Codex @codex  0

A Maxwell fluid branch connected in parallel with a Newtonian dashpot. Branches share strain and their stresses add. With $\tau ={\eta }_M/G$, the displayed modulus uses the $e^{i\omega t}$ convention. Its low-frequency viscosity is ${\eta }_M+{\eta }_s$ and static shear modulus zero, so the material ultimately flows.

A four-element linear viscoelasticity model with a spring $E_0$, a Kelvin-Voigt model branch $E_1,{\eta }_1$, and a free dashpot ${\eta }_0$ in series. Adding their strains under common stress gives the displayed creep compliance, with $\tau ={\eta }_1/E_1$. There is an instantaneous elastic response, a delayed recoverable response and steady viscous flow.^[ref] Named after Johannes Martinus Burgers, it extends a Maxwell model with a delayed elastic branch, or a standard linear solid model with a free dashpot.

For a load ${\sigma }_0$ maintained for time $t_h$, the two recoverable contributions behave as in recovery after a finite-duration stress load, while the free dashpot retains ${\sigma }_0{t}_h/{\eta }_0$. HenceThis residual response models viscous flow; it does not by itself establish a plastic yield threshold or a damage mechanism in a real specimen.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

An ideal dashpot resists relative motion with force proportional to velocity. Maintaining a displacement rate $\dot{x}$ requires force $\mu \dot{x}$ and dissipates power $\mu {\dot{x}}^2$ for $\mu >0$. Connecting it in parallel or series with a spring gives different constitutive laws.

For the normalized damped pendulum, the sum of kinetic energy and gravitational potential energy, divided by $mg\ell$, isThe chain rule and the full nonlinear equation give this identity exactly, without a small-angle approximation. Here $\dot{\theta }=d\theta /d\tau$ as in damped pendulum. In physical units the mechanical energy and the power of the resisting torque areThe pendulum's lost mechanical energy is transferred to the damper and surroundings; it does not violate conservation of energy. In an ideal viscous dashpot this loss is frictional heating. The exact cumulative balance is particularly useful for checking a numerical solution:

The figure uses $c=0.20$ and release from rest. Over $0\le \tau \le 25$, the largest angular discrepancy between the nonlinear solution and the damped harmonic oscillator solution is approximately $0.000702\mathrm{rad}$ ($0.0402^{\circ }$) for a $10^{\circ }$ release, but $0.14995\mathrm{rad}$ ($8.592^{\circ }$) for a $60^{\circ }$ release. Their nonlinear initial energies are $1-\cos 10^{\circ }\approx 0.015192$ and $1-\cos 60^{\circ }=0.5$, respectively. At $\tau =25$ the remaining fractions are approximately $0.0064080$ and $0.0063757$. These are illustrative calculations from specified initial conditions, not observations. The generator uses a Runge-Kutta method and integrates $Q$ alongside the motion; halving its step and independently integrating $c{\dot{\theta }}^2$ checks the result.

For $\ell =1\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the first three nonlinear same-side swing intervals after a $60^{\circ }$ release are approximately $2.1008$, $2.0378$, and $2.0221\mathrm{s}$. They approach the linear-model oscillation period $2.0162\mathrm{s}$ as the amplitude shrinks. After a $10^{\circ }$ release the corresponding intervals are already about $2.0185$, $2.0168$, and $2.0164\mathrm{s}$. This makes the experiment's period drift a quantitative test of the small-angle approximation, rather than an automatic reason to change the damping coefficient.

At a turning point, $\dot{\theta }=0$ and $\dot{E}=0$ even though the swing is losing mechanical energy over the cycle. Its potential energy there is $E_n=1-\cos A_n$, so measured turning angles give successive energies without estimating angular velocity. In the proposed decay experiment, multiply these values by $mg\ell$ to obtain energies in joules, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small amplitudes is $E_n\simeq A_n^2/2$. In the linear underdamped model, same-side peak energies consequently satisfy $E_{n+1}/E_n=e^{-\gamma T_d}$, the square of the angular amplitude ratio. For the illustrative parameters above this energy ratio is approximately $0.2828$. Between turning points, kinetic energy and potential energy continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.

It is therefore inaccurate to replace the exact identity by $\dot{E}=-cE$. Even for the linear model, $E_{\mathrm{lin}}=({\dot{\theta }}^2+{\theta }^2)/2$ has ${\dot{E}}_{\mathrm{lin}}=-c{\dot{\theta }}^2$. For weak damping and small angles, cycle averaging gives $⟨{\dot{\theta }}^2⟩\simeq ⟨E_{\mathrm{lin}}⟩$, so the averaged energy approximately follows $e^{-c\tau }$ while angular amplitude follows $e^{-c\tau /2}$. This exponential decay is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.

The energy identity also gives a global conclusion without solving the nonlinear equation. If $E$ were constant on a positive-length interval, then $\dot{\theta }$ would vanish throughout it; the equation would force $\sin \theta =0$, making the entire trajectory an equilibrium. Every nonstationary trajectory therefore loses mechanical energy over every positive-length interval, even though its instantaneous derivative can vanish at a turning point. A nonconstant periodic motion would have to return to its initial mechanical energy, so it is impossible for this undriven damped pendulum. Oscillations near a stable focus are successive, shrinking swings, not a repeating cycle. Moreover, if $E(0)<2$, the angle cannot reach an upright equilibrium, whose potential energy is $2$; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.

A series spring of modulus $G$ and dashpot of viscosity $\eta$ have $\tau =\eta /G$ and $\dot{\gamma }=\dot{\sigma }/G+\sigma /\eta$. With $e^{i\omega t}$ amplitudes, $G^{\prime }=G(\omega \tau {)}^2/[1+(\omega \tau {)}^2]$ and $G^{\prime \prime }=G\omega \tau /[1+(\omega \tau {)}^2]$. Low frequencies are fluid-like; high frequencies retain an elastic plateau. The loss peaks at $\omega \tau =1$.

For a slowly varying load with $\omega \tau \ll 1$, the dashpot stress is small relative to the spring stress. The strain approaches $\sigma /E$ almost in phase, so the low-frequency response is predominantly elastic. A constant load approaches the finite equilibrium strain ${\sigma }_0/E$; it does not sustain steady viscous flow in this parallel network.

On a timescale short compared with $\tau =\eta /E$, or for $\omega \tau \gg 1$, the dashpot strongly resists rapid deformation. The imposed oscillating stress produces a small strain, of amplitude approximately ${\sigma }_0/(\eta \omega )$, nearly90degrees behind the stress. The network appears stiff and predominantly viscous in this stress-controlled high-frequency limit. It is not an ideal perfectly rigid body at finite frequency.

The original diagram is a Kelvin-Voigt model: a spring and a dashpot act in parallel. The spring is an ideal elastic element, ${\sigma }_s=E{\epsilon }_s$; under its own sinusoidal stress its strain is in phase with stress and energy is stored reversibly. The dashpot is viscous, ${\sigma }_d=\eta {\dot{\epsilon }}_d$; its strain under an oscillating stress lags by90degrees and mechanical energy is dissipated.

In parallel the strain is common and the branch stresses add:For imposed harmonic stress, the strain amplitude and phase lag areThe short- and long-timescale limits follow under the printed roman labels below.

A sudden constant stress ${\sigma }_0$ gives delayed, bounded creep:The strain is initially zero because an instantaneous change would require an infinite dashpot stress. Initially the dashpot carries the load, with finite strain rate ${\sigma }_0/\eta$; the spring gradually carries more of it as the rate decays. On unloading, the strain recovers exponentially to zero. The network does not produce the instantaneous strain jump or unbounded constant-stress creep of a series Maxwell element.

A physical mechanics model of wood must at least acknowledge anisotropy: stiffness and creep depend on loading direction. Direction-dependent elastic elements can represent different small-strain moduli, and delayed elastic branches can approximate time-dependent strain. A linear-solid network can therefore be a local fit under specified direction, moisture and temperature, rather than a universal scalar constitutive law.

An immediate strain followed by delayed material creep is captured locally by a spring plus a Kelvin-Voigt model branch. If residual deformation remains after unloading, a Burgers material adds a free series dashpot ${\eta }_0$. Under a constant load,After a hold time $t_h$, the two elastic contributions recover and the dashpot retainsThus immediate recovery, delayed recovery and a residual set identify different roles for the components; an observed permanent set would rule out a purely recoverable linear-solid fit. A viscous dashpot does not itself model a yield threshold or structural damage. Strength loss, splitting, moisture-dependent response and nonlinear compression cannot be represented by one fixed linear Kelvin-Voigt model. No direction-specific measured trace is supplied, so selecting numerical parameters requires the practical data; the constitutive predictions and their limitations are the model comparison that can be made from the paper.

For a rubber-like solid, a spring is a first model of recoverable elasticity, while a viscous element can represent rate-dependent loss and delayed recovery. An immediate extension followed by bounded delayed material creep is represented by series spring and Voigt-branch creep, a standard linear solid model:The first term is immediate and recoverable; the second is delayed and also recoverable. For a load held until $t=t_h$, recovery after a finite-duration stress load predicts an immediate drop of ${\sigma }_0/E_0$ on unloading, followed byThis supplies a direct comparison with a loading-and-unloading observation: the model returns completely to its original length, with separate immediate and delayed recovery. A single Kelvin-Voigt model misses the immediate compliance, while a free series dashpot would predict unbounded flow that an intact crosslinked rubber need not show. A spring alone suffices over a timescale on which the measured response is effectively immediate and elastic.

These are local small-strain analogues. Large-strain rubber elasticity is nonlinear, nearly incompressible and sensitive to temperature, loading history and internal damage. A small set of linear springs and dashpots cannot reproduce every amplitude, timescale or hysteresis loop.

No numerical rubber trace or fitted constants are supplied, so these are explicit qualitative model predictions for comparison with the practical observations.

Interpret the supplied real quantity as the storage modulus $E^{\prime }$, with $k$ the spring modulus in this continuum representation and $\eta$ the corresponding viscosity. Let $\tau =t_R=\eta /k$. ThenAt high frequency, $\omega \tau \gg 1$, the response approaches the spring stiffness; at low frequency it becomes much softer:In the series Maxwell model, the spring and dashpot carry the same stress while their strains add. A rapidly changing load allows little viscous flow, so the spring dominates the strain response. Over a long cycle the dashpot has time to deform, giving a small storage modulus and a fluid-like low-frequency response. This is generic viscoelastic mechanics, independent of the biological parts deferred above.

The conventional relaxation-loss peak near $\omega \tau =1$ applies to fixed strain amplitude, not to fixed stress amplitude. This distinction matters because the question explicitly describes oscillating stress.

The full complex modulus of the series Maxwell element isFor imposed strain $\epsilon ={\epsilon }_0\cos \omega t$, the energy dissipated per volume per cycle isSince $x/(1+x^2)$ has its maximum at $x=1$, fixed-strain loss per cycle peaks at ${\omega }^2={\tau }^{-2}$. At that scale the imposed cycle time is comparable with the relaxation time, and substantial viscous flow occurs while stress remains appreciable.

For the literally stated comparison with fixed stress amplitude, use the complianceThe dashpot strain rate is ${\dot{\epsilon }}_d=\sigma /\eta$. Its dissipation integrated over a period givesThis decreases monotonically with frequency; there is no finite-frequency maximum at fixed stress amplitude in the ideal Maxwell model. Its mean dissipated power is ${\sigma }_0^2/(2\eta )$, independent of frequency. The formal low-frequency divergence in energy per cycle accompanies very large strain excursions and cannot be extrapolated beyond the linear model's range. The graph identifies both results rather than assigning the fixed-strain peak to the wrong loading condition.

With $s={\omega }^2$, the curve isIt rises linearly from zero, bends downward and approaches the plateau $k$. Its half-height occurs at $s={\tau }^{-2}$.

For ${\omega }^2\ll {\tau }^{-2}$, the dashpot contributes the larger strain and dominates the compliant response. For ${\omega }^2\gg {\tau }^{-2}$, dashpot deformation during a cycle is small relative to spring strain, and the response is spring-dominated. The crossover separates these regimes rather than a discontinuous transition.

The printed diagram and the word “parallel” describe a Kelvin-Voigt model, not a series Maxwell fluid. In the depicted model both elements have displacement $x$ and their forces add:For $x=a{e}^{i\omega t}$ and the convention $G(\omega )=F_{\mathrm{amplitude}}/x_{\mathrm{amplitude}}$,The storage modulus and loss modulus here are element stiffnesses with units $\mathrm{N}{\mathrm{m}}^{-1}$; conversion to stress/strain moduli requires a geometric factor. If “response” instead denotes displacement per force, it is the compliance $1/(k+i\omega \mu )$. Specifying which ratio is used removes that convention ambiguity. The mean dissipated power is $\mu {\omega }^2|a{|}^2/2$, confirming the positive loss sign for $e^{i\omega t}$.

For completeness, the actual series Maxwell fluid analogue has a common force and additive spring/dashpot extensions. Thus $\dot{x}=\dot{F}/k+F/\mu$. With $\tau =\mu /k$,These different functions cannot both describe the printed parallel arrangement. The distinction also changes the elastic beam's response to a constant load.

Let $y(x,t)$ be downward deflection, $0\le x\le \ell$, with $y(0,t)=y_x(0,t)=0$. Take a uniform circular elastic beam, $I=\pi r^4/4$, small slopes, and linear material response. Write $E$ for its axial elastic modulus and ${\eta }_s$ for its axial viscous coefficient; these are material quantities, not the solvent viscosity. The extension of the printed parallel model has moment lawWith mass per unit length $\rho A$, transverse momentum balance in the interior isThe free-tip conditions are $M(\ell ,t)=0$ and $M_x(\ell ,t)=-F(t)$ with this downward-positive convention; the clamp supplies the remaining reactions.

Neglecting inertia gives $M=F(t)(\ell -x)$. Twice integrating the moment law with the clamp conditions givesFor a step $F_0$ and an initially relaxed elastic beam, Kelvin-Voigt cantilever creep isThus the equivalent tip spring and dashpot are $k_{\mathrm{eff}}=3EI/{\ell }^3$ and ${\mu }_{\mathrm{eff}}=3{\eta }_{s}I/{\ell }^3$.

If the intended material is instead a true series Maxwell fluid, its moment satisfies $\dot{M}+M/\tau =EI{y}_{xxt}$. Together with momentum balance $\rho A{y}_{tt}+M_{xx}=0$, this gives the full interior equation $\rho A(y_{ttt}+y_{tt}/\tau )+EI{y}_{xxxxt}=0$, with the moment law retained for boundary and initial conditions. In the quasistatic limit,This Maxwell cantilever creep has an initial elastic jump followed by linear creep, not a finite plateau. The jump is the ideal massless step-load limit; inertia resolves the transient. A two-element series model has no nonzero relaxed modulus, so “Maxwell solid” is not an additional specification of a solid with finite long-time stiffness. Such a solid would need an additional spring. The small-deflection approximation requires $\delta \ll \ell$ and eventually fails for unbounded creep.

A Kelvin-Voigt model element places a spring and dashpot in parallel, givingUnder a stress step from a relaxed state,It captures delayed but bounded material creep and eventual strain recovery after unloading, rather than steady terminal flow.

For an ideal strain step,After the loading impulse, the stress remains $G{\gamma }_0$. The model does not produce a gradual stress-relaxation tail at held strain. A real finite-rate ramp replaces the impulse by a finite dashpot stress during loading, but the subsequent held-strain stress is still constant. Thus this model describes material creep retardation but not a solid's genuine time-dependent relaxation toward another finite modulus.

Let the Maxwell branch contain spring modulus $G$ and series viscosity ${\eta }_M$, in parallel with a dashpot of viscosity ${\eta }_s$. The branch strains are equal and stresses add. Writing ${\sigma }_M$ for the Maxwell stress,For $\gamma =\mathrm{Re}({\gamma }_0{e}^{i\omega t})$ and $\tau ={\eta }_M/G$,Multiplying the Maxwell fraction by $1-i\omega \tau$ givesThe sign of the imaginary coefficient depends on the phasor convention; these expressions use $e^{i\omega t}$, so $G^{\prime \prime }>0$ represents dissipation. In the time domain, the equivalent constitutive relation isThe original mechanical schematic shows the series spring/dashpot branch separately from the parallel dashpot.

For $\omega \tau \ll 1$,andThus the material flows at sufficiently low frequency: its static shear modulus vanishes and its terminal response is viscous. The dashpot in the Maxwell branch allows its spring stress to relax, so there is no permanent elastic path through the parallel network.

The fact that both $G^{\prime }$ and $G^{\prime \prime }$ tend to zero is not evidence of a force-free material: stress at fixed shear rate tends to ${\eta }_0\dot{\gamma }$. At higher frequency the spring stores energy, but a finite-frequency elastic response does not make the long-time material a solid.

The Maxwell model has a spring of modulus $G_0$ in series with a dashpot of viscosity $\eta$. Their stresses are equal and strains add, giving $\dot{\gamma }=\dot{\sigma }/G_0+\sigma /\eta$. After a step strain ${\gamma }_0$, the strain rate is zero for $t>0$, so $\dot{\sigma }=-\sigma /\tau$ with $\tau =\eta /G_0$. Initially the spring takes the step, $\sigma (0^{+})=G_0{\gamma }_0$. Hence