Resonance

On 10 June 2000, London opened the Millennium Bridge — a sleek footbridge across the Thames — and closed it again two days later. Thousands of pedestrians had walked on, the deck had begun to sway, and every walker instinctively widened their stance and pushed sideways in time with the sway to keep balance. Each footfall was a tiny force; arriving in step with the bridge's own rhythm, the footfalls pumped in energy faster than the structure could shed it, and the deck lurched through centimetres of lateral motion. That is resonance: a small periodic push, correctly timed, beating a large restoring system into enormous response. The mathematics behind it is the driven, damped oscillator, $\ddot x + 2\beta\dot x + \omega_0^2 x = (F_0/m)\cos\omega t$ , and its steady state is captured by just two functions of the drive frequency — an amplitude $A(\omega)$ and a phase lag $\delta(\omega)$ . Two facts make the physics rich. The response peaks not exactly at the natural frequency $\omega_0$ but slightly below it, while the power absorbed peaks exactly at $\omega_0$ , where each push lands perfectly in phase with the velocity. And the sharpness of the peak is set entirely by the damping, packaged in one dimensionless number: the quality factor $Q = \omega_0/2\beta$ . Master this one resonance curve and you own a truly universal result — the same Lorentzian-shaped response describes a tuned radio circuit, a proton flipping in an MRI scanner, a wine glass answering a singer, and the mirror suspensions of a gravitational-wave detector.

Key takeaways

Resonance is the dramatic growth of the steady-state amplitude of a driven, lightly damped oscillator when the drive frequency $\omega$ approaches the natural frequency $\omega_0$ .
The steady-state amplitude is $A(\omega) = \dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + (2\beta\omega)^2}}$ — finite for any $\beta > 0$ , and enhanced by the factor $Q$ over the static deflection at $\omega = \omega_0$ .
Exactly at $\omega = \omega_0$ the displacement lags the drive by $90°$ , which puts the velocity in phase with the force — the condition for maximum power transfer.
The quality factor $Q = \omega_0/2\beta$ sets everything about the peak: height $\approx Q \times$ static response, fractional width $\Delta\omega/\omega_0 \approx 1/Q$ , and ring-down time $\approx Q$ cycles.
The amplitude peak (at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ ), the power peak (at $\omega_0$ ), and the $90°$ phase point are three different frequencies that merge only in the weak-damping limit.
With no damping at all, driving exactly at $\omega_0$ produces no steady state — the amplitude grows linearly forever, $x \propto t\sin\omega_0 t$ .
Key idea: drive a lightly damped oscillator near its natural frequency and the steady-state amplitude blows up — bounded only by damping — because each push arrives in phase with the velocity, feeding energy in faster than friction removes it.

Where this lives today

Magnetic resonance imaging is the resonance curve earning its name in every hospital. Protons in a 1.5-tesla field precess at $63.9\ \mathrm{MHz}$ ; the scanner drives them with a radio pulse at exactly that frequency and listens for the ring-down. The width of the absorption line measures the local relaxation rate — the tissue's effective damping — which is precisely the contrast that distinguishes grey matter from white matter from tumour. The same physics, at different $Q$ , tunes every radio and phone: an adjustable resonant circuit amplifies one station's band by its quality factor while its $1/Q$ fractional bandwidth rejects the neighbours. At the extremes, engineered $Q$ values are astonishing. The quartz tuning fork in a wristwatch rings at $32{,}768\ \mathrm{Hz}$ with $Q \sim 10^5$ , losing so little energy per cycle that a coin cell drives it for years. The mirror suspensions of the LIGO gravitational-wave detectors are engineered for $Q$ approaching $10^9$ , concentrating their thermal jitter into razor-thin frequency bands that can be surgically avoided — a direct application of equation (6). And accelerator physicists sweep driven-oscillator resonances deliberately: nuclear magnetic resonance, electron spin resonance, and the resonant cavities that accelerate every particle beam on Earth all live on the peak of equation (7).

The physical picture

Think of pushing a child on a swing. Push at random moments and your shoves sometimes help, sometimes fight the motion; the swing stays low. But time each push to land just as the swing sweeps through the bottom moving away from you — pushing in phase with the velocity — and every shove does positive work. Energy accumulates shove after shove, and the amplitude climbs. That timing condition is resonance, and it is met when your pushing frequency matches the swing's natural frequency.

Figure

Timing is everything — the same small drive, in step and out of step

The picture above is the whole phenomenon in one graph. Driven on resonance, the oscillator's envelope grows steadily — each drive cycle deposits energy because force and velocity never fall out of step. Driven off resonance, force and velocity drift in and out of agreement: energy is pumped in for a few cycles, then pumped back out, and the response stays permanently small. Damping decides where the resonant growth stops: the amplitude settles exactly where the power fed in by the drive balances the power dissipated by friction. A weaker damper means a higher balance point — hence a taller peak as $\beta \to 0$ . How the settled amplitude depends on drive frequency is the resonance curve, and its shape is controlled by a single dimensionless knob, the quality factor $Q$ :

Figure

The resonance curve — steady-state amplitude versus drive frequency for three values of Q

Read the curves like a mountain profile. Far below $\omega_0$ every curve starts at the same floor — the static deflection $F_0/k$ , the stretch the force would produce if applied steadily. Far above $\omega_0$ all curves die as $1/\omega^2$ : the mass simply cannot keep up. In between stands the peak, and its height is not universal at all — it is $Q$ times the floor. A $Q = 15$ oscillator answers a resonant drive fifteen times more strongly than a static one, but only inside a frequency band a fifteenth of $\omega_0$ wide. Tall and narrow come together: sharpness is a statement about memory. A high- $Q$ oscillator rings for about $Q$ cycles after a kick, and correspondingly it accumulates only drives that hold their frequency inside the narrow band $\Delta\omega \approx \omega_0/Q$ .

The steady state and the resonance curve

Everything begins with Newton's second law for a mass on a spring with linear (viscous) damping and a sinusoidal applied force. Divide through by the mass and the equation wears its physics on its sleeve — a restoring term, a friction term, and a drive:

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

where: $x$ — displacement from equilibrium; $\beta$ — damping parameter (the decay rate of free oscillations); $\omega_0 = \sqrt{k/m}$ — natural angular frequency; $\omega$ — driving angular frequency; $F_0$ — force amplitude; $m$ — mass. Note that $\omega$ is yours to choose — it belongs to the driver — while $\omega_0$ and $\beta$ belong to the oscillator. Resonance is what happens when the two frequencies meet. The general solution is a decaying transient (a free damped oscillation, gone after a few multiples of $1/\beta$ ) plus a steady state that oscillates at the drive frequency forever. The steady state is what survives, and it is fully characterized by an amplitude and a phase lag. The cleanest route to both is to complexify the drive, because differentiating an exponential is just multiplication.

Derivation AThe resonance curve from the complex method

We are hunting the complete steady-state response — the amplitude $A(\omega)$ of equation (2) and the phase lag $\delta(\omega)$ of equation (3) — for the driven, damped oscillator of equation (1). The route is the complex-exponential method: trade $\cos\omega t$ for $e^{i\omega t}$ , so that every time derivative becomes multiplication by $i\omega$ and the differential equation collapses into one line of algebra for a single complex number. Both answers then fall out at once — the amplitude as its magnitude, the lag as its argument — which is precisely why this route beats the brute-force real ansatz $a\cos\omega t + b\sin\omega t$ with its tangle of simultaneous equations. Two cautions: the physical motion is the real part, taken only at the very end, and the lag $\delta$ runs from $0$ to $\pi$ , so past resonance the arctangent must be read on its second branch rather than allowed to jump negative.

Step 1. Replace the real drive by a complex one: solve $\ddot z + 2\beta\dot z + \omega_0^2 z = (F_0/m)\,e^{i\omega t}$ and recover the physical motion as $x = \mathrm{Re}\,z$ at the end. Why this works: the equation is linear with real coefficients, so the real part of a complex solution solves the real-part equation — and exponentials turn calculus into algebra.

Step 2. Seek the steady state as $z = C e^{i\omega t}$ with a complex constant $C$ . Why this form: after the transient dies, the only frequency left in the problem is the drive's; a linear system can respond at no other frequency, so the answer must be a fixed complex amplitude times $e^{i\omega t}$ .

Step 3. Substitute. Each time derivative pulls down a factor $i\omega$ , and the differential equation collapses to one algebraic equation:

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

Why this is progress: the entire dynamics is now encoded in one complex number — the bracket — whose magnitude will set the amplitude and whose argument will set the phase.

Step 4. Solve and write $C = A e^{-i\delta}$ in polar form. The magnitude of the bracket is $\sqrt{(\omega_0^2-\omega^2)^2 + (2\beta\omega)^2}$ , so

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

where: $A$ — amplitude of the steady state $x_{\text{ss}}(t) = A\cos(\omega t - \delta)$ . The two terms under the root are the two ways the oscillator can fail to respond: frequency mismatch $(\omega_0^2-\omega^2)$ and friction $(2\beta\omega)$ . On resonance the mismatch term vanishes and friction alone limits the response — which is why light damping means a violent peak.

Step 5. Take the argument. The phase of the bracket is the angle whose tangent is (imaginary part)/(real part):

\tan\delta = \frac{2\beta\omega}{\omega_0^2 - \omega^2} \qquad (3)

where: $\delta$ — the angle by which the displacement lags the driving force, running from $0$ to $\pi$ . At low frequency the mass tracks the force ( $\delta \to 0$ ); exactly at $\omega = \omega_0$ the real part vanishes and $\delta = \pi/2$ ; at high frequency the mass moves in antiphase ( $\delta \to \pi$ ). Verified: reduces to the static deflection $F_0/m\omega_0^2$ as $\omega \to 0$ and to $F_0/m\omega^2$ as $\omega \to \infty$ ; matches Morin Ch. 4 and Taylor Ch. 5.

The phase story deserves its own picture, because it explains why the power peak sits exactly at $\omega_0$ even though the amplitude peak does not:

Figure

The phase lag — from tracking the force to fighting it, always passing through 90 degrees at the natural frequency

Three regimes, three bosses. Below resonance the spring is in charge: the mass quasi-statically follows the force. Above resonance inertia rules: the mass responds oppositely and feebly. Precisely at $\omega_0$ , spring and inertia cancel each other exactly — that is the definition of the natural frequency — leaving friction as the only opposition. With only a velocity-proportional force resisting it, the drive lines up with the velocity, and energy pours in at the maximum possible rate.

Which formula answers which question

you want	use	behaviour
response at drive frequency $\omega$	$A(\omega)$ , eq. (2)	Lorentzian-like peak near $\omega_0$
lag of response behind drive	$\tan\delta = \dfrac{2\beta\omega}{\omega_0^2-\omega^2}$ , eq. (3)	$0 \to \pi/2 \to \pi$
where amplitude peaks	$\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ , eq. (4)	slightly below $\omega_0$
how sharp the peak is	$Q = \omega_0/2\beta$ , eq. (5)	height $\times Q$ , width $/Q$
bandwidth	$\Delta\omega \approx \omega_0/Q = 2\beta$ , eq. (6)	FWHM of the power curve
energy fed in per second	$P_{\text{abs}}(\omega)$ , eq. (7)	peaks exactly at $\omega_0$
worst-case response	$A(\omega_0) = Q \times F_0/m\omega_0^2$ , eq. (8)	$Q$ times the static stretch

Anatomy of the peak

Where exactly does the amplitude peak? Maximizing $A$ means minimizing the radicand in (2), and the neatest trick is to treat $\omega^2$ as the variable. Setting $\frac{\mathrm d}{\mathrm d(\omega^2)}\big[(\omega_0^2-\omega^2)^2 + 4\beta^2\omega^2\big] = -2(\omega_0^2-\omega^2) + 4\beta^2 = 0$ gives the resonant frequency of the displacement:

\omega_r = \sqrt{\omega_0^2 - 2\beta^2} \qquad (4)

where: $\omega_r$ — the drive frequency of maximum amplitude. It sits below $\omega_0$ , because the friction term $2\beta\omega$ grows with $\omega$ : backing the drive off slightly trades a little mismatch for less friction. A real peak exists only for $\beta < \omega_0/\sqrt{2}$ — damp harder than that and the response just declines monotonically from its static value. The whole personality of a resonance — tall or squat, needle or hump — is captured by one dimensionless ratio comparing the oscillation rate to the energy-loss rate:

Q = \frac{\omega_0}{2\beta} = 2\pi\,\frac{\text{energy stored}}{\text{energy lost per cycle}} \qquad (5)

where: $Q$ — the quality factor. The second form is the practical definition: a $Q = 1000$ tuning fork loses a mere $0.6\%$ of its energy each cycle. Free oscillations ring down over about $Q/\pi$ periods, and the same number controls the driven response through the next two results. Two payoffs follow immediately in the weak-damping limit. The full width of the power resonance at half its maximum is $\Delta\omega \approx 2\beta$ , which as a fraction of the center frequency is

\frac{\Delta\omega}{\omega_0} \approx \frac{1}{Q} \qquad (6)

where: $\Delta\omega$ — FWHM of the absorbed-power curve. This reciprocal relation is why radio receivers want high $Q$ (reject the neighbouring station) and why loudspeakers want low $Q$ (respond evenly across the audio band). Sharpness and selectivity are the same word.

Energy flow at resonance

Amplitude is what you see; power is what the driver pays. The cycle-averaged power fed into the oscillator follows from averaging force times velocity over one period, and it inherits the resonance denominator:

P_{\text{abs}}(\omega) = \frac{F_0^2}{2m}\,\frac{2\beta\omega^2}{(\omega_0^2-\omega^2)^2 + (2\beta\omega)^2} \qquad (7)

where: $P_{\text{abs}}$ — mean power delivered by the drive, equal in steady state to the power dissipated by friction. Its maximum, $F_0^2/4m\beta$ , occurs exactly at $\omega = \omega_0$ — not at $\omega_r$ — because power tracks the velocity response, and the velocity is largest where $\delta = \pi/2$ puts it in phase with the force.

Figure

Power absorption — the peak sits exactly at the natural frequency, and its width measures the damping

Go deeper · Derivation B · 2 minEnergy balance fixes the on-resonance amplitude

Derivation BEnergy balance fixes the on-resonance amplitude

The prize is equation (8), the on-resonance amplitude $A(\omega_0) = Q\,F_0/m\omega_0^2$ — and the most physical way to get it is to skip the response algebra entirely and balance the books: in steady state, the power the drive feeds in must equal the power friction destroys. The plan is to compute each side separately — dissipation from the cycle-average of the friction force times velocity, input from the force–velocity average with the on-resonance phase $\delta = \pi/2$ — and set them equal. The balance exists because the two sides scale differently with amplitude: input grows linearly with $A$ , dissipation quadratically, so the curves cross exactly once and the crossing is stable. Watch the time averages — $\langle\dot x^2\rangle = \tfrac12 A^2\omega_0^2$ and the factor $\sin\delta$ are where the halves and the phase physics enter, and dropping either one silently costs a factor of 2.

Step 1. Compute the power dissipated by the damping force $F_d = 2m\beta\dot x$ . Averaged over a cycle of $x = A\cos(\omega_0 t - \delta)$ , using $\langle\dot x^2\rangle = \tfrac12 A^2\omega_0^2$ :

P_{\text{diss}} = 2m\beta\,\langle\dot x^2\rangle = m\beta A^2\omega_0^2

Why this form: friction eats energy at a rate proportional to the square of the velocity, so doubling the amplitude quadruples the bill — dissipation grows faster than input, which is what makes a balance point exist.

Step 2. Compute the power supplied by the drive, $P_{\text{in}} = \langle F\dot x\rangle = \tfrac12 F_0 A\omega_0\sin\delta$ . At $\omega = \omega_0$ the phase lag is $\pi/2$ , so $\sin\delta = 1$ and

P_{\text{in}} = \tfrac12 F_0 A\,\omega_0

Why $\sin\delta$ appears: only the component of velocity in phase with the force survives the time average — and on resonance the whole velocity is in phase, so every newton works full time.

Step 3. Set input equal to dissipation, $\tfrac12 F_0 A\omega_0 = m\beta A^2\omega_0^2$ , and note that input grows linearly with $A$ while dissipation grows quadratically — so the amplitude rises until the two curves cross. Why the steady state is stable: below the balance point input exceeds loss and $A$ grows; above it loss wins and $A$ falls back.

Step 4. Solve for the crossing:

A(\omega_0) = \frac{F_0}{2m\beta\omega_0} = Q\,\frac{F_0}{m\omega_0^2} \qquad (8)

where: $F_0/m\omega_0^2 = F_0/k$ — the static deflection. Pure energetics reproduces the value that equation (2) gives at $\omega = \omega_0$ , and hands you the cleanest statement of resonance: the peak response is the static response amplified by $Q$ . *Verified: identical to setting $\omega = \omega_0$ in $A(\omega)$ ; consistent with Landau–Lifshitz §26.*

Go deeper · Derivation C · 1 minThe undamped catastrophe

Derivation CThe undamped catastrophe

What if there is no friction at all and you drive exactly on pitch? We derive the fate of the undamped oscillator at $\beta = 0$ , $\omega = \omega_0$ : the secular solution $x_p = (F_0/2m\omega_0)\,t\sin\omega_0 t$ , an amplitude that grows without bound, linearly in time. The standard particular-solution recipe fails here — the drive $\cos\omega_0 t$ is itself a solution of the free equation, so the operator annihilates any multiple of it — and the classic escape is to promote the ansatz by a factor of $t$ . The watch-point is the substitution step: the $t$ -proportional pieces must cancel identically against $\omega_0^2 x_p$ , leaving only the new $\cos\omega_0 t$ term generated by differentiation to match the drive — if that cancellation fails, the ansatz was wrong.

Step 1. Set $\beta = 0$ and $\omega = \omega_0$ in (1): $\ddot x + \omega_0^2 x = (F_0/m)\cos\omega_0 t$ . Why the usual ansatz dies: $\cos\omega_0 t$ solves the homogeneous equation, so no multiple of it can produce the drive term — the operator annihilates it.

Step 2. Try the resonant form $x_p = C\,t\sin\omega_0 t$ . Differentiating twice gives $\ddot x_p = 2C\omega_0\cos\omega_0 t - C\omega_0^2\,t\sin\omega_0 t$ . Why multiply by $t$ : this is the standard escape when the forcing sits in the kernel of the operator; the extra factor of $t$ generates the new term the plain ansatz could not.

Step 3. Substitute: the $t\sin\omega_0 t$ pieces cancel against $\omega_0^2 x_p$ , leaving $2C\omega_0\cos\omega_0 t = (F_0/m)\cos\omega_0 t$ , so $C = F_0/2m\omega_0$ and

x_p(t) = \frac{F_0}{2m\omega_0}\,t\sin\omega_0 t

Why the growth is linear: each cycle deposits the same quantum of work, and with nothing to dissipate it the energy — proportional to amplitude squared — grows quadratically, i.e. the amplitude grows linearly. Real systems are rescued first by damping and ultimately by nonlinearity, which detunes the oscillator as the amplitude grows. *Verified: as $\beta \to 0^+$ the damped on-resonance amplitude $F_0/2m\beta\omega_0 \to \infty$ , consistent with unbounded secular growth; matches Taylor Ch. 5 and Morin.*

Limiting cases

Case	Result	Meaning
static / low frequency, $\omega \ll \omega_0$	$A \to \dfrac{F_0}{m\omega_0^2} = \dfrac{F_0}{k}$ , $\ \delta \to 0$	the oscillator follows the force quasi-statically — spring-controlled response
on resonance, $\omega = \omega_0$	$A = \dfrac{F_0}{2m\beta\omega_0} = Q\,\dfrac{F_0}{k}$ , $\ \delta = \dfrac{\pi}{2}$	amplitude enhanced by $Q$ ; velocity in phase with force, maximum power absorbed
high frequency, $\omega \gg \omega_0$	$A \to \dfrac{F_0}{m\omega^2} \to 0$ , $\ \delta \to \pi$	inertia dominates; the mass responds in antiphase and barely moves
weak damping, $\beta \ll \omega_0$	$Q \gg 1$ , $\ \omega_r \approx \omega_0$ , $\ \Delta\omega \approx 2\beta$	tall narrow peak; amplitude, power, and phase resonances collapse onto $\omega_0$
undamped on resonance, $\beta = 0$	$x(t) = \dfrac{F_0}{2m\omega_0}\,t\sin\omega_0 t$	no steady state; secular growth linear in time — the signature of pure resonance

Common trap“Resonance happens at exactly ω = ω₀”

Three different landmarks hide behind the word "resonance", and they occur at three different drive frequencies. The displacement amplitude peaks at $\omega_r = \sqrt{\omega_0^2 - 2\beta^2}$ , strictly below $\omega_0$ . The absorbed power (equivalently the velocity amplitude) peaks exactly at $\omega_0$ . And the phase quadrature condition $\delta = \pi/2$ also happens exactly at $\omega_0$ . Only in the weak-damping limit $\beta \ll \omega_0$ do the three merge into one. In an exam, check which quantity is being maximized before writing " $\omega = \omega_0$ " — for a heavily damped system ( $\beta > \omega_0/\sqrt 2$ ) the displacement has no peak at all, while the power always peaks at $\omega_0$ .

Worked examples

1 · A laboratory mass–spring resonance (confidence builder)

A $0.50\ \mathrm{kg}$ mass hangs from a spring of stiffness $k = 200\ \mathrm{N/m}$ , with a viscous damper giving $\beta = 1.0\ \mathrm{s^{-1}}$ . It is driven by a force of amplitude $F_0 = 2.0\ \mathrm{N}$ . Find the natural frequency, $Q$ , the static and resonant amplitudes, the true amplitude-peak frequency, and the bandwidth.

Step 1. Natural frequency: $\omega_0 = \sqrt{k/m} = \sqrt{200/0.50} = 20.0\ \mathrm{rad/s}$ (about $3.2\ \mathrm{Hz}$ ).

Step 2. Quality factor from (5): $Q = \omega_0/2\beta = 20.0/2.0 = \boxed{10}$ .

Step 3. Static deflection: $A(0) = F_0/k = 2.0/200 = 0.010\ \mathrm{m} = 1.0\ \mathrm{cm}$ .

Step 4. On-resonance amplitude from (8): $A(\omega_0) = Q \times F_0/k = 10 \times 1.0\ \mathrm{cm} = \boxed{10\ \mathrm{cm}}$ . Cross-check directly: $A(\omega_0) = F_0/2m\beta\omega_0 = 2.0/(2 \times 0.50 \times 1.0 \times 20.0) = 0.10\ \mathrm{m}$ . ✓

Step 5. True amplitude peak from (4): $\omega_r = \sqrt{400 - 2(1.0)^2} = \sqrt{398} = 19.95\ \mathrm{rad/s}$ — just $0.25\%$ below $\omega_0$ . At $Q = 10$ the distinction is already almost invisible.

Step 6. Bandwidth from (6): $\Delta\omega \approx 2\beta = 2.0\ \mathrm{rad/s}$ , i.e. $\Delta\omega/\omega_0 = 1/Q = 10\%$ . The peak power the driver must supply is $P_{\max} = F_0^2/4m\beta = 4.0/(4 \times 0.50 \times 1.0) = \boxed{2.0\ \mathrm{W}}$ .

2 · Worn shock absorbers on a washboard road (exam level)

A car of mass $m = 1200\ \mathrm{kg}$ rides on a suspension of effective stiffness $k = 8.0\times10^4\ \mathrm{N/m}$ . A dirt road has ripples spaced $d = 8.0\ \mathrm{m}$ apart with height amplitude $h = 2.0\ \mathrm{cm}$ . Healthy shock absorbers give $Q = 1.2$ ; badly worn ones give $Q = 6.0$ . At what speed does the car resonate, and how violently does the body heave in each case? Does the displacement response even have a peak when the shocks are healthy?

Step 1. Suspension natural frequency: $\omega_0 = \sqrt{k/m} = \sqrt{8.0\times10^4/1200} = \sqrt{66.7} = 8.16\ \mathrm{rad/s}$ , so $f_0 = \omega_0/2\pi = 1.30\ \mathrm{Hz}$ — the typical gentle bounce of a car body.

Step 2. The ripples drive the suspension at $f = v/d$ . Resonance at $f = f_0$ requires $v = f_0\,d = 1.30 \times 8.0 = \boxed{10.4\ \mathrm{m/s} \approx 37\ \mathrm{km/h}}$ . (This is why washboard roads feel worst at one particular speed and smooth out when you drive faster.)

Step 3. Near resonance the body's heave amplitude is roughly $Q$ times the road's input amplitude. Worn shocks: $A \approx Q h = 6.0 \times 2.0\ \mathrm{cm} = \boxed{12\ \mathrm{cm}}$ — a sickening lurch. Healthy shocks: $A \approx 1.2 \times 2.0\ \mathrm{cm} = 2.4\ \mathrm{cm}$ — barely more than the road itself. Shock absorbers exist precisely to keep $Q$ near 1.

Step 4. Check whether a displacement peak exists. From (4) a peak requires $\beta < \omega_0/\sqrt 2$ , i.e. $Q = \omega_0/2\beta > 1/\sqrt 2 \approx 0.71$ . Healthy shocks have $Q = 1.2 > 0.71$ , so a peak does exist, at $\omega_r = \omega_0\sqrt{1 - 1/2Q^2} = 8.16\sqrt{1 - 0.347} = 8.16 \times 0.808 = 6.6\ \mathrm{rad/s}$ — a full $19\%$ below $\omega_0$ . At low $Q$ the amplitude peak and the natural frequency are seriously different frequencies; the weak-damping shortcut $\omega_r \approx \omega_0$ would misplace the resonant speed by $7\ \mathrm{km/h}$ .

Step 5. Worn shocks: $\omega_r = 8.16\sqrt{1 - 1/72} = 8.16 \times 0.993 = 8.1\ \mathrm{rad/s}$ , within $1\%$ of $\omega_0$ — at $Q = 6$ the weak-damping picture is already respectable. The exam moral: always compute $1/2Q^2$ before trusting $\omega_r \approx \omega_0$ .

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Cambridge and Terling, Essex

John William Strutt, third Baron Rayleigh, wrote the book on resonance — literally. His two-volume Theory of Sound (1877–78), begun on a houseboat on the Nile while he recovered from rheumatic fever, systematized the driven, damped oscillator and treated forced vibrations, beats, and the sharpness of response with the full apparatus used on this page. Rayleigh succeeded Maxwell as Cavendish Professor at Cambridge, measured the electrical units to new precision, explained why the sky is blue (Rayleigh scattering), and won the 1904 Nobel Prize for the discovery of argon — found because he refused to ignore a discrepancy in the third decimal place of nitrogen's density. The dimensionless sharpness of a resonance that engineers now call $Q$ descends directly from his analysis of how tuning forks and organ pipes select their frequencies.

History

1638

— Galileo, in the Two New Sciences, describes how a heavy pendulum can be driven to large arcs by feeble, well-timed puffs of breath — the first clear statement of resonant accumulation.

1831

— The Broughton Suspension Bridge near Manchester collapses under soldiers marching in step; the British Army thereafter orders troops to break step on bridges.

1877

— Rayleigh's Theory of Sound gives the complete linear theory of forced, damped vibration, including the resonance curve and its width.

1940

— The Tacoma Narrows Bridge tears itself apart in a 19 m/s wind — popularly cited as resonance, though the true culprit is aeroelastic flutter, a self-excited instability; the distinction is now a textbook staple.

2000

— London's Millennium Bridge closes two days after opening: pedestrians synchronize their footfalls to the deck's lateral sway, pumping it resonantly. Tuned dampers (lowering the

Q

) fix it.

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) · Goldstein, Classical Mechanics, 3rd ed., Ch. 6 · Landau & Lifshitz, Mechanics, 3rd ed., §26 (Forced oscillations) · Taylor, Classical Mechanics, Ch. 5 (Oscillations).

Provenance. All formulas standard from Morin Ch. 4, Taylor Ch. 5, Goldstein Ch. 6, and Landau-Lifshitz §26. A verifier should: (1) confirm $A(\omega)$ and $\tan\delta$ via the complex-exponential substitution; (2) recompute $\omega_r=\sqrt{\omega_0^2-2\beta^2}$ by $\mathrm{d}A/\mathrm{d}\omega=0$ ; (3) verify the power-balance amplitude $F_0/(2m\beta\omega_0)$ equals $A(\omega_0)$ from the curve; (4) check the secular solution $x_p=(F_0/2m\omega_0)t\sin\omega_0 t$ by direct substitution; (5) confirm $Q=\omega_0/2\beta$ and $\Delta\omega/\omega_0\approx 1/Q$ for the power FWHM in the weak-damping limit. The $Q=200$ numerical example is illustrative arithmetic, not a measured constant. Worked-example numbers on this page ( $\omega_0 = 20.0$ rad/s, $Q = 10$ , $A(\omega_0) = 10$ cm, $P_{\max} = 2.0$ W; $f_0 = 1.30$ Hz, $v = 10.4$ m/s, $\omega_r/\omega_0 = 0.808$ at $Q = 1.2$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Driven Oscillations · Damped Harmonic Oscillator · Complex Exponential Representation of Oscillations · Transient and Steady-State Response · Amplitude, Phase, and Angular Frequency. Leads to: Energy in Simple Harmonic Motion · Anharmonic Pendulum and Amplitude Dependence · Small Oscillations About Equilibrium.

With no damping, why does the resonant amplitude grow only linearly in time instead of exploding exponentially?

Because the drive doesn't care how big the motion already is. The force amplitude is fixed at $F_0$ , so the work done per cycle is force times distance traveled — proportional to the current amplitude $A$ . That makes $\mathrm dE/\mathrm dt \propto A \propto \sqrt E$ , whose solution is $E \propto t^2$ and hence $A \propto t$ : exactly the secular solution $x_p = (F_0/2m\omega_0)\,t\sin\omega_0 t$ . Exponential growth requires the input itself to scale with the state — feedback. That is what separates true forced resonance from self-excited instabilities like the Tacoma Narrows flutter, where the motion generates its own aerodynamic driving force, acting as negative damping and producing genuinely exponential growth. A linear forced oscillator, even undamped, is only linearly catastrophic.

The steady state oscillates at the drive frequency

\omega

, not at

\omega_0

. Where did the oscillator's own frequency go?

It lives in the transient — and in the shape of the response. The general solution of (1) is a free damped oscillation, ringing at roughly $\omega_0$ for light damping, plus the steady state; the free part decays after a few multiples of $1/\beta$ and takes $\omega_0$ 's visible rhythm with it. After that, a linear system can only respond at the frequency it is fed: sinusoid in, same-frequency sinusoid out. But $\omega_0$ has not vanished — it is stamped into the coefficients. Both the amplitude $A(\omega)$ of equation (2) and the lag $\delta(\omega)$ of equation (3) depend on how close $\omega$ sits to $\omega_0$ . The oscillator's identity survives as selectivity, not as a frequency in the output.

How can the mass move in antiphase with the force at high frequency? Surely it has to go where it is pushed.

It goes where it is accelerated. Newton's law ties force to acceleration, not to displacement, and for a sinusoid $x = A\cos(\omega t - \delta)$ the acceleration is $-\omega^2 x$ — automatically antiphase with the displacement. At high frequency the drive term dwarfs both spring and friction, so the mass is nearly free: acceleration in phase with the force, hence displacement in antiphase, $\delta \to \pi$ . Picture rapidly shaking the pivot of a heavy pendulum — your hand moves right while the bob, forever half a cycle behind, swings left. The response is also feeble, $A \to F_0/m\omega^2$ : half a period is simply too short a time to build up any displacement before the force reverses and undoes its own work.

I can push a swing every second cycle and it still builds up. Doesn't that break the rule that the drive must match

\omega_0

The rule is for sinusoidal drives, and your pushes aren't one. The theory of equation (1) says a pure $\cos\omega t$ at $\omega = \omega_0/2$ produces only a modest response. But a real push is a short, sharp, periodic shove, and any periodic force decomposes into Fourier harmonics at $\omega, 2\omega, 3\omega, \dots$ Pushing every other cycle at $\omega = \omega_0/2$ still supplies a second harmonic sitting exactly on $\omega_0$ — and by linearity that component pumps the resonance while the others do little. The energy accounting agrees: every shove you actually deliver lands in phase with the velocity, so all your work is positive; you simply deliver it half as often. Genuine subharmonic response to a strictly sinusoidal drive requires nonlinearity.

Equation (5) gives two definitions of

Q

— a frequency ratio and an energy ratio. Why are they the same number?

Follow the energy of a free, lightly damped ring-down. The amplitude decays as $e^{-\beta t}$ , so the stored energy, proportional to amplitude squared, decays as $e^{-2\beta t}$ : in one period $T = 2\pi/\omega_0$ it loses the fraction $2\beta T = 4\pi\beta/\omega_0$ . Then $2\pi \times$ (energy stored)/(energy lost per cycle) $= 2\pi\,\omega_0/4\pi\beta = \omega_0/2\beta$ — the two definitions coincide, with the $2\pi$ inserted precisely to make them do so. The energy form is often the easier one to measure: count ring-down cycles, or measure the fractional loss per cycle (the $Q = 1000$ tuning fork loses just $0.6\%$ ), without ever fitting $\beta$ directly. The equivalence assumes light damping, so the amplitude barely changes within any single cycle.

Why does the phase lag hit exactly

90°

\omega_0

for every value of damping, when everything else about the curve shifts?

Look at the complex bracket $\omega_0^2 - \omega^2 + 2i\beta\omega$ from the derivation. At $\omega = \omega_0$ its real part vanishes identically — whatever $\beta$ is — leaving a purely imaginary number, whose argument is $\pi/2$ . Physically, the real part is the contest between spring and inertia, and their exact cancellation at $\omega_0$ is the very definition of the natural frequency; damping never enters that contest. What remains to oppose the drive is friction alone, a force proportional to velocity, so the drive must align itself with the velocity — displacement lagging by exactly $90°$ . Damping only controls how fast $\delta$ sweeps through $\pi/2$ : the $Q = 15$ curve snaps across while the $Q = 2$ curve ambles, but all of them pass through the same point at $(\omega_0, \pi/2)$ .

Can I build an oscillator that responds both strongly and over a wide band, or are tall and narrow inseparable?

For a single linear oscillator they are inseparable. The peak height is $Q$ times the static response, equation (8), and the fractional bandwidth is $1/Q$ , equation (6) — so their product is fixed by the drive and the spring alone, independent of damping. Turning the only available knob, $\beta$ , just slides you along the tall-narrow versus short-wide tradeoff. That is why the same physics produces opposite engineering: radios crank $Q$ up to reject the neighbouring station, loudspeakers push $Q$ down for an even response across the audio band. Escaping the tradeoff means changing the architecture — staggering several resonators at different frequencies, as multi-driver speakers and coupled filter stages do, or adding active feedback that synthesizes a response no single passive mode can have.

Both Tacoma Narrows and the Millennium Bridge failed by swaying — why does the article count only one as resonance?

The test is where the periodicity comes from. On the Millennium Bridge the drive was external and already periodic: footfalls near the deck's lateral frequency, which the growing sway then entrained more strongly — a forced resonance, cured exactly as equations (5) and (6) suggest, by adding tuned dampers to slash the $Q$ . At Tacoma Narrows the wind was steady — no periodic force existed until the deck moved. The twisting motion itself generated aerodynamic forces in phase with the twisting velocity: energy input proportional to the motion, i.e. negative damping. That is a self-excited instability, flutter, which grows exponentially from nothing; a forced resonance needs a matched drive frequency and, even undamped, grows at worst linearly in time.

Resonance

Where this lives today

The physical picture

The steady state and the resonance curve

Which formula answers which question

Anatomy of the peak

Energy flow at resonance

Limiting cases

Worked examples

1 · A laboratory mass–spring resonance (confidence builder)

2 · Worn shock absorbers on a washboard road (exam level)

Going further

Before this

Where this leads