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Adding a load $M$ to a spring-supported table shifts its static downward position by $\Delta =Mg/k$. If the load is deposited with negligible impulse and moves with the table, the moving mass is $m+M$. Starting at the former equilibrium with zero velocity, an underdamped response isIts transient envelope decays over $1/\beta$. Critical damping $b=2\sqrt{k(m+M)}$ gives $x=\Delta [1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. A nonzero impact impulse changes the initial velocity and therefore the transient.

Critical damping is the repeated-root case of the viscous damping model: its two characteristic roots coincide at $r=-{\omega }_0$. The solution with arbitrary initial conditions isThe extra factor $t$ is essential: one exponential function alone cannot fit arbitrary initial displacement and velocity when the characteristic equation has a double polynomial root. Substitution, or taking the limit of the underdamped formula as its angular frequency tends to zero, gives the displayed expression.

The precise fastest-return statement is this: hold $m$, $k$, $x_0>0$ and $v_0=0$ fixed, and compare all $\zeta \ge 1$. Every response decreases positively toward equilibrium; the critically damped responseis smaller than every overdamped response at each $t>0$. It therefore reaches and stays within any specified band $|x|\le ϵx_0$, $0<ϵ<1$, sooner than those responses. Exact arrival at rest takes infinite time. This is a comparison among nonoscillatory regimes for these particular initial conditions, not a universal minimum of every possible settling-time criterion.

One short comparison argument uses $q=\partial X/\partial \zeta$. Its initial conditions are zero and $q^{\prime \prime }+2\zeta q^{\prime }+q=-2X^{\prime }$. For release from rest, $X^{\prime }<0$ at $\tau >0$; the causal Green function of a damped oscillator is positive at positive time for $\zeta \ge 1$. Its convolution with $-2X^{\prime }$ gives $q>0$: increasing damping beyond the critical value increases the remaining displacement.

There is no repeated oscillation at critical damping, but there can be overshoot once. For $x_0>0$, the factor in square brackets crosses zero precisely when $v_0<-{\omega }_0{x}_0$, atFor example, $v_0=-2{\omega }_0{x}_0$ gives $x=x_0(1-{\omega }_{0}t)e^{-{\omega }_{0}t}$, crossing equilibrium at $t=1/{\omega }_0$ as shown above. An overdamped response can also cross once for suitable initial conditions. Conversely, specially chosen initial conditions can remove its slow exponential decay entirely. Thus the generic slow-root argument must not be applied to a response whose slow-root coefficient is zero.

A mass-spring-damper subject to a constant load obeys $\ddot{z}+2\gamma \dot{z}+{\gamma }^{2}z={\gamma }^2{z}_{\mathrm{eq}}$ at critical damping. With $z(0)=0$ and $\dot{z}(0)=v_0$,For $z_{\mathrm{eq}}>0$, the displacement exceeds its final value at a positive time exactly when $v_0>\gamma z_{\mathrm{eq}}$. Thus a critically damped step response released from rest is non-overshooting, but critical damping alone does not forbid overshoot from arbitrary initial momentum. For a vertical scale with downward coordinate, $z_{\mathrm{eq}}=mg/k$ and $\gamma =\sqrt{k/m}$, making the threshold $v_0>g\sqrt{m/k}$.

With $e^{i\omega t}$ phasors, critical damping gives $\Theta (\omega )=F_0/({\omega }_0+i\omega {)}^2$, amplitude $F_0/({\omega }_0^2+{\omega }^2)$ and phase $-2\arctan (\omega /{\omega }_0)$. The response strongly suppresses alternating currents well above the mechanical frequency.

A critically damped RLC response is the boundary between underdamped RLC response and overdamped RLC response. The free series RLC circuit has a repeated characteristic root $-{\omega }_0$, with general solutionFor initial conditions $q(0)=q_0$, $i(0)=0$,In the plotted normalization, $q(1)=2/e\approx 0.735759$ and $i(1)=-1/e\approx -0.367879$. The electric charge returns without ringing; the polynomial factor $1+{\omega }_{0}t$ matters, so a single exponential function does not give the whole response.

For fixed $L,C$ and these initial conditions, critical damping avoids ringing while giving a faster asymptotic return than any larger $R$: the overdamped RLC response has a root closer to zero. This makes critical damping a useful tuning target for settling after switching. An MIT laboratory exercise makes the distinction observable by asking students to compare all three cases with $L=47\mathrm{mH}$ and $C=47\mathrm{nF}$. Those values give $R_{\mathrm{crit}}=2\sqrt{L/C}=2000\Omega$ and ${\omega }_0\approx 21276.6{\mathrm{s}}^{-1}$. ^[ref].

For $\ddot{\theta }+2{\omega }_0\dot{\theta }+{\omega }_0^2\theta =F$ switched on from rest, $\theta (t)=(F/{\omega }_0^2)[1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. critical damping is the fastest nonoscillatory step response at fixed natural frequency within the ordinary second-order model.

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.

An experimentally useful question is whether the disappearing oscillation follows viscous damping, and whether increasing the brake changes $b$ rather than $m$ or $k$. A published teaching protocol uses a cart on an air track, attached between two springs, with ceramic magnets providing eddy-current damping and an ultrasonic motion sensor recording position against time. Replace the magnets with equal masses in the baseline trial so the moving mass stays fixed. Obtain each spring constant from the slope of applied force versus extension, then use $k=k_1+k_2$. Time many baseline cycles to check ${\omega }_0\simeq \sqrt{k/m}$. Compare one-magnet and two-magnet free decays; the protocol calls for a fit of the full trace including its offset. ^[ref]

The following additional controls make the inference testable. Level the air track and hold its air supply fixed. Check the ultrasonic motion sensor at several stationary flag positions against a ruler, including the full planned travel range; record the offset and scale correction. Check elapsed acquisition time against an independent clock, and choose a sampling rate that resolves many samples within the shortest expected oscillation period. Record the resting equilibrium before and after each run. Release the cart from a small measured displacement with minimal initial velocity, with the driving apparatus inactive. Repeat at several release sizes and magnet settings while keeping the springs and total mass unchanged. These are proposed calibration and repeatability checks, not reported measurements from the source.

For underdamping, fit the recorded position to $y(t)=y_{\mathrm{eq}}+A{e}^{-\alpha t}\cos ({\omega }_{\mathrm{d}}t+\varphi )$ by nonlinear regression. The readout to inspect is a sequence of alternating extrema about $y_{\mathrm{eq}}$ with shrinking amplitudes, and the deviations of the fitted curve from the recorded points. The model predictsCompare the right-hand side, determined independently from mass and spring constants, with the fitted left-hand side. A good fit supports a constant-coefficient single normal mode over the observed range; it does not establish a universal microscopic force law. The published protocol denotes the decay rate by $b$ and the velocity-proportional coefficient by $c$; here they are $\alpha$ and $b$, respectively.

At critical damping or overdamping, fit the corresponding nonoscillatory response including its offset and both initial conditions, rather than forcing a damped cosine onto the data. Adjusting the brake can explore the transition, but the cited protocol does not establish that its one-magnet or two-magnet settings attain critical damping. A near-critical trace over a short interval can be hard to distinguish from neighbouring damping ratios.

Important limitations are finite sensor range, missed or reflected acoustic wave pulses, measurement noise near equilibrium, nonzero release velocity, and position-dependent magnetic braking. Changing the release size tests for quadratic drag or nonlinear spring response; inspecting deviations near turning points can expose kinetic friction. Slow drift of the offset can imitate a long overdamped tail. Report uncertainties in fitted rates and independently measured $m,k$; absence of a visible crossing only bounds the behaviour within the instrument's sensitivity and recording interval. No numerical experimental results are asserted here.

The damping ratio is the dimensionless ratio $b/b_{\mathrm{crit}}$. Substituting the exponential function $x=e^{rt}$ into the viscous damping model gives the characteristic equation and its characteristic roots:For $0<\zeta <1$, underdamping gives a pair of complex conjugate roots with negative real part. For $\zeta =1$, critical damping gives the repeated negative real root. For $\zeta >1$, overdamping gives two distinct negative real roots. At $\zeta =0$ the roots are purely imaginary and the harmonic oscillator is undamped; this boundary case has no energy loss. With $m,k>0$, negative $b$ would inject energy and lies outside the dissipative model.

The left panel locates $r/{\omega }_0$ in the complex plane; the imaginary part gives the damped angular frequency, while the negative real part gives the exponential decay rate. The right panel shows the positive rates $-\mathrm{Re}(r_{\pm })/{\omega }_0$. Beyond critical damping, one root moves toward zero, explaining slow late motion for generic initial conditions. The repeated-root solution also has a polynomial prefactor, so these rates alone do not describe its complete response.

For underdamping, let $x_n,x_{n+1}$ be successive same-sign extrema of displacement relative to equilibrium, one full oscillation period apart. The logarithmic decrement isComparing extrema $N$ full cycles apart instead gives $\delta =N^{-1}\ln (|x_n|/|x_{n+N}|)$. Adjacent extrema of opposite sign are only half a cycle apart and give half this decrement. This distinction matters when inferring a damping ratio. Near critical damping, a full second same-sign extremum may be too small to resolve above measurement noise; a full-trace fit with uncertainty is then more informative than a peak ratio.

At critical damping, the characteristic polynomial has the repeated root $r=-{\omega }_0$. Its independent solutions are $e^{-{\omega }_{0}t}$ and $t{e}^{-{\omega }_{0}t}$, givingThe factor $t$ supplies the second independent solution that would be lost by using just one exponential. There is no oscillatory factor at this boundary between the underdamped and overdamped regimes.

Assume the impulse point is in the interior, $0<a<l$. Expand the displacement in the fixed-end Fourier sine seriesOrthogonality gives the initial velocity coefficient by applying the Dirac delta to the sine function:Substitution in the wave equation yields $T_n^{\prime \prime }+2k{T}_n^{\prime }+{\alpha }_n^2{T}_n=0$. For $k<{\alpha }_n$, its roots are $-k\pm i{\omega }_n$, where ${\omega }_n=\sqrt{{\alpha }_n^2-k^2}$. The two initial conditions determineHence the normalized velocity-impulse Green function for a damped string isFor a mode with critical damping, use the limit $e^{-kt}t$ in place of the time factor. For $k>{\alpha }_n$, put ${\eta }_n=\sqrt{k^2-{\alpha }_n^2}$ and use $e^{-kt}\sinh ({\eta }_{n}t)/{\eta }_n$. These choices solve the same modal ordinary differential equation and initial conditions, so the formula covers arbitrary $k>0$. The impulse initial velocity is a distribution, so initial differentiation of the series is interpreted distributionally.

The printed formula requires correction: the factor depending on $n$ must lie inside the sum, and sine-mode normalization requires $2/l$, rather than $2$. The printed slash after the final sine is also stray. Direct differentiation at $t=0$ verifies the corrected normalization: the initial sine coefficients are exactly those of $\delta (x-a)$.

A mode has dimensionless damping ratio $k/{\alpha }_n$, and characteristic rootsIf $k/{\alpha }_n\ll 1$, the normal mode oscillates at ${\omega }_n={\alpha }_n+O(k^2/{\alpha }_n)$ with exponentially decaying envelope $e^{-kt}$. As $k\to 0$, the solution approaches the undamped wave equation response.

If $k/{\alpha }_n\gg 1$, the roots are real and negative, withThe mode has a fast transient and a slow nonoscillatory relaxation. After the fast transient, neglecting $y_{tt}$ gives the diffusion equation $y_t\simeq c^2{y}_{xx}/(2k)$ for these long-wavelength modes. At $k={\alpha }_n$, the repeated root produces $t{e}^{-kt}$: this is critical damping.

Thus the critical parameter is $k/{\alpha }_n=1$ for each mode, and the fundamental threshold isWhen $k<\pi c/l$ every mode is oscillatory. For any fixed, finite $k$, sufficiently large $n$ still gives oscillatory modes, so large damping of the lowest modes does not mean that every mode of the full string is overdamped.

The small-angle approximation replaces $\sin \theta$ by $\theta$, giving the damped harmonic oscillator equation $\ddot{\theta }+c\dot{\theta }+\theta =0$. Its characteristic roots satisfy $r^2+cr+1=0$.

For $0<c<2$, put $\nu =\sqrt{1-c^2/4}$; the motion is underdamped:It oscillates with decaying amplitude. For $c=2$, the repeated root gives critical damping:For $c>2$, the motion is overdamped:Both real exponents are negative, so there is no oscillatory factor. The two constants are arbitrary in each case; these approximations require the displacement to remain small.

With gravity omitted, the initial data contain only the first fixed-end normal mode, so writeThenBecause $k=\pi c/l$, this oscillator is critically damped, andTherefore

Gas in hydrostatic balance has zero vertical velocity. A dust grain subject to linear drag with aerodynamic stopping time $\tau$ therefore obeysThis is a damped harmonic oscillator. For ${\Omega }_z\tau >1/2$, the motion is underdamped, with angular frequencyand envelope $e^{-t/(2\tau )}$. Critical damping occurs at ${\Omega }_z\tau =1/2$. For ${\Omega }_z\tau <1/2$, the motion is overdamped. In the strong-drag limit ${\Omega }_z\tau \ll 1$, a rapid transient on timescale $\tau$ leaves slow dust settling in an astrophysical disk at rate ${\Omega }_z^2\tau$; in the weak-drag limit the grain makes many damped vertical oscillations.

An even function satisfying the spatial equation has $X=C\cos (kx)$. The Dirichlet boundary conditions require $\cos (kL)=0$, so the allowed wavenumbers areThere is no nonzero zero-wavenumber mode. With $T=e^{-\alpha t/2}S$, the time equation becomes $S^{\prime \prime }+{\omega }_n^{2}S=0$, where ${\omega }_n^2=c^2{k}_n^2-{\alpha }^2/4$. Thus superposition of the even normal modes givesThis uses completeness of the even Fourier series modes. Imaginary ${\omega }_n$ describe overdamping. At critical damping, ${\omega }_n=0$, the two displayed exponentials coincide: the complete time factor must instead be $e^{-\alpha t/2}(C_n+D_{n}t)$. This repeated-root exception is necessary for a genuinely general solution.

For each sine normal mode, the temporal characteristic equation is $s^2+2\lambda s+{\alpha }_n^2=0$, with roots $s=-\lambda \pm \sqrt{{\lambda }^2-{\alpha }_n^2}$. Since $\lambda L/\pi =M+1/2$, the first $M$ modes are overdamped and all higher modes are underdamped. No integer $n$ gives critical damping.

The zero initial displacement excludes the hyperbolic cosine and cosine terms. HencewhereTo impose the initial velocity, let $f_n=\frac{2}{L}{\int }_0^{L}f(x)\sin ({\alpha }_{n}x)dx$ be its Fourier sine series coefficients. Differentiating the expansion at $t=0$ gives $A_n{\Omega }_n=f_n$ for the overdamped modes and $A_n{\omega }_n=f_n$ for the underdamped modes. These determine all coefficients. Despite the hyperbolic sine, the first modes decay: ${\Omega }_n<\lambda$, so both temporal exponential rates are negative.

Let $Q(t)$ denote the electric charge on the initially positive capacitor plate and define the discharge electric current by $I=-\dot{Q}$. Kirchhoff's voltage law for the series RLC circuit gives the damped harmonic oscillator equationThe initial conditions are $Q(0)=Q_0$ and $\dot{Q}(0)=0$: the open switch initially prevents electric current, and finite inductance prevents an instantaneous jump in it. Put $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The two characteristic roots are $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. An oscillatory electric current requires a nonzero imaginary part, henceFor positive $L,C$ and nonnegative $R$, this is the underdamped regime. The initial conditions giveAt equality there is critical damping, and above it there is an overdamped response, with no repeated changes of sign of the discharge electric current. A nonzero initial electric charge is understood; if $Q_0=0$, the stated initial conditions give no electric current at all.