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A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

For a uniform rigid beam of mass $m$ and two identical vertical end springs of stiffness $k$, small vertical end displacements have kinetic energy $m({\dot{x}}_1^2+{\dot{x}}_1{\dot{x}}_2+{\dot{x}}_2^2)/6$ and displacement-dependent potential energy $k(x_1^2+x_2^2)/2$. The mass and stiffness matrices are $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. Their generalized eigenvalue problem gives symmetric translation $(1,1)$ and antisymmetric rocking $(1,-1)$ at the displayed angular frequencies. The centre coordinate contains only the symmetric mode. Neglect spring inertia, damping and finite-angle geometric corrections.

Optical molasses surrounds slow chemical atoms with opposing red-detuned laser beams, usually three pairs along perpendicular axes. In the elementary Doppler cooling picture, each pair damps one component of velocity. The result resembles viscous drag, even though the atoms are moving through a vacuum rather than a liquid.

Uniform balanced optical molasses supplies velocity damping, without a restoring force towards a chosen position. An atom at rest has zero mean force wherever it sits within the uniform illuminated region. Without noise or external forces, $m\dot{v}=-\alpha v$ givesThe atom approaches a position determined by its initial condition, not the beam intersection. Recoil-driven diffusion and gravity eventually remove atoms from finite beams. A long residence time in optical molasses is therefore compatible with the absence of position trapping. An optical trap provides spatial confinement by a different force mechanism; a magneto-optical trap adds a position-dependent magnetic field and suitable beam electromagnetic polarization to create a restoring force.

For a loop of area $A$ and self-inductance $L$, negligible resistance conserves $LI+{\Phi }_{\mathrm{ext}}$. With zero trapped total flux and ${\Phi }_{\mathrm{ext}}=A{B}_0\sin (\omega t)\cos \theta$, the electric current is $I=-A{B}_0\sin (\omega t)\cos \theta /L$. The magnetic torque generalized to the angle between loop normal and field is ${\tau }_{\theta }=-IAB\sin \theta$. For an orientation varying slowly over a field cycle, averaging gives the displayed torque and effective potential $U_{\mathrm{av}}=A^2{B}_0^2{\cos }^2\theta /(4L)$. Its stable orientations have loop normal perpendicular to the applied field. Without damping, a freely released loop librates around such an orientation instead of settling there.

An oscillation is a repeated variation about an equilibrium or central value. A normal mode describes a single frequency of oscillation.

An overdamped RLC response is a free series RLC circuit response with $R>2\sqrt{L/C}$, equivalently $\alpha >{\omega }_0$. Put $\beta =\sqrt{{\alpha }^2-{\omega }_0^2}$. The distinct negative characteristic roots are $r_{\pm }=-\alpha \pm \beta$, so the general electric charge isFor initial conditions $q(0)=q_0$, $i(0)=0$,For $q_0>0$, this electric charge decreases to zero without reversing sign. In the dimensionless example ${\omega }_0=1$, $\alpha =2$, the characteristic roots are $-2\pm \sqrt{3}\approx -0.268,-3.732$. The slow term dominates late times. Stronger damping can therefore make the final return slower: for $\alpha \gg {\omega }_0$, $r_{+}\sim -{\omega }_0^2/(2\alpha )=-1/(RC)$. Nonoscillation alone does not imply a fast response, and monotonicity here uses the specified initial conditions.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.

The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.

Seek the steady forced displacement as $x=x_0{e}^{-i\omega t}$. The oscillator equation givesFor $n$ ion pairs per volume, the ionic polarization is $P_{\mathrm{ion}}=nqx$. Add the electronic polarization ${ϵ}_0({ϵ}_{\infty }-1)E$, assumed frequency-independent over this infrared range. The complex relative permittivity is thereforeFor a longitudinal free oscillation without external free charge, $\mathbf{k}\cdot \mathbf{D}=0$ and the longitudinal electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$ in the lossless limit. HenceEliminating the microscopic oscillator strength givesHere ${\omega }_{\mathrm{LO}}$ denotes the undamped longitudinal frequency; with finite damping the zero is at a complex frequency. The sign of the imaginary term follows the stated $e^{-i\omega t}$ convention and gives positive absorptive $\mathrm{Im}{ϵ}_r$ for $\omega >0$, $\gamma >0$. At zero frequency the same model yields the Lyddane-Sachs-Teller relation ${ϵ}_r(0)/{ϵ}_{\infty }={\omega }_{\mathrm{LO}}^2/{\omega }_{\mathrm{TO}}^2$. The high-frequency limit here is above the phonon resonance but below electronic resonances, not arbitrarily high photon energy.

At normal incidence from vacuum the reflectivity iswith the passive branch of the complex refractive index. Below the optical resonance the real permittivity rises towards its pole and reflectance increases. Between the TO and LO frequencies, the lossless permittivity is negative; the wave is evanescent and the reflectivity is unity in that idealization. Finite damping rounds and reduces this reststrahlen band. Above the LO edge, positive permittivity returns and reflectivity can become very small when the refractive index is near one, before rising towards its electronic-background value.

Reading the surviving source graph to its available precision givesThese are band-edge estimates, not a high-precision damping fit. The minimum immediately above the band is not exactly the LO zero: for zero damping ${ϵ}_r=1$ occurs at ${\omega }^2=({ϵ}_{\infty }{\omega }_{\mathrm{LO}}^2-{\omega }_{\mathrm{TO}}^2)/({ϵ}_{\infty }-1)>{\omega }_{\mathrm{LO}}^2$.

The high-energy plotted tail near $90\mathrm{meV}$ has $R\simeq 0.22$. Ignoring residual phonon dispersion gives $n\simeq (1+\sqrt{R})/(1-\sqrt{R})=2.77$ and $ϵ\simeq 7.65$. Correcting with the finite-energy oscillator ratio givesThus a background of order $8$–$9$ is supported; equating the still-dispersive tail exactly to the asymptotic background would slightly underestimate it. The original model curves below use these approximate energies, without copying or fitting individual source data points.

For a transverse electromagnetic plane wave in a nonmagnetic medium, Maxwell's equations give $c^2{k}^2={\omega }^2{ϵ}_r(\omega )$. Set damping to zero and define $u=c^2{k}^2/{ϵ}_{\infty }$. Substitution yieldsso ${\omega }^4-({\omega }_{\mathrm{LO}}^2+u){\omega }^2+u{\omega }_{\mathrm{TO}}^2=0$. Therefore the two phonon polariton branches areTake the nonnegative square root for the positive physical frequencies. At $k=0$, ${\omega }_{-}=0$ and ${\omega }_{+}={\omega }_{\mathrm{LO}}$. At small wavevector the lower branch is photon-like with velocity $c/\sqrt{{ϵ}_r(0)}$; at large wavevector it approaches ${\omega }_{\mathrm{TO}}$ from below. The upper branch ultimately becomes photon-like with velocity $c/\sqrt{{ϵ}_{\infty }}$. There is no real propagating transverse branch with frequency strictly between ${\omega }_{\mathrm{TO}}$ and ${\omega }_{\mathrm{LO}}$ in this lossless bulk model. Coupling changes the eigenmodes from a bare photon crossing a flat oscillator into mixed normal modes.

For zero damping the conserved mechanical energy isFor $\gamma <0$, the potential grows to infinity in both directions, so every nonstationary finite-energy trajectory is trapped between two turning points and is periodic; the zero-energy solution is stationary. The case $\gamma =0$ is the harmonic oscillator. The negative-$\gamma$ comparison extends the parameter range initially stated in the source.

For $\gamma >0$, the origin lies in a local potential well, with barriers at $x=\pm {\gamma }^{-1/2}$ of height $1/(4\gamma )$. A trajectory launched from the origin oscillates precisely when its initial energy is below that barrier:At equality the trajectory approaches a barrier equilibrium asymptotically, rather than being periodic; for example the positive branch is $x={\gamma }^{-1/2}\tanh (t/\sqrt{2})$. Above the barrier the trajectory escapes the well. The source's inequality without the absolute value tacitly takes $v\ge 0$; for signed initial velocity the energy criterion requires $|v|$.

Next set $\beta =\gamma =0$ and assume $\alpha >0$. The energy $H=({\dot{x}}^2+x^2)/2$ satisfies $\dot{H}=-\alpha x^2{\dot{x}}^2\le 0$. This bounds the complete trajectory. The zero-dissipation set is the union of $x=0$ and $\dot{x}=0$, and its only invariant trajectory is the origin: a nonzero point on either axis immediately leaves it under the differential equation. The LaSalle invariance principle therefore gives $\boxed{x(t)\to 0,\dot{x}(t)\to 0}$. If $\alpha =0$ as well, damping is absent and that conclusion fails for nonzero $v$; the positive order-one damping assumption is essential.

For the weakly nonlinear periodic motion, write $\gamma =\alpha \kappa$, set $\tau =\omega t$, $\omega =1+\alpha {\omega }_1+\cdots$, and expand $x=A\cos \tau +\alpha x_1+\cdots$. The first correction obeysResonant forcing at the fundamental frequency would generate secular terms. Eliminating both coefficients yieldsConsequently the maximum displacement is $\boxed{2\sqrt{\beta }+O(\alpha )}$ for fixed positive $\beta$. A possible third-harmonic correction is $x_1=-A^3\sin 3\tau /32-\kappa A^3\cos 3\tau /32$, up to homogeneous amplitude and phase choices. The leading amplitude also balances averaged damping: its slow evolution is $\dot{A}=\alpha A(\beta /2-A^2/8)$, whose nonzero fixed point is attracting. The case $\beta =0$ is a degenerate zero-amplitude limit, rather than a nonzero limit cycle.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

There is no Landau damping of this branch within the completely degenerate collisionless model. Its phase velocity obeys $\omega /k>V_e>V_i$, while each parallel marginal distribution vanishes outside $|u|\le V_{\alpha }$. There are no particles at the resonant velocity, and $F_{\alpha }^{\prime }(\omega /k)=0$. The Landau contour consequently contributes no imaginary part to the dielectric function; a real root is allowed with $\boxed{\gamma =0}$.

If a small nonzero temperature supplied a tail beyond the Fermi surface, the conclusion would change: the resonant derivative would generally be nonzero. For the convention $p=\gamma -i\omega$ and a simple weakly damped root, expanding the continued dielectric function would give $\gamma =-{ϵ}_I(\omega ,k)/{\partial }_{\omega }{ϵ}_R(\omega ,k)$. This conditional formula is not needed to manufacture damping for the exact cutoff. Neither collisional attenuation nor nonlinear evolution is included in the given dispersion relation.

The two real branches start at $\omega =\pm {\omega }_{\mathrm{p}e}$, have zero initial slope, and approach the lines $\omega =\pm k{V}_e$ from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient $3V_e^2/5$ in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has $d{P}_e/d{n}_e=m_e{V}_e^2/3$, suggesting, at long wavelengths,Both species are degenerate here, so $V_i=(m_e/m_i)V_e$. The hierarchy $V_i\ll c_s\ll V_e$ explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at $u=c_s$, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order $\sqrt{m_e/m_i}$, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to $c_s^2$. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

Take an ideal flexible mechanical string with constant tension $T>0$ and uniform linear mass density $\mu >0$. Let $x$ measure distance along its straight equilibrium position and let $y(x,t)$ be its small transverse displacement. Neglect damping, bending stiffness and external distributed forces, and assume $|y_x|\ll 1$.

A segment of reference length $\Delta x$ has mass $\mu \Delta x$. The endpoint tension forces have transverse resultantsince $\sin \theta \simeq \tan \theta =y_x$ for small slopes. Newton's second law, divided by $\Delta x$ and taken to the limit, givesHere $t$ is time, the subscripts denote partial derivatives, and $c$ is the wave speed. The transverse acceleration is $y_{tt}$; $y_{xx}$ describes the small-slope curvature responsible for the net transverse force.

A normal mode of a stable undamped linear coupled oscillator has a single angular frequency and a fixed ratio of component displacements. All components share a sinusoidal time factor, with relative oscillation phases of zero or $\pi$ for real mode shapes. A general free motion is a superposition of the normal modes. For this system the symmetric stiffness matrix gives real eigenvalues and orthogonal eigenvectors; positive spring constants make the squared angular frequencies positive.

Write $m=m_1=m_2>0$ and $\kappa =k_{12}$. Assume ideal massless linear springs, negligible damping and positive $k_1,k_2,\kappa$. By Hooke's law and Newton's second law,The coupling spring extends by $x_2-x_1$, giving equal and opposite coupling forces.

For a normal mode, take $\mathbf{x}=\mathrm{Re}(\mathbf{a}{e}^{i\omega t})$. Nonzero eigenvectors requireSolving this quadratic equation in $m{\omega }^2$ givesA corresponding mode shape has ratioand the general real solution is $\mathbf{x}=C_{-}{\mathbf{a}}_{-}\cos ({\omega }_{-}t+{\varphi }_{-})+C_{+}{\mathbf{a}}_{+}\cos ({\omega }_{+}t+{\varphi }_{+})$. The arbitrary amplitudes and oscillation phases account for four independent initial data.

Weak coupling alone does not guarantee large beats. A useful representative sketch assumes nearly equal uncoupled angular frequencies. In the resonant case $k_1=k_2=k$ with $\kappa \ll k$, the mode shapes are $(1,1)$ and $(1,-1)$, andFor $x_1(0)=A$, $x_2(0)={\dot{x}}_1(0)={\dot{x}}_2(0)=0$, both normal modes have equal coefficients. With $\bar{\omega }=({\omega }_{+}+{\omega }_{-})/2$ and $\Delta \omega ={\omega }_{+}-{\omega }_{-}$,These rapid oscillations have slowly varying complementary wave envelopes. The initial displacement shifts almost entirely to the other mass at $t=\pi /\Delta \omega$, and returns after $2\pi /\Delta \omega$. Arbitrary initial data give a different superposition; if $|k_1-k_2|\gg \kappa$, the normal modes are localized mainly on individual masses, and strong exchange is suppressed.

Holding one mass fixed leaves the coupling spring attached and stretched by the other mass. Therefore the two constrained angular frequencies areThese differ from the angular frequencies obtained by physically removing the coupling spring. Ordinary frequencies in hertz are each corresponding $\omega /(2\pi )$.

Finally, for $k_1>k_2$ and $\kappa >0$, put $D=\sqrt{(k_1-k_2{)}^2+4{\kappa }^2}>k_1-k_2$. ThenSince ${\omega }_1>{\omega }_2$, taking positive square roots provesAt $\kappa =0$, the outer inequalities become equalities.

Use the same difference $w$ and wave energy as in part (i). The damped wave equation now gives the energy dissipation identity for a linearly damped stringBecause $E(0)=0$ and $E(t)\ge 0$, the energy remains zero. Hence $w_t=w_x=0$, and the homogeneous Dirichlet boundary conditions force $w=0$. Therefore the solution of a differential equation is unique also for $\boxed{\alpha >0}$; positive damping can only decrease this energy.

Put $C=q^2/(4\pi {ϵ}_0)>0$. The spring exerts the force $k(\ell -x)$ by Hooke's law, while Coulomb's law gives the attractive force $-C/x^2$. Newton's second law therefore gives the equation of motionFor $x>0$, the derivative is $f^{\prime }(x)=-k+2C/x^3$. Thus $f$ increases up to $x_{\ast }=(2C/k{)}^{1/3}$ and decreases thereafter. Also $f(x)\to -\infty$ at both ends of the positive half-axis. Its unique maximum isThere are therefore two equilibrium points precisely whenWriting them $x_{-}<x_{\ast }<x_{+}$, the linearization $m\ddot{\eta }=f^{\prime }(x_{\pm })\eta$ has exponentially growing solutions at $x_{-}$ and oscillatory solutions at $x_{+}$. Equivalently, the potential energy has a strict local maximum at $x_{-}$ and a strict local minimum at $x_{+}$. Hence the smaller equilibrium point is an unstable equilibrium; the larger is a stable equilibrium, with no damping to make it attracting.

At equality $\ell ={\ell }_1$ there is one degenerate equilibrium point $x_{\ast }$; it is unstable, since a displacement to its left gives a leftward force. For $\ell <{\ell }_1$ there are no equilibrium points.