Transient and Steady-State Response

Set up the classic lab demonstration: a mass on a spring, a signal generator, a shaker, an oscilloscope. Switch on a sinusoidal drive and watch the trace. It does not immediately oscillate at a fixed size. For the first seconds the motion writhes — two rhythms fighting, the amplitude swelling and shrinking as if the system cannot make up its mind. Then, gradually, the struggle fades and the trace locks into a clean, constant sinusoid at the drive frequency, and stays there forever. Run it again with a completely different launch — pull the mass aside first, or fling it — and the early writhing changes completely, but the final oscillation is identical. That two-act structure is the subject of this article. The general solution of the driven, damped oscillator is the sum of two pieces with utterly different jobs: a steady state that oscillates at the drive frequency with amplitude and phase the system imposes, and a transient — a free, damped oscillation at the system's own ringing frequency — that decays as $e^{-\beta t}$ and carries every trace of the initial conditions to the grave. The steady state is where the motion is going; the transient is how it gets there, and it is the only part of the solution you are allowed to adjust when matching how the motion started. Mastering the split buys you real laboratory power: you will compute exactly how a resonator builds up from rest, why the startup shows beat-like interference, and why a high- $Q$ oscillator — a quartz fork, a superconducting readout cavity — takes roughly $Q/\pi$ cycles to settle, which is the hidden speed limit on every precision resonant measurement.

Key takeaways

The general solution of $\ddot x + 2\beta\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ is $x = x_{\text{p}} + x_{\text{h}}$ : a particular (steady-state) piece plus a homogeneous (transient) piece.
The steady state $x_{\text{p}} = A\cos(\omega t - \delta)$ is fixed entirely by the system and the drive — its $A(\omega)$ and $\delta(\omega)$ are derived in Driven Oscillations and contain no free constants.
The transient $x_{\text{h}} = e^{-\beta t}(C_1\cos\omega_1 t + C_2\sin\omega_1 t)$ decays as $e^{-\beta t}$ and its two constants $C_1, C_2$ absorb the initial position and velocity — never touch $A$ or $\delta$ to fit initial conditions.
Started from rest, the amplitude builds toward the steady state as $1 - e^{-\beta t}$ on resonance; off resonance the transient (at $\omega_1$ ) interferes with the drive (at $\omega$ ), producing decaying beat-like ringing.
Settling takes a time $\sim 1/\beta$ , which is about $Q/\pi$ oscillation cycles to reach 63% — high- $Q$ systems respond sharply but slowly, and every resonant instrument must wait out its own memory.
Every launch converges to the same steady state: dissipation erases initial-condition memory exponentially.
Key idea: Linearity splits the driven motion into transient + steady state: $x(t)=x_h(t)+x_p(t)$ , where $x_h$ decays and absorbs the initial conditions, and $x_p$ is the persistent drive-locked response whose amplitude and phase are independent of how the motion began.

Where this lives today

The settling rule $N \approx Q/\pi$ is a live engineering constraint at the frontier of quantum computing. Reading out a superconducting qubit means driving a microwave resonator coupled to it and watching the reflected steady state; the resonator cannot report faster than it can settle, so designers deliberately spoil its quality factor — internal $Q$ values above a million are throttled to a few thousand of external coupling — trading sensitivity for a ring-up of tens of nanoseconds. Every 2026-era processor's measurement chain is a negotiated compromise on exactly equation (8): too high a readout $Q$ and the computer idles waiting for photons to build up; too low and the signal drowns. The same waiting game runs at every scale. A lock-in amplifier hunting a sharp mechanical resonance must dwell $\sim Q$ cycles per frequency point, which is why characterizing a $Q = 10^7$ microresonator can take hours of automated sweeping. Concert-hall acousticians measure the room's transient directly — the reverberation time RT60 is the ring-down of the hall's modes — and tune absorbers (damping) so music's steady states bloom and decay on musical timescales. And car suspension engineers still buy Kelvin's bargain in reverse: damping near critical, $Q \sim 1$ , so the transient from each pothole is forgotten within a single bounce.

The physical picture

Think of the steady state as a moving target that the motion is falling toward. The drive and the damping between them define one — and only one — self-consistent perpetual motion: the oscillation at the drive frequency in which the energy fed in per cycle exactly equals the energy friction destroys per cycle. Start the system anywhere else and it is not on that motion; the difference between where it is and where it must end up is precisely the transient. And the transient is nothing new — it is a free, damped oscillation, exactly the ring-down studied in the Damped Harmonic Oscillator article, obeying the homogeneous equation because the drive is already fully "spent" sustaining the steady state. Free damped oscillations die as $e^{-\beta t}$ ; therefore memory dies as $e^{-\beta t}$ .

Figure

Startup from rest on resonance — the amplitude climbs the envelope toward the steady state

The figure above is the cleanest startup there is: drive switched on at $t = 0$ with the oscillator at rest, drive frequency right on $\omega_0$ . At $t = 0$ the steady state alone would say the mass is already moving — so the transient must start with exactly the opposite motion to cancel it. As the transient decays, the cancellation fails at the rate $e^{-\beta t}$ , and what is left grows along the envelope $A(1 - e^{-\beta t})$ : the oscillator does not so much "build up" as "stop cancelling." One damping time $1/\beta$ brings the amplitude to 63% of its final value; three damping times bring it to 95%. Off resonance the startup is more dramatic, because the two pieces now run at different frequencies — the transient rings at the damped natural frequency $\omega_1$ , the steady state at the drive's $\omega$ — and their sum shows interference:

Figure

The anatomy of a startup — steady state plus decaying transient equals the observed motion

Read the third trace against the first two. Early on, the observed motion is genuinely two-frequency: it is not "the oscillator hesitating," it is the arithmetic sum of a clean drive-frequency sinusoid and a clean decaying ring at $\omega_1$ . Where the two agree in sign the motion is big; where they oppose it is small. All the apparent complexity of startup is this superposition — and since one partner is doomed, the complexity is temporary. The initial conditions choose only $C_1$ and $C_2$ , the size and phase of the doomed partner.

The equations, assembled

The system is the same driven, damped oscillator whose steady state is derived in Driven Oscillations; this article is about the complete solution, so let us fix notation once:

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

where: $\omega_0 = \sqrt{k/m}$ — natural frequency; $\beta = b/2m$ — damping parameter; $F_0, \omega$ — drive amplitude and frequency. (Several references — and this article's provenance note — use $\gamma \equiv 2\beta$ , so their $e^{-\gamma t/2}$ decay is our $e^{-\beta t}$ : same physics, different packing of the 2.) We assume the underdamped case $\beta < \omega_0$ , the one with visible ringing; the overdamped and critical cases swap the transient's sinusoids for decaying exponentials but change nothing structural. The theory of linear equations dictates the shape of every solution before any physics is done. One particular solution plus the general homogeneous solution is the general solution:

x(t) = x_{\text{p}}(t) + x_{\text{h}}(t) \qquad (2)

where: $x_{\text{p}}$ — any one motion that satisfies the driven equation (we use the steady state); $x_{\text{h}}$ — the general solution of the free equation, carrying the two arbitrary constants a second-order equation owes us. This bland-looking line is the whole conceptual content of the article: the driven problem = the free problem + one forced motion. The homogeneous piece is the free damped oscillator, solved in its own article; for $\beta < \omega_0$ it rings at the damped frequency and dies exponentially:

x_{\text{h}}(t) = e^{-\beta t}\big(C_1\cos\omega_1 t + C_2\sin\omega_1 t\big), \qquad \omega_1 = \sqrt{\omega_0^2 - \beta^2} \qquad (3)

where: $\omega_1$ — the damped natural frequency, a bit below $\omega_0$ ; $C_1, C_2$ — the two free constants, soon to be fixed by initial conditions. Note the three distinct frequencies now in play: the drive's $\omega$ (chosen by you), the natural $\omega_0$ (property of the system), and the ringing $\omega_1$ (property of the system with its damping). The transient rings at $\omega_1$ — neither of the other two. The particular piece we take straight from the Driven Oscillations article, where the complex-exponential method produces it in five lines:

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

where: $A(\omega), \delta(\omega)$ — the steady-state amplitude and phase lag, functions fixed by the system and drive with no adjustable freedom whatsoever ( $0 \le \delta \le \pi$ ). This is the piece that survives, and the peak it develops near $\omega_0$ is the Resonance article's territory. Here its only role is: the destination.

Derivation AWhy every solution is transient plus steady state

We are establishing the fundamental decomposition, equation (2), with its three claims: that the sum solves the driven equation, that every solution has this form, and that all solutions converge to the same steady state — the system forgets its initial conditions. The strategy is pure linearity: work with the operator $L[x] = \ddot x + 2\beta\dot x + \omega_0^2 x$ and let superposition do everything; no formula-grinding is needed. The subtle claim is the second one — not just that transient-plus-steady-state is a solution, but that there are no others — and the trick for it is to look at the difference of two solutions. Watch that step; it is where uniqueness of the long-time motion comes from.

Step 1. Check the sum solves the driven equation. With $L[x_{\text{p}}] = (F_0/m)\cos\omega t$ and $L[x_{\text{h}}] = 0$ , linearity gives

L[x_{\text{p}} + x_{\text{h}}] = L[x_{\text{p}}] + L[x_{\text{h}}] = \frac{F_0}{m}\cos\omega t + 0

Why linearity holds: every term of $L$ is proportional to $x$ or a derivative of $x$ — no squares, no products — so $L$ distributes over sums. This is the same structural fact that powers the superposition-of-drives result in the Driven Oscillations article.

Step 2. Show there are no other solutions. Let $x$ be any solution of the driven equation and consider its difference from our particular one, $u = x - x_{\text{p}}$ . Then

L[u] = L[x] - L[x_{\text{p}}] = \frac{F_0}{m}\cos\omega t - \frac{F_0}{m}\cos\omega t = 0

so $u$ solves the free equation, i.e. $u$ is one of the motions (3). Every solution is therefore $x_{\text{p}}$ plus some homogeneous motion — the form (2) is exhaustive. Why this matters: it forbids exotic alternatives. There is no special launch that locks the oscillator into some other perpetual motion; whatever you do, you are on the steady state plus a doomed free ring.

Step 3. Count the freedom. The constants $C_1, C_2$ in (3) are exactly two — the number a second-order equation needs to match an arbitrary initial position and velocity. The steady state contributes none: $A$ and $\delta$ in (4) are already determined functions of $\omega$ . Why the bookkeeping is satisfying: the drive fixes where you end up; the two-dimensional space of initial conditions maps one-to-one onto the two-dimensional space of transients. Nothing is over- or under-determined.

Step 4. Take the long-time limit. Since $|x_{\text{h}}| \le e^{-\beta t}\sqrt{C_1^2 + C_2^2} \to 0$ for any $C_1, C_2$ ,

\lim_{t\to\infty}\big[x(t) - x_{\text{p}}(t)\big] = 0

— every solution, from every launch, converges to the same steady state. Why this is memory loss: the only place the initial data lived was in $C_1, C_2$ , and both ride the dying exponential. Dissipation is the eraser; without it ( $\beta = 0$ ) the free part never dies and the system remembers its launch forever. *Verified: setting $F_0 = 0$ collapses (2) to the free damped oscillator, and setting $t \gg 1/\beta$ collapses it to the steady state of the Driven Oscillations article; structure as in Morin Ch. 4 and Taylor Ch. 5.*

Matching initial conditions — the transient's one job

Here is where students lose the most points, so let us do it completely and then numerically. Suppose at $t = 0$ you know the position $x(0) = x_0$ and velocity $\dot x(0) = v_0$ . The temptation is to bend $A$ or $\delta$ to fit — resist it; they are not yours. Evaluate the full solution (2) and its derivative at $t = 0$ :

x_0 = A\cos\delta + C_1, \qquad v_0 = A\omega\sin\delta - \beta C_1 + \omega_1 C_2 \qquad (5)

where: the $A$ -terms are the steady state's position and velocity at $t = 0$ — already fixed numbers once you know $\omega$ — and $C_1, C_2$ are the unknowns. Two linear equations, two unknowns, always solvable: $C_1 = x_0 - A\cos\delta$ , then $C_2 = (v_0 + \beta C_1 - A\omega\sin\delta)/\omega_1$ . The transient is literally the difference between how you started and how the steady state wanted you to start, packaged as a free damped oscillation.

Go deeper · Derivation B · 2 minThe start from rest and the shape of the ringing

Derivation BThe start from rest and the shape of the ringing

We are deriving the explicit motion for the most common laboratory launch — oscillator at rest, drive switched on at $t = 0$ — and reading off why the startup shows beat-like ringing when $\omega \ne \omega_0$ . The strategy is to feed $x_0 = 0, v_0 = 0$ into the matching equations (5), solve for $C_1, C_2$ , and assemble the full $x(t)$ ; the interference interpretation then falls out by comparing the two pieces' frequencies and initial sizes. Watch the signs: both constants come out negative multiples of the steady state's coefficients, which is the algebra saying the transient begins life opposing the steady state exactly.

Step 1. Impose $x(0) = 0$ in (5): $C_1 = -A\cos\delta$ . Why the minus sign: at $t = 0$ the steady state alone sits at $A\cos\delta$ , but the mass is actually at zero — the transient must start exactly opposite to cancel it.

Step 2. Impose $\dot x(0) = 0$ : $0 = A\omega\sin\delta - \beta C_1 + \omega_1 C_2$ , so with $C_1$ from Step 1,

C_2 = -\frac{A\,(\beta\cos\delta + \omega\sin\delta)}{\omega_1}

Why $\beta$ appears here: the transient's velocity has two parts — the ring's own swing and the envelope's shrinkage — and the matching must cancel the steady state's initial velocity with both included.

Step 3. Assemble the complete rest-start motion:

x(t) = A\cos(\omega t - \delta) - A e^{-\beta t}\left[\cos\delta\,\cos\omega_1 t + \frac{\beta\cos\delta + \omega\sin\delta}{\omega_1}\,\sin\omega_1 t\right] \qquad (6)

where: the first term is the destination, the bracket is the memory. Both terms have amplitude of order $A$ at $t = 0$ — the transient starts as big as the steady state itself — and the bracket rides the dying exponential.

Step 4. Read off the interference. For light damping the two pieces are comparable in size for the first few $1/\beta$ , one oscillating at $\omega$ , the other at $\omega_1 \approx \omega_0$ . Their sum drifts periodically in and out of phase at the difference frequency: the envelope swells and shrinks with period $2\pi/|\omega - \omega_1|$ , fading as $e^{-\beta t}$ . Why this is "beats with a doomed partner": the arithmetic is identical to two-tone beats — the topic of the Beats article, where both tones persist — but here one tone decays, so the interference pattern is transient ringing rather than a steady throb. When $\omega$ is close to $\omega_0$ the beat period is long and the startup looks like slow, dying surges; far from $\omega_0$ the wiggles are fast and vanish within a few drive cycles. *Verified: direct substitution confirms $x(0) = 0$ exactly, and $\dot x(0) = 0$ using $\frac{\mathrm d}{\mathrm dt}$ of each term at $t=0$ ; equation (6) reduces to the textbook rest-start response (Taylor §5.6) with $\gamma = 2\beta$ .*

Figure

Transient ringing — the dying transient beats against the drive before the steady state wins

The settling time — why high Q means slow

How long must you wait for the steady state? Equation (6) answers with one number: the transient's envelope is $e^{-\beta t}$ , so the only timescale of forgetting is

\tau = \frac{1}{\beta} \qquad (7)

where: $\tau$ — the settling (memory) time: amplitude within 63% of final after $\tau$ , 95% after $3\tau$ , 99% after $\ln(100)\,\tau \approx 4.6\tau$ . Note what $\tau$ does not depend on: the drive. Loud or soft, on-resonance or off, the approach rate to the steady state is the system's own $\beta$ — the drive sets the destination, damping sets the travel time. Counting in cycles instead of seconds turns this into the practical rule of resonant instrumentation. Using $Q = \omega_0/2\beta$ (taken as a given here; the definitions live in the Quality Factor article), the number of oscillation periods in one settling time is

N_{\text{settle}} = \frac{\tau}{T} = \frac{\omega_0}{2\pi\beta} = \frac{Q}{\pi} \qquad (8)

where: $N_{\text{settle}}$ — cycles to reach 63% of the final amplitude; roughly $1.5\,Q$ cycles gets you to 99%. This is the flip side of everything that makes high $Q$ desirable: the same weak damping that gives a tall, narrow response makes the system slow to change its mind. A resonator that rings for $Q$ cycles must also be excited for $Q$ cycles.

Go deeper · Derivation C · 2 minThe resonant build-up envelope and the Q-cycle settling rule

Derivation CThe resonant build-up envelope and the Q-cycle settling rule

We are deriving the clean on-resonance limit of the rest-start solution — the envelope $x \approx A(1 - e^{-\beta t})\sin\omega_0 t$ behind the hero figure — and from it the settling rule of equations (7) and (8). The strategy is to specialize equation (6) to $\omega = \omega_0$ , where the phase lag is exactly $\pi/2$ , and then use light damping ( $\beta \ll \omega_0$ ) to identify $\omega_1$ with $\omega_0$ . The step to watch is the ratio $\omega_0/\omega_1$ multiplying the transient: it is what keeps the cancellation at $t = 0$ exact before any approximation is made, and only after the envelope is extracted do we set it to one.

Step 1. Specialize to resonance. At $\omega = \omega_0$ the steady state has $\delta = \pi/2$ (from equation (4): the mismatch $\omega_0^2 - \omega^2$ vanishes), so $\cos\delta = 0$ , $\sin\delta = 1$ , and the rest-start constants collapse to $C_1 = 0$ , $C_2 = -A\,\omega_0/\omega_1$ . Why $C_1$ dies: with a $90°$ lag the steady state passes through zero displacement at $t = 0$ — the position needs no cancelling, only the velocity does.

Step 2. Write the motion with these constants:

x(t) = A\sin\omega_0 t - A\,\frac{\omega_0}{\omega_1}\,e^{-\beta t}\sin\omega_1 t

Why both terms are sines: on resonance the response lags the cosine drive by a quarter cycle, and the transient must be a sine too, to cancel the response's initial velocity with its own.

Step 3. Apply light damping, $\beta \ll \omega_0$ : then $\omega_1 = \omega_0\sqrt{1 - \beta^2/\omega_0^2} \approx \omega_0$ (the difference is second order in $\beta/\omega_0$ ), and the two sines merge:

x(t) \approx A\big(1 - e^{-\beta t}\big)\sin\omega_0 t \qquad (9)

where: the oscillation is a pure tone at $\omega_0$ from the first cycle; only its envelope evolves, rising as $1 - e^{-\beta t}$ . This is the trace on the oscilloscope in every driven-resonance lab: no beats on resonance (the two frequencies coincide), just a smooth climb.

Step 4. Extract the settling rule. The envelope reaches $1 - e^{-1} \approx 63\%$ at $t = 1/\beta$ , which spans $t/T = \omega_0/2\pi\beta = Q/\pi$ oscillation periods — equation (8) — and $99\%$ at $4.6/\beta \approx 1.5\,Q$ periods. Why the rule is worth memorizing: it converts a quoted $Q$ directly into waiting time. It is also the energy story in disguise: near the start the drive pumps at its full rate while friction, which bills as amplitude squared, barely collects — the gap closes only as the amplitude nears the balance point. *Verified: equation (9) satisfies $x(0) = 0$ , $\dot x(0) = 0$ to first order in $\beta/\omega_0$ ; differentiating and substituting into (1) at $\omega = \omega_0$ leaves only terms of order $\beta^2/\omega_0^2$ ; the $t \to \infty$ limit reproduces $A(\omega_0)$ and the $\beta \to 0$ limit recovers linear secular growth $x \to (F_0/2m\omega_0)\,t\sin\omega_0 t$ , matching the undamped-catastrophe result quoted in the Resonance article.*

Figure

Settling is measured in cycles — and the price of a cycle is set by Q

Limiting cases

Case	Result	Meaning
long times, $t \gg 1/\beta$	$x \to A\cos(\omega t - \delta)$	the transient is gone; motion is the pure steady state, independent of launch
high $Q$ (weak damping)	$\tau = 1/\beta$ long; settle $\approx Q/\pi$ cycles	slow forgetting: long beat-like ringing off resonance, slow build-up on resonance
low $Q$ (strong damping)	transient dies within $\sim$ 1 cycle	steady state established almost immediately; phasor analysis valid at once
undamped, $\beta = 0$ , $\omega \ne \omega_0$	transient never decays	free and forced parts coexist forever as a permanent two-tone superposition — no true "steady state"
undamped on resonance, $\beta = 0$ , $\omega = \omega_0$	$x \propto t\sin\omega_0 t$	secular growth, no settling at all — the limit the Resonance article dissects

Common trapYou can meet the initial conditions by adjusting the steady-state amplitude

You cannot — and trying is the single most common error in driven-oscillator problems. The steady-state amplitude $A$ and lag $\delta$ are outputs of the system, computed functions of $\omega_0$ , $\beta$ , $\omega$ , $F_0$ with zero adjustable freedom; they are the same whether you release the mass from rest, from a stretch, or with a fling. All initial-condition information goes into the transient constants $C_1$ and $C_2$ via equations (5) — first write down the fixed steady state, then solve for the transient that makes $x(0)$ and $\dot x(0)$ come out right. The give-away symptom of the error: a "solution" that behaves correctly at early times but predicts a launch-dependent long-time amplitude. Long-time behaviour never depends on the launch — dissipation guarantees it.

Worked examples

1 · Matching a launch, number by number (confidence builder)

A cart of mass $m = 0.50\ \mathrm{kg}$ on a spring $k = 200\ \mathrm{N/m}$ with damping $\beta = 2.0\ \mathrm{s^{-1}}$ is driven by $F(t) = (2.0\ \mathrm{N})\cos(10\,t)$ . At $t = 0$ the cart is held at $x_0 = 20\ \mathrm{mm}$ and released from rest. Find the complete motion.

Step 1. The fixed part first. $\omega_0 = \sqrt{200/0.50} = 20.0\ \mathrm{rad/s}$ , $\omega_1 = \sqrt{\omega_0^2 - \beta^2} = \sqrt{400 - 4} = 19.90\ \mathrm{rad/s}$ . From equation (4) at $\omega = 10\ \mathrm{rad/s}$ : mismatch $= 400 - 100 = 300$ , friction term $= 2\beta\omega = 40$ , so $A = 4.0/\sqrt{300^2 + 40^2} = 13.2\ \mathrm{mm}$ and $\delta = \arctan(40/300) = 0.133\ \mathrm{rad}$ ( $7.6°$ ). These numbers are now frozen — the launch cannot touch them.

Step 2. First matching equation (5): $C_1 = x_0 - A\cos\delta = 20.0 - 13.2\cos(0.133) = 20.0 - 13.1 = \boxed{6.9\ \mathrm{mm}}$ . The steady state already wants to start at 13.1 mm; the transient supplies the missing 6.9.

Step 3. Second matching equation. The steady state's initial velocity is $A\omega\sin\delta = 13.2 \times 10 \times 0.1325 = 17.5\ \mathrm{mm/s}$ , so with $v_0 = 0$ : $C_2 = (v_0 + \beta C_1 - A\omega\sin\delta)/\omega_1 = (0 + 2.0 \times 6.9 - 17.5)/19.90 = (13.8 - 17.5)/19.90 = \boxed{-0.18\ \mathrm{mm}}$ .

Step 4. Assemble and check:

x(t) = 13.2\cos(10t - 0.133) + e^{-2.0t}\big[6.9\cos(19.90\,t) - 0.18\sin(19.90\,t)\big] \ \mathrm{mm}

At $t = 0$ : $13.1 + 6.9 = 20.0\ \mathrm{mm}$ ✓, and the initial velocity terms cancel by construction ✓. The transient starts at $\sqrt{6.9^2 + 0.18^2} = 6.9\ \mathrm{mm}$ and is down to 5% of that by $t = 3/\beta = 1.5\ \mathrm{s}$ — after which the cart oscillates at $10\ \mathrm{rad/s}$ with 13.2 mm amplitude, all memory of the 20 mm launch erased.

2 · How long before you can trust a resonant instrument? (exam level)

A quartz tuning-fork resonator runs at $f_0 = 32{,}768\ \mathrm{Hz}$ with $Q = 1.0\times10^5$ . It is driven on resonance from rest. How long, in seconds and in cycles, until its amplitude reaches 63% of the steady state? Until 99%? Compare with a loudspeaker cone ( $f_0 = 50\ \mathrm{Hz}$ , $Q = 0.7$ ).

Step 1. Convert $Q$ to a damping rate. $\omega_0 = 2\pi f_0 = 2.06\times10^5\ \mathrm{rad/s}$ and $\beta = \omega_0/2Q = 2.06\times10^5/(2\times10^5) = 1.03\ \mathrm{s^{-1}}$ . Remarkable in itself: a resonator vibrating two hundred thousand radians per second forgets at roughly one per second.

Step 2. Settling time from equation (7): $\tau = 1/\beta = \boxed{0.97\ \mathrm{s}}$ to 63%. In cycles, equation (8): $N = Q/\pi = 1.0\times10^5/\pi \approx \boxed{32{,}000\ \mathrm{cycles}}$ .

Step 3. The 99% mark needs $\ln(100)\,\tau = 4.6 \times 0.97 = \boxed{4.5\ \mathrm{s}}$ — about $147{,}000$ cycles, or $1.5\,Q$ . Any measurement scheme that reads this resonator's amplitude must wait seconds per data point; this, not electronics, is the bandwidth bottleneck of high- $Q$ sensors, and it is why a lock-in measurement of a sharp resonance is inherently slow.

Step 4. The loudspeaker: $\beta = \omega_0/2Q = 2\pi(50)/1.4 = 224\ \mathrm{s^{-1}}$ , so $\tau = 4.5\ \mathrm{ms}$ and 99% arrives in $20\ \mathrm{ms}$ — about one cycle at 50 Hz. The design moral is the contrast: $Q = 0.7$ is chosen for loudspeakers precisely so the cone settles within a cycle and can track music's transients; $Q = 10^5$ is chosen for the fork precisely so it forgets nothing and keeps time. Same equations, opposite corners.

Origins · 1 minLord Kelvin and how this was discovered

Lord Kelvin

1824–1907 · Glasgow

William Thomson — Lord Kelvin from 1892 — met the transient response of a driven system as a matter of intercontinental commerce. Consulting for the Atlantic telegraph in the 1850s, he showed that a pulse fed into a long undersea cable does not arrive as a pulse: the cable's distributed capacitance and resistance smear each symbol into a slow rise toward its steady value, so the operator must wait out a settling time before the next symbol is readable — his "law of squares" predicted that this waiting time grows as the cable length squared. His mirror galvanometer, sensitive enough to read the transient's earliest rise rather than its final level, is what made the 1866 cable commercially workable. Kelvin lived in the world this article describes: steady states you must wait for, transients that carry the message, and instruments designed around the settling time. His harmonic analyser — built with his brother James — even mechanized Fourier superposition, summing sinusoids by machine to predict tides decades before electronics.

History

1855

— William Thomson derives the telegraph equation for the Atlantic cable and the "law of squares": the settling time of a cable's response grows as the square of its length.

1858

— The first transatlantic cable carries faint, transient-smeared pulses; Thomson's mirror galvanometer reads the rising edge of each symbol before the steady level arrives.

1866

— The Great Eastern lays the first permanently successful Atlantic cable with Thomson aboard; he is knighted for making the transient readable.

1877

— Rayleigh's Theory of Sound formalizes the general solution of forced vibration as decaying free motion plus persistent forced motion.

1893

— Charles Steinmetz's phasor method lets engineers compute only the steady state — legitimate precisely because transients die exponentially.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §3.x and small-oscillation chapter · Landau & Lifshitz, Mechanics (Vol. 1), §22, §26 (Forced oscillations) · Taylor, Classical Mechanics, Ch. 5 (Oscillations)

Provenance. Verify the amplitude $A(\omega)=(F_0/m)/\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}$ , phase $\tan\delta=\gamma\omega/(\omega_0^2-\omega^2)$ , damped frequency $\omega_1=\sqrt{\omega_0^2-\gamma^2/4}$ , amplitude-resonance frequency $\omega_{\text{res}}=\sqrt{\omega_0^2-\gamma^2/2}$ , and the power-balance result against Morin Ch. 4, Taylor Ch. 5, Goldstein small-oscillations chapter, and Landau–Lifshitz §26. The rest-start solution can be checked by direct substitution into $x(0)=\dot x(0)=0$ . Quality factor $Q=\omega_0/\gamma$ is the standard convention (some texts use $Q=\omega_0/\Delta\omega$ which agrees at high $Q$ ). This page writes $\beta = \gamma/2$ throughout, matching its sibling articles; all formulas map under $\gamma = 2\beta$ . Worked-example numbers ( $A = 13.2$ mm, $\delta = 0.133$ rad, $C_1 = 6.9$ mm, $C_2 = -0.18$ mm; fork $\beta = 1.03\ \mathrm{s^{-1}}$ , $\tau = 0.97$ s, $Q/\pi \approx 32{,}000$ cycles, 99% at 4.5 s) are illustrative arithmetic recomputed from equations (4)–(8).

Builds on: Driven Oscillations · Damped Harmonic Oscillator · Underdamped, Critically Damped, and Overdamped Regimes · Complex Exponential Representation of Oscillations · Simple Harmonic Motion Leads to: Resonance · Amplitude, Phase, and Angular Frequency · Energy in Simple Harmonic Motion

The transient amplitude decays as e^(−βt), but I have also seen e^(−2βt). Which is right?

Both — they track different quantities. The amplitude (and any displacement or velocity coordinate) decays as $e^{-\beta t}$ : that is the factor multiplying the transient in equation (3). The energy is quadratic in amplitude, so it decays as $(e^{-\beta t})^2 = e^{-2\beta t}$ — twice the rate. Confusing the two silently costs factors of 2 in settling estimates: in a ring-down the amplitude takes $1/\beta$ to fall to 37%, while the stored energy gets there in half that time — and a power meter and a displacement sensor will report different-looking curves for the same event. When quoting a settling time, always say which quantity — and if you use the $\gamma = 2\beta$ convention, keep track of which symbol your textbook decays with.

Is the transient a real motion, or just a mathematical correction term?

Completely real — for the first few damping times it can dominate what you see. In the rest-start solution (6) the transient begins exactly as large as the steady state; the wild early oscilloscope trace, the surging envelope, the first violent swing of a pendulum you start pushing — all of that is mostly transient. Engineers fight real transients daily: inrush currents when circuits switch on are transients, the lurch of a car hitting a pothole is a transient, the "thump" a loudspeaker makes at power-on is a transient. What makes the transient special is not unreality but mortality: it is the only part of the motion with an expiry date, and the only part that knows how the motion started.

Why does the transient ring at ω₁ rather than at the natural frequency ω₀ or the drive frequency ω?

Because the transient is a solution of the free damped equation — the drive does not appear in it at all, so $\omega$ is irrelevant to its rhythm; and damping slows a free oscillator's swing slightly below $\omega_0$ , to $\omega_1 = \sqrt{\omega_0^2 - \beta^2}$ . Physically, friction steals a little energy each swing, so the oscillator takes marginally longer per cycle than it would undamped. For light damping the correction is tiny — $\beta = 0.05\,\omega_0$ shifts the frequency by barely 0.1% — which is why one loosely says the transient "rings at the natural frequency." In precision work the distinction matters: the beat frequency in a startup measurement is $|\omega - \omega_1|$ , not $|\omega - \omega_0|$ , and fitting ring-downs with $\omega_0$ instead of $\omega_1$ biases damping estimates.

The startup wiggles look exactly like beats. Are they the same phenomenon as two-tuning-fork beats?

The interference arithmetic is identical; the cast is different. Classic beats — the throbbing of two tuning forks slightly out of tune, covered in the Beats article — involve two persistent tones, so the envelope oscillates at the difference frequency forever. In a driven startup the two tones are the steady state at $\omega$ and the transient at $\omega_1$ , and the second is dying as $e^{-\beta t}$ : you get the same difference-frequency envelope, but each surge is weaker than the last, and after a few $1/\beta$ the pattern is gone. A useful diagnostic reading: if you observe decaying beats after switching on a drive, you are watching the system forget; the beat rate tells you $|\omega - \omega_1|$ and the decay of the surges tells you $\beta$ — two system parameters from one scope trace.

Why exactly do high-Q systems take so long to settle? Is it just a formula, or is there a physical reason?

There is a clean energy reason. To sit at the steady-state amplitude, a resonator must hold a certain stored energy, and the drive can deliver energy only so fast — at a rate limited by the force and the (growing) velocity. A high- $Q$ system on resonance settles at a large amplitude ( $Q$ times the static response, as the Resonance article shows), so there is a great deal of energy to pour in, while the drive's per-cycle deposit stays modest. Equivalently, in the frequency domain: building a response that distinguishes $\omega_0$ from frequencies $\Delta\omega \sim \omega_0/Q$ away requires observing the drive for at least $\sim Q$ cycles — you cannot resolve a narrow line quickly. Sharpness in frequency and sluggishness in time are the same property, seen from two sides.

Does it matter at what phase of the force cycle I switch the drive on?

For the destination, no; for the journey, yes. Switching on at the force's peak versus its zero is equivalent to changing the drive's phase constant, which shifts the steady state's phase accordingly — but its amplitude $A$ and lag $\delta$ relative to the force are untouched. What changes is the mismatch between the actual initial conditions and what the steady state "wants" at the switch-on instant, i.e. the constants $C_1, C_2$ : a different switch-on phase means a different-sized, differently-phased transient and a visibly different startup wiggle. The settling rate is always $\beta$ regardless. Pulse-sequence designers exploit this: choosing the switch-on phase can minimize (or deliberately maximize) the transient — a trick used to ring up resonators faster than $1 - e^{-\beta t}$ by briefly overdriving.

What does the transient look like if the system is critically damped or overdamped?

The decomposition survives untouched; only the transient's costume changes. For overdamping ( $\beta > \omega_0$ ) the homogeneous solution is a sum of two decaying exponentials, $C_1 e^{-(\beta - \sqrt{\beta^2 - \omega_0^2})t} + C_2 e^{-(\beta + \sqrt{\beta^2-\omega_0^2})t}$ — no ringing, just a slouch toward the steady state; at critical damping it is $(C_1 + C_2 t)e^{-\beta t}$ . Two constants remain in every case, so initial conditions are matched the same way. One subtlety: overdamping does not settle faster — the slower of the two exponentials has a decay rate below $\beta$ , so heavy damping actually prolongs the memory. The fastest possible forgetting is at critical damping, which is exactly why measuring instruments and vehicle suspensions are tuned near it.

During the transient, is the energy balance of the steady state violated?

Not violated — just not yet in balance, and that is the whole point. In steady state the drive's average power input equals the dissipated power exactly, which is why the amplitude holds constant. During startup from rest the amplitude is below the balance point: friction (billing as amplitude squared) collects less than the drive deposits, and the surplus accumulates as growing stored energy. As the amplitude nears the steady value the books close and growth stops. Run it the other way — switch the drive off — and the input drops to zero while dissipation continues: the stored energy drains and you get the mirror-image transient, a free ring-down at $\omega_1$ . The detailed power bookkeeping of the settled state lives in the Power Absorption and Phase Lag article.

Transient and Steady-State Response

Where this lives today

The physical picture

The equations, assembled

Matching initial conditions — the transient's one job

The settling time — why high Q means slow

Limiting cases

Worked examples

1 · Matching a launch, number by number (confidence builder)

2 · How long before you can trust a resonant instrument? (exam level)

Going further

Before this

Where this leads