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For $m\ddot{x}+b\dot{x}+kx=F_0\cos \omega t$, define ${\omega }_0=\sqrt{k/m}$ and $\beta =b/(2m)$. The parameter relating quality factor and viscous damping regimes isAt weak damping it also equals $2\pi$ times stored energy divided by energy lost per cycle. The stored-energy convention and frequency matter: this cycle-based interpretation should not be extended to free motions that never complete a cycle.

For $\beta <{\omega }_0$ the free motion is $e^{-\beta t}[A\cos {\omega }_{d}t+B\sin {\omega }_{d}t]$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. Thus underdamping means $Q>1/2$. High $Q$ gives many oscillations per amplitude-decay time, narrow resonance and strong frequency selectivity; a lightly loaded tuning fork, optical cavity or low-loss electrical resonator provides examples. The amplitude decay time is ${\tau }_A=1/\beta =2Q/{\omega }_0$, while the slowly averaged energy decays with ${\tau }_E=1/(2\beta )=Q/{\omega }_0$. These times differ by a factor of two.

At critical damping, $\beta ={\omega }_0$ and $Q=1/2$, with $x=(A+Bt)e^{-{\omega }_{0}t}$. There is no ringing; a simple critically damped position indicator can settle without oscillation. For $\beta >{\omega }_0$ and $Q<1/2$, there are two real negative decay rates $-\beta \pm \sqrt{{\beta }^2-{\omega }_0^2}$. Increasing very strong drag slows the long-time return, whose rate approaches ${\omega }_0^2/(2\beta )$. A strongly viscously damped mechanical indicator illustrates this regime. These classifications concern the linear homogeneous equation, not every possible nonlinear damping law.

The driven displacement amplitude isThe displacement-resonance threshold of a viscous oscillator requires $\beta <{\omega }_0/\sqrt{2}$, or $Q>1/\sqrt{2}$; it lies at $\sqrt{{\omega }_0^2-2{\beta }^2}$. Some underdamped systems therefore have no such maximum. Average absorbed power instead peaks at ${\omega }_0$ for this viscous model. Its half-power frequencies are $\sqrt{{\omega }_0^2+{\beta }^2}\pm \beta$, so their separation is $2\beta$ and $Q={\omega }_0/\Delta \omega$. For broad response this ratio remains a useful parameter but not a claim of many free oscillations. The familiar displacement-linewidth version is a high-$Q$ approximation.

A series RLC circuit has ${\omega }_0=1/\sqrt{LC}$ and $Q={\omega }_{0}L/R$ for the charge oscillation. Resistance dissipates energy, inductance provides the inertial analogue and capacitance the restoring analogue. Ringdown and a calibrated half-power scan independently estimate loss; parasitic resistance, loading and radiation change the measured value. A large $Q$ increases frequency selectivity but also lengthens settling time, so it is not universally the best instrumental choice.

For $b^2=4mk$, the repeated characteristic root requires a second linearly independent solution $t{e}^{-\gamma t}$:This is critical damping. If $x(0)=x_0$ and $\dot{x}(0)=0$, then $x=x_0(1+\gamma t)e^{-\gamma t}$, demonstrating the monotone return used in the suspension discussion.

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.

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. First take $\lambda =e\gamma >0$. Since $|e^{-i\omega \tau }|=e^{\tau \mathrm{Im}\omega }$, close the contour below for $\tau >0$, and above for $\tau <0$. On the chosen semicircle the exponential is bounded and the denominator grows as $R^2$, so the arc integral tends to zero. The upper contour encloses no pole, giving $G=0$ for $t<t^{\prime }$. The lower contour is clockwise, hence the real-axis integral divided by $2\pi$ is $-i$ times the sum of residues. Since $D^{\prime }({\omega }_{+})=-2\Omega$ and $D^{\prime }({\omega }_{-})=2\Omega$, that sum isThusThis is the causal Green function of a damped oscillator. Its value is continuous through $\tau =0$, and its derivative jumps from zero to one; integrating the differential equation through zero verifies the delta-source normalization and the residue sign. The limits are $H(\tau )\tau e^{-\lambda \tau }$ at critical damping and $H(\tau )e^{-\lambda \tau }\sinh (\kappa \tau )/\kappa$ in the overdamped regime.

For completeness, if the unspecified charge has $e<0$, the same real-axis integral instead encloses upper poles for $\tau <0$ and gives $-H(-\tau )e^{-\lambda \tau }\sin (\Omega \tau )/\Omega$, an advanced Green function. A causal growing response would require an inverse-Laplace contour above those poles, not the stated real-axis Fourier integral. At $e=0$ the real-axis integral needs a pole prescription; shifting both poles below it gives the retarded zero-damping limit. These distinctions are necessary because $\gamma >0$ alone does not fix the sign of damping.

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 $y_{tt}+2k{y}_t=c^2{y}_{xx}$ with fixed endpoints, initial displacement zero and initial velocity a Dirac delta at $a\in (0,l)$, the Fourier sine series coefficient of that velocity is $2\sin (n\pi a/l)/l$. Each coefficient evolves by $T_n^{\prime \prime }+2k{T}_n^{\prime }+(n\pi c/l{)}^2{T}_n=0$. Here ${\omega }_n^2=(n\pi c/l{)}^2-k^2$. At critical damping, replace $\sin ({\omega }_{n}t)/{\omega }_n$ by $t$; for an overdamped mode replace it by $\sinh (\sqrt{k^2-(n\pi c/l{)}^2}t)/\sqrt{k^2-(n\pi c/l{)}^2}$. The initial conditions hold in the sense of distributions.

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.