — so noise moves the crossing hardly at all. A lock-in amplifier measuring the quadrature signal turns this into a routine: sweep, watch the in-phase signal cross zero, read $\omega_0$ .

If input equals dissipation exactly in steady state, how did the oscillator acquire its stored energy in the first place?

During the transient — the audit of equation (10) is a statement about the settled state only. While the amplitude is still building, the velocity is small, so the friction bill $2m\beta\langle\dot x^2\rangle$ runs below the drive's deposit and the surplus accumulates as stored energy, roughly $\tfrac12 m\omega^2 A^2$ worth by the time it settles. The books record this honestly: $\langle\mathrm dE/\mathrm dt\rangle > 0$ throughout the ramp-up, approaching zero as $e^{-2\beta t}$ over a few multiples of $1/\beta$ . Run the movie backward at switch-off and the same store drains through friction as the free oscillation rings down. Steady state is not "no energy in the oscillator" — it is "no change in the ledger," a full reservoir with matched inflow and outflow.

Is this Lorentzian really the same one measured in absorption spectroscopy, or just an analogy?

The same, with translated symbols — that is the content of the Lorentz oscillator model. An electron bound in an atom responds to the oscillating electric field of light as a damped driven oscillator: $F_0 \to eE_0$ , $\omega_0 \to$ the transition frequency, $2\beta \to$ the decay rate $\Gamma$ of the excited state. The power an atom removes from a light beam is then equation (6) verbatim, and the measured absorption line is a Lorentzian of FWHM $\Gamma$ — for the sodium D line, $\Gamma/2\pi \approx 9.8$ MHz on a carrier of $5.1\times10^{14}$ Hz, an effective $Q$ of five times $10^7$ . Quantum mechanics changes how you compute $\omega_0$ and $\Gamma$ , but the response shape survives because the atom-field system is still linear for weak driving. When the article calls $A_{\text{abs}}$ a precursor of $\chi''$ , that is a literal ancestry, not a metaphor.

Power Absorption and Phase Lag in Driven Systems

Where this lives today

The physical picture

The steady state, inherited

The energy ledger, instant by instant

The journey of the phase lag

Elastic and absorptive: the two response curves

The absorption curve across the whole spectrum

The steady-state energy audit

Which form answers which question

Limiting cases

Worked examples

1 · The power bill of a lab resonance (confidence builder)

2 · Reading a detuned drive like a spectroscopist (exam level)

Going further

Before this

Where this leads