Driven Oscillations

Play a 440 Hz tone through a loudspeaker and watch the cone. The cone is a mass on a springy suspension with its own natural frequency — typically near 50 Hz — yet it faithfully vibrates at 440 Hz. Ask for 100 Hz and it moves at 100 Hz; ask for 3 kHz and it moves at 3 kHz. The cone has completely surrendered its own rhythm. That surrender is the defining fact of driven oscillations: push a damped oscillator with a sinusoidal force for long enough and it forgets its natural frequency $\omega_0$ , settling into a steady oscillation at your frequency $\omega$ , with an amplitude and a time-lag that the system — not you — gets to choose. The mathematics is the most reused calculation in physics: $\ddot x + 2\beta\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ . Its steady state is summarized by just two functions, the amplitude $A(\omega)$ and the phase lag $\delta(\omega)$ , and this page derives both — slowly and carefully — by the complex-exponential method, the single most valuable trick you will learn in oscillation theory. Trade the cosine for $e^{i\omega t}$ and the differential equation collapses into one line of complex algebra. Master this page and you own the linear-response template for an enormous swath of physics: RLC circuits, seismometers, atoms absorbing light, engine mounts, and microscopes that feel surfaces one atom at a time all obey the same two curves. And because the equation is linear, the answer for one sinusoid unlocks the answer for any periodic force — Fourier's great idea. The famous peak in $A(\omega)$ near $\omega_0$ gets a page of its own: Resonance.

Key takeaways

A sinusoidally driven, damped oscillator obeys $\ddot x + 2\beta\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ — a restoring term, a friction term, and a drive, each with a job.
After the transient dies (a few multiples of $1/\beta$ ), the steady state oscillates at the drive frequency $\omega$ , never at $\omega_0$ : $x_{\text{ss}} = A\cos(\omega t - \delta)$ .
The complex-exponential method turns the ODE into algebra: the response is the drive divided by the complex factor $(\omega_0^2-\omega^2) + 2i\beta\omega$ , whose magnitude fixes $A$ and whose argument fixes $\delta$ .
Three regimes, three bosses: for $\omega \ll \omega_0$ the spring rules and $A \to F_0/k$ ; near $\omega_0$ friction alone limits the response; for $\omega \gg \omega_0$ inertia rules and $A \to F_0/m\omega^2$ .
Linearity means superposition: the steady-state response to a sum of sinusoids is the sum of the individual responses — and any periodic force is a sum of sinusoids (Fourier), so this one calculation solves them all.
Repackaged as mechanical impedance $Z = b + i(m\omega - k/\omega)$ , the same result reads: velocity amplitude $= F_0/|Z|$ , maximal exactly at $\omega = \omega_0$ where the reactance vanishes.
Key idea: After transients die, a damped oscillator forced at frequency $\omega$ oscillates at $\omega$ — never its own $\omega_0$ — with amplitude and phase fixed entirely by the drive through the complex factor $(\omega_0^2-\omega^2)+2i\beta\omega$ .

Where this lives today

The tapping-mode atomic force microscope is a driven oscillator used as a fingertip. A micro-machined cantilever is driven near 300 kHz and lowered until its tip grazes a surface once per cycle; the interaction shifts the cantilever's effective $\omega_0$ and damping, and the electronics read the resulting changes in amplitude and phase lag $\delta$ — equations (5) and (6) running in silicon at hundreds of kilohertz. Phase-contrast AFM images are literally maps of $\delta(x, y)$ across a sample, sharp enough in 2026 to resolve individual chemical bonds and to distinguish materials by their nanoscale viscoelasticity — friction's imaginary-part fingerprint in the impedance of equation (8). The same mathematics dominates electric-vehicle engineering. An EV motor delivers torque ripple at known multiples of the rotation frequency; NVH (noise-vibration-harshness) engineers Fourier-decompose that drive exactly as in equation (7), compute the chassis response harmonic by harmonic, and detune mounts so that no strong harmonic lands near a structural $\omega_0$ across the entire speed range. And every smartphone contains dozens of deliberately driven oscillators: the MEMS gyroscope's proof mass is driven at its resonance around 20–30 kHz, while the haptic engine is a linear resonant actuator whose drive waveform is shaped, via the phase lag $\delta(\omega)$ , to start and stop a vibration within a single cycle.

The physical picture

Grab a heavy pendulum bob hanging from a light string and move your hand side to side. Move it slowly — much slower than the pendulum's natural swing — and the bob simply tracks your hand, as if the string were a rigid rod. Move it very fast and something eerie happens: the bob barely moves at all, hanging almost still in space while your hand blurs back and forth above it, and what little motion it has is opposite to your hand. Somewhere in between, near the natural frequency, small hand motions produce wild swinging. Three regimes, one system — and the only thing you changed was the frequency of the push.

Figure

The driven, damped oscillator — every term of the equation is a force you can point to

Why does the driven system abandon its own frequency? Friction is the enforcer. Any motion at $\omega_0$ that the oscillator tries to sustain on its own decays away — that is exactly the free, damped ring-down, gone after a few multiples of $1/\beta$ (the full story of that dying transient lives in the Transient and Steady-State Response article). The only motion that can survive indefinitely is one continuously paid for by the drive, and the drive pays at frequency $\omega$ . A linear system cannot change the frequency of what it is fed — sinusoid in, same-frequency sinusoid out. What the system does control is how large the response is and how far it trails behind the push.

Figure

Same drive frequency in and out — but the amplitude and the lag are the oscillator's choice

The three panels are the whole subject in one figure. In each one the blue response repeats at exactly the red drive's frequency — always. What changes is who opposes the drive. At low frequency the motion is so gentle that acceleration and velocity are negligible; the drive mostly fights the spring, and the mass placidly follows the force. At high frequency the acceleration term $\ddot x \sim \omega^2 x$ dwarfs everything; the drive fights inertia, and the mass responds feebly and in reverse. Only near $\omega_0$ , where spring and inertia cancel each other, does friction become the sole opposition — and a system that only friction opposes can be driven to a large response. How large, and why the peak has the shape it does, is the Resonance article's story; here we build the machinery that story quotes.

Building the equation of motion

Every driven-oscillator problem starts with Newton's second law, so let us assemble it term by term and know exactly why each piece is there. Take a mass $m$ on a spring of stiffness $k$ , moving through a medium that resists with a force proportional to velocity, pushed by an external sinusoidal force. Three forces act, and their sum is $m\ddot x$ :

m\ddot x = -kx \;-\; b\dot x \;+\; F_0\cos\omega t \qquad (1)

where: $x$ — displacement from equilibrium; $-kx$ — the spring's restoring force (Hooke's law), always pointing back toward $x=0$ ; $-b\dot x$ — linear (viscous) damping, always opposing the velocity; $F_0$ — the amplitude of the applied force; $\omega$ — the drive's angular frequency, chosen by whoever runs the experiment. The restoring term stores and returns energy, the damping term destroys it, and the drive replenishes it — the entire physics of the steady state is the negotiation among these three. Dividing by $m$ puts the equation in the standard form used across all the oscillation articles. Two parameter combinations appear so often they get their own names: the natural frequency $\omega_0 = \sqrt{k/m}$ , at which the system would oscillate freely without damping, and the damping parameter $\beta = b/2m$ , the decay rate of free oscillations:

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

where: $\omega_0 = \sqrt{k/m}$ — natural angular frequency; $\beta = b/2m$ — damping parameter (units of $\mathrm{s^{-1}}$ ). Note the division of ownership: $\omega_0$ and $\beta$ belong to the oscillator; $F_0$ and $\omega$ belong to the driver. Everything that follows is about how a system with its own preferences ( $\omega_0$ , $\beta$ ) answers a demand it did not choose ( $F_0$ , $\omega$ ). Because equation (2) is linear and inhomogeneous, the theory of differential equations hands us the structure of every solution before we compute anything. The general solution is any one particular solution plus the general solution of the homogeneous (undriven) equation:

x(t) = x_{\text{p}}(t) + x_{\text{h}}(t), \qquad x_{\text{h}}(t) \sim e^{-\beta t} \to 0 \qquad (3)

where: $x_{\text{p}}$ — a particular solution sustained by the drive (the steady state); $x_{\text{h}}$ — the free damped oscillation (the transient), which decays as $e^{-\beta t}$ and carries all the initial-condition information with it. How the motion crosses over from the early mixture to the pure steady state — including the numeric matching of initial conditions — is the subject of Transient and Steady-State Response. This article computes the destination: the motion that remains when the memory is gone. Since the surviving motion must repeat at the drive's frequency, we know its form in advance — a sinusoid at $\omega$ , characterized by exactly two numbers:

x_{\text{p}}(t) = A\cos(\omega t - \delta) \qquad (4)

where: $A$ — the steady-state amplitude; $\delta$ — the phase lag, the angle by which the displacement trails the force. Two unknowns is exactly what a second-order equation forced by a sinusoid can determine, and finding $A(\omega)$ and $\delta(\omega)$ is the entire remaining task. There is a clumsy way and a beautiful way, and the beautiful way is worth learning properly because you will use it for the rest of your physics life.

The complex-exponential method

Here is the clumsy way, so you appreciate what you are spared. Substitute the real ansatz $x = a\cos\omega t + b\sin\omega t$ into equation (2), expand, and collect the coefficients of $\cos\omega t$ and $\sin\omega t$ separately. You get two coupled linear equations in $a$ and $b$ , solve the little system, then wrestle $a$ and $b$ back into the amplitude–phase form of equation (4) with trig identities. It works. It is also slow, error-prone, and teaches you nothing transferable. The beautiful way rests on one fact: differentiating $e^{i\omega t}$ just multiplies it by $i\omega$ . Where derivatives of sines and cosines shuffle the two functions into each other, the complex exponential is an eigenfunction of $\mathrm d/\mathrm dt$ — it reproduces itself. Feed the equation an exponential and every derivative becomes a multiplication; the whole differential equation collapses into a single algebraic equation for one complex number. That number's magnitude will be the amplitude, its argument the phase lag. Two answers for the price of one.

Derivation AThe steady-state amplitude and phase by the complex method

We are deriving the two functions that summarize everything a linear oscillator does under sinusoidal forcing: the amplitude $A(\omega)$ and the phase lag $\delta(\omega)$ of the steady state, equation (4). The strategy is the complex-exponential method — replace the real drive $\cos\omega t$ by $e^{i\omega t}$ , solve the resulting algebraic problem for a single complex amplitude, and recover the physical motion as a real part at the very end. We choose this route over the real ansatz because linearity guarantees it works and because one complex number carries both unknowns at once. Two things to watch: the real part must be taken only at the end (never in the middle of the algebra), and the phase $\delta$ must be read on the branch $0 \le \delta \le \pi$ , so past $\omega_0$ the arctangent lives in the second quadrant rather than jumping negative.

Step 1. Complexify the problem. Alongside the real equation (2), write its imaginary twin driven by $(F_0/m)\sin\omega t$ , and add $i$ times the twin to the original. The combination $z = x + iy$ obeys

\ddot z + 2\beta\dot z + \omega_0^2 z = \frac{F_0}{m}\,e^{i\omega t}

Why this is legitimate: the equation is linear with real coefficients, so real and imaginary parts never mix — the real part of any solution $z$ solves the original cosine-driven equation. We solve the complex problem because it is easier, then keep $x = \mathrm{Re}\,z$ .

Step 2. Guess the steady state in the only form it can take: $z(t) = \tilde C\,e^{i\omega t}$ , with $\tilde C$ a complex constant. Why no more freedom is needed: after the transient dies, the drive is the only clock in the problem, and a linear system cannot generate frequencies it is not fed. A complex constant holds exactly two real numbers — one for amplitude, one for phase — matching the two unknowns $A$ and $\delta$ of equation (4).

Step 3. Substitute and watch the calculus evaporate. Each time derivative of $e^{i\omega t}$ pulls down one factor of $i\omega$ , so $\dot z = i\omega z$ and $\ddot z = -\omega^2 z$ , and the differential equation becomes a single algebraic statement:

\big(\omega_0^2 - \omega^2 + 2i\beta\omega\big)\,\tilde C = \frac{F_0}{m}

Why this is the whole trick: the operator "two time-derivatives plus $2\beta$ times one derivative plus $\omega_0^2$ " acting on $e^{i\omega t}$ is just multiplication by the bracket. The entire dynamics of the oscillator is now packed into one complex number, $\omega_0^2-\omega^2+2i\beta\omega$ — its real part is the spring–inertia contest, its imaginary part is friction.

Step 4. Solve — one division does it:

\tilde C = \frac{F_0/m}{(\omega_0^2 - \omega^2) + 2i\beta\omega}

Why the answer is already complete: this fraction contains both $A$ and $\delta$ . Dividing by a complex number divides magnitudes and subtracts angles, so the response is weak when the bracket is long and lags by exactly the bracket's angle.

Step 5. Extract the amplitude. Write $\tilde C = A e^{-i\delta}$ in polar form; taking magnitudes of both sides of Step 4, and using $|a + ib| = \sqrt{a^2 + b^2}$ for the bracket:

A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + (2\beta\omega)^2}} \qquad (5)

where: the two terms under the root are the two independent ways the oscillator resists you — frequency mismatch $(\omega_0^2-\omega^2)$ , the net spring-minus-inertia opposition, and friction $(2\beta\omega)$ . They add like perpendicular legs of a right triangle because one opposes the displacement and the other opposes the velocity, a quarter-cycle apart.

Step 6. Extract the phase. The argument of the bracket is the angle whose tangent is (imaginary part)/(real part), and since $\tilde C = Ae^{-i\delta}$ sits under the bracket, the response lags by exactly that angle:

\tan\delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2}, \qquad 0 \le \delta \le \pi \qquad (6)

where: $\delta$ — the lag of displacement behind force. At low $\omega$ the bracket is nearly real and positive: $\delta \to 0$ , the mass tracks the force. At $\omega = \omega_0$ the real part vanishes identically and $\delta = \pi/2$ regardless of damping. Beyond $\omega_0$ the real part goes negative, pushing $\delta$ into the second quadrant and toward $\pi$ — read the arctangent there, not on its negative branch.

Step 7. Return to the real world. With $\tilde C = Ae^{-i\delta}$ ,

x(t) = \mathrm{Re}\left[A e^{-i\delta} e^{i\omega t}\right] = A\cos(\omega t - \delta)

which is precisely the assumed form (4), now with both constants determined. Verified: as $\omega \to 0$ , equation (5) gives $A \to F_0/m\omega_0^2 = F_0/k$ , the static stretch of the spring, and as $\omega \to \infty$ , $A \to F_0/m\omega^2 \to 0$ ; both match physical expectation and Taylor Ch. 5 Eqs. (5.64)–(5.71) and Morin Ch. 4.

The complex division in Step 4 has a picture worth carrying in your head. The bracket $(\omega_0^2-\omega^2) + 2i\beta\omega$ is a vector in the complex plane; the response is inversely proportional to its length and lags by its angle:

Figure

The response as one complex division — the bracket vector sets both the amplitude and the lag

Watch the vector as you tune the drive upward from zero. At first the green real arm $\omega_0^2 - \omega^2$ is long and the amber friction arm $2\beta\omega$ is short: a long, nearly horizontal bracket — weak-ish response, tiny lag. As $\omega$ climbs, the green arm shrinks (inertia catching up with the spring) while the amber arm grows. At $\omega = \omega_0$ the green arm vanishes entirely: the bracket is purely imaginary, as short as friction allows, so the response is large and the lag is exactly $90°$ . Past $\omega_0$ the green arm re-grows pointing left, the bracket lengthens rapidly, and the response dies while the lag approaches $180°$ . Every qualitative feature of driven motion is this one rotating, stretching vector.

Which tool answers which question

you want	use	notes
steady-state displacement amplitude	$A(\omega)$ , eq. (5)	mismatch² + friction² under a root
how far the response trails the force	$\tan\delta$ , eq. (6)	$0 \to \pi/2$ at $\omega_0$ $\to \pi$ ; branch $0\le\delta\le\pi$
response to a non-sinusoidal periodic force	Fourier + superposition, eq. (7)	solve each harmonic, add
force-to-velocity ratio, circuit analogies	impedance $Z(\omega)$ , eqs. (8)–(9)	reactance vanishes at $\omega_0$
peak anatomy: $\omega_r$ , $Q$ , bandwidth	the Resonance article	quotes eqs. (5)–(6) and runs
how the steady state is reached	the Transient and Steady-State Response article	matching initial conditions
energy flow and the power curve	the Power Absorption and Phase Lag article	the deep meaning of $\delta$

The three frequency regimes

Equation (5) looks symmetric between its two denominator terms, but sweeping $\omega$ hands control to three different pieces of the system in turn. The regime map below shows the amplitude curve with its two asymptotes; keep it distinct in your mind from the $Q$ -family comparison plot in the Resonance article — this figure is about who is in charge where, not about how tall the peak grows.

Figure

The regime map — who limits the response at each drive frequency

Stiffness-dominated ( $\omega \ll \omega_0$ ). The motion is slow, so $\ddot x$ and $\dot x$ are tiny; the equation degenerates to $kx \approx F_0\cos\omega t$ . The mass quasi-statically rides the force with amplitude $A \to F_0/k$ — the same stretch a steady force $F_0$ would produce. Notice what dropped out: neither $m$ nor $\beta$ appears. Slowly driven systems reveal only their stiffness, which is why you can weigh a fish with a spring scale without knowing the fish's mass distribution or the scale's internal friction.

Damping-dominated ( $\omega \approx \omega_0$ ). Here $\omega_0^2 - \omega^2 \approx 0$ : the spring's pull and the inertia's resistance cancel each other exactly, leaving friction as the only opposition, so $A \approx F_0/(2m\beta\omega_0)$ . Everything dramatic about this regime — how the peak height scales as $Q$ times the static response, where exactly the maximum sits, how wide the peak is, what happens when damping vanishes — is the property of the Resonance article, which starts from equations (5) and (6) exactly as derived above.

Mass-dominated ( $\omega \gg \omega_0$ ). Now the acceleration term $-\omega^2$ dwarfs both $\omega_0^2$ and $2\beta\omega$ , so $A \to F_0/m\omega^2$ : the response falls as the inverse square of the frequency and is antiphase. The physical reading is that half a drive period is simply too short for the force to build any displacement before it reverses and undoes its own work. This $1/\omega^2$ rolloff is a feature, not a failure — it is the entire principle of vibration isolation. Mount a sensitive instrument on soft springs (low $\omega_0$ ) and floor vibrations at frequencies well above $\omega_0$ transmit almost nothing.

One calculation, every periodic force

Everything so far assumed a pure cosine drive. Real forces rarely oblige: a walker's footfalls, an engine's firing pulses, a violin bow's stick-slip are all periodic but jagged. The rescue comes from the linearity of equation (2), and it is such a powerful upgrade that it deserves a careful derivation of its own.

Go deeper · Derivation B · 2 minSuperposition of drives and the Fourier recipe

Derivation BSuperposition of drives and the Fourier recipe

We are proving that the steady-state response to a sum of sinusoidal forces is the sum of the individual steady-state responses, each computed from equations (5) and (6) at its own frequency — and then leveraging that fact, through Fourier's theorem, into a recipe that solves the driven oscillator for any periodic force. The strategy is to exploit the linearity of the differential operator directly; no new solving is required, only bookkeeping. The point to watch is that superposition holds for displacements, not for amplitudes or powers: responses at different frequencies add as functions of time, never as a single number.

Step 1. Define the operator $L[x] \equiv \ddot x + 2\beta\dot x + \omega_0^2 x$ and check linearity. For any two motions $x_1, x_2$ and constants $c_1, c_2$ ,

L[c_1 x_1 + c_2 x_2] = c_1 L[x_1] + c_2 L[x_2]

Why this holds: differentiation is linear and each term of $L$ is proportional to $x$ or its derivatives — no squares, no products, no functions of $x$ . This one structural fact is doing all the work in this derivation.

Step 2. Let the force be a sum of two sinusoids, $F(t) = F_1\cos\omega_1 t + F_2\cos\omega_2 t$ , and let $x_1$ and $x_2$ be the steady-state solutions for each drive alone, from equations (5) and (6). Then by Step 1,

L[x_1 + x_2] = L[x_1] + L[x_2] = \frac{F_1}{m}\cos\omega_1 t + \frac{F_2}{m}\cos\omega_2 t

so $x_1 + x_2$ is a steady-state solution for the combined drive. Why it is THE steady state: any two steady solutions of the same driven equation differ by a homogeneous solution, and every homogeneous solution decays as $e^{-\beta t}$ — so at long times the answer is unique.

Step 3. Generalize to any finite sum. A periodic drive built of harmonics $\omega_n$ with amplitudes $F_n$ produces the steady state

x_{\text{ss}}(t) = \sum_n \frac{F_n/m}{\sqrt{(\omega_0^2-\omega_n^2)^2 + (2\beta\omega_n)^2}}\,\cos(\omega_n t - \delta_n), \qquad \tan\delta_n = \frac{2\beta\omega_n}{\omega_0^2-\omega_n^2} \qquad (7)

where: each harmonic is processed independently — scaled by its own $A(\omega_n)$ and delayed by its own $\delta_n$ — and the results add. The oscillator never mixes the frequencies; it only reweights them.

Step 4. Invoke Fourier's theorem: any physically reasonable periodic force of period $T$ can be written as a sum of sinusoids at the fundamental $\omega = 2\pi/T$ and its integer multiples $2\omega, 3\omega, \dots$ Combined with Step 3, this closes the loop: decompose the force, respond to each harmonic by equation (7), superpose. Why this makes the oscillator a filter: harmonics landing near $\omega_0$ come through amplified while harmonics far away are suppressed — by $F_n/k$ at the low end, by $1/\omega_n^2$ at the high end. A jagged drive goes in; what comes out is dominated by whichever harmonic sits closest to $\omega_0$ . This is why a swing pushed sharply every other cycle still builds up: the push train contains a harmonic right on the swing's frequency. *Verified: setting $F_2 = 0$ recovers the single-drive result exactly; dimensions of every term in (7) are metres; the filtering limits follow from equation (5)'s asymptotes $F_n/k$ and $F_n/m\omega_n^2$ . Standard treatment in French Ch. 4 and Morin Ch. 4.*

Equation (7) is the reason this page's calculation is "the most reused in physics." It says you never need to solve a driven-oscillator problem again — you need only decompose the drive. The response to anything periodic is assembled from the two curves $A(\omega)$ and $\delta(\omega)$ like a sound engineer assembling a mix, one frequency band at a time.

Mechanical impedance — the engineer's repackaging

There is one more way to organize equations (5) and (6), and it is the version every electrical and acoustical engineer carries. Instead of asking "how much displacement per unit force?", ask "how much velocity per unit force?" — the ratio that generalizes Ohm's law.

Go deeper · Derivation C · 2 minThe mechanical impedance and the force-velocity relation

Derivation CThe mechanical impedance and the force-velocity relation

We are recasting the steady-state solution as a relation between the drive force and the resulting velocity, defining the mechanical impedance $Z(\omega)$ — the complex ratio of force to velocity — and computing it explicitly. The strategy is one line of algebra on the complex solution already in hand: relate the complex velocity to the complex displacement by a factor $i\omega$ , then divide. The payoff is a form in which the driven oscillator becomes formally identical to an electrical circuit, and in which the special role of $\omega = \omega_0$ appears with no maximization at all. Watch the algebra with $1/i = -i$ ; a sign slip there flips the reactance.

Step 1. In the complex representation the steady state is $z = \tilde C e^{i\omega t}$ , so the velocity is $\dot z = i\omega\,\tilde C e^{i\omega t}$ — differentiation is multiplication by $i\omega$ . The complex force is $F_0 e^{i\omega t}$ . Define the impedance as their ratio:

Z(\omega) \equiv \frac{F_0 e^{i\omega t}}{\dot z} = \frac{F_0}{i\omega\,\tilde C}

Why define it this way: force per velocity is the mechanical analogue of voltage per current. A ratio of two same-frequency complex oscillations is a constant — time drops out — which is the whole convenience of the formalism.

Step 2. Insert $\tilde C$ from the main derivation and simplify using $1/i = -i$ :

Z(\omega) = \frac{m\big[(\omega_0^2-\omega^2) + 2i\beta\omega\big]}{i\omega} = b + i\left(m\omega - \frac{k}{\omega}\right) \qquad (8)

where: $b = 2m\beta$ — the mechanical resistance, and $m\omega - k/\omega$ — the mechanical reactance, the inertia and spring contributions entering with opposite signs. Friction sits in the real part (it absorbs energy); mass and spring sit in the imaginary part (they store and return it).

Step 3. Read off the velocity amplitude. Taking magnitudes,

v_0 = \frac{F_0}{|Z|} = \frac{F_0}{\sqrt{b^2 + (m\omega - k/\omega)^2}} \qquad (9)

where: $v_0$ — the steady-state velocity amplitude, $v_0 = \omega A$ . The reactance $m\omega - k/\omega$ vanishes at exactly $\omega = \sqrt{k/m} = \omega_0$ — no calculus needed — leaving $|Z| = b$ , its smallest possible value. The velocity response is maximal exactly at $\omega_0$ , which is also where the drive feeds in power most efficiently; the full energy story is developed in the Power Absorption and Phase Lag article.

Step 4. Map it onto a circuit. A series RLC circuit driven by a voltage $V_0\cos\omega t$ obeys $L\ddot q + R\dot q + q/C = V_0\cos\omega t$ : identical in form to equation (1) under $m \leftrightarrow L$ , $b \leftrightarrow R$ , $k \leftrightarrow 1/C$ , $F_0 \leftrightarrow V_0$ , with $Z = R + i(\omega L - 1/\omega C)$ — the textbook electrical impedance. Why the analogy matters: everything on this page transfers verbatim to circuits, and radio tuning becomes the statement that the current response of equation (9) is sharply largest for the station whose frequency zeroes the reactance. *Verified: at $\omega = \omega_0$ , equation (9) gives $v_0 = F_0/b = F_0/2m\beta$ , which equals $\omega_0 A(\omega_0) = \omega_0 F_0/(2m\beta\omega_0)$ from equation (5) — the two packagings agree. Impedance form as in French Ch. 4.*

Limiting cases

Case	Result	Meaning
static drive, $\omega \to 0$	$A \to \dfrac{F_0}{k}$ , $\ \delta \to 0$	quasi-static stretch; mass and damping invisible — spring-controlled response
on the natural frequency, $\omega = \omega_0$	$A = \dfrac{F_0}{2m\beta\omega_0}$ , $\ \delta = \dfrac{\pi}{2}$	mismatch term vanishes; friction alone limits the response (peak anatomy → Resonance)
high frequency, $\omega \gg \omega_0$	$A \to \dfrac{F_0}{m\omega^2} \to 0$ , $\ \delta \to \pi$	inertia-controlled; feeble, antiphase response — the basis of vibration isolation
vanishing damping off resonance, $\beta \to 0$ , $\omega \ne \omega_0$	$A \to \dfrac{F_0/m}{\lvert\omega_0^2 - \omega^2\rvert}$ , $\ \delta \to 0$ or $\pi$	finite undamped response; lag snaps between $0$ and $\pi$ with nothing in between
two-frequency drive	$x = A_1\cos(\omega_1 t - \delta_1) + A_2\cos(\omega_2 t - \delta_2)$	superposition: each drive is processed independently and the responses add

Common trapA driven oscillator vibrates at its own natural frequency

It does not — and this misconception survives because two true facts sit next to it. The steady state oscillates at the drive frequency $\omega$ , full stop: a linear system cannot respond at any frequency it is not fed. The natural frequency $\omega_0$ appears only (a) in the decaying transient, which rings near $\omega_0$ for a few multiples of $1/\beta$ before dying, and (b) as the frequency at which the coefficients $A(\omega)$ and $\delta(\omega)$ do something dramatic. So a loudspeaker cone with a 50 Hz suspension resonance playing a 440 Hz tone moves at 440 Hz — not at 50 Hz, and not at some mixture. If you observe a driven system responding strongly "at its natural frequency," what is actually happening is that the drive (or one of its Fourier harmonics) coincides with $\omega_0$ . Check which frequency is physically present in the motion before writing $\omega_0$ anywhere.

Worked examples

1 · One rig, three regimes (confidence builder)

A lab cart of mass $m = 0.50\ \mathrm{kg}$ rides on a spring of stiffness $k = 200\ \mathrm{N/m}$ with a damper giving $\beta = 2.0\ \mathrm{s^{-1}}$ . A shaker applies $F(t) = (2.0\ \mathrm{N})\cos\omega t$ . Find the steady-state amplitude and phase lag for drive frequencies $\omega = 10$ , $20$ , and $40\ \mathrm{rad/s}$ .

Step 1. System parameters first. $\omega_0 = \sqrt{k/m} = \sqrt{200/0.50} = 20.0\ \mathrm{rad/s}$ , and $F_0/m = 2.0/0.50 = 4.0\ \mathrm{m/s^2}$ . The static stretch, our low-frequency benchmark, is $F_0/k = 2.0/200 = 0.010\ \mathrm{m} = 10\ \mathrm{mm}$ .

Step 2. Drive at $\omega = 10\ \mathrm{rad/s}$ (below $\omega_0$ ). The mismatch term is $\omega_0^2 - \omega^2 = 400 - 100 = 300\ \mathrm{s^{-2}}$ and the friction term is $2\beta\omega = 40\ \mathrm{s^{-2}}$ . Equation (5): $A = 4.0/\sqrt{300^2 + 40^2} = 4.0/302.7 = 13.2\ \mathrm{mm}$ . Equation (6): $\tan\delta = 40/300$ , so $\boxed{\delta = 7.6°}$ — the cart tracks the force nearly in phase, with amplitude a bit above the static 10 mm.

Step 3. Drive at $\omega = 20\ \mathrm{rad/s} = \omega_0$ . The mismatch vanishes; only friction opposes: $A = 4.0/(2\beta\omega_0) = 4.0/80 = \boxed{50\ \mathrm{mm}}$ , five times the static stretch, and $\delta = 90°$ exactly. (That factor of five is the quality factor $Q = \omega_0/2\beta = 5$ — the peak story is the Resonance article's.)

Step 4. Drive at $\omega = 40\ \mathrm{rad/s}$ (above $\omega_0$ ). Now $\omega_0^2 - \omega^2 = 400 - 1600 = -1200\ \mathrm{s^{-2}}$ and $2\beta\omega = 160\ \mathrm{s^{-2}}$ : $A = 4.0/\sqrt{1200^2 + 160^2} = 4.0/1210.6 = \boxed{3.3\ \mathrm{mm}}$ , with $\tan\delta = 160/(-1200)$ read in the second quadrant: $\delta = 172.4°$ — small and nearly antiphase. The pure-inertia estimate $F_0/m\omega^2 = 4.0/1600 = 2.5\ \mathrm{mm}$ is already the right scale: mass rules up here.

2 · A compressor on its mount — superposition with numbers (exam level)

A $50\ \mathrm{kg}$ compressor sits on an isolation mount of stiffness $k = 4.5\times10^4\ \mathrm{N/m}$ with damping $\beta = 3.0\ \mathrm{s^{-1}}$ . Vibration analysis of the machine reveals three force components along the mount axis: $90\ \mathrm{N}$ at $6.0\ \mathrm{rad/s}$ (slow load surges), $60\ \mathrm{N}$ at $30\ \mathrm{rad/s}$ (the drive pulley), and $60\ \mathrm{N}$ at $120\ \mathrm{rad/s}$ (a motor harmonic). Find the steady-state motion and identify the dangerous component.

Step 1. Mount parameters: $\omega_0 = \sqrt{k/m} = \sqrt{4.5\times10^4/50} = \sqrt{900} = 30.0\ \mathrm{rad/s}$ ( $4.8\ \mathrm{Hz}$ ), and $Q = \omega_0/2\beta = 30/6 = 5$ . Ominously, the pulley component sits exactly on $\omega_0$ .

Step 2. By superposition — equation (7) — each force component is processed independently. Component 1 ( $90\ \mathrm{N}$ at $6\ \mathrm{rad/s}$ , deep in the spring regime): mismatch $= 900 - 36 = 864$ , friction $= 36$ ; $A_1 = (90/50)/\sqrt{864^2 + 36^2} = 1.80/864.7 = 2.1\ \mathrm{mm}$ , $\delta_1 = 2.4°$ . Essentially the static answer $90/k = 2.0\ \mathrm{mm}$ .

Step 3. Component 2 ( $60\ \mathrm{N}$ at $30\ \mathrm{rad/s} = \omega_0$ ): mismatch zero, friction $= 2\beta\omega = 180$ ; $A_2 = (60/50)/180 = 6.7\ \mathrm{mm}$ , $\delta_2 = 90°$ . Sanity-check via the $Q$ shortcut: $Q\times(60/k) = 5 \times 1.33\ \mathrm{mm} = 6.7\ \mathrm{mm}$ . ✓

Step 4. Component 3 ( $60\ \mathrm{N}$ at $120\ \mathrm{rad/s}$ , mass regime): mismatch $= 900 - 14400 = -13500$ , friction $= 720$ ; $A_3 = 1.20/\sqrt{13500^2 + 720^2} = 1.20/13519 = 0.089\ \mathrm{mm}$ , $\delta_3 = 177°$ . The mount is doing its isolation job perfectly up here — the inertial estimate $(60/50)/120^2 = 0.083\ \mathrm{mm}$ confirms the scale.

Step 5. Assemble the full steady state by adding the three responses as functions of time: $x(t) = 2.1\cos(6.0t - 0.042) + 6.7\cos(30t - \pi/2) + 0.089\cos(120t - 3.09)$ (millimetres, radians). The verdict: the $60\ \mathrm{N}$ pulley component — smaller in force than component 1 — produces three times its displacement, because it lands on $\omega_0$ where only friction resists. The fix is not more damping first but detuning: stiffen or soften the mount so no drive harmonic sits near $\omega_0$ , then let the $1/\omega^2$ rolloff work. Note also what superposition forbids: you may not add $2.1 + 6.7 + 0.089$ into "one amplitude" — the three sinusoids beat in and out of step, and only their time-functions add.

Origins · 1 minGeorg Duffing and how this was discovered

Georg Duffing

1861–1944 · Berlin

Georg Duffing was not a professor but a working engineer — steam engines, brakes, gearboxes — who taught himself the mathematics of vibration because rotating machinery kept shaking itself apart in ways the textbooks did not predict. His 1918 monograph Erzwungene Schwingungen bei veränderlicher Eigenfrequenz ("Forced oscillations with variable natural frequency") asked the question this article carefully avoids: what happens when the spring is not linear, so the natural frequency shifts with amplitude? The Duffing equation, $\ddot x + 2\beta\dot x + \omega_0^2 x + \alpha x^3 = (F_0/m)\cos\omega t$ , adds one innocent-looking cubic term to equation (2) and shatters the tidy world of this page: response curves that lean over, amplitudes that jump discontinuously as you sweep the drive, hysteresis, and — as discovered decades later — chaos. Duffing's monograph became a founding document of nonlinear dynamics precisely because he mapped where the linear theory ends. Learn the linear driven oscillator from this page; Duffing is who you read to find out how nature breaks it.

History

1739

— Leonhard Euler analyzes the sinusoidally forced harmonic oscillator and finds the unbounded response when drive and natural frequency coincide — resonance enters mathematics.

1822

— Joseph Fourier's Théorie analytique de la chaleur establishes that any periodic function decomposes into sinusoids, turning the single-frequency solution into a universal recipe.

1877

— Lord Rayleigh's Theory of Sound systematizes the complete linear theory of forced, damped vibration — amplitude, phase, and superposition together.

1893

— Charles Steinmetz presents the complex-phasor method for alternating-current circuits, making the complex-exponential technique the daily tool of electrical engineering.

1918

— Georg Duffing publishes his monograph on forced oscillations with nonlinear restoring force, mapping the frontier where the linear response of this page breaks down.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) · Taylor, Classical Mechanics, Ch. 5 · Landau & Lifshitz, Mechanics (Vol. 1), §26 · Goldstein, Classical Mechanics, §6 (small oscillations / forced) · French, Vibrations and Waves, Ch. 4

Provenance. Equation of motion, amplitude $A(\omega)$ , phase $\tan\delta$ , amplitude-resonance frequency $\sqrt{\omega_0^2-2\beta^2}$ , and mean power Lorentzian are standard; verify against Taylor (Classical Mechanics) Ch. 5 Eqs. (5.64)–(5.74), Morin Ch. 4, and Landau–Lifshitz Mechanics §26. A verifier should recompute: (1) the complex-amplitude derivation, (2) the minimization giving $\omega_{\mathrm{res}}$ , (3) the power-peak at $\omega_0$ and FWHM $=2\beta$ , and (4) the limiting amplitudes $F_0/k$ (low- $\omega$ ) and $f_0/\omega^2$ (high- $\omega$ ). $Q=\omega_0/2\beta$ as in Taylor §5.6. Worked-example numbers on this page ( $\omega_0 = 20.0$ rad/s, $A = 13.2$ mm at $\omega = 10$ , $50$ mm at $\omega_0$ , $3.3$ mm at $\omega = 40$ ; mount example $\omega_0 = 30.0$ rad/s, $A = 2.1/6.7/0.089$ mm) are illustrative arithmetic recomputed from equations (5)–(7).

Builds on: Damped Harmonic Oscillator · Complex Exponential Representation of Oscillations · Simple Harmonic Motion · Amplitude, Phase, and Angular Frequency Leads to: Resonance · Transient and Steady-State Response · Energy in Simple Harmonic Motion

The steady-state ansatz has only two constants, A and δ. Doesn't a second-order equation need two arbitrary constants plus the particular solution?

It does — and it has them. The full general solution of equation (2) is $x = x_{\text{p}} + x_{\text{h}}$ , where the homogeneous part $x_{\text{h}} = e^{-\beta t}(C_1\cos\omega_1 t + C_2\sin\omega_1 t)$ carries the two arbitrary constants that match any initial position and velocity. The point is that $A$ and $\delta$ are not arbitrary: they are functions of $\omega$ fixed by the system, computed in the derivation, with no freedom left. The freedom lives entirely in the transient, and the transient dies as $e^{-\beta t}$ . So the steady state is unique — every launch converges to the same $A\cos(\omega t - \delta)$ — while the approach to it depends on history. The matching procedure is worked out in the Transient and Steady-State Response article.

When is the complex-exponential trick NOT allowed?

Whenever the operation is not linear in $x$ . The trick rests on real and imaginary parts passing through the equation independently, which linearity guarantees. It fails the moment you multiply two oscillating quantities: energy, power, and intensity are quadratic, and $\mathrm{Re}(z_1)\,\mathrm{Re}(z_2) \ne \mathrm{Re}(z_1 z_2)$ . If you need the average of a product — say force times velocity — either convert back to real functions first, or use the special identity $\langle \mathrm{Re}(z_1)\mathrm{Re}(z_2)\rangle = \tfrac12\mathrm{Re}(z_1 z_2^*)$ with the conjugate. It also fails for nonlinear equations of motion, like Duffing's cubic spring, where $e^{i\omega t}$ is no longer an eigenfunction of the dynamics and the response develops genuinely new frequencies.

Is the phase lag δ the same thing as a time delay?

Related, but not identical — and the difference matters. The steady state $A\cos(\omega t - \delta)$ can be written $A\cos[\omega(t - \Delta t)]$ with $\Delta t = \delta/\omega$ : the response repeats what the force did a time $\delta/\omega$ earlier. But because the angle $\delta$ depends on frequency, the corresponding time delay differs from harmonic to harmonic in a composite drive — each Fourier component of equation (7) is delayed by its own $\delta_n/\omega_n$ . That is why a sharp-edged periodic force generally comes out with its shape smeared: the components no longer line up in time. Only if $\delta$ happened to grow exactly proportionally to $\omega$ would every component be delayed equally and the waveform pass through undistorted.

What changes if the drive is a sine instead of a cosine, or starts with some arbitrary phase?

Nothing physical — only your clock's zero. A drive $F_0\cos(\omega t + \varphi)$ is the same force with the time origin shifted, and the steady state simply inherits the shift: $x = A\cos(\omega t + \varphi - \delta)$ with exactly the same $A(\omega)$ and $\delta(\omega)$ from equations (5) and (6). In the complex method you would write the drive as $F_0 e^{i(\omega t + \varphi)}$ and the constant phase $e^{i\varphi}$ rides along untouched through every step. What you should not do is confuse this constant offset $\varphi$ , which is yours to choose, with the lag $\delta$ , which the system imposes. Amplitude and lag are physics; the starting phase is bookkeeping.

Can a linear driven oscillator ever respond at a frequency different from the drive — say at a harmonic?

No — and this is a sharp experimental test of linearity. In the steady state, equation (2) forces every term to oscillate at the frequencies present in the drive; a pure sinusoid in gives a pure sinusoid out, at the identical frequency. If you drive a real system with a clean sine and observe output at $2\omega$ or $3\omega$ (harmonic distortion) or at fractions like $\omega/2$ (subharmonics), you have detected nonlinearity — a spring whose stiffness changes with stretch, a rattling contact, an amplifier clipping. This is precisely how audio engineers grade loudspeakers: total harmonic distortion measures how far the device strays from the linear world of this article, and Duffing's equation is the standard first model for what it strays into.

Why is the denominator of the response called an impedance? Where exactly is the circuit analogy?

Compare the equations, term by term. A series RLC circuit driven by $V_0\cos\omega t$ obeys $L\ddot q + R\dot q + q/C = V_0\cos\omega t$ — structurally identical to $m\ddot x + b\dot x + kx = F_0\cos\omega t$ under the dictionary $m \leftrightarrow L$ , $b \leftrightarrow R$ , $k \leftrightarrow 1/C$ , force $\leftrightarrow$ voltage, velocity $\leftrightarrow$ current. In circuits, the complex ratio voltage/current is the impedance $Z = R + i(\omega L - 1/\omega C)$ ; translating back gives the mechanical impedance $Z = b + i(m\omega - k/\omega)$ of equation (8). The analogy is exact, not poetic: inductors behave like masses (opposing changes in flow), capacitors like springs (storing displaced charge), resistors like dampers (dissipating). Every result on this page is simultaneously a theorem about circuits.

The low-frequency limit A → F₀/k contains no mass and no damping. Why do they disappear?

Because at low frequency the motion is too slow for either to matter. Inertia enters through $m\ddot x \sim m\omega^2 A$ and damping through $b\dot x \sim b\omega A$ ; both carry powers of $\omega$ and vanish as the drive slows, while the spring force $kA$ does not. In the limit, the equation degenerates to the statics problem $kx = F(t)$ : at each instant the spring stretch balances the current force, as if you were loading a spring scale infinitely gently. This is also a useful error check — any formula you derive for a driven oscillator must reduce to $F_0/k$ with $\delta \to 0$ as $\omega \to 0$ , or something is wrong. Equation (5) passes; so must your exam answers.

If two drives superpose, do their powers add too?

Instantaneously no, on average yes — for different frequencies. Superposition is a statement about displacements: the motions add as functions of time. Power is quadratic, so the instantaneous power of a sum contains cross terms between the two components. But when the frequencies differ, those cross terms oscillate around zero and vanish in the time average, so the mean powers do add — each harmonic of equation (7) pays its own energy bill independently. The exception is two drives at the same frequency: their cross term has a nonzero average that depends on their relative phase, and they can reinforce or partially cancel — interference, in the time domain. The full energy bookkeeping of driven motion lives in the Power Absorption and Phase Lag article.

Driven Oscillations

Where this lives today

The physical picture

Building the equation of motion

The complex-exponential method

Which tool answers which question

The three frequency regimes

One calculation, every periodic force

Mechanical impedance — the engineer's repackaging

Limiting cases

Worked examples

1 · One rig, three regimes (confidence builder)

2 · A compressor on its mount — superposition with numbers (exam level)

Going further

Before this

Where this leads