Quality Factor and Exponential Energy Decay

Derivation

Quality Factor and Exponential Energy Decay

D-060 Home PU-103 Threads energy Depends on The Damped Harmonic Oscillator, Energy Exchange and Conservation in SHM

Statement

For a lightly damped linear oscillator m ẍ + b ẋ + k x = 0, the cycle-averaged mechanical energy obeys d⟨E⟩/dt = −γ⟨E⟩ with γ = b/m, so ⟨E⟩(t) = E₀ e^−γt. The quality factor Q = ω₀/γ (with ω₀ = √(k/m)) equals the number of radians of oscillation completed while the stored energy falls by one factor of e.

Why it matters

Almost every resonator you meet — a pendulum, a quartz crystal, an RLC tank, an optical cavity, a swinging bridge, an atomic transition — is a lightly damped oscillator, and the single dimensionless number Q characterises how well it stores energy. It sets ringdown times, resonance sharpness, and the linewidth of any spectral feature.

The result also shows why energy is the natural variable for damping: while the amplitude and the position oscillate, the envelope of the energy decays in a clean, memoryless, exponential way, and its decay constant is exactly what a decibel-per-second or a "Q of the mode" measurement reports.

Assumptions

Linear (viscous) damping, force −bẋ with constant b.If drag is nonlinear (∝ ẋ² for turbulent air, or a constant Coulomb friction for dry sliding) the loss per cycle is amplitude-dependent and the decay ceases to be exponential — it becomes algebraic or linear-in-time instead. Light damping, γ ≪ 2ω₀ (equivalently Q ≫ ½).If dropped the system is over- or critically damped, there is no oscillation to average over, and "radians per e-fold" and the cycle-average both lose meaning. Constant parameters m, b, k over many periods.If the mass, stiffness or damping drift on the timescale of an oscillation the ODE is no longer autonomous, γ becomes time-dependent, and the exponential envelope acquires a slowly varying rate γ(t). Free (undriven) motion — the cycle-average uses equipartition of the SHM solution.If a drive is present, in steady state the input power balances dissipation and ⟨E⟩ is constant, not decaying; the exponential law describes only the transient after the drive is removed.

Derivation

m ẍ + b ẋ + k x = 0

Equation of motion of the damped oscillator (prior result: damped-harmonic-oscillator). A

E = ½ m ẋ² + ½ k x²

Total mechanical energy = kinetic + spring potential (prior result: shm-energy-conservation defines this E). A

dE/dt = m ẋ ẍ + k x ẋ

Differentiate term by term in time and apply the chain rule; m, k constant. A

m ẍ = −b ẋ − k x

Solve the equation of motion in step 1 for the highest derivative, ready to substitute. A

dE/dt = ẋ(−b ẋ − k x) + k x ẋ = −b ẋ²

Insert step 4 into step 3; the +k x ẋ and −k x ẋ cancel exactly. This is the instantaneous power dissipated — exact, and ≤ 0. B

x(t) ≈ A₀ e^−γt/2 cos(ω_d t + φ), ω_d = √(ω₀² − γ²/4)

Light-damping solution: over one period the motion is nearly SHM with a slowly shrinking envelope, so a cycle-average is well defined (prior result: damped-harmonic-oscillator). C

⟨½ m ẋ²⟩ = ⟨½ k x²⟩ = ½⟨E⟩ ⟹ ⟨ẋ²⟩ = ⟨E⟩ / m

Over one cycle of near-SHM the kinetic and potential energies share equally (equipartition / virial averaging — shm-energy-conservation); their sum is ⟨E⟩. C

⟨dE/dt⟩ = −b ⟨ẋ²⟩ = −(b/m) ⟨E⟩ ≡ −γ⟨E⟩, γ = b/m

Cycle-average step 5 and substitute step 7; define the energy decay rate γ = b/m. B

⟨E⟩(t) = E₀ e^−γt

Integrate the linear first-order ODE d⟨E⟩/dt = −γ⟨E⟩ with initial energy E₀. A

Q ≡ ω₀/γ = ω₀ τ_E, τ_E = 1/γ

Define the quality factor as the ratio of the natural angular frequency to the energy decay rate; in one energy e-fold time τ_E the phase advances by ω₀τ_E = Q radians. B

Result

⟨E⟩(t) = E₀ e^−γt, Q = ω₀/γ

Reading. The stored energy decays exponentially with rate γ = b/m; the e-fold time is τ_E = 1/γ. Because energy ∝ (amplitude)², the amplitude envelope decays at half that rate, e^−γt/2, with e-fold time 2/γ = 2τ_E. The quality factor counts radians per energy e-fold: Q = ω₀τ_E, equivalently Q/2π full oscillations before E falls to E₀/e. The fractional energy lost per cycle is 1 − e^−γT ≈ γT = 2π/Q, so Q = 2π × (energy stored)/(energy lost per cycle).

Units check. γ = b/m has units (kg s⁻¹)/(kg) = s⁻¹, so γt is dimensionless and e^−γt is well posed. ω₀ = √(k/m) has units √((N m⁻¹)/kg) = √(s⁻²) = s⁻¹. Hence Q = ω₀/γ is (s⁻¹)/(s⁻¹) = dimensionless, as a pure count must be.

Limiting cases

Undamped, γ → 0: Q → ∞, τ_E → ∞, energy is conserved and the oscillator rings forever — recovers shm-energy-conservation.
High Q (Q ≫ 1): hundreds to millions of oscillations per e-fold; resonance is sharp, with fractional bandwidth Δω/ω₀ ≈ 1/Q.
Approaching critical damping, γ → 2ω₀: ω_d → 0 and Q → ½; the "per cycle" picture collapses because there is barely one oscillation.
One e-fold in cycles: the number of full periods to lose 1/e of the energy is τ_E/T = ω₀/(2πγ) = Q/2π.

Breaks when

Over- or critically damped (γ ≳ 2ω₀, Q ≲ ½). There is no periodic motion; the cycle-average is undefined and the energy relaxes as a sum of two real exponentials, not a single e^−γt ringing envelope.
Nonlinear damping. Quadratic (turbulent) drag ∝ ẋ² gives an amplitude that decays roughly as 1/(1+t) (algebraic); dry Coulomb friction gives an amplitude that falls linearly in time and stops dead — neither is exponential, and Q is not constant.
Driven steady state. With a sustained drive the energy is constant on average; the exponential law applies only to the free transient once the drive is switched off.
Slowly varying parameters. If b, k or m change over many periods, γ and Q become time-dependent and the envelope is no longer a single exponential.

Failure modes

Amplitude/energy rate mix-up. Writing energy as e^−γt/2 or amplitude as e^−γt — the energy rate is γ, the amplitude rate is γ/2, a factor of 2 that changes every ringdown time.
Convention clash. Many texts write ẍ + 2β ẋ + ω₀²x = 0 with 2β = b/m, so their β = γ/2. Reusing Q = ω₀/(2β) is right but Q = ω₀/β is wrong — check whether the symbol multiplies the first or the amplitude derivative.
Instantaneous vs averaged. Reading dE/dt = −bẋ² as itself exponential. It is exact but oscillatory (it dips to zero at every turning point); only its cycle-average is −γ⟨E⟩.
Missing the 2π. Quoting Q = E/ΔE_cycle instead of 2π E/ΔE_cycle — Q is radians per e-fold, and one cycle is 2π radians.
ω_d for ω₀. Using the damped frequency ω_d in Q; for light damping the difference is O(1/Q²) and negligible, but stating Q = ω_d/γ as exact is wrong.

Discussion

The heart of the result is step 5: energy leaves the oscillator at a rate −bẋ² that is always non-negative and vanishes only at the turning points. This is Joule heating in the dashpot — the mechanical analogue of I²R in a resistor — and it is the microscopic reason the second law shows up in a "frictionless" Newtonian problem. The oscillator does not lose energy uniformly in time; it loses it fastest as it whips through equilibrium and not at all at the extremes, yet the envelope smooths this into a clean exponential.

Why exponential, and why memoryless? Because linear damping makes the loss per cycle a fixed fraction 2π/Q of whatever energy is present, independent of amplitude. A constant fractional loss per unit phase is the defining property of an exponential — the same logic behind radioactive decay and RC discharge. The quality factor is then just the phase-space "half-life" expressed in radians, which is why the same Q governs the ringdown time and the resonance linewidth: a Fourier transform of e^−γt/2cos ω₀t is a Lorentzian of full width γ centred at ω₀, so Δω/ω₀ = γ/ω₀ = 1/Q. Ringdown in time and sharpness in frequency are one fact seen two ways.

The energy method also generalises where the explicit solution does not. For any weakly perturbed conservative system, dE/dt = (∂L/∂t)-type accounting plus averaging over the fast phase — the method of averaging, or adiabatic invariance of the action J = ∮ p\,dq = 2πE/ω — reproduces ⟨E⟩ ∝ e^−γt without ever writing down ω_d. In the quantum oscillator the same Q reappears as the ratio of a level's frequency to its linewidth, and the number of coherent oscillations before decoherence is again ∼Q; the classical ringdown and the quantum coherence time are the same quantity in different clothing.

Common misconceptions. Q is not "how much energy the oscillator holds" — a tiny tuning fork and a massive flywheel can share the same Q. It measures the ratio of stored to per-cycle-lost energy, i.e. how many oscillations survive, not their size. And a high Q does not mean slow oscillation; it means slow decay relative to fast oscillation — a 5 MHz quartz crystal with Q ≈ 10⁶ rings for only ∼0.2 s yet completes a million cycles per e-fold.

Worked examples

Mass–spring ringdown: m = 0.20 kg, k = 80 N/m, b = 0.050 kg/s, E₀ = 1.0 J

Find ω₀, γ, Q, the energy e-fold time, and the energy remaining after 10 s. Work symbolically first. A

ω₀ = √(k/m) = √(80/0.20) = √400 = 20 rad/s

Natural angular frequency from stiffness and mass. A

γ = b/m = 0.050/0.20 = 0.25 s⁻¹, τ_E = 1/γ = 4.0 s

Energy decay rate and its e-fold time. Check γ = 0.25 ≪ 2ω₀ = 40, so light damping holds. A

Q = ω₀/γ = 20/0.25 = 80

Quality factor; the fork rings for about Q/2π ≈ 13 cycles per e-fold. A

E(10 s) = E₀ e^−γt = 1.0 · e^−0.25·10 = e^−2.5 = 0.082 J

Evaluate the exponential envelope at t = 10 s = 2.5 τ_E. A

ω₀ = 20 rad/s, γ = 0.25 s⁻¹, Q = 80, E(10 s) ≈ 0.082 J

Reading. After 2.5 energy e-folds only 8% of the initial joule survives; the amplitude, decaying at γ/2, is down to √0.082 ≈ 29% of its start.

Tuning fork by decay: f = 512 Hz; amplitude halves in t_½ = 2.0 s

Infer Q and the number of oscillations per energy e-fold from a stopwatch measurement of the amplitude. B

A(t) = A₀ e^−γt/2 ⟹ e^−γt_½/2 = ½ ⟹ γ = 2 ln 2 / t_½

Amplitude decays at half the energy rate; set the envelope to ½ and solve for γ. B

γ = 2(0.693)/2.0 = 0.693 s⁻¹

Numeric energy decay rate. A

ω₀ = 2πf = 2π(512) = 3217 rad/s

Convert cyclic frequency to angular frequency. A

Q = ω₀/γ = 3217/0.693 ≈ 4.6 × 10³; Q/2π ≈ 7.4 × 10²

Quality factor and cycles per energy e-fold. A

γ ≈ 0.69 s⁻¹, Q ≈ 4.6 × 10³, ≈ 740 oscillations per e-fold

Reading. A clean audible ring of a few seconds already implies a Q of thousands — the fork completes hundreds of oscillations before the stored energy drops by a single factor of e.

Problems

An oscillator has ω₀ = 100 rad/s and Q = 250. Find γ, the energy e-fold time τ_E, and the amplitude e-fold time.
Solution
γ = ω₀/Q = 100/250 = 0.40 s⁻¹. Energy e-fold time τ_E = 1/γ = 2.5 s. Amplitude decays at γ/2 = 0.20 s⁻¹, so its e-fold time is 2/γ = 5.0 s (twice the energy time).
Show that the fraction of energy lost in one period is 1 − e^−γT, and that for light damping this is ≈ 2π/Q. Evaluate for Q = 50.
Solution
Over one period ⟨E⟩ goes from E to E e^−γT, so the fractional loss is ΔE/E = 1 − e^−γT. With T = 2π/ω_d ≈ 2π/ω₀ and γT ≪ 1, expand: 1 − e^−γT ≈ γT = γ·2π/ω₀ = 2π/Q. For Q = 50: ΔE/E ≈ 2π/50 = 0.126, about 13% per cycle.
A 0.50 kg mass on a spring of k = 200 N/m is observed to have Q = 40. Find the damping coefficient b.
Solution
ω₀ = √(k/m) = √(200/0.50) = √400 = 20 rad/s. From Q = ω₀/γ, γ = ω₀/Q = 20/40 = 0.50 s⁻¹. Then b = γ m = 0.50 × 0.50 = 0.25 kg/s (i.e. 0.25 N·s/m).
An RLC-like mode has Q = 1.0 × 10⁴ at f = 1.0 MHz. How many oscillations occur before the stored energy falls to 1/e, and what is τ_E in real time?
Solution
Cycles per energy e-fold: Q/2π = 10⁴/2π ≈ 1.6 × 10³ oscillations. ω₀ = 2π(10⁶) = 6.28 × 10⁶ rad/s, so γ = ω₀/Q = 6.28×10⁶/10⁴ = 628 s⁻¹ and τ_E = 1/γ = 1.6 × 10⁻³ s ≈ 1.6 ms. (Consistency: 1.6 ms × 10⁶ Hz ≈ 1.6×10³ cycles, matching Q/2π.)
A pendulum starts with E₀ = 2.0 J and γ = 0.10 s⁻¹. (a) When does the energy reach 0.10 J? (b) What is the amplitude then, as a fraction of the initial amplitude? (c) State one physical condition under which this exponential answer would be invalid.
Solution
(a) E(t) = E₀e^−γt = 0.10 ⟹ e^−0.10t = 0.10/2.0 = 0.05 ⟹ t = −ln(0.05)/0.10 = 3.00/0.10 = 30 s. (b) Amplitude ∝ √E, so A/A₀ = √(0.10/2.0) = √0.05 ≈ 0.22 (about 22%). (c) If the damping were dominated by dry (Coulomb) friction or turbulent (quadratic) drag, or if the amplitude grew large enough for the pendulum to be nonlinear, the decay would not be exponential and constant γ would fail; likewise if the pendulum were being driven, E would settle to a steady value rather than decay.