Resonance — physical consequences, design implications: Question bank

Before you start, four anchors you'll lean on repeatedly. Do not skip these — every trap below reuses one of them.

Definition The four symbols this page uses

==Spring constant $k$ == — the stiffness of the restoring pull: the force $- k x$ grows in proportion to how far $x$ you displace the mass. Big $k$ = stiff (snaps back hard), small $k$ = floppy. Its units are newtons per metre.
Natural frequency $ω_{0} = k / m$ — the rhythm the system swings at with no pushing and no friction. Stiffer spring (bigger $k$ ) or lighter mass (smaller $m$ ) means faster swinging. Built in Simple Harmonic Motion.
Damping $b$ — the friction term $- b \overset{x}{˙}$ that removes energy; it is the only thing that stops resonance blowing up to infinity.
Drive $F_{0} cos ω t$ — the outside push at frequency $ω$ . Resonance is about matching $ω$ to $ω_{0}$ .
==Phase lag $ϕ$ == — in the steady solution $x (t) = A cos (ω t - ϕ)$ , the number $ϕ$ says how far behind the driver the mass lags, measured in the angle of one full cycle ( $360°$ = one whole cycle). $ϕ = 0$ means the mass moves exactly with the push; $ϕ = 90°$ means it lags by a quarter cycle; $ϕ = 180°$ means it moves exactly opposite the push.

Now let's see the resonance curve these symbols describe.

Figure — Resonance — physical consequences, design implications

The blue curve is the amplitude $A (ω)$ . Notice its peak sits just left of $ω_{0}$ (dashed line) — that offset is what the "displacement resonance" traps are about. The pale-yellow band, labelled $Δ ω$ directly on the figure, spans the two half-power points: where $A$ has fallen to $1/ 2$ of its peak, so the delivered power (∝ $A^{2}$ ) is exactly halved.

Formula The half-power condition ⇒ the bandwidth

$Δ ω = ω_{0} / Q$ What "half-power" means algebraically. Power delivered scales as $A^{2}$ , so a half-power point is any $ω$ where $A (ω) = \frac{A _{ma x}}{2} ⟺ A (ω)^{2} = \frac{1}{2} A_{ma x}^{2} .$ Turn that into an equation. With $A_{ma x}^{2} = (F_{0} / m)^{2} / (b ω_{0} / m)^{2}$ , squaring $A (ω)$ and demanding it equal half of that gives, after cancelling $(F_{0} / m)^{2}$ , $(ω_{0}^{2} - ω^{2})^{2} + (\frac{bω}{m})^{2} = 2 (\frac{b ω _{0}}{m})^{2} .$ Solve near the peak. For light damping the two solutions sit close to $ω_{0}$ ; write $ω = ω_{0} \pm \frac{δ}{2}$ and keep only leading terms. The condition reduces to $(ω_{0}^{2} - ω^{2})^{2} \approx (b ω_{0} / m)^{2}$ , i.e. $∣ ω_{0}^{2} - ω^{2} ∣ \approx b ω_{0} / m$ . Since $ω_{0}^{2} - ω^{2} \approx 2 ω_{0} (ω_{0} - ω)$ , each half-power point sits a distance $∣ ω_{0} - ω ∣ \approx b / (2 m)$ from $ω_{0}$ . The full width between them is therefore $Δ ω \approx \frac{b}{m} = \frac{ω _{0}}{Q}, using Q = \frac{m ω _{0}}{b} .$ That is where the clean rule $Δ ω = ω_{0} / Q$ comes from — and it is the yellow band on the figure.

Finally, the phase. Watch how $ϕ$ swings from $0$ to $180°$ as you sweep the drive through resonance, crossing exactly $90°$ at $ω_{0}$ :

At $ϕ = 90°$ (the crossing point) the force is a quarter-cycle ahead of displacement, which makes it line up perfectly with velocity — that is why energy pours in most efficiently there.

True or false — justify

True or false: at resonance the driving force is in phase with the displacement.

False. At resonance the force leads displacement by 90° (

ϕ = 90°

), which puts it in phase with velocity — that is exactly why every push adds energy and the swing keeps growing.

True or false: with zero damping the steady-state amplitude at

ω_{0}

would be infinite.

True (mathematically). The denominator becomes

0 + 0 = 0

, so

A \to \infty

. Physically impossible — which is the whole point that real systems must have damping to cap it.

True or false: the displacement amplitude peaks exactly at

ω_{0}

False. The displacement peak sits just below, at

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2}

. Only the velocity/power resonance lands exactly on

ω_{0}

True or false: a heavier oscillator is safe from resonance because it is sluggish.

False. Extra mass merely lowers

ω_{0} = k / m

; it relocates the resonance rather than removing it. You could accidentally move

ω_{0}

onto a driver you were previously safe from.

True or false: a microwave oven works by resonating water molecules.

False. It uses dielectric (dipole-relaxation) heating — dipoles lag as they try to follow the field, dumping heat over a broad band. If it were a sharp resonance, the frequency would need to be precise, which it deliberately is not.

True or false: a high-

Q

system has a tall, narrow resonance peak.

True. High

Q = m ω_{0} / b

means low damping, so energy leaks slowly, the peak is sharp and the bandwidth

Δ ω = ω_{0} / Q

is narrow — great for selectivity (radio tuning, quartz clocks).

True or false: increasing damping always lowers the peak amplitude.

True. A larger

b

increases the denominator's minimum value

bω / m

, so

A_{ma x} = F_{0} / (b ω_{0})

falls. This is precisely the engineer's lever.

True or false: soldiers break step because their combined weight is too heavy for the bridge.

False. It is not the static weight but the coherent periodic force at ~2 Hz. Breaking step randomises the timing so no single frequency dominates the drive.

True or false: at the half-power points the amplitude has dropped to half its peak.

False. Amplitude drops to

1/ 2 \approx 0.707

of peak; power (∝ amplitude²) drops to a half, since

(1/ 2)^{2} = \frac{1}{2}

True or false: a tuned mass damper works by adding stiffness to make the building rigid.

False. It adds a second oscillator tuned to the building's frequency; it absorbs and dissipates energy, splitting and lowering the resonance peak rather than stiffening the structure.

Spot the error

"Resonance is dangerous, so good engineering means avoiding it entirely." — find the flaw.

Resonance is often wanted: radios, MRI, and quartz watches exploit sharp resonance. The goal is to place it deliberately — high

Q

where selectivity helps, damped and detuned where amplitude threatens structures.

"Since

A_{ma x} = F_{0} / (b ω_{0})

, doubling the drive force

F_{0}

always doubles danger." — find the flaw.

The amplitude does scale linearly with

F_{0}

, but "danger" also depends on whether the system is even at its resonant frequency. A huge force far from

ω_{0}

can be harmless; a tiny one at

ω_{0}

can be catastrophic.

"The Tacoma Narrows bridge failed because steady wind pushed it over." — find the flaw.

Steady wind is not periodic and cannot drive resonance. The rhythmic driver was vortex shedding (a Kármán vortex street) whose frequency matched a torsional mode. The relevant force oscillated even though the wind speed was roughly constant.

"Both power resonance and displacement resonance occur at

ω_{0}

." — find the flaw.

Only power/velocity resonance is exactly at

ω_{0}

. Displacement resonance is at

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2}

, a little lower — the two coincide only in the limit

b \to 0

"A low-

Q

system is just a badly built high-

Q

system." — find the flaw.

Low

Q

is often the design goal: shock absorbers, door dampers, and the microwave oven all want broad, robust response so behaviour barely changes with small frequency drift. "Good" depends entirely on the job.

"Because

ϕ = 90°

at resonance, the force does no work — it's perpendicular to displacement." — find the flaw.

Work depends on force aligned with velocity, not displacement. At

ϕ = 90°

the force is exactly in phase with velocity, so it does maximum work per cycle — the opposite conclusion.

Why questions

Why does even a tiny periodic push build a huge amplitude at resonance?

Because each push arrives in step with the motion (force in phase with velocity), so every push adds energy coherently instead of some adding and some subtracting. Growth is limited only when energy fed in per cycle equals energy lost to damping.

Why is damping the quantity that "saves the day" rather than mass or stiffness?

Damping is the only term that removes energy. Mass and stiffness merely set where

ω_{0}

sits; without any

b

the balance between energy in and out never closes, so amplitude runs away.

Why does higher

Q

give a radio better station selectivity?

High

Q

means a narrow bandwidth

Δ ω = ω_{0} / Q

, so only frequencies extremely close to

ω_{0}

are amplified strongly and neighbouring stations are rejected. See LC Circuits & AC Resonance.

Why is the displacement peak at

ω_{r es}

below

ω_{0}

rather than on it?

Because the damping term

(bω / m)^{2}

in the denominator grows with

ω

; it penalises higher frequencies, nudging the smallest-denominator (largest-amplitude) point to the left of the spring–inertia balance at

ω_{0}

Why must a sharp driver excite a matching normal mode strongly?

A single-frequency drive pours energy almost entirely into the mode whose

ω_{0}

matches it; via Fourier Analysis, a mismatched or broadband drive spreads its energy thinly across many modes, exciting none strongly.

Why does phase lag

ϕ

pass through

90°

as you sweep through resonance?

ϕ = arctan [(bω / m) / (ω_{0}^{2} - ω^{2})]

the denominator is positive below

ω_{0}

(

ϕ

small), zero at

ω_{0}

(

arctan \infty = 90°

), and negative above (

ϕ \to 180°

). The crossover at

90°

marks the frequency of perfect energy transfer.

Edge cases

ω = 0

(a constant push, no oscillation) what is the amplitude?

The formula gives

A = F_{0} / k

— just the static spring stretch. No resonance, no damping effect: a steady force simply displaces the spring by Hooke's law.

ω \to \infty

(driving far faster than natural) what happens to

A

Amplitude

\to 0

. The mass's inertia cannot keep up with the rapid reversals, so the

m ω^{2}

term dominates the denominator and swamps everything — the oscillator barely moves.

What happens to

ω_{r es}

as damping grows toward

b^{2} = 2 m^{2} ω_{0}^{2}

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2} \to 0

: the displacement-resonance peak slides all the way to zero frequency. Beyond that damping level there is no peak at all — the response just falls monotonically from its static value (see the overdamped curve on the figure below).

In the light-damping limit

b \to 0

, where does the peak go and how tall?

ω_{r es} \to ω_{0}

and

A_{ma x} = F_{0} / (b ω_{0}) \to \infty

. This is the idealised undamped picture that makes resonance look infinitely dangerous — a limit, never reality.

For a perfectly rigid, undriven, undamped mass on a spring, is "resonance" even defined?

No drive means no resonance to speak of — you only have free Simple Harmonic Motion at

ω_{0}

. Resonance is inherently a forced-oscillation phenomenon: it needs an external periodic push to match against

ω_{0}

Connections

Forced Oscillations — the steady-state amplitude and phase these traps probe.
Damped Oscillations — where $b$ , $Q$ , and the amplitude cap come from.
Simple Harmonic Motion — the source of $ω_{0} = k / m$ .
LC Circuits & AC Resonance — the selectivity questions in electrical form.
Standing Waves & Normal Modes — mode-matching behind the "why" section.
Fourier Analysis — why sharp drivers excite matching modes.
Resonance — physical consequences, design implications — parent topic.