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An air track supports a cart on a cushion of air to reduce contact friction. It provides a nearly one-dimensional platform for a damped harmonic oscillator; residual drag remains part of the baseline damping.

antidamping is a velocity-proportional term that supplies rather than removes mechanical energy. For the displayed damped harmonic oscillator with negative damping coefficient, $d[({\dot{x}}^2+{\omega }_0^2{x}^2)/2]/dt=-2\lambda {\dot{x}}^2\ge 0$. causality of a driven solution and its long-time stability are different properties.

Adding a velocity-proportional force $-b\dot{y}$ produces a damped harmonic oscillator. The fastest non-oscillatory release from rest is critical damping, with the repeated-root solution in the title. Reaching fraction $\epsilon$ of the initial displacement requires $(1+s)e^{-s}=\epsilon$, $s={\omega }_{0}t$. At one percent, $s=6.638352\dots$, rather than $\log 100$. Exact arrival at equilibrium takes infinite time; the tolerance reference must be stated.

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.

A damped torsional oscillation is rotation about a fiber's fixed reference orientation with a restoring torque and dissipative torque. In the ideal viscous model, damping contributes $-b\dot{\theta }$ with constant $b\ge 0$. This is a rotational damped harmonic oscillator. For known constant moment of inertia $I$ and torsion constant $\kappa$, define $\gamma =b/(2I)$. In the oscillatory regime ${\gamma }^2<\kappa /I$,$A$ and $\phi$ are set by the initial state. Successive positive peaks measured from the same equilibrium have magnitudes $p_n$ and $p_{n+1}$; their ratio gives $\delta =\ln (p_n/p_{n+1})=\gamma T_d$. Consequently a measured damped oscillation period and decay together recoverUsing only $4{\pi }^{2}I/T_d^2$ underestimates the torsion constant in this model. The toy curve in torsion constant sets $I=2.0\times 10^{-6}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, and $\gamma =0.020{\mathrm{s}}^{-1}$. It gives $T_0\simeq 88.86\mathrm{s}$, $T_d\simeq 92.64\mathrm{s}$, and $\delta \simeq 1.853$. Ignoring the decay recovers $9.2\times 10^{-9}\mathrm{N}m/rad$, an $8\%$ underestimate; including it recovers the stipulated torsion constant.

These equations assume a single rotational mode, a stationary zero, linear elasticity, and viscous damping. Internal fiber losses need not behave as a constant viscous coefficient: Fleischer and colleagues model their precision apparatus using a loss angle. Their response is evidence for specifying a damping model, not for applying the viscous correction universally. ^[ref]

A wave equation with a linear damping term and positive wave speed $c$ separates into spatial normal modes and damped harmonic oscillators. For a spatial eigenvalue $k^2$, the time factor obeys $T^{\prime \prime }+\alpha T^{\prime }+c^2{k}^{2}T=0$. Its characteristic roots are $-\alpha /2\pm \sqrt{{\alpha }^2/4-c^2{k}^2}$. Thus underdamping requires $|\alpha |<2c|k|$, critical damping gives a repeated root and the second solution $t{e}^{-\alpha t/2}$, and overdamping gives two real exponential rates. Negative $\alpha$ amplifies rather than damps the motion.

Damping dissipates the energy of an oscillation, reducing its amplitude. A linearly damped string applies this mechanism to each normal mode.

Why can stronger damping make a return to equilibrium slower? A weak brake lets a spring exchange potential energy with kinetic energy many times; a very strong brake makes the final motion creep. Critical damping marks the boundary between these behaviours. The distinction depends on the force law: the classification below concerns a single damped harmonic oscillator with viscous damping, rather than every mechanism that removes energy.

Let $x(t)$ be displacement from equilibrium, $m>0$ the moving mass, $k>0$ the spring constant, and $b\ge 0$ the constant coefficient of viscous damping. With no driving force, Newton's second law givesThe two initial conditions specify displacement and velocity. Define the undamped eigenfrequency as an angular frequency ${\omega }_0=\sqrt{k/m}$ and the damping ratio as $\zeta =b/(2\sqrt{mk})$. The ordinary differential equation becomes $\ddot{x}+2\zeta {\omega }_0\dot{x}+{\omega }_0^{2}x=0$. A constant weight in a vertical spring merely shifts equilibrium; it is already absorbed in the definition of $x$.^[ref]

For example, the ideal parameters $m=1\mathrm{kg}$ and $k=4\mathrm{N}{\mathrm{m}}^{-1}$ give ${\omega }_0=2{\mathrm{s}}^{-1}$ and critical damping at $b=4\mathrm{kg}{\mathrm{s}}^{-1}$. Values $b=1,4,8\mathrm{kg}{\mathrm{s}}^{-1}$ give damping ratios $0.25,1,2$, respectively. These are model choices, not measured values. For release from rest with $x_0>0$, set $X=x/x_0$ and $\tau ={\omega }_{0}t$; then $X^{\prime \prime }+2\zeta X^{\prime }+X=0$, where primes now denote derivatives with respect to $\tau$.

The left panel uses identical initial conditions $X(0)=1$, $X^{\prime }(0)=0$. Underdamping produces repeated crossings of equilibrium; critical damping and overdamping give positive decreasing responses. The right panel keeps critical damping but changes initial velocity: $X^{\prime }(0)=-2$ gives one crossing, although there is no oscillation. Both panels are exact model predictions.

For a damped harmonic oscillator $\ddot{y}+\gamma \dot{y}+{\omega }_0^{2}y=0$, differentiation of the mechanical energy $E=m({\dot{y}}^2+{\omega }_0^2{y}^2)/2$ gives $\dot{E}=-m\gamma {\dot{y}}^2$. This identity is exact. In the weak-damping regime $\gamma \ll {\omega }_0$, the leading period average has equal kinetic energy and potential energy, so $\overline{{\dot{y}}^2}\simeq \overline{E}/m$ and $\dot{\overline{E}}\simeq -\gamma \overline{E}$. Thus the averaged energy has exponential decay, while the instantaneous energy has a small oscillatory correction. Its derivative vanishes at every turning point, ruling out an exact pointwise nonconstant exponential. The quality factor is $Q\simeq {\omega }_0/\gamma$ in this approximation.

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 small-angle galvanometer model combines its moment of inertia $J_g$, damping coefficient $b_g$ and restoring coefficient ${\kappa }_g$ with a torque per electric current $K_g$. The resulting damped harmonic oscillator responds to the current pulse through the displayed equation. Its angular deflection $\theta$ can continue after the driving electric current has ceased. Inferring the current's detailed time dependence requires the instrument response, rather than equating needle angle with instantaneous current. This is a generic readout model, not a calibration of Faraday's historical instrument.

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

When $k^{(p)}\gg k_J^{(p)}$, omit the gravitational term and define $\omega (t)=c_s(t)k/a(t)$. ThenThis is an acoustic damped harmonic oscillator with time-dependent coefficients. Freezing the background locally at $t_0$, its amplitude damping rate is $H_0$, soThe oscillation-averaged quadratic oscillator energy decays twice as fast in this frozen-frequency model, with time $1/(2H_0)$. Stating which quantity is damped avoids a factor-of-two convention error. Locally the solution isFor evolution over many expansion times, retain the varying frequency. Put ${\delta }_{\mathbf{k}}=\chi /a$. Direct substitution givesWhen $\omega \gg H$ and $|\dot{\omega }|\ll {\omega }^2$, the WKB approximation yieldsThese are rapidly oscillating sound waves in comoving density, diluted by the expanding background. For constant $c_s$ their envelope scales as $a^{-1/2}$, not universally as $a^{-1}$, because $\omega \propto a^{-1}$. In the Einstein-de Sitter universe, introducing conformal time $d\eta =dt/a$ gives the exact constant-$c_s$ pressure-only solution ${\delta }_{\mathbf{k}}=[C\sin (k{c}_s\eta )+D\cos (k{c}_s\eta )]/\eta$, consistent with $a\propto {\eta }^2$. For that background the frozen-coefficient damping time is $3t/2$, while the slowly varying envelope has instantaneous decay time $2/H$.

After the force is removed, the damped harmonic oscillator evolves freely. There is no impulsive force at zero, so both displacement and velocity are continuous. Matching the values in part (i) givesEquivalently,This is bounded in positive time, completing the unique globally bounded solution. The formulas also include $\omega =0$, for which the forcing and solution both vanish.

Dividing Newton's equation by the mass gives the stated model: $\ddot{z}$ is the acceleration, $\gamma \dot{z}$ represents viscous damping, $(k/m)z$ is the spring's restoring acceleration, and the right side is the applied driving force per unit mass. The complex notation is a convenience; the physical displacement is the real part. Put ${\omega }_0^2=k/m$. The homogeneous characteristic roots areFor $m,k>0$ and damping $\gamma \ge 0$, oscillations occur when ${\gamma }^2<4{\omega }_0^2$, the underdamped case of the damped harmonic oscillator.

For a steady response $z=Z{e}^{i\omega t}$, substitution givesTaking the modulus and phase givesThe phase is between zero and $\pi$ for positive damping and frequency; an unqualified arctangent loses the correct quadrant above resonance.

With the real drive $F_0\cos \omega t$ and real response $r\cos (\omega t-\varphi )$, the average driving power equals the dissipated viscous power:For $\gamma >0$, write this asIt is largest when the squared term vanishes, givingThis power peak is at ${\omega }_0$ even when the displacement-amplitude peak is shifted. The frequency bandwidth of viscous oscillator power absorption follows from the half-power condition $|{\omega }_0^2-{\omega }^2|=\gamma \omega$:In fact the bandwidth identity is exact in this model. Sharp resonance additionally gives ${\omega }_{1,2}\simeq {\omega }_0\mp \gamma /2$. With zero damping at exact resonance, there is no bounded steady harmonic response to which these finite-power formulas apply.

Set ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}>0$. The real solution of the damped harmonic oscillator equation is $x-x_0=e^{-\gamma t/2}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$. The initial displacement fixes $A=a-x_0$, and release from rest gives $B=\gamma A/(2{\omega }_d)$. ThereforeThe exponential decay rate of the amplitude is $\gamma /2$, while the weak-damping energy decay rate is $\gamma$.

For the damped harmonic oscillator, the characteristic polynomial is $r^2+2\gamma r+{\omega }_0^2$. When $0\le \gamma <{\omega }_0$, its roots are $-\gamma \pm i\Omega$, with $\Omega =\sqrt{{\omega }_0^2-{\gamma }^2}$. ThusHere $A$ and $B$ are arbitrary real constants. The frequency is $\Omega$, and the amplitude envelope decays with time constant $1/\gamma$ when $\gamma >0$; the zero-damping limit recovers simple harmonic motion.

The characteristic equation of the damped harmonic oscillator is $r^2+\gamma r+{\omega }_0^2=0$. In the underdamped case its roots are $-\gamma /2\pm i{\omega }_d$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}$. Thus the exact real solution is $y=A{e}^{-\gamma t/2}\cos ({\omega }_{d}t+\alpha )$. For $\gamma \ll {\omega }_0$, ${\omega }_d={\omega }_0+O({\gamma }^2/{\omega }_0)$, giving the requested leading approximationFor mass $m$ and spring constant $m{\omega }_0^2$, the mechanical energy is $E=\frac{1}{2}m({\dot{y}}^2+{\omega }_0^2{y}^2)$. Differentiating and using the differential equation gives $\dot{E}=-m\gamma {\dot{y}}^2$. In particular, energy decay of a viscously damped harmonic oscillator is not an exact pointwise exponential decay: $\dot{E}=0$ at a turning point. To leading order in $\gamma /{\omega }_0$, the period average isThe quality factor is $Q=2\pi \overline{E}/(\text{energy lost in one cycle})$. Using the period $T\simeq 2\pi /{\omega }_0$ and the small fractional loss $\gamma T$ gives $\boxed{Q\simeq {\omega }_0/\gamma }$. Equivalently, the energy-decay definition $Q={\omega }_0\overline{E}/{\overline{P}}_{\mathrm{loss}}$ yields the same leading result. Replacing ${\omega }_d$ by ${\omega }_0$ in the oscillatory phase is not uniform over arbitrarily long times; its phase error is $O({\gamma }^{2}t/{\omega }_0)$.

Assume a physical mass $m>0$, spring constant $k>0$, and nonnegative viscous damping parameter $\gamma$. The homogeneous damped harmonic oscillator has characteristic equationIts roots are $r=-\gamma /2\pm \sqrt{{\gamma }^2/4-{\omega }_0^2}$. They have nonzero imaginary parts precisely when ${\gamma }^2<4k/m$. Defining $\Omega =\sqrt{k/m-{\gamma }^2/4}$, the real oscillatory solutions areFor strictly decaying oscillations also require $\gamma >0$. The limiting $\gamma =0$ case is an undamped harmonic oscillator. Equality gives critical damping with $z_h=(C+Dt)e^{-\gamma t/2}$, and a larger positive $\gamma$ gives two real decaying exponentials rather than oscillations.

At fixed $t$, expand the exact damped harmonic oscillator solution using $e^{-ϵt}=1-ϵt+O({ϵ}^2)$, $\omega =1+O({ϵ}^2)$ and $1/\omega =1+O({ϵ}^2)$. Its Poincaré expansion through first order isThe correction $-ϵt\sin t$ is a secular term. When $ϵt$ is order one it is not small, and the truncated amplitude eventually changes sign and grows in magnitude, instead of reproducing positive damping. This loss of ordering is why a fixed-time regular perturbation is unsuitable on the damping timescale.

The linear damped harmonic oscillator has characteristic exponents $-ϵ\pm i\sqrt{1-{ϵ}^2}$. Imposing both initial conditions gives, for $|ϵ|<1$,For completeness, at $|ϵ|=1$ its continuous limiting form is $t{e}^{-ϵt}$; for $|ϵ|>1$ replace $\sin (\omega t)/\omega$ by $\sinh (\sqrt{{ϵ}^2-1}t)/\sqrt{{ϵ}^2-1}$. The initial derivative is one, as the original PDF specifies; the converted TeX loses this derivative and corrupts the time derivatives in both oscillator equations.

The separate nonlinear oscillator has quadratic drag. Introduce the slow time $T=ϵt$ and use the method of multiple scales,The order-$ϵ$ equation isFor positive amplitude, write $\vartheta =t+\varphi$. Its right side is $-2A_T\cos \vartheta +2A{\varphi }_T\sin \vartheta -A^2|\cos \vartheta |\cos \vartheta$. The cosine coefficient of the last angular function iswhile its sine coefficient is zero. The solvability condition in the method of multiple scales removes the resonant harmonics, givingThe leading initial conditions select $A(0)=1$, $\varphi (0)=0$. Therefore the quadratically damped oscillator hasThe same sign follows from the exact energy balance $d[({\dot{x}}^2+x^2)/2]/dt=-ϵ|\dot{x}{|}^3$. Averaging $|\cos \vartheta {|}^3$ gives $4/(3\pi )$, consistent with the amplitude equation.

For $ϵ>0$, the amplitude decays algebraically and stays at most one. On each fixed interval $0\le ϵt\le T_{\ast }$, the omitted displacement correction is $O(ϵ)$, so this approximation is valid through times of order ${ϵ}^{-1}$. The effective drag strength $ϵA$ decreases rather than increases, and the envelope predicts $A\sim 3\pi /(4ϵt)$ at later times. There is no finite-time pole in the positive-drag approximation. A uniform error assertion for arbitrarily long intervals is stronger than the finite-slow-time expansion just derived; the stated $O(ϵ)$ estimate applies to the specified $ϵt=O(1)$ range.

For $ϵ<0$, write $\delta =1-4|ϵ|t/(3\pi )$. The amplitude grows as $1/\delta$, with a formal pole atOn fixed slow-time intervals strictly before that pole, the displacement error remains $O(|ϵ|)$. This estimate is not uniform as $\delta \to 0$: nonresonant corrections have size comparable to $|ϵ|A^2$, and the required weak-drag condition is $|ϵ|A=|ϵ|/\delta \ll 1$. Thus the approximation necessarily fails when $\delta =O(|ϵ|)$, corresponding to a distance of order one in fast time from the formal pole, and may require additional error control when approaching it. The plot therefore stops short of the pole. Its position is a prediction of the leading amplitude equation, not an exact blow-up time for the original nonlinear equation; continuing the formula past it is invalid.