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 obeying \( \ddot{x} + \gamma\dot{x} + \omega_0^2 x = 0 \) with \( \gamma \ll \omega_0 \), the total mechanical energy averaged over one cycle decays exponentially, \( \langle E \rangle(t) = E_0\, e^{-\gamma t} \), so the energy e-folding time is \( \tau_E = 1/\gamma \). The dimensionless quality factor \( Q \equiv \omega_0/\gamma \) equals \( 2\pi \) times the energy stored divided by the energy dissipated per cycle, and counts the number of radians of oscillation phase accumulated while the stored energy falls by one factor of \( e \).

Why it matters

The quality factor is the single number that characterises how "good" a resonator is, from a pendulum to a quartz crystal to the LIGO test masses to an atomic transition. It fixes the sharpness of the resonance peak, the ringdown time after the drive is removed, and the fundamental phase-noise floor of clocks and oscillators, so it recurs identically across mechanics, circuits, optics and quantum systems.

Working in terms of energy rather than amplitude is the natural language: energy is the conserved quantity of the undamped problem (from shm-energy-conservation), and turning damping on converts that conservation law into a controlled leak. The rate of that leak, \( \gamma \), and the phase clock, \( \omega_0 \), combine into the one ratio \( Q \) that survives when we non-dimensionalise the whole problem.

Assumptions

Linear viscous damping.The dissipative force is \( -b\dot{x} = -m\gamma\dot{x} \), strictly proportional to velocity; if dropped (dry/Coulomb friction, quadratic drag, hysteretic loss) the decay is no longer a single exponential and \( Q \) becomes amplitude-dependent.

Light damping, \( \gamma \ll \omega_0 \).Guarantees the system is underdamped and oscillatory with \( \omega_d = \sqrt{\omega_0^2-\gamma^2/4}\approx\omega_0 \); if dropped the motion is over- or critically damped, no cycles exist, and "energy per cycle" is undefined.

Cycle-averaging is meaningful.The amplitude changes negligibly over one period, \( \gamma T \ll 1 \), so averaging kinetic and potential energy over a cycle is well defined; if dropped the average is over an interval during which the envelope itself moves appreciably and the clean \( e^{-\gamma t} \) separation of envelope from oscillation breaks.

Constant coefficients.\( m \), \( k=m\omega_0^2 \) and \( b \) are time-independent; if dropped (aging, temperature drift, nonlinear stiffness) \( \omega_0 \) and \( \gamma \) drift and the constant-rate exponential is replaced by \( \exp\!\big(-\int\gamma\,dt\big) \).

Derivation

\[ \ddot{x} + \gamma\dot{x} + \omega_0^2 x = 0, \qquad \gamma=\frac{b}{m},\ \ \omega_0^2=\frac{k}{m} \]

Starting point: the equation of the damped harmonic oscillator (prior result damped-harmonic-oscillator), written in standard normalised form. A

\[ x(t) = A\,e^{-\gamma t/2}\cos(\omega_d t + \varphi), \qquad \omega_d=\sqrt{\omega_0^2-\tfrac{\gamma^2}{4}} \]

Underdamped solution of the linear constant-coefficient ODE, quoted from damped-harmonic-oscillator. The envelope decays at rate \( \gamma/2 \) in amplitude. A

\[ E(t) = \tfrac{1}{2}m\dot{x}^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}m\dot{x}^2 + \tfrac{1}{2}m\omega_0^2 x^2 \]

Definition of total mechanical energy (kinetic plus potential), using \( k=m\omega_0^2 \). This is the quantity conserved when \( \gamma=0 \) (from shm-energy-conservation). A

\[ \dot{x} = -A\,e^{-\gamma t/2}\!\left[\tfrac{\gamma}{2}\cos(\omega_d t+\varphi) + \omega_d\sin(\omega_d t+\varphi)\right] \]

Differentiate the step-2 envelope-times-cosine by the product rule, one operation. A

\[ \dot{x} \simeq -A\,\omega_0\,e^{-\gamma t/2}\sin(\omega_0 t+\varphi) \]

Light-damping approximation: \( \tfrac{\gamma}{2}\ll\omega_d \) drops the first bracketed term, and \( \omega_d\to\omega_0 \). Legal because we assumed \( \gamma\ll\omega_0 \). B

\[ E(t) \simeq \tfrac{1}{2}mA^2\omega_0^2 e^{-\gamma t}\Big[\sin^2(\omega_0 t+\varphi) + \cos^2(\omega_0 t+\varphi)\Big] \]

Substitute step 5 and step 2 into step 3. Kinetic and potential terms each carry the same prefactor \( \tfrac{1}{2}mA^2\omega_0^2 e^{-\gamma t} \) at this order. B

\[ E(t) \simeq \tfrac{1}{2}mA^2\omega_0^2\, e^{-\gamma t} \equiv E_0\, e^{-\gamma t} \]

Apply the Pythagorean identity \( \sin^2+\cos^2=1 \). The oscillatory ripple cancels exactly at leading order, leaving a pure exponential envelope with \( E_0=\tfrac{1}{2}mA^2\omega_0^2 \). A

\[ \frac{dE}{dt} = -\,b\,\dot{x}^2 = -m\gamma\dot{x}^2 \le 0 \]

Independent check via the work–energy theorem: differentiating \( E \) and using the equation of motion, the only surviving term is the power removed by the damping force. This confirms energy is strictly non-increasing and is the rigorous origin of the leak. C

\[ \left\langle\frac{dE}{dt}\right\rangle = -m\gamma\langle\dot{x}^2\rangle = -m\gamma\cdot\frac{E}{m} = -\gamma E \;\Rightarrow\; \langle E\rangle = E_0 e^{-\gamma t} \]

Cycle-average step 8 using virial equipartition \( \langle \tfrac12 m\dot{x}^2\rangle=\tfrac12 E \), i.e. \( m\langle\dot x^2\rangle=E \). This reproduces step 7 from the dissipation side and closes the loop: the two routes agree. C

\[ Q \equiv \frac{\omega_0}{\gamma} = 2\pi\,\frac{E}{\Delta E_{\text{cycle}}}, \qquad \Delta E_{\text{cycle}} = E\big(1-e^{-\gamma T}\big)\approx E\gamma T \]

Define \( Q \) and evaluate energy lost per period \( T=2\pi/\omega_0 \). Since \( \gamma T=2\pi\gamma/\omega_0\ll1 \), \( 1-e^{-\gamma T}\approx\gamma T \), so \( E/\Delta E_{\text{cycle}}=1/(\gamma T)=\omega_0/(2\pi\gamma) \), giving \( 2\pi E/\Delta E=\omega_0/\gamma=Q \). B

Result

\[ \langle E\rangle(t)=E_0\,e^{-\gamma t}, \qquad Q=\frac{\omega_0}{\gamma}=2\pi\,\frac{E_{\text{stored}}}{E_{\text{lost per cycle}}} \]

Reading. The stored energy falls by a factor \( e \) every time \( t=1/\gamma \), i.e. after \( Q/(2\pi) \) full oscillations, or equivalently after the phase \( \omega_0 t \) advances by \( Q \) radians. A high-\( Q \) resonator therefore rings for many cycles: \( Q \) literally counts radians of oscillation per energy e-fold. Amplitude decays half as fast as energy (rate \( \gamma/2 \)), so the amplitude e-folding time is \( 2/\gamma \) and amplitude falls by \( e \) in \( Q/\pi \) cycles.

Units check. \( \gamma=b/m \) has units \( (\text{kg·s}^{-1})/\text{kg}=\text{s}^{-1} \), so \( \gamma t \) is dimensionless and \( e^{-\gamma t} \) is a pure number — good. \( Q=\omega_0/\gamma=(\text{rad·s}^{-1})/(\text{s}^{-1}) \) is dimensionless, as required of a figure of merit. \( E_0=\tfrac12 mA^2\omega_0^2 \) has units \( \text{kg·m}^2\text{·s}^{-2}=\text{J} \). Consistent.

Limiting cases

\( \gamma\to0 \) (no damping): \( Q\to\infty \), \( \tau_E\to\infty \); \( E\to E_0 \) constant, recovering exact energy conservation of SHM.
\( \gamma\to2\omega_0 \) (critical damping): \( \omega_d\to0 \), \( Q\to\tfrac12 \); the "cycle" and cycle-averaged-energy picture breaks down — no oscillation to average over.
Very high \( Q \) (e.g. atomic clock, \( Q\sim10^{15} \)): \( \gamma\ll\omega_0 \) is superbly satisfied; the exponential is essentially exact over enormous numbers of cycles and the ripple correction is negligible.
Short times, \( t\ll1/\gamma \): \( e^{-\gamma t}\approx1-\gamma t \); energy decays linearly at first with slope \( -\gamma E_0 \), matching the instantaneous dissipation rate.
One half-life of energy: \( E=E_0/2 \) at \( t=\ln2/\gamma\approx0.693/\gamma \), i.e. after \( \approx0.11\,Q \) oscillations.

Breaks when

Overdamped/critical regime \( \gamma\gtrsim2\omega_0 \): the motion is a sum of real exponentials with no oscillation, so \( \omega_d \) is imaginary or zero, "energy per cycle" is undefined, and \( Q=\omega_0/\gamma \) no longer measures ringdown in cycles.
Nonlinear damping (quadratic drag \( \propto\dot{x}^2 \), dry friction \( \propto\operatorname{sgn}\dot{x} \)): the loss per cycle is not proportional to \( E \), so the decay is not exponential — dry friction gives a linear-in-time amplitude decay that stops abruptly, and \( Q \) becomes amplitude-dependent.
Strong drive / large amplitude (anharmonic potential): \( \omega_0 \) becomes amplitude-dependent, energy is exchanged with higher harmonics, and the clean envelope/oscillation factorisation \( E=E_0 e^{-\gamma t} \) fails.
Time-varying parameters: if \( \gamma(t) \) drifts, the constant-rate exponential is replaced by \( \exp\!\big(-\int_0^t\gamma\,dt'\big) \) and a single \( Q \) is ill-defined.

Failure modes

Confusing the two rates: writing energy decay as \( e^{-\gamma t/2} \). Amplitude decays at \( \gamma/2 \); energy \( \propto \) amplitude\( ^2 \) decays at \( \gamma \). Squaring \( e^{-\gamma t/2} \) gives \( e^{-\gamma t} \).
Using \( Q=\omega_0/(2\gamma) \): mixing conventions. With the EOM written as \( \ddot x+\gamma\dot x+\omega_0^2x=0 \), the standard result is \( Q=\omega_0/\gamma \). Errors come from an inconsistent definition of \( \gamma \) versus \( b \), or from confusing amplitude and energy rates.
Forgetting the \( 2\pi \): stating \( Q=E/\Delta E_{\text{cycle}} \). The correct energy definition carries a factor \( 2\pi \): \( Q=2\pi E/\Delta E_{\text{cycle}} \).
Cycle-averaging too early: replacing \( \sin^2\to\tfrac12 \) before combining kinetic and potential terms, then wrongly concluding energy ripples at \( 2\omega_0 \) at leading order. The ripple cancels once both terms are summed.
Treating \( Q \) as the number of cycles to a full stop: the oscillation never fully stops in finite time; \( Q \) counts radians per e-fold, not cycles to rest.
Applying \( \omega_d\approx\omega_0 \) when \( Q\sim1 \): the light-damping approximations used in steps 5–7 are invalid near critical damping.

Discussion

The deep content of this result is a change of description. The undamped oscillator has one conserved scalar, its energy; damping does not destroy that scalar but makes it leak at a rate set entirely by \( \gamma \). Because the loss mechanism (viscous force \( \propto\dot{x} \)) removes power \( \propto\dot{x}^2\propto E \), the leak is proportional to what remains, and any quantity whose loss rate is proportional to itself decays exponentially. The exponential is thus not an accident of the sinusoid but a direct consequence of the loss being linear and the energy being quadratic in the state.

The factorisation \( x=(\text{envelope})\times(\text{oscillation}) \) reflects a separation of timescales: the fast clock \( \omega_0 \) and the slow envelope \( \gamma/2 \). Their ratio \( Q=\omega_0/\gamma \) is the only dimensionless group in the problem, which is why every damped resonator — mechanical, electrical (\( Q=\tfrac{1}{R}\sqrt{L/C} \)), optical (cavity finesse), or quantum (linewidth \( \Delta\omega=\gamma \)) — is fully characterised by \( Q \) once amplitude and phase are scaled out. The frequency-domain face of the same number is the resonance: a driven oscillator has a Lorentzian response of fractional width \( \Delta\omega/\omega_0=1/Q \) (power FWHM \( =\gamma \)), so the ringdown time \( 1/\gamma \) and the linewidth \( \gamma \) are Fourier conjugates — a time–bandwidth statement.

Physically \( Q \) is a memory: it is (up to \( 2\pi \)) the number of oscillations the system performs before it forgets its initial energy. This is why high \( Q \) is prized in clocks (a quartz watch crystal has \( Q\sim10^{4}\text{–}10^{5} \), a hydrogen maser \( Q\sim10^{9} \), an optical clock transition \( Q\gtrsim10^{15} \)): a narrow line and a long coherence are the same property viewed in frequency or in time. The LIGO mirrors are suspended with \( Q\sim10^{8} \) precisely so thermal noise stays confined to a narrow band away from the measurement.

At the rigorous level, the "energy" that decays cleanly as \( e^{-\gamma t} \) is the cycle-averaged energy; the instantaneous energy also carries a small ripple at \( 2\omega_d \) with relative amplitude \( \mathcal{O}(\gamma/\omega_0)=\mathcal{O}(1/Q) \), from the cross term dropped in step 5. Systematically, one can pass to the complex amplitude \( a=\sqrt{m\omega_0/2}\,\big(x+i\dot{x}/\omega_0\big) \), whereupon the EOM becomes \( \dot{a}=\big(-i\omega_0-\gamma/2\big)a \) and \( |a|^2=E \) obeys \( \frac{d}{dt}|a|^2=-\gamma|a|^2 \) exactly at this order — the same slowly-varying-envelope object that reappears as the field amplitude in laser physics and as the annihilation operator whose expectation decays at \( \gamma/2 \) in the quantum-optical master equation, with \( Q \) setting the number of coherent cavity cycles before decoherence.

Common misconceptions. \( Q \) is not a measure of how much energy the oscillator holds (that is \( E_0 \)); it is a rate ratio. A more massive or more energetic oscillator with the same \( \gamma \) and \( \omega_0 \) has the identical \( Q \). Nor does "high \( Q \)" mean "small friction force" in absolute terms — a very stiff, fast resonator can have large absolute damping yet enormous \( Q \) because \( \omega_0 \) is large.

Worked examples

Tuning fork: \( f_0=440\ \text{Hz} \), and its amplitude falls to \( 1/e \) of its initial value after \( t^{\text{amp}}_{1/e}=3.0\ \text{s} \). Find \( \gamma \), \( Q \), the energy half-life, and the number of oscillations per energy e-fold.

Identify what is given: amplitude e-folding, not energy. Amplitude decays at \( \gamma/2 \), so \( t^{\text{amp}}_{1/e}=2/\gamma \). A

\[ \gamma=\frac{2}{t^{\text{amp}}_{1/e}}=\frac{2}{3.0\ \text{s}}=0.667\ \text{s}^{-1} \]

Solve for the energy decay constant from the amplitude e-folding time. A

\[ \omega_0=2\pi f_0=2\pi(440)=2765\ \text{rad·s}^{-1},\qquad Q=\frac{\omega_0}{\gamma}=\frac{2765}{0.667}=4.15\times10^{3} \]

Convert frequency to angular frequency and form the ratio. A

\[ t^{E}_{1/2}=\frac{\ln2}{\gamma}=\frac{0.693}{0.667}=1.04\ \text{s};\qquad N_{e\text{-fold}}=\frac{Q}{2\pi}=\frac{4150}{2\pi}=6.6\times10^{2}\ \text{cycles} \]

Energy half-life from \( E=E_0/2 \); cycles per energy e-fold from the reading of \( Q \). B

\[ \gamma=0.67\ \text{s}^{-1},\quad Q\approx4.2\times10^{3},\quad t^{E}_{1/2}\approx1.0\ \text{s},\quad \approx660\ \text{cycles per energy }e\text{-fold} \]

Reading. A modest, easily heard 3-second ring already corresponds to \( Q\sim4000 \) — audible-length rings imply high \( Q \) because \( \omega_0 \) is large. Since \( \gamma/\omega_0\sim2\times10^{-4}\ll1 \), light damping is amply justified.

Units check. \( \gamma \) in \( \text{s}^{-1} \), \( \omega_0 \) in \( \text{rad·s}^{-1} \), ratio dimensionless. Times in seconds. Consistent.

RLC circuit: \( L=10\ \text{mH} \), \( C=100\ \text{nF} \), \( R=5.0\ \Omega \). Find \( \omega_0 \), \( Q \), and the time for the stored energy to fall to \( 1\% \) of its initial value.

Map the circuit to the canonical oscillator: charge \( q \) plays the role of \( x \), with \( \ddot q+\tfrac{R}{L}\dot q+\tfrac{1}{LC}q=0 \), so \( \omega_0^2=1/(LC) \), \( \gamma=R/L \). A

\[ \omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(10^{-2})(10^{-7})}}=\frac{1}{\sqrt{10^{-9}}}=3.16\times10^{4}\ \text{rad·s}^{-1} \]

Compute the natural angular frequency; \( LC=10^{-9}\ \text{s}^2 \). A

\[ \gamma=\frac{R}{L}=\frac{5.0}{10^{-2}}=500\ \text{s}^{-1},\qquad Q=\frac{\omega_0}{\gamma}=\frac{3.16\times10^{4}}{500}=63.2 \]

Damping constant and quality factor. Equivalently \( Q=\tfrac1R\sqrt{L/C}=\tfrac15\sqrt{10^{-2}/10^{-7}}=\tfrac15\sqrt{10^{5}}=63.2 \), confirming. B

\[ E=E_0 e^{-\gamma t}=0.01\,E_0 \;\Rightarrow\; \gamma t=\ln(100)=4.605 \;\Rightarrow\; t=\frac{4.605}{500}=9.2\ \text{ms} \]

Invert the exponential energy decay for the \( 1\% \) threshold. A

\[ \omega_0\approx3.2\times10^{4}\ \text{rad·s}^{-1},\quad Q\approx63,\quad t_{1\%}\approx9.2\ \text{ms} \]

Reading. In 9.2 ms the circuit completes \( \omega_0 t/2\pi\approx46 \) oscillations while losing \( 99\% \) of its energy — consistent with \( Q/(2\pi)\approx10 \) cycles per e-fold and \( \ln100\approx4.6 \) e-folds. Light damping holds since \( \gamma/\omega_0=1/Q\approx0.016\ll1 \).

Units check. \( \sqrt{LC} \) has units \( \sqrt{\text{H·F}}=\sqrt{\text{s}^2}=\text{s} \), so \( \omega_0 \) is \( \text{s}^{-1} \). \( R/L \) is \( \Omega/\text{H}=\text{s}^{-1} \). \( \tfrac1R\sqrt{L/C} \) is \( \tfrac{1}{\Omega}\sqrt{\text{H/F}}=\tfrac{1}{\Omega}\cdot\Omega \), dimensionless. Consistent.

Problems

A pendulum of natural frequency \( f_0=0.50\ \text{Hz} \) has \( Q=200 \). Find the damping constant \( \gamma \) and the time for its energy to fall to \( 1/e \).
Solution
\( \omega_0=2\pi f_0=\pi=3.14\ \text{rad·s}^{-1} \). \( \gamma=\omega_0/Q=3.14/200=0.0157\ \text{s}^{-1} \). Energy e-folding time \( \tau_E=1/\gamma=63.7\ \text{s} \). (Equivalently \( \tau_E=Q/\omega_0=200/3.14=63.7\ \text{s} \), or \( Q/(2\pi)\approx32 \) cycles, \( =32/0.5=64\ \text{s} \).)
The amplitude of a damped oscillator drops to half its initial value after 10 complete cycles. Estimate \( Q \).
Solution
Amplitude decays as \( e^{-\gamma t/2} \). After \( n=10 \) cycles, \( t=nT=n\cdot2\pi/\omega_0 \). Half-amplitude: \( \gamma t/2=\ln2 \Rightarrow \gamma t=2\ln2=1.386 \). So \( \gamma\,(20\pi/\omega_0)=1.386 \Rightarrow \gamma/\omega_0=1.386/(20\pi)=0.02206 \). Thus \( Q=\omega_0/\gamma=1/0.02206=45.3 \). (Rule of thumb: amplitude halves in \( \approx0.11Q \) cycles; \( 0.11Q=10\Rightarrow Q\approx45 \).)
An RLC circuit has \( L=2.0\ \text{mH} \), \( C=50\ \text{nF} \) and quality factor \( Q=80 \). Find the resistance \( R \) and the resonant frequency \( f_0 \).
Solution
\( \omega_0=1/\sqrt{LC}=1/\sqrt{(2\times10^{-3})(5\times10^{-8})}=1/\sqrt{10^{-10}}=1.0\times10^{5}\ \text{rad·s}^{-1} \), so \( f_0=\omega_0/2\pi=1.59\times10^{4}\ \text{Hz}=15.9\ \text{kHz} \). From \( Q=\tfrac1R\sqrt{L/C} \): \( \sqrt{L/C}=\sqrt{(2\times10^{-3})/(5\times10^{-8})}=\sqrt{4\times10^{4}}=200\ \Omega \). So \( R=200/Q=200/80=2.5\ \Omega \).
A quartz crystal resonator operates at \( f_0=32.768\ \text{kHz} \) with \( Q=1.0\times10^{5} \). How long does the stored energy take to decay to \( 1/e \), and how many oscillations occur in that time?
Solution
\( \omega_0=2\pi f_0=2\pi(32768)=2.059\times10^{5}\ \text{rad·s}^{-1} \). \( \gamma=\omega_0/Q=2.059\times10^{5}/10^{5}=2.059\ \text{s}^{-1} \). Energy e-folding time \( \tau_E=1/\gamma=0.486\ \text{s} \). Cycles in that time \( =f_0\tau_E=32768\times0.486=1.59\times10^{4} \), which equals \( Q/(2\pi)=10^5/6.283=1.59\times10^{4} \). Consistent.
A mass–spring system has \( m=0.20\ \text{kg} \), \( k=80\ \text{N·m}^{-1} \), and damping coefficient \( b=0.10\ \text{kg·s}^{-1} \). Its initial amplitude is \( A=5.0\ \text{cm} \). Find (a) \( \omega_0 \), (b) \( Q \), (c) the initial stored energy, and (d) the average power dissipated at \( t=0 \).
Solution
(a) \( \omega_0=\sqrt{k/m}=\sqrt{80/0.20}=\sqrt{400}=20\ \text{rad·s}^{-1} \). (b) \( \gamma=b/m=0.10/0.20=0.50\ \text{s}^{-1} \); \( Q=\omega_0/\gamma=20/0.50=40 \). (c) \( E_0=\tfrac12 kA^2=\tfrac12(80)(0.050)^2=\tfrac12(80)(2.5\times10^{-3})=0.10\ \text{J} \) (equivalently \( \tfrac12 mA^2\omega_0^2=\tfrac12(0.20)(0.0025)(400)=0.10\ \text{J} \)). (d) Average dissipated power at \( t=0 \): \( \langle P\rangle=\gamma E_0=0.50\times0.10=0.050\ \text{W}=50\ \text{mW} \). Light damping holds since \( \gamma/\omega_0=1/Q=0.025\ll1 \).