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Assume $\omega >0$. The characteristic roots of the damped harmonic oscillator are $-k\pm \sqrt{k^2-{\omega }^2}$. Its general homogeneous solution isThese are respectively the underdamped, critically damped and overdamped regimes.

For the velocity-switched driven oscillator, the initial acceleration is $-{\omega }^2{x}_1<0$. Its velocity is therefore negative on the first half-cycle, and the forcing is zero:Indeed $\dot{x}=-({\omega }^2{x}_1/p)e^{-kt}\sin pt<0$ in that open interval. Set $r=e^{-k\pi /p}$; at the first turning point, $x_2=-r{x}_1$ and $\dot{x}=0$.

Put $\tau =t-\pi /p$. On the second half-cycle the velocity is positive, so the particular solution is $a/{\omega }^2$. Matching position and velocity givesIts derivative is positive for $0<\tau <\pi /p$. At the next turning point,The turning-point return map therefore repeats the initial state exactly whenThis value exceeds $a/{\omega }^2$, so the subsequent acceleration initiates the required negative-velocity half-cycle again. The switched solution is continuous with continuous velocity, and is piecewise twice differentiable; acceleration has one-sided jumps where the forcing switches.

For $k>\omega$, the specified positive initial value givesThere is no next turning point: the forcing stays zero and $x$ approaches zero monotonically. The motion from the specified initial state cannot be periodic. More generally, a solution started at any positive maximum has this same obstruction; a nonconstant periodic orbit would require such a maximum. The zero equilibrium is a separate constant solution.

For the damped harmonic oscillator, the characteristic roots are $-k\pm i\omega$. For $\omega \ne 0$, the real general solution isIf the frequency parameter is zero, the characteristic root is repeated and the complete limiting basis is $e^{-kt},t{e}^{-kt}$, so $y_1=e^{-kt}(A+Bt)$. No sign restriction on $k$ is needed for these formulas, although $k>0$ corresponds to decay.

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.

Every solution is the resonant series RLC response $I_0(t)=R^{-1}\sin ({\omega }_{0}t)$ plus a solution of the homogeneous damped harmonic oscillator. Define $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The three possibilities areThese are respectively the underdamped RLC response, critically damped RLC response and overdamped RLC response. Both exponents in the last case are negative because $0<\kappa <\gamma$. All three homogeneous contributions tend to zero as $t\to \infty$, so every response approaches the unique periodic steady current $I_0(t)$. No initial conditions were specified in this clause; they determine $A,B$.

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.

The characteristic equation is $r^2+\lambda r+k=0$, whose roots are $r_{\pm }=(-\lambda \pm \sqrt{{\lambda }^2-4k})/2$. For distinct real roots, $y=C_1{e}^{r_{+}t}+C_2{e}^{r_{-}t}$; at the repeated-root threshold ${\lambda }^2=4k$, $y=(C_1+C_{2}t)e^{-\lambda t/2}$. In the complex-root case put $\Omega =\sqrt{k-{\lambda }^2/4}$ to obtainThis is precisely the oscillatory regime of the damped harmonic oscillator. The zero solution is of course nonoscillatory; the regime describes its nontrivial motions.

Apply the Fourier transform of a derivative twice to the damped harmonic oscillator equation. Under the corresponding convergence assumptions,ThusSince $\kappa ,\Omega >0$, this denominator has no real zero: its imaginary part could vanish only at $\omega =0$, where its real part is positive. The transform therefore defines a regular response on the real frequency axis.

The convolution theorem gives the causal responseDefine the integrated impulse response, for $s\ge 0$, byDirect differentiation gives $F^{\prime }(s)=e^{-ps}\sin (qs)/q$ and $F(0)=F^{\prime }(0)=0$. ThusIn particular, after switch-off the answer isThe response and its first derivative are continuous at both switching times, and the post-pulse motion is a freely decaying damped harmonic oscillator. The expression described as valid for $t<1$ in the paper is understood for $0<t<1$; before the pulse the causal output is zero.

Model the AFM cantilever as a driven damped harmonic oscillator. If the drive stays at the free-space resonance and the spring constant $k$ and effective mass $m$ are unchanged, its steady amplitude isHere ${\gamma }_0$ is intrinsic damping and ${\gamma }_{\mathrm{ts}}$ the added dissipative interaction with the surface. At large average separation there is no tapping, so ${\gamma }_{\mathrm{ts}}\simeq 0$ and $A=A_{\mathrm{free}}$. As the lower turning point reaches the surface, intermittent contacts dissipate energy and the amplitude decreases. With further approach the oscillation becomes strongly constrained, eventually approaching a continuously contacting, nearly quenched response. The damping-only sketch has a free-amplitude plateau followed by a descending amplitude on approach. The rounded transition in the sketch is illustrative; adhesion and actual contact mechanics determine its shape. The deliberately omitted force-gradient-induced resonance shift can make real approach curves more complicated.

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.

The transient is the homogeneous part of a damped harmonic oscillator, $m\ddot{x}+b\dot{x}+kx=0$. Define $\alpha =b/(2m)$ and ${\omega }_0=\sqrt{k/m}$. Its characteristic roots are $-\alpha \pm \sqrt{{\alpha }^2-{\omega }_0^2}$.

Initial position and velocity determine A,B. For an imposed force or support motion, the full solution is a particular response plus this transient, which decays for b>0 and k>0. The mechanical energy obeys $d(K+U)/dt=-b{\dot{x}}^2$. In light damping its cycle-averaged envelope decays as $e^{-2\alpha t}$, with $Q\simeq {\omega }_0/(2\alpha )$. Initial conditions can still produce an overshoot in a nonoscillatory regime; the damping category describes the roots rather than every possible trajectory.

Substituting $x=A{e}^{pt}$ into the damped harmonic oscillator equation givesFor nonnegative viscous damping, the cases areThe PDF's request to express these conditions in terms of $\gamma /m$ is dimensionally inconsistent with its definition $\gamma =b/m$: $\gamma$ already has units ${\mathrm{s}}^{-1}$. With the unnormalised equation $m\ddot{x}+b\dot{x}+kx=0$, the equivalent comparisons are $b/m\gtrless 2{\omega }_0$, or $b^2\gtrless 4mk$. Also ${\omega }_0$ is the undamped natural angular frequency, not in general the driven displacement peak.

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

Write $\gamma =b/(2m)$ and $\nu =\sqrt{{\omega }^2-{\gamma }^2}$, where $\omega =\sqrt{k/m}$ is the undamped angular frequency. The linear drag gives the vector damped harmonic oscillator equationFor the original circular initial conditions, ${\mathbf{r}}_0=(c,c)$ and ${\mathbf{v}}_0=\omega (-c,c)$, the exact underdamped solution isUsing $|{\mathbf{r}}_0{|}^2=2c^2$, $|{\mathbf{v}}_0{|}^2=2{\omega }^2{c}^2$ and ${\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$ givesThis exact potential energy is useful because it satisfies the prescribed initial conditions, including $\dot{U}(0)=k{\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$. In the stated weak-damping limit $\gamma \ll \omega$, $\nu \simeq \omega$ and the square bracket differs from one by a small oscillatory term. The leading approximation is thereforeThe radius is approximately $\sqrt{2}c{e}^{-bt/(2m)}$, so the trajectory is approximately a shrinking logarithmic spiral. The exactly exponential result would require slightly adjusted initial velocity; with the specified undamped circular velocity, the small modulation above is real. This is the initially circular damped planar oscillator.

In the Lorentz oscillator model, each participating bound electron is displaced by an applied electric field against a restoring force. The model replaces complicated interband transitions by a driven damped harmonic oscillator. If $u$ is the electron displacement along the field and the electron charge is $-e$, its equation of motion isHere ${\omega }_0$ describes binding and $\gamma >0$ describes loss of coherent motion. For $E(t)=\mathrm{Re}(E_0{e}^{-i\omega t})$, the complex amplitude isThe induced electric dipole moment is $-eu$, so the electric polarization amplitude for number density $n$ isUsing $D={ϵ}_{0}E+P={ϵ}_0{ϵ}_{r}E$ givesThe paper's $ϵ$ is this dimensionless relative permittivity, rather than the absolute permittivity ${ϵ}_0{ϵ}_r$. Independent classes of bound electrons contribute additive susceptibilities, usually with different resonance frequencies and optical oscillator strengths.

Writing $D_{\omega }=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$ separates the response intoAt zero frequency the real part is $1+{\omega }_p^2/{\omega }_0^2$ and the imaginary part vanishes. Near resonance the imaginary part has a positive photon absorption peak and the real part changes rapidly, passing through $1$ at ${\omega }_0$. Above resonance the real part can become negative when damping is sufficiently weak; at very high frequency it approaches $1$ from below while the imaginary part tends to zero. Finite damping removes the undamped divergence.

For a nonmagnetic material take the passive square-root branch $N=\eta +i\kappa =\sqrt{{ϵ}_r}$, with $\kappa \ge 0$. A plane wave then has an intensity attenuation coefficient $\alpha =2\omega \kappa /c$. Thus the imaginary permittivity expresses dissipation, whereas the actual photon absorption coefficient is determined by the imaginary refractive index. The normal-incidence power reflectance isA large refractive-index mismatch produces wave reflection even without loss. In the undamped model ${ϵ}_r<0$ between ${\omega }_0$ and $\sqrt{{\omega }_0^2+{\omega }_p^2}$, so the field is evanescent and wave reflection is total in a semi-infinite sample. With damping the response includes both wave reflection and photon absorption; a peak in the imaginary permittivity should not simply be identified with a reflectance peak.

To infer the short-wavelength background from $R=0.35$, assume weak photon absorption and take the physical semiconductor branch $\eta >1$. ThensoThis is an estimate of the valence-electron background in a frequency window above the doped-carrier response and below the effective bound-electron resonance. One reflectance value alone does not determine both parts of a complex dielectric constant. Even for real ${ϵ}_r$, a reciprocal-index branch gives the same $R$; the ordinary background index greater than unity selects the branch used here.

Using the supplied conventional cubic lattice parameter, the participating valence-electron number density isUsing $m=m_e$ for the stipulated free-electron comparison givesAn angular frequency “in eV” means its quantum energy $\hslash \omega$, not $h\omega$. No doped-carrier effective mass should be inserted into this valence-electron count.

The low-frequency limit of the bound-electron contribution obeysConsequentlyThis effective resonance is much larger than the quoted $0.2\mathrm{eV}$ band gap. There is no contradiction: the single oscillator has been assigned the strength of all sixteen participating valence electrons per conventional cell, whereas the band gap is only the lowest interband excitation threshold. A real spectrum distributes optical oscillator strength over many transitions. Equating the oscillator resonance to the lowest threshold would misrepresent the assumed background response. Here “low frequency” means low relative to ${\omega }_0$; the static response of the doped conducting sample also contains its free carriers.

The additional mobile carriers in doped indium antimonide contribute a Drude frequency response, obtained by setting their restoring frequency to zero. At infrared frequencies well below the effective valence resonance, the sum is approximatelyAt the shorter wavelengths the carrier term is small and $R$ approaches the background value. As wavelength increases, the negative free-carrier contribution lowers the real permittivity towards unity. For weak damping, refractive-index matching to vacuum gives the Drude-Lorentz reflectivity minimum atThis minimum is near ${ϵ}_r^{\prime }=1$, not its zero. At the slightly lower screened plasma frequency ${\omega }_{pc}/\sqrt{{ϵ}_b}$, the real permittivity crosses zero; at longer wavelengths it becomes negative and the material becomes strongly reflective. Finite damping prevents a perfectly vanishing minimum and rounds the edge. At still lower frequency the large conducting response also drives $R$ towards unity.

For fixed carrier effective mass and background permittivity, ${\omega }_{pc}\propto \sqrt{n_c}$ and ${\lambda }_{\min }\propto n_c^{-1/2}$. The observed shift of the minimum towards shorter wavelength as doping increases is therefore the expected free-carrier trend. The figure is an original model illustration, not a digitization or quantitative fit of the supplied curves; the carrier effective mass, damping and their density dependence have not been specified.

Put $y^{(0)}=y$ and $y^{(1)}=\dot{y}$. ThenThe second row is the original damped harmonic oscillator equation and the first row defines the auxiliary velocity. This converts a second-order ordinary differential equation into two first-order equations without changing the required two initial values.

Ringdown is the freely decaying motion of a resonator after an applied drive is removed. For an underdamped damped harmonic oscillator, the amplitude decays as $e^{-\gamma t}$ while the angular frequency is $\nu =\sqrt{{\Omega }^2-{\gamma }^2}$. In the weak-damping limit, the cycle-averaged stored energy decays as $e^{-2\gamma t}$, giving the quality factor from ringdown $Q\simeq \Omega /(2\gamma )$.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].