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The angular frequency of an underdamped mode isFor ordinary damping, $\alpha \ge 0$, mode $n$ oscillates precisely when $\alpha <cn\pi /L$. In particular, all modes, including the fundamental, oscillate when $\alpha <c\pi /L$. There are some oscillatory modes for every finite $\alpha$, since the allowed $n$ are unbounded; no finite damping coefficient overdamps the entire ideal string spectrum.

If $\alpha <0$, the same mode has negative damping: oscillatory amplitudes grow as $e^{|\alpha |t/2}$. For modes with real characteristic roots, both roots are positive, since $|\alpha |/2>\sqrt{{\alpha }^2/4-c^2{k}_n^2}$. Hence those modes also grow. Negative damping makes the equilibrium unstable.

Put ${\omega }_0^2=k/m$, where $k$ is the binding spring constant rather than an optical wavenumber. The steady complex amplitude solution of the forced Lorentz oscillator model isThe acceleration complex amplitude is $-{\omega }^2{\mathbf{x}}_0$. HenceThis is the scattering cross-section of a harmonically bound charge. The displacement and induced electric dipole moment are parallel to the incident linear polarization, and integrating their dipole radiation over all directions gives the total cross section above.

For $\omega \ll {\omega }_0$, binding dominates andrecovering Rayleigh scattering with static electric polarizability $q^2/k$. For $\omega \gg {\omega }_0$, inertia dominates: ${\mathbf{x}}_0\simeq -q{\mathbf{E}}_0/(m{\omega }^2)$, so its second time derivative becomes frequency independent andThis is a classical Thomson scattering regime only while $\hslash \omega \ll m{c}^2$. Exactly at $\omega ={\omega }_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite damping and radiation reaction regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.

The deductive-nomological model represents an explanation as a valid deduction of an event description from scientific laws and particular circumstances. In the schemathe premises are the explanans and $E$ the explanandum. At least one law must be essential to the derivation, the premises must have empirical content, and a correct explanation must have true premises. Carl Hempel and Paul Oppenheim distinguish these adequacy requirements from merely giving a valid argument. Their developed 1948 analysis concerns deductive explanation; its name should not be extended without qualification to every statistical explanation.^[ref]

The model captures something valuable. For an ideal simple harmonic oscillator, the equation of motion $m\ddot{x}=-kx$, a positive mass $m$, a positive spring constant $k$ and non-equilibrium initial conditions determine a periodThe derivation explains why the period changes with those parameters within the model. It is not just the report that an oscillation occurred. Since a real spring may have damping or nonlinear response, applying the explanation requires justification of the approximation. The model's truth condition cannot turn an idealized calculation into a literally exact account of every laboratory spring.

The account is not sufficient to settle explanatory direction. For a vertical pole of height $h$ on level ground, a solar elevation angle $0<\theta <\pi /2$ and parallel light rays give a shadow length $\ell =h\cot \theta$. Given $h$ and $\theta$, one can deduce $\ell$; given $\ell$ and $\theta$, one can equally deduce $h$. Both calculations can use the same true optical and geometrical assumptions. Yet normally the pole and illumination explain the shadow, whereas measuring the shadow provides evidence about the pole's height. Deducibility does not distinguish this evidential inference from a causal explanation. This is the problem of explanatory asymmetry.

Nor is strict deduction necessary for every good physical explanation. A radioactive decay model can explain a sample's decay rate and the chance of a nucleus decaying in a specified interval, while not entailing which particular nucleus decays first. The exponential distribution supplies a controlled statistical account. Adding the observed result as a premise would permit a deduction but would not supply the missing explanation. An inductive-statistical model is a different proposal, and even that needs an account of which statistical information is relevant.

These objections do not show that every explanation must instead recount a sequence of causes. Explanations based on conservation laws or symmetry can constrain a phenomenon without specifying a detailed mechanism. The important point is that the role of the premises, the question being asked and the relevant dependence matter, in addition to logical consequence. The deductive-nomological model is a useful pattern for some physical explanations, but it is not an adequate necessary-and-sufficient account of explanation throughout the physical sciences.

As $\omega$ approaches ${\omega }_0=\sqrt{g/l}$, the driving motion couples resonantly to the pendulum. The particular-solution amplitude has denominator ${\omega }_0^2-{\omega }^2$, so a steady harmonic-response formula alone is singular. For a specified finite initial state, the near-resonant motion instead develops beats, and its exact resonant limit has a growing envelope.

For example, with the vertical zero-angular-velocity initial condition above, taking the limit givesThe ideal undamped linear theory predicts amplitude growing linearly with time, not an infinite displacement at the first instant. In reality the motion leaves the small-angle regime, and damping and nonlinear changes of the pendulum's oscillation period alter the response. Below and above resonance, the sign of the harmonic response coefficient changes, corresponding to a phase reversal.

For weak damping, the ringdown comparison bound for two resonator modes gives $I(t)=I(0)e^{-\omega t/Q}$. Interpreting the stated loudness comparison as an intensity comparison,Therefore the information gives an upper bound, not a unique value:Equality would describe equal intensities at two seconds; the stated strict inequality allows every smaller positive $Q_L$ consistent with the weak-damping model.

The restoring, damping and electric forces on the electron give $m_e\ddot{x}=-kx-b\dot{x}-eE(t)$. Dividing by $m_e$ gives a Lorentz oscillator model withFor the $e^{i\omega t}$ convention, substitute $x=\mathrm{Re}(x_0{e}^{i\omega t})$. Then $x_0=-e{E}_0/[m_e({\omega }_0^2-{\omega }^2+i\gamma \omega )]$ andThe time-averaged dissipated power isFor positive frequency the squared term is minimized to zero only at $\omega ={\omega }_0$. Thus the power maximum is exactly at the undamped natural frequency, even though the displacement maximum is slightly shifted. Here $P_{\max }=e^2|E_0{|}^2/(2b)$. The half-power equation is $|{\omega }_0^2-{\omega }^2|=\gamma \omega$, whose two positive roots areThis is the absorbed-power bandwidth of a viscously damped oscillator, exact within the model. Near resonance the profile is approximately a Lorentzian line shape. With $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$,For this wavelength linewidth of a damped-oscillator power resonance we used ${\omega }_{+}{\omega }_{-}={\omega }_0^2$. The half-power wavelength endpoints are not exactly symmetric around ${\lambda }_{\mathrm{peak}}$, although that distinction is tiny here.

In a constant field the settled displacement is $x_s=-e{E}_{\mathrm{dc}}/k$. On sudden removal, with zero initial velocity, the free motion isThe electron oscillates about its original equilibrium, losing the previously stored spring energy. The period is $3.52\mathrm{fs}$, the amplitude decay time is $2/\gamma =0.319\mathrm{ps}$, and the cycle-averaged energy decay time is $1/\gamma =0.160\mathrm{ps}$. This is the effective classical model supplied, not a prediction of the full hydrogen line radiation spectrum.

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.

For $b^2<4mk$, put ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. Complex-conjugate roots give the real solutionThis is underdamping: oscillation at ${\omega }_d$ persists within a decaying exponential envelope. damping reduces the oscillation frequency below ${\omega }_0$. Constants in all three cases are determined by two initial conditions.

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.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

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:

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.

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.

For underdamping, the general response is a free transient plus a particular driven response. Use the paper's $e^{i\omega t}$ convention. The steady displacement complex amplitude isThusHere $\gamma =b/(2m)$, ${\omega }_d=\sqrt{k/m-{\gamma }^2}$ and the real constants $C,D$ enforce the initial conditions. For a real positive $F_0$, the steady term is $|x_0|\cos (\omega t-\varphi )$, whereFor light damping, ${\omega }_d\simeq {\omega }_0$ and the transient decays slowly compared with an oscillation. The two-argument phase convention makes the displacement lag evolve from approximately zero below resonance through $\pi /2$ near ${\omega }_0$ to approximately $\pi$ far above it.

The response is quasistatic at low frequency, resonantly enhanced when the spring and inertia nearly cancel, and suppressed as $1/{\omega }^2$ at high frequency. The original graph gives approximate readings $x_s=|x_0(0)|\simeq 0.10\mathrm{mm}$, $x_p\simeq 1.0\mathrm{mm}$ and $f_r\simeq 150\mathrm{Hz}$. These are graph estimates, not exact supplied measurements.

Use oscillator parameters from static and resonant displacement. The static spring constant isFor light damping, ${\omega }_0\simeq {\omega }_r=2\pi (150)\simeq 9.42\times 10^2{\mathrm{s}}^{-1}$, givingNear resonance, $x_p\simeq F_0/(b{\omega }_0)$, soIndependently, the resonance enhancement $x_p/x_s\simeq 10$ estimates the same quality factor. The half-power displacement level is $x_p/\sqrt{2}\simeq 0.71\mathrm{mm}$; its intersections are approximately $142$ and $157\mathrm{Hz}$, giving bandwidth $\Delta f\simeq 15\mathrm{Hz}$ and $Q\simeq f_0/\Delta f\simeq 10$.

The exact displacement-peak frequency is ${\omega }_r^2={\omega }_0^2-b^2/(2m^2)$. At $Q\simeq 10$ its difference from ${\omega }_0$ is only about $0.25\%$, well below the precision justified by this printed curve. Greater numerical precision for the inferred parameters would be misleading.

Choose positive electric current in the initial discharge direction, since the source diagram does not label the initial plate polarity or a current arrow. Let $q$ be the charge remaining on the initially positive plate, so $I=-\dot{q}$. Kirchhoff's voltage law for the series RLC circuit givesThe natural angular frequency, damping coefficient and quality factor areSince $\alpha <{\omega }_0$ (equivalently $Q>1/2$), this is an underdamped RLC response. With ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}=815.82\mathrm{rad}{\mathrm{s}}^{-1}$, satisfying the initial conditions givesand differentiating givesThe electric current starts at zero with $\dot{I}(0)=V_0/L=100\mathrm{A}{\mathrm{s}}^{-1}$. Its wave period is $T_d=2\pi /{\omega }_d=7.702\mathrm{ms}$, its zeros occur every $3.851\mathrm{ms}$, and its positive and negative wave envelopes decay with time constant $30.0\mathrm{ms}$. The first maximum obeys $\tan {\omega }_{d}t={\omega }_d/\alpha$ and is $0.1151\mathrm{A}$ at $1.875\mathrm{ms}$. The electric current tends to zero as the initial capacitor energy is dissipated in the resistor.

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.

A normal mode is a motion in which all displacement coordinates oscillate at one angular frequency with fixed amplitude ratios. For small conservative oscillations, the symmetric stiffness and mass matrices give the generalized eigenvalue problem for small oscillations, $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. For positive mass and a stable equilibrium, its mass-orthogonal eigenvectors span the motion. A superposition of these modes, with phases and amplitudes fixed by initial displacements and velocities, gives the general solution; different modes need not share a phase.

Use upward end displacements. The spring compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The potential energy isStatic force balance gives $2k{x}_0=mg$, cancelling the linear terms and leavingFor a uniform beam, its central moment of inertia in the vertical plane is $m{l}^2/12$. To first order in the tilt, $\theta =(x_2-x_1)/l$. Its translational and rotational kinetic energies areThus $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. The normal modes of a rigid beam on two end springs have eigenvectorsThe first translates without tilt; the second rocks about a stationary centre. Ordinary frequencies are ${\omega }_s/(2\pi )$ and ${\omega }_a/(2\pi )$.

Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static spring load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial velocities vanish because no impulse is imparted. Decomposing into the two normal modes givesOnly the symmetric mode contributes to the centre position:The datum $z_{\mathrm{eq}}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll 1$, ideal light vertical springs and negligible damping.