Damped, Driven Oscillators and Resonance

Derivation

Damped, Driven Oscillators and Resonance

D-032 Home PU-101 Threads force · energy · waves Depends on Small Oscillations and Simple Harmonic Motion

Statement

For a linear oscillator of mass m, restoring constant k and linear drag b, driven by F(t) = F₀ cos ωt, the long-time (steady-state) motion is x(t) = A(ω) cos(ωt − δ) with amplitude A(ω) = (F₀/m) / √[(ω₀² − ω²)² + 4β²ω²], phase lag tan δ = 2βω / (ω₀² − ω²), an amplitude-resonance peak at ω_r = √(ω₀² − 2β²), and a sharpness set by the quality factor Q = ω₀/2β, where ω₀² = k/m and β = b/2m.

Why it matters

Almost every measuring instrument, filter, and structural element that responds to a periodic input is, near one of its normal modes, a driven damped oscillator. The three outputs derived here — how large the response is, how much it lags the drive, and how sharply it peaks — are exactly what an engineer tunes or an experimenter exploits, from a seismometer and a radio front-end to nuclear magnetic resonance and the mirrors of a gravitational-wave detector.

The single dimensionless number Q compresses the whole story: it is simultaneously the resonant gain over the static response, the ratio of centre frequency to bandwidth, and 2π times the energy stored per energy lost per cycle. Learning to read Q off the equation of motion is one of the highest-leverage skills in applied physics.

Assumptions

The restoring force is linear (Hooke's law), F_rest = −kx.If dropped, the equation becomes nonlinear; the response amplitude curve bends over (Duffing behaviour), resonance becomes multivalued, and jump/hysteresis phenomena appear instead of a single-valued A(ω). The damping is linear and velocity-proportional, F_drag = −bẋ.If dropped (e.g. dry Coulomb friction or turbulent ẋ² drag), the superposition/complex method fails, the decay is no longer exponential, and the Lorentzian resonance shape is lost. A pure single-frequency, constant-amplitude drive has acted long enough that the homogeneous transient has decayed.If dropped, the observed motion is the steady state plus a beating transient ∝ e^−βt; the clean formulas describe only the t ≫ 1/β asymptote. The system is underdamped, β < ω₀.If dropped, there is no real amplitude-resonance peak (ω_r becomes imaginary) and Q < ½; the response is monotonic in ω and the notion of a sharp resonance disappears. Parameters m, b, k, F₀, ω are constant in time.If dropped, the coefficients become time-dependent (parametric or swept driving) and one must solve Mathieu-type or chirp equations rather than an algebraic amplitude relation.

Derivation

mẍ = −kx − bẋ + F₀ cos ωt

Newton's second law: sum the Hooke restoring force, the linear drag, and the external drive. A

ẍ + 2βẋ + ω₀²x = (F₀/m) cos ωt

Divide by m and define ω₀² ≡ k/m, β ≡ b/2m. Pure symbol rearrangement into standard form. A

ẍ + 2βẋ + ω₀²z = (F₀/m) e^iωt, x = Re z

The operator is linear with real coefficients, so if complex z solves the complex-drive equation then Re z solves the cos-drive equation (cos ωt = Re e^iωt). Complexifying turns derivatives into multiplications. B

z(t) = X e^iωt, X ∈ ℂ

Steady-state ansatz: after the homogeneous part (∝ e^−βt) has died away, the system oscillates at the drive frequency. The complex constant X carries both amplitude and phase. B

(−ω² + 2iβω + ω₀²) X e^iωt = (F₀/m) e^iωt

Substitute the ansatz: ẋ = iωXe^iωt and ẍ = −ω²Xe^iωt. Differentiation has become algebra. B

X = (F₀/m) / (ω₀² − ω² + 2iβω)

Cancel the common factor e^iωt (never zero) and solve the resulting algebraic equation for X. B

A = |X| = (F₀/m) / √[(ω₀² − ω²)² + 4β²ω²]

The physical amplitude is the modulus of X: |quotient| = |numerator| / |denominator|, and |a + ic| = √(a² + c²). B

X = A e^−iδ ⇒ x(t) = A cos(ωt − δ), tan δ = 2βω / (ω₀² − ω²)

The phase lag δ = −arg X equals the argument of the denominator; taking Re(Xe^iωt) gives the shifted cosine. δ is taken in [0, π] so the response always lags the drive. B

minimise D(ω) = (ω₀² − ω²)² + 4β²ω²: dD/du = −2(ω₀² − u) + 4β² = 0, u = ω²

A is largest where its denominator D is smallest. Substituting u = ω² makes D a quadratic; setting the derivative to zero locates the minimum. C

ω_r² = ω₀² − 2β², A_max = (F₀/m) / [2β√(ω₀² − β²)]

Solve for the minimising frequency, then substitute back: D_min = 4β⁴ + 4β²(ω₀² − 2β²) = 4β²(ω₀² − β²), and take the square root. C

Q ≡ ω₀ / 2β; β ≪ ω₀: A_max ≈ Q · (F₀/k), Q = ω₀ / Δω

Define the quality factor. For light damping ω_r ≈ ω₀ and the peak reduces to Q times the static deflection F₀/k; the power-absorption curve is a Lorentzian of full width at half-maximum Δω = 2β. C

Result

A(ω) = (F₀/m) / √[(ω₀² − ω²)² + 4β²ω²], tan δ = 2βω/(ω₀² − ω²)

ω_r = √(ω₀² − 2β²), A_max = (F₀/m)/[2β√(ω₀² − β²)], Q = ω₀/2β

Reading. At low drive frequency the mass follows the force nearly in phase with the static deflection F₀/k (δ → 0). As ω approaches ω₀ the amplitude swells to roughly Q times that static value and the response crosses δ = 90° exactly at ω = ω₀. Well above resonance the mass can no longer keep up: the amplitude falls as F₀/mω² and the motion swings toward antiphase (δ → 180°). A larger Q (weaker damping) means a taller, narrower peak sitting closer to ω₀.

Units check. F₀/m has units N·kg⁻¹ = m·s⁻². Every term under the root is a squared frequency: (s⁻²)² = s⁻⁴, so the root is s⁻². Hence A = (m·s⁻²)/(s⁻²) = m ✓. The argument of tan δ is (s⁻²)/(s⁻²), dimensionless, and Q = ω₀/2β is (s⁻¹)/(s⁻¹), dimensionless ✓.

Limiting cases

Static / DC limit (ω → 0): A → (F₀/m)/ω₀² = F₀/k and δ → 0: the spring simply balances a slowly varying force.
High-frequency / mass-controlled limit (ω ≫ ω₀): A → F₀/mω² and δ → 180°: inertia dominates and the mass moves opposite to the force.
Exactly at ω = ω₀: A = F₀/(2mβω₀) = Q F₀/k and δ = 90° — the velocity is in phase with the drive, so power transfer is maximal.
Weak damping (β ≪ ω₀, Q ≫ 1): ω_r ≈ ω₀(1 − β²/ω₀²) ≈ ω₀ and the resonance is a sharp Lorentzian of width Δω = ω₀/Q.
Critical / heavy damping (β ≥ ω₀/√2): ω_r² ≤ 0, the amplitude peak vanishes, and A(ω) falls monotonically from its DC value.
Zero damping (β → 0): A diverges at ω = ω₀ and δ jumps discontinuously from 0 to 180° — the idealised undamped resonance catastrophe.

Breaks when

Large amplitude / nonlinear spring. When the drive pushes the displacement beyond the Hooke regime, the effective stiffness becomes amplitude-dependent (Duffing). The resonance curve leans over, folds back on itself, and shows amplitude jumps and hysteresis as ω is swept up versus down — the single-valued Lorentzian no longer exists.
Non-viscous damping. Dry Coulomb friction (constant magnitude) or quadratic aerodynamic drag (∝ ẋ²) breaks linearity: superposition and the complex-X method fail, the amplitude does not diverge as β → 0, and the phase relation is no longer arctan-shaped.
Transient not yet decayed. For times t ≲ 1/β (which is Q cycles) after the drive is switched on, the motion is steady state plus a decaying homogeneous solution, producing ring-up and beats; the steady-state formulas apply only asymptotically.
Overdamped / undefined Q. Once β ≥ ω₀ the resonant frequency ω_r is imaginary and Q ≤ ½; talk of a "resonance peak" or "bandwidth" becomes meaningless.

Failure modes

Confusing the three "resonant" frequencies. Amplitude resonance is at ω_r = √(ω₀² − 2β²), velocity/power resonance is exactly at ω₀, and the free-decay frequency is ω_d = √(ω₀² − β²) — three different numbers, equal only in the β → 0 limit.
Dropping the factor of 2 in the damping term. Writing β = b/m instead of b/2m propagates into Q, the linewidth, and every peak height.
Getting the sign / quadrant of δ wrong. Above resonance ω₀² − ω² < 0, so tan δ is negative but δ lies in the second quadrant (between 90° and 180°); blindly taking arctan gives a spurious negative lag.
Assuming the peak amplitude occurs where velocity or power is maximal. Maximum energy dissipation is at ω₀, not at ω_r; students who maximise the wrong quantity misplace the peak.
Using the amplitude formula during switch-on transients. Applying steady-state A(ω) before the system has rung up for ∼Q cycles gives the wrong instantaneous displacement.
Reading Q off the amplitude FWHM instead of the power FWHM. The bandwidth relation Q = ω₀/Δω uses the half-power (1/√2 amplitude) points, not the half-amplitude points.

Discussion

The physics hides three competing agents in the denominator of A(ω). At low frequency the stiffness term ω₀² wins and the spring sets the response; at high frequency the inertial term ω² wins and the mass sets it; these two cancel at ω = ω₀, leaving only the damping term 2βω to limit the amplitude. Resonance is precisely this cancellation of the reactive (spring and mass) contributions, exposing the small dissipative one — which is why the peak height scales as 1/β and the phase passes through 90° there.

The 90° phase crossing is the deepest single feature. When the displacement lags the force by a quarter cycle, the velocity is exactly in phase with the drive, so the instantaneous power Fẋ is always positive on average and energy is pumped in most efficiently. This is why the sharp, symmetric Lorentzian belongs to the power-absorption curve and peaks at ω₀, whereas the amplitude peak sits slightly below at ω_r. The quality factor unifies these views: Q = ω₀/2β equals the resonant gain, equals the number of radians of free oscillation before the energy falls by e, and equals 2π × (energy stored)/(energy dissipated per cycle).

The same algebra reappears verbatim across physics with a change of names: the series RLC circuit maps m → L, b → R, k → 1/C, giving Q = (1/R)√(L/C) and the identical Lorentzian used for radio tuning. In wave and quantum contexts the driven-oscillator response function becomes the complex susceptibility or the Breit–Wigner cross-section, where ω₀ is a resonant energy and 2β a decay width Γ = ħ/τ — the finite lifetime of an excited state is literally the damping of a driven oscillator.

Viewed as a linear system, the steady-state relation X = χ(ω)(F₀/m) defines a complex response function χ(ω) = 1/(ω₀² − ω² + 2iβω) whose poles sit at ω = iβ ± ω_d in the upper half-plane. Their upper-half-plane location is exactly the statement of causality (the response cannot precede the force), and it forces the real and imaginary parts of χ — the in-phase elastic response and the out-of-phase absorptive response — to obey the Kramers–Kronig dispersion relations. The single damped oscillator is thus the archetype linear-response system, and the fluctuation–dissipation theorem ties the same 2β that damps the driven motion to the thermal noise the oscillator emits at equilibrium.

Common misconceptions. Resonance does not mean the amplitude becomes infinite — that only happens for the unphysical β = 0 case; real damping always caps it at A_max = QF₀/k. And "driving at the natural frequency" gives maximum power absorption (at ω₀) but not quite maximum amplitude (at the slightly lower ω_r); the distinction only matters when damping is not negligible.

Worked examples

Peak amplitude of a driven spring. m = 0.50 kg, k = 200 N·m⁻¹, b = 0.50 kg·s⁻¹, F₀ = 2.0 N. Find ω₀, β, Q, ω_r, and A_max.

ω₀ = √(k/m) = √(200/0.50) = √400 = 20 rad·s⁻¹

Natural frequency from stiffness and mass. A

β = b/2m = 0.50/(2·0.50) = 0.50 s⁻¹ ⇒ Q = ω₀/2β = 20/1.0 = 20

Damping constant and quality factor; Q = 20 confirms light damping. A

ω_r = √(ω₀² − 2β²) = √(400 − 0.50) = √399.5 = 19.99 rad·s⁻¹

Amplitude-resonance frequency, essentially ω₀ because β ≪ ω₀. B

A_max = (F₀/m)/[2β√(ω₀² − β²)] = (2.0/0.50)/[1.0·√(400 − 0.25)] = 4.0/19.994

Substitute into the peak formula; the numerator is F₀/m = 4.0 m·s⁻². B

A_max = 0.20 m (= Q · F₀/k = 20 × 0.010 m ✓)

Reading. The static deflection F₀/k = 10 mm is amplified twentyfold at resonance, exactly the Q = 20 gain. Units check. (m·s⁻²)/(s⁻¹·s⁻¹) = m ✓.

Amplitude and phase off resonance. Same system (ω₀ = 20 s⁻¹, β = 0.50 s⁻¹, F₀/m = 4.0 m·s⁻²), now driven at ω = 25 rad·s⁻¹, above resonance. Find A and the phase lag δ.

ω₀² − ω² = 400 − 625 = −225 s⁻², 2βω = 2·0.50·25 = 25 s⁻²

Evaluate the two pieces of the complex denominator. A

A = 4.0/√[(−225)² + 25²] = 4.0/√(50625 + 625) = 4.0/226.4

Amplitude is F₀/m divided by the modulus of the denominator. B

tan δ = 2βω/(ω₀² − ω²) = 25/(−225) = −0.1111

Numerator positive, denominator negative ⇒ δ is in the second quadrant. B

δ = 180° − arctan(0.1111) = 180° − 6.3° = 173.7°

Take the correct quadrant: above resonance the response lags by nearly half a cycle. B

A = 1.8 × 10⁻² m ≈ 18 mm, δ ≈ 174°

Reading. Driven above ω₀ the amplitude has already dropped from its 200 mm peak to 18 mm, and the mass moves almost exactly opposite to the force — inertia dominates. Units check. (m·s⁻²)/(s⁻²) = m ✓; δ dimensionless ✓.

Problems

A driven damped oscillator has m = 0.20 kg, k = 80 N·m⁻¹, and b = 0.16 kg·s⁻¹. Find ω₀ and Q.
Solution
ω₀ = √(k/m) = √(80/0.20) = √400 = 20 rad·s⁻¹. β = b/2m = 0.16/0.40 = 0.40 s⁻¹. Q = ω₀/2β = 20/0.80 = 25.
For the oscillator of Problem 1, driven by F₀ = 1.0 N, find the static deflection and the peak amplitude A_max.
Solution
Static: A₀ = F₀/k = 1.0/80 = 0.0125 m = 12.5 mm. Weak damping (Q = 25), so A_max ≈ Q A₀ = 25 × 0.0125 = 0.31 m. (Exact: A_max = (F₀/m)/[2β√(ω₀²−β²)] = 5.0/[0.80·√399.84] = 5.0/15.997 = 0.3126 m, confirming 0.31 m.)
A tuning fork behaves as a lightly damped oscillator at f₀ = 440 Hz. After being struck, its free vibration amplitude falls to 1/e of its initial value in 2.0 s. Estimate Q.
Solution
Free amplitude decays as e^−βt, so 1/e occurs at t = 1/β = 2.0 s ⇒ β = 0.50 s⁻¹. ω₀ = 2πf₀ = 2π(440) = 2765 rad·s⁻¹. Q = ω₀/2β = 2765/1.0 = ≈ 2.8 × 10³. (Equivalently Q = ω₀τ/2 with τ = 2.0 s.)
Show that at the half-power points the amplitude has fallen to 1/√2 of its resonant value, and that for weak damping these occur at ω ≈ ω₀ ± β, giving bandwidth Δω = 2β and Q = ω₀/Δω.
Solution
Near resonance write ω = ω₀ + ε with |ε| ≪ ω₀. Then ω₀² − ω² = (ω₀−ω)(ω₀+ω) ≈ (−ε)(2ω₀) = −2ω₀ε, and 2βω ≈ 2βω₀. The denominator squared is D ≈ 4ω₀²(ε² + β²), so A² ∝ 1/(ε² + β²) — a Lorentzian in ε. At ε = 0, A² is maximal; the power (∝ A²) halves when ε² + β² = 2β², i.e. ε = ±β, where A = A_peak/√2. The two points are ω = ω₀ ± β, so Δω = 2β and Q = ω₀/2β = ω₀/Δω. ✓
A series RLC circuit has L = 10 mH, C = 1.0 µF, R = 5.0 Ω, driven by an AC source. Find the resonant (angular) frequency, Q, and the −3 dB bandwidth in Hz.
Solution
Map m→L, b→R, k→1/C. ω₀ = 1/√(LC) = 1/√(10×10⁻³ · 1.0×10⁻⁶) = 1/√(10⁻⁸) = 1.0×10⁴ rad·s⁻¹. Q = ω₀L/R = (10⁴)(0.010)/5.0 = 100/5.0 = 20. Bandwidth Δω = ω₀/Q = 10⁴/20 = 500 rad·s⁻¹, so Δf = Δω/2π = 500/6.283 = 80 Hz (centre frequency f₀ = ω₀/2π ≈ 1.6 kHz).