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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 the initially circular damped planar oscillator with overdamping $\gamma >\omega$, put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}$. The exact solution replaces $\cos \nu t,\sin \nu t/\nu$ by $\cosh \alpha t,\sinh \alpha t/\alpha$. Resolving along the starting radius and its anticlockwise tangent yieldsIts limiting angle is the title value. Both coordinates remain positive when $c>0$, so the trajectory returns to the origin without repeated revolutions. It initially moves tangentially and bends toward its slow decaying eigenmode, rather than following a radial straight line from the start.

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 $b^2>4mk$, define $\beta =\sqrt{{\gamma }^2-{\omega }_0^2}$. The two characteristic roots are real and negative:This is overdamping: there is no repeated oscillation, and the slow root controls late-time return. A generic initial velocity can still cause a single crossing of equilibrium, so “nonoscillatory” need not mean monotone for all initial data. For large $b$, the slow root approaches $-k/b$.

Take $m,k,b>0$, and define ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. Substitution of $e^{rt}$ into the unforced equation gives $r^2+2\gamma r+{\omega }_0^2=0$. The roots determine whether the return is oscillatory.

For return after a displacement in this ideal model, critical damping, case (b), is the usual comfort choice: it removes repeated bouncing while avoiding the slow relaxation mode of strong overdamping. For release from rest it gives a monotone return with the fastest asymptotic decay available without oscillation. This criterion does not prove optimal comfort for every road profile or initial velocity; actual comfort also depends on acceleration, road forcing and suspension travel.

For the forced response, set $\Omega =2\pi u/d$ and use $x_{\mathrm{ss}}=\mathrm{Re}(\hat{x}{e}^{i\Omega t})$. ThenThe general forced solution adds the appropriate homogeneous transient from (a), (b) or (c). Initial data were not specified, so its constants cannot be fixed. Positive damping removes this transient at long times. This solution uses the particular driving force in the supplied equation; it does not silently replace it by a different moving-support model.

Maximising $X$ is minimising its squared denominator $D$. Differentiation givesA nonzero displacement resonance exists only if $b^2<2mk$. Here ${\Omega }_r^2=0.175{\mathrm{s}}^{-2}$, and thereforeThe low-frequency limit is $X(0)=A$, whereas $X\to 0$ at large $\Omega$. The parameters give damping ratio $\zeta =b/(2\sqrt{mk})=0.456$, consistent with an underdamped system and the existence of this amplitude maximum.

Put ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. The characteristic roots are $-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. For release from rest at displacement $x_0>0$, the three regimes areunderdamping produces decaying oscillations; critical damping is the repeated-root boundary; overdamping has two real decay rates. The critical and overdamped curves from rest approach equilibrium without crossing it. All three initially have zero slope and negative acceleration. A sufficiently strong initial kick can cause a single crossing even without an oscillatory mode.

For the critical oscillator with initial velocity $-v_0$, the repeated-root solution isIts exponential is positive at finite time. Therefore the bracket has a positive-time zero precisely whenEquality gives a pure positive exponential and no finite-time crossing.

For the specified numerical oscillator, ${\omega }_0=1{\mathrm{s}}^{-1}$ and $\gamma =2/\sqrt{3}{\mathrm{s}}^{-1}>1{\mathrm{s}}^{-1}$, verifying overdamping. Its two roots are $-1/\sqrt{3}$ and $-\sqrt{3}$ in inverse seconds. The initial conditions givewith $t$ in seconds. The slow component has a negative coefficient; the fast positive component initially dominates.

The values for the requested plot areThe curve crosses near $0.56\mathrm{s}$ and then returns toward zero from below. The exact expression provides a check on the graphical estimate:

To settle quickly after loading, critical damping is preferable to the supplied weak damper:Using the strict-bound mass instead changes this by less than $1\%$. This eliminates free ringing while avoiding the slow tail of severe overdamping. In the stipulated external-force model, greater damping also reduces the steady forced amplitude; there is no finite optimum based solely on that amplitude. A critical or near-critical value is a practical compromise for rapid settling. If the disturbance were specified as floor displacement transmitted through the legs, damping would introduce an additional transmission tradeoff, but that is a different forcing model.

Set the drive to zero and try $x=e^{st}$. The characteristic equation isFor underdamping, $\boxed{b^2<4mk}$, the roots are $-\gamma \pm i{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The displacement oscillates with an exponentially decaying envelope:For critical damping, $\boxed{b^2=4mk}$, there is a repeated root $-{\omega }_0$, giving $x=(C+Dt)e^{-{\omega }_{0}t}$. The motion is nonoscillatory and is the limiting case between oscillatory and overdamped response.

For overdamping, $\boxed{b^2>4mk}$, the roots are distinct negative real numbers and $x=C{e}^{s_{+}t}+D{e}^{s_{-}t}$. Both components decay without sustained oscillation; strong damping gives a particularly slow late-time relaxation. Constants $C,D$ depend on the initial conditions. A nonoscillatory solution can cross equilibrium once for some initial data, so “nonoscillatory” does not mean monotone for every possible initial condition.

Let the initial conditions be $x(0)=0$ and $\dot{x}(0^{+})=v_0>0$. The velocity-impulse response of a damped oscillator isAll curves start at the origin with the same positive slope $v_0$. Underdamping gives a decaying sequence of positive and negative lobes. Critical damping gives one positive lobe followed by return to zero. Overdamping also remains positive for these initial data, but has a longer tail. The initial impulse changes velocity abruptly, not displacement.

A mass $M$ on a spring of stiffness $K$, subject to linear viscous damping and a sinusoidal drive, obeys the forced harmonic oscillator equationHere $x$ is displacement, $2M\gamma$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, and $F_0$ is a force in newtons. A series RLC circuit is an electrical analogue, with $M\leftrightarrow L$, $2M\gamma \leftrightarrow R$, $K\leftrightarrow 1/C$ and $x\leftrightarrow q$.

For free motion set $F_0=0$ and solve the characteristic equation $s^2+2\gamma s+{\omega }_0^2=0$. The roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$ distinguish three cases:

The constants are set by initial position and velocity. The mechanical energy and its rate of change in free motion areFor weak damping, the cycle-averaged energy decays approximately as $e^{-2\gamma t}$. The quality factor is $Q={\omega }_0/(2\gamma )$; for $Q\gg 1$ it equals $2\pi$ times stored energy divided by the loss per cycle. Logarithmic decrement of successive free-decay peaks determines $\gamma$ experimentally.

For driven motion the complete answer is the decaying free transient plus a steady response at the driving angular frequency:At low frequency, $X\to F_0/K$ and the displacement is nearly in phase with the drive. At $\Omega ={\omega }_0$, the lag is $\pi /2$ and $X=F_0/(2M\gamma {\omega }_0)$. At high frequency, $X\simeq F_0/(M{\Omega }^2)$ and the lag tends to $\pi$. The displacement resonance maximum lies atonly when $\gamma <{\omega }_0/\sqrt{2}$. Thus an underdamped oscillator need not have a nonzero-frequency displacement resonance peak.

In the steady state the mean input power equals the mean dissipated power:This absorption peaks exactly at $\Omega ={\omega }_0$, rather than at ${\Omega }_r$. Its half-power points have bandwidth $\Delta \Omega =2\gamma$, so $Q={\omega }_0/\Delta \Omega$ for this linear model. For weak damping, a high quality factor therefore means a narrow strong resonance and a long free-decay time. Exactly zero damping at resonant driving has no finite steady response: a secular amplitude grows with time.

Use the diagram's downward displacement $z$, measured from the unloaded plate position, and take $z(0)=0$. The mechanical energy, with gravitational potential zero at that reference, isThe loaded equilibrium follows from $k{z}_{\mathrm{eq}}=mg$, so its downward shift is $\boxed{z_{\mathrm{eq}}=mg/k}$.

The shock absorber's force is $-b\dot{z}$, hence its power is force times velocity: $dE/dt=-b{\dot{z}}^2\le 0$. Differentiating the energy gives $\dot{z}(m\ddot{z}+kz-mg)=-b{\dot{z}}^2$, which yields, with continuity through instants of zero velocity,Set ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$.

For release from rest, the three step response of a damped oscillator regimes are easy to compare after shifting to $z-z_{\mathrm{eq}}$. With light damping $0<\gamma <{\omega }_0$, write ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$:It overshoots and oscillates around the loaded position with a decaying envelope. At critical damping, $\gamma ={\omega }_0$, it becomes $z/z_{\mathrm{eq}}=1-(1+\gamma t)e^{-\gamma t}$, approaching monotonically from below. For heavy damping $\gamma >{\omega }_0$, set $\eta =\sqrt{{\gamma }^2-{\omega }_0^2}$ and replace the trigonometric bracket by $\cosh \eta t+(\gamma /\eta )\sinh \eta t$. This is also monotonic, but its slow exponential delays settling.

Critical damping gives the fastest nonoscillatory settling from the unloaded rest condition, avoiding a repeatedly swinging reading without the sluggishness of overdamping. Its required coefficient isAt critical damping the repeated characteristic root is $-\gamma$. Thus the general displacement under constant load is $(A+Bt)e^{-\gamma t}+C$. The equilibrium and initial conditions give $C=z_{\mathrm{eq}}$, $A=-z_{\mathrm{eq}}$ and $B-\gamma A=v_0$, henceSubtracting the final position gives $z-z_{\mathrm{eq}}=[-z_{\mathrm{eq}}+(v_0-\gamma z_{\mathrm{eq}})t]e^{-\gamma t}$. The exponential is positive and the bracket starts negative. It becomes positive at some $t>0$ if and only if its slope is positive. Therefore critical-damping overshoot from initial velocity requireswith $v_0$ positive downwards. At equality the scale still approaches from below. Critical damping prevents oscillatory modes, but does not guarantee a non-overshooting response to an arbitrarily large impact velocity.

For overdamping, $\gamma =b/(2m)>\omega$. Put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}>0$. Both characteristic roots $-\gamma \pm \alpha$ are negative real numbers. Replacing trigonometric functions by hyperbolic functions gives the exact trajectoryIt starts from $(c,c)$ with tangent $(-\omega c,\omega c)$, bends anticlockwise toward the origin, and approaches it without repeated revolutions. The right panel gives a representative choice $\gamma =1.5\omega$; the paper does not specify one particular value of $b$ in this regime.

One can make the qualitative shape precise. In radial and tangential directions relative to the initial radius, the angle turned from the starting direction obeysFor $c>0$, both coordinates stay positive, and the approach is along the fixed ray with $y/x\to (\gamma +\alpha +\omega )/(\gamma +\alpha -\omega )$. The initial velocity is tangential, so an immediately straight radial return would be incorrect. The absence of oscillatory characteristic roots prevents repeated revolutions. This is a non-spiralling overdamped central oscillator trajectory.

Apply separation of variables to each mode of the damped wave equation $u_{tt}+2k^2{u}_t=c^2{u}_{xx}$, assuming $L,c>0$. The spatial ordinary differential equation is $X_n^{\prime \prime }+({\lambda }_n/c{)}^2{X}_n=0$. Its general solution is $C\sin ({\lambda }_{n}x/c)+D\cos ({\lambda }_{n}x/c)$. The boundary condition at $0$ gives $D=0$; a nonzero mode at $L$ requiresOnly odd $n=2m+1$ occur in the initial Fourier sine series. Put ${\omega }_m=(2m+1)\pi c/L$ and $d_m=\sqrt{k^4-{\omega }_m^2}$. The temporal equation $T_m^{\prime \prime }+2k^2{T}_m^{\prime }+{\omega }_m^2{T}_m=0$ has roots $-k^2\pm d_m$. DefineFor distinct roots, let the full amplitude be $q_m(t)=A_m{e}^{-{\alpha }_{m}t}+B_m{e}^{-{\beta }_{m}t}$. The initial displacement and zero initial velocity impose $A_m+B_m=F_m$ and ${\alpha }_m{A}_m+{\beta }_m{B}_m=0$. Solving givesThus the physically consistent superposition isFor ${\omega }_m<k^2$ there is overdamping, with two real decaying exponentials. For ${\omega }_m>k^2$ there is underdamping; set ${\nu }_m=\sqrt{{\omega }_m^2-k^4}$; the equivalent real expression isAt critical damping, the repeated-root case ${\omega }_m=k^2$, take the limit, giving $q_m(t)=F_m(1+k^{2}t)e^{-k^{2}t}$. These formulas all satisfy $q_m(0)=F_m$ and $q_m^{\prime }(0)=0$.

Three source qualifications matter. First, the sum must start at $m=0$, for the same reason as in (b). Second, the printed $X(x)=u_0(x)$ cannot hold for a single separated mode: the tent has $X^{\prime \prime }=0$ on each open linear segment but $X\ne 0$, contradicting $X^{\prime \prime }=-({\lambda }_n/c{)}^{2}X$ for positive ${\lambda }_n$, and its peak is not twice differentiable. Separation applies mode by mode, with the initial tent recovered by superposition. Third, no finite “large” $k$ makes every mode overdamped, because ${\omega }_m\to \infty$. The two-exponential expression permits complex conjugate roots, while a repeated root requires its displayed limiting form. The tent initial data are naturally interpreted through the convergent modal solution, rather than as a single smooth separated product.

For $\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0$, an initial velocity $v_0$ at zero displacement givesAll three cases begin with slope $v_0$. Only underdamping produces repeated crossings through equilibrium; critical damping and overdamping remain on the initial side of equilibrium for this particular initial condition. The overdamped long-time tail decays at rate $\gamma -s$, which becomes slow for strong damping.