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For the linear three-reservoir carbon model, with concentrations as a column vector, the linear ordinary differential equation isA homogeneous normal mode has $\mathbf{x}=\mathbf{v}{e}^{\lambda t}$, where $J\mathbf{v}=\lambda \mathbf{v}$. For negative real eigenvalues the e-folding relaxation time is $\tau =-1/\lambda$. The row sums vanish, so ${\mathbf{v}}_3=(1,1,1{)}^T$ has ${\lambda }_3=0$ andEqual effective concentrations generate no exchange. Summing the reservoir budgets givesThus injected carbon dioxide is conserved after emissions cease: the model redistributes it but contains no permanent removal. The zero normal mode records this conserved excess mass, rather than a long but finite decay.

To check the other normal modes are damped, set $D=\mathrm{diag}(R_a,R_b,R_d)$ and $L=-DJ$. ThenThe similarity transformation $D^{1/2}J{D}^{-1/2}=-D^{-1/2}L{D}^{-1/2}$ is a symmetric negative-semidefinite matrix with a one-dimensional kernel, so the remaining two eigenvalues are strictly negative.

Define the two concentration differences $u=C_a-C_b$ and $v=C_b-C_d$. For zero emissions and $R_a=R_b=R$, their ordinary differential equation isWriting the positive decay rates as ${\mu }_i=1/{\tau }_i$, their characteristic equation isConsequentlyFor $r\gg s$, the first normal mode rapidly equalizes the fast reservoirs, giving ${\mu }_1\simeq 2r/R$. The slower evolution treats them as a combined capacity $2R$ exchanging with $R_d$, giving ${\mu }_2\simeq s[1/(2R)+1/R_d]$. ThereforeThe timescale ratio is ${\tau }_2/{\tau }_1\simeq 4r/s\gg 1$ in the joint limit. Since the conserved normal mode has ${\tau }_3=\infty$, the hierarchy is ${\tau }_1\ll {\tau }_2\ll {\tau }_3$.

Here $R=7.8$, $R_d=78$, $r=0.4$, and $s=0.04$, in the stated reservoir units. The exact characteristic coefficients are $P=0.10820513{\mathrm{yr}}^{-1}$ and $Q=0.000315582{\mathrm{yr}}^{-2}$, givingThe leading separation estimates give $9.75$ yr and $325$ yr if finite $R_d$ is retained, or $390$ yr if $R_d/R$ is also sent to infinity. Thus ${\tau }_1\ll 100\mathrm{yr}<{\tau }_2$; a ratio of ten warrants using the exact values for the requested sketch.

Assume initially zero anomalies in all reservoirs. The atmospheric impulse response, obtained by decomposing $e^{Jt}$ into the three normal modes, iswith $w_0=1/93.6=0.01068376$, $w_1=0.06087812$, and $w_2=0.05664324$, in ppmv/GtCO${}_2$. The checks $w_0+w_1+w_2=1/R_a$ and $w_1/{\tau }_1+w_2/{\tau }_2=r/R_a^2$ fix the instantaneous injection response and its initial decay.

The concentration is the convolution $C_a(t)={\int }_0^{t}G(t-t^{\prime })E(t^{\prime })d{t}^{\prime }$. To evaluate it transparently, define $H_0(q)=q^2/2$ for $q>0$, andsetting every $H_i(q)=0$ when $q\le 0$. Expressing the triangular emission history as $E(t)=0.5[t_{+}-2(t-100{)}_{+}+(t-200{)}_{+}]$ givesHere times are in years and $0.5$ has the corresponding emission-ramp units. Differentiating or evaluating the convolution givesThe maximum concentration occurs earlier than the end of emissions, at approximately year $172$, with $C_a\simeq 280.7$ ppmv. Initially there is a short adjustment over ${\tau }_1$; the curve then rises, turns over as emissions fall, and continues declining after year 200 as the deep ocean takes up carbon dioxide.

The total emitted mass is the area of the emission triangle, $\frac{1}{2}(200)(50)=5000$ GtCO${}_2$. All reservoirs eventually share one concentration, so conservation of mass gives

Use the flat two-layer model, with thickness $h=30\mathrm{km}$, crustal shear speed ${\beta }_1=3.5\mathrm{km}{\mathrm{s}}^{-1}$ and mantle speed ${\beta }_2=4.5\mathrm{km}{\mathrm{s}}^{-1}$. For SH-wave travel times in a crustal waveguide, unfolding each reflected path gives all reflection orders at once:with $x$ in kilometres and times in seconds. The direct wave begins at the origin; reflected orders have intercepts $60m/3.5=17.14m\mathrm{s}$ and approach the direct-wave slope at large distance.

The critical angle, measured from the vertical, is $i_c=\arcsin (3.5/4.5)\simeq 51.1^{\circ }$. Associated critical refraction and its multiples giveEach head-wave line is tangent to its reflection curve at the critical distance. In particular the primary head wave exists before it becomes the first arrival; it overtakes the direct wave only around $170\mathrm{km}$.

The slope of a reflected travel-time curve is $p=d{t}_m/dx=\sin i/{\beta }_1$. For $x<x_{c,m}$, $p<1/{\beta }_2$, so the mantle supports outgoing shear radiation and successive reverberations leak energy. For $x>x_{c,m}$, $1/{\beta }_2<p<1/{\beta }_1$: the mantle field is evanescent and ideal interface reflection is total. These leaking and trapped SH-wave reverberations explain the annotations in the original travel-time drawing below. A reflected ray is not itself a separate Love mode: constructive interference of the trapped contributions, including reflection phase, selects the normal modes.

For a horizontally propagating mode with ${\beta }_1<c<{\beta }_2$, define vertical wavenumbersThe free upper surface and welded interface impose the Love-wave dispersion conditionwhere ${\mu }_j={\rho }_j{\beta }_j^2$ are shear moduli. Each phase-velocity branch falls towards $3.5\mathrm{km}{\mathrm{s}}^{-1}$ with increasing frequency. Higher branches join $4.5\mathrm{km}{\mathrm{s}}^{-1}$ at their low-frequency cutoffs. At cutoff $s\to 0$, soThus the first three overtones have maximum periods about $10.77$, $5.39$ and $3.59\mathrm{s}$. The fundamental mode has $n=0$ and no positive-frequency cutoff in this flat layer-over-half-space model: it approaches $c={\beta }_2$ as $f\to 0$, with ever-deeper penetration. The phase-velocity sketch uses an explicitly illustrative equal-density choice, because the question gives speeds but no density or rigidity ratio. That missing ratio changes detailed branch curvature, not the cutoff values or the requested branch topology.

For the actual finite spherical Earth, the very-long-period continuation must instead be described by discrete toroidal normal modes, not a flat infinite waveguide at arbitrary wavelength. The flat-model answer therefore does not assert a continuum of arbitrarily low-frequency global nonrigid oscillations. A global lowest oscillation frequency cannot be computed from the supplied crust-and-half-space data alone.^[ref]

First-motion source mechanisms. Locate an earthquake from arrival times and use a seismic velocity model to calculate each station's source takeoff direction. Measure the initial direct P wave polarity after checking instrument sign, component orientation, timing and phase identification. An upward trace at an instrument is not automatically a source compression without the propagation and component convention. Plot the inferred source radiation signs on a focal sphere, commonly using a lower-hemisphere equal-area projection, and seek nodal planes separating positive and negative observations.

For a shear dislocation with orthogonal unit slip and fault-normal vectors $\mathbf{s},\mathbf{n}$, the double-couple fault source has $M=M_0(\mathbf{s}{\mathbf{n}}^{\mathsf{T}}+\mathbf{n}{\mathbf{s}}^{\mathsf{T}})$. Its far-field P radiation sign is that ofThe two zeros are orthogonal nodal planes; the four quadrants alternate between outward and inward initial radiation. Their orientations and quadrant signs determine candidate strike, dip and rake. Normal, reverse and strike-slip mechanisms give different projected patterns. The earthquake pressure and tension axes are the negative and positive moment-tensor principal directions, respectively; they do not by themselves measure the complete regional stress tensor.

There is a fundamental fault-plane ambiguity: exchanging $\mathbf{s}$ and $\mathbf{n}$ leaves the radiation tensor unchanged. First motions therefore give a fault plane and an auxiliary plane without saying which actually slipped. Use mapped rupture, aftershock geometry or independent geodetic displacement to select the physical fault where possible. Polarity signs alone do not measure scalar seismic moment, source duration or rupture extent. A poorly sampled focal sphere or many near-nodal observations can admit several mechanisms rather than one defensible beachball.

Propagate uncertainty in event depth, velocity gradients and takeoff angles, allow for mispicked or reversed polarities, and report the range of mechanisms consistent with the observations. A primary method did this by testing alternative locations and velocity models and validated its uncertainty estimates using noisy synthetic data and clusters of events with similar waveforms.^[ref] Broadband waveforms and amplitude ratios add information, especially for non-double-couple sources, but they are additional measurements rather than a way to extract amplitude information from a sign-only data set.

Radial density structure. In an isotropic adiabatic material, the seismic parameter isFor a chemically homogeneous region whose background stratification is adiabatic, $d\rho /dP=\rho /K_S=1/\Phi$. Combine this with spherical hydrostatic balance and mass conservation to obtain the Adams-Williamson equationsThe inward density increase follows compressibility under the increasing pressure. A specified starting density and measured $\Phi (r)$ allow integration within a homogeneous adiabatic layer. Seismic velocities alone, however, do not determine that density normalization, since they contain ratios of moduli to density. Chemical boundaries, phase transitions, nonadiabatic thermal gradients and anisotropy require additional modeling; a composition jump is not obtained by blindly integrating the smooth relation across it.

Use travel times and surface waves to constrain elastic velocities, reflected/converted phases to locate interfaces, and normal-mode frequencies to constrain density as well as elastic structure through their sensitivity kernels. Impose independently known total mass and moment of inertia:Because these weight depth differently, they help distinguish admissible profiles, but two integrals alone do not specify an arbitrary radial function. Laboratory equations of state and mineral physics restrict composition, compressibility and density jumps. pressure and enclosed mass are matched across interfaces, while density may jump. In a liquid region $\beta =0$ and $\Phi ={\alpha }^2$, so the shear-wave absence constrains phase state without itself providing a direct density measurement.

The Preliminary Reference Earth Model is a concrete joint inversion of seismic travel times, normal modes, mass and inertia, including density and attenuation along with elastic structure. The original analysis combined these complementary data rather than deriving density uniquely from P wave speed.^[ref] A defensible radial density model combines hydrostatic/thermodynamic relations with seismic and global integral constraints, and states where homogeneity and adiabaticity fail.

For small oscillations about equilibrium, put $M_{ij}=T_{ij}(0)$ and $K_{ij}={\partial }_i{\partial }_{j}V(0)$. The potential has no linear term at equilibrium; to quadratic order the Lagrangian is $L_2={\dot{x}}^{\mathsf{T}}M\dot{x}/2-x^{\mathsf{T}}Kx/2$. Its equations are $M\ddot{x}+Kx=0$. A normal mode $x=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem $(K-{\omega }^{2}M)v=0$. For a positive-definite mass matrix, diagonalization $M^{-1/2}K{M}^{-1/2}$ gives the squared frequencies and independent normal coordinates. A strictly positive Hessian gives positive frequencies; zero eigenvalues require separate higher-order stability analysis.

Here the spring extensions are $x_1$, $x_2-x_1$, and $-x_2$. ThusThe stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and $M=mI$. Its eigenvectors $(1,1)$ and $(1,-1)$ have stiffness eigenvalues $k$ and $3k$, givingIn the first mode the masses move together with equal displacements and the middle spring keeps its length. In the second they move oppositely with equal amplitudes, stretching and compressing the middle spring twice as much as either end spring.

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.

In the three-dimensional Debye model, the lattice has $3N$ normal modes for $N$ atoms. Replacing their dispersion by a linear acoustic dispersion up to a cutoff gives the Debye heat capacityAt low temperature the upper limit can be replaced by infinity. Since the integral then equals $4{\pi }^4/15$,At high temperature each harmonic mode contributes $k_B$ by the equipartition theorem, giving the Dulong-Petit lawFor one mole replace $N{k}_B$ by $R$. The curve starts with zero slope, rises as $T^3$, and levels off toward the classical limit. This is the lattice contribution; a metal's electronic heat capacity is a separate low-temperature term.

The Debye wavevector follows by counting three polarizations inside the cutoff sphere:With the supplied speed of sound treated as the effective Debye speed, ${\omega }_D=v{k}_D$ andHere $k_D\simeq 1.52\times 10^{10}{\mathrm{m}}^{-1}$. Real crystals have different longitudinal and transverse sound speeds; the one-speed estimate uses the stated value, rather than inferring those unprovided branches.

A primitive cell containing two atoms has two independent displacement coordinates for each chosen polarization. Their coupled normal modes split into an acoustic phonon branch, which translates the cell at long wavelength, and an optical phonon branch, which moves its constituents against one another. The nonzero restoring force for the relative displacement gives a nonzero optical frequency even at the zone centre.

For an explicit scalar model, take alternating masses $M,m$, equal nearest-neighbor spring constant $K$, and cell length $a$. If $u_n,v_n$ are the displacements of the two atoms in cell $n$, Newton's second law givesA mode proportional to $e^{i(kna-\omega t)}$ has a nonzero amplitude only when the two-by-two dynamical determinant vanishes. The resulting diatomic-chain dispersion relation with equal springs, in this cell-length convention, isThe minus branch is acoustic, with ${\omega }_{-}(0)=0$ and a linear small-$|k|$ dispersion. The plus branch is optical, withAt $k=0$, the acoustic eigenvector has $u=v$. The optical eigenvector instead satisfies $Mu+mv=0$: its cell centre of mass stays fixed while the atoms move oppositely. At the zone edge, for $M>m$, the frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, giving a gap between branches. Both slopes vanish there. The dispersion and displacement patterns are shown in the preceding original figure.

If the two atoms carry opposite effective charges, the relative motion creates an oscillating electric dipole moment, enabling infrared absorption by long-wavelength light. The photon momentum is tiny compared with the zone size, so ordinary optical absorption principally excites modes close to $k=0$. Raman activity instead depends on modulation of polarizability; the name optical does not imply that every such mode is infrared active. Three-dimensional crystals have separate longitudinal and transverse polarizations, and long-range Coulomb forces can split their optical frequencies. A monatomic primitive cell has no internal relative coordinate and therefore no optical branch of this type.

Let the nearest-neighbour spacing be $d$, so the primitive unit cell has length $\ell =2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K^{\prime }$ between cells. Hooke's law and Newton's second law giveFor normal modes $u_n=u{e}^{i(qn\ell -\omega t)}$ and $v_n=v{e}^{i(qn\ell -\omega t)}$, the two amplitudes satisfyIts vanishing determinant gives the alternating-spring chain dispersion relationThe minus sign is the acoustic phonon branch, with ${\omega }_{-}(0)=0$ and $u=v$ at the origin. The plus sign is the optical phonon branch, with $u=-v$ at the origin. For positive springs the acoustic phonon branch rises from zero to $\sqrt{2\min (K,K^{\prime })/m}$ at the Brillouin zone edge; the optical phonon branch falls from $\sqrt{2(K+K^{\prime })/m}$ to $\sqrt{2\max (K,K^{\prime })/m}$. Their long-wavelength limits areThus the leading frequency in the strong-bond limit of the alternating-spring chain is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.

Let adjacent equal masses $m$ be separated by $a$, with alternating strong and weak spring constants $K$ and $\kappa$, $K\gg \kappa >0$. The pattern repeats after $2a$, even though the masses are identical. Thus the first Brillouin zone isChoose the strong bond inside each two-atom primitive unit cell. With displacement amplitudes $(u,v)e^{i(2naq-\omega t)}$, the normal mode equations areThe alternating-spring chain dispersion relation is consequentlyThe lower acoustic phonon branch has in-phase motion within the strongly bound pair and ${\omega }_{-}(0)=0$. Its small-$q$ slope is the speed of sound,The upper optical phonon branch has opposite motion within each pair, with ${\omega }_{+}(0)=\sqrt{2(K+\kappa )/m}$. Here “optical” names the gapped branch; equal neutral masses need not couple to light through an electric dipole.

In the strong-bond limit of the alternating-spring chain, the pairs act almost as rigid dimers in the low-frequency branch, while their internal stretch gives a high, nearly flat branch:At the zone edge the two frequencies are $\sqrt{2\kappa /m}$ and $\sqrt{2K/m}$, separated by a large forbidden frequency interval. Neighbouring cells are out of phase there. Both branches have horizontal tangents at the edge.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

Set ${\lambda }^{\prime }=0$. A decaying normal mode $e^{-kt}$ obeysFor $C_d\gg C_s$, its sum and product of roots giveInitially $T_s=T_d=T_0$, so $a_1+a_2=1$ and $a_1{k}_1+a_2{k}_2=\lambda /C_s$. To leading order,On the slow normal mode, the first equation gives $T_d/T_s=(\lambda +\gamma -C_s{k}_2)/\gamma \simeq (\lambda +\gamma )/\gamma$, verifying proportionality after the fast transient has decayed. The small parameter must also make $k_2/k_1$ small; the supplied positive coefficients and $C_d\gg C_s$ satisfy this.

The Euler-Lagrange equations from the quadratic Lagrangian areFor a normal mode proportional to $e^{i\omega t}$, let $\lambda ={\omega }^{2}a/g$. Nonzero amplitudes requireHenceThe corresponding amplitude ratios are $\Phi /\Theta =2\lambda /(1-\lambda )$. The lower-frequency mode has $\Phi /\Theta =2(\sqrt{2}-1)$, with the angles in phase; the higher-frequency mode has $\Phi /\Theta =-2(1+\sqrt{2})$, with opposite signs. The small hoop's spin remains zero in both modes.

With endpoints held at zero displacement but free to rotate, variation permits arbitrary endpoint $\delta {\psi }_x$ and gives the natural zero-moment condition ${\psi }_{xx}=0$ there. These are simply supported endpoints. The normal modes areThe first mode becomes unstable when $F+B(\pi /l{)}^2<0$, soEquivalently the compressive load exceeds $|F{|}_{\mathrm{crit}}={\pi }^{2}B/l^2$. At equality the linear mode is neutral and nonlinear terms are needed to determine its post-buckling amplitude. The quadratic energy of $A\sin (\pi x/l)$ is $A^{2}l[F(\pi /l{)}^2+B(\pi /l{)}^4]/4$, giving the same sign test.

The endpoint displacement condition alone does not impose clamping. The quoted threshold uses the natural free-slope boundary condition of the stated variational problem. If rotations were also fixed, the admissible modes and threshold would change; one must not import a clamped-end formula into this result.

For the horizontal generalized coordinate $\eta$,The corresponding Euler-Lagrange equation isFor $\theta$, the terms proportional to $\dot{\eta }\dot{\theta }\sin \theta$ cancel between $d(\partial L/\partial \dot{\theta })/dt$ and $\partial L/\partial \theta$. Hence the other exact equation isKeeping only terms linear in small coordinates and their derivatives givesThe equilibrium is $\theta =\eta =0$. These characteristic frequencies label gravitational and spring restoring scales; they are not two independent normal mode frequencies of this massless-support system.