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For small horizontal displacement $x_i\simeq l{\theta }_i$, the gravitational restoring force is $-mg{\theta }_i\simeq -(mg/l)x_i$. A spring between adjacent bobs has extension equal to their displacement difference; Hooke's law pulls each bob towards its neighbour's displaced position. Thus comparison with the stated equations givesThe small-angle approximation makes the bob motion horizontal to first order; its vertical shift changes spring geometry only at higher order. Multiplying all force equations by minus one puts the positive restoring coefficients on the displayed right-hand side.

The central spring extension is $x_2-x_1$. Applying Hooke's law and Newton's second law to each mass givesIn matrix form $\ddot{x}+(k/m)\left(\begin{array}{cc}1+ϵ& -ϵ\\ -ϵ& 1+ϵ\end{array}\right)x=0$. The positive off-diagonal coupling in the force equations means that displacing one mass pulls the other in the same direction.

Let $x_1,x_2,x_3$ be longitudinal displacements from equilibrium, with positive direction to the right. The fixed walls have zero displacement. The first mass feels forces $-k{x}_1$ and $k(x_2-x_1)$; the central mass feels $k(x_1-x_2)$ and $k(x_3-x_2)$; the last mass feels $k(x_2-x_3)$ and $-k{x}_3$. Hooke's law and Newton's second law therefore giveEquivalently,All four springs contribute: the wall springs supply one of the two diagonal terms in the first and third rows. This is a fixed-end chain, not a freely ended chain with a zero-frequency translation mode.

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 downward-positive displacement and the filament bending modulus $B$ and load density $w$ defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is $-\delta E/\delta h=-B{h}_{xxxx}+w$. In resistive-force theory, transverse motion has drag per unit length $-{\zeta }_{\perp }{h}_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.

At steady state, $B{h}_s^{\prime \prime \prime \prime }=w$. Integrating first from the free end gives $h_s^{\prime \prime \prime }=(w/B)(x-L)$ and $h_s^{\prime \prime }=(w/2B)(x-L{)}^2$. Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is $\delta =h_s(L)=w{L}^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses $A$ as Young's modulus; if $A$ itself is flexural rigidity, the same result reads $k_{\mathrm{eff}}=8A/L^3$. This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.

For dimensional analysis, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[{\zeta }_{\perp }]=$ force times time/length squared makes ${\zeta }_{\perp }{h}_t$, $B{h}_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $w{L}^3/B\ll 1$, since $h_s^{\prime }(L)=w{L}^3/(6B)$.

Because increasing the length by $dL$ requires work $fdL$ on the chain, its first law of thermodynamics isAll configurations have the same energy, so at fixed $E$ and $N$,Differentiating the entropy found above givesand hence the entropic Hooke law for a one-dimensional chain isWhen $L\ll Na$, the Taylor expansion of $\mathrm{artanh}x$ giveswhich is Hooke's law. Conversely, at fixed force,The extension decreases as the temperature rises and tends to zero as $T\to \infty$: the rubber contracts on heating at fixed tension.

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.

Measure $x_1,x_2$ as horizontal displacements from the respective downward equilibrium positions; the absolute horizontal coordinates are $x_1$ and $L+x_2$. Each vertical displacement is of second order in its small angle. Thus the spring is horizontal to first order, its extension is $x_2-x_1$, and Hooke's law gives horizontal forces $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized simple pendulum restoring force yieldsFor the normal modes of two spring-coupled equal pendulums, add and subtract these ordinary differential equations. The normal mode coordinates $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$ satisfyHence the two natural angular frequencies and displacement patterns areOrdinary frequencies are ${\omega }_{+}/(2\pi )$ and ${\omega }_{-}/(2\pi )$. In the first normal mode, both particles swing together with equal amplitudes and the spring is unextended to first order. In the second normal mode, the particles swing in opposite directions with equal amplitudes, so the spring adds a restoring force and increases the angular frequency.

Label the equal masses $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi /3,4\pi /3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe first is rigid rotation of all three beads: every arc length stays fixed, giving a zero mode. In the second, the equal masses move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal masses move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero normal modes have zero total angular momentum. These are the normal modes of three beads connected by springs on a circular hoop with $M=2m$.

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.

A molecule is an electrically neutral bound assembly of more than one atom. It can contain identical atoms, as in ${\mathrm{O}}_2$, or different atoms, as in ${\mathrm{H}}_{2}O$. Discrete molecules are not a universal description of matter: an extended ionic crystal does not consist of separate neutral molecules corresponding to each empirical formula unit. The IUPAC definition also specifies binding sufficient to support a vibrational state.^[ref]

Elasticity is a body's tendency to recover its original configuration when the deforming force is removed. Within a linear regime, Hooke's law makes stress proportional to strain, but elasticity does not require exact linearity. The elastic limit matters: permanent plastic deformation is a different response.

Inertia is the property expressed by Newton's first law: without a resultant external force, a body's velocity remains constant in an inertial frame of reference. Mass measures translational inertia. It is not a force acting against motion, and a moving body does not require a continuing forward force merely to keep moving.

A force describes an interaction that changes momentum. In Newtonian mechanics,A force has magnitude and direction; its point of application also matters for the torque and hence the body's rotational response.

Take $x$ positive downward and put zero gravitational potential energy at the spring's support. The mass is then a distance $l+x$ below that reference, so $V_g=-mg(l+x)$. Hooke's law gives the elastic potential energy $k{x}^2/2$. ThereforeThe constant term $mgl$ fixes the requested reference but does not affect the motion. The Euler-Lagrange equation isThis is exactly Newton's second law with gravity downward and spring force $kx$ upward for positive extension. The static extension is $x_0=mg/k$.

Let $s$ denote distance from the neutral axis within a cross section. Bending to radius $R$ gives longitudinal strain $s/R$, so Hooke's law gives stress $Es/R$. Its moment isHere $E$ is Young's modulus and $I$ is the second moment of area, not a mass moment of inertia. For the stated square orientation, $I=a^4/12$.

For self-weight deflection of a square cantilever, the downward load per unit length is $q=\rho a^{2}g$. A free-end segment from $x$ to $l$ has shear-force magnitude $F(x)=q(l-x)$ and bending-moment magnitude $B(x)=q(l-x{)}^2/2$. Thus, with the given sign conventions,Using the small-slope curvature $1/R\simeq y^{\prime \prime }$ for downward deflection,The clamp fixes position and slope, and the unloaded free end has zero moment and shear:Four integrations yieldThe linear beam model requires small slopes and strains; the strong $l^4$ dependence eventually invalidates that approximation if other dimensions and material properties are fixed.

Constitutive law and Poisson ratio. Use positive stress for tension. In an isotropic linearly elastic medium, Hooke's law givesThe analogous equations follow by cyclic permutation. Under uniaxial tension in direction 1, $e_1={\tau }_1/E$ and $e_2=e_3=-\sigma {\tau }_1/E$. Thus Poisson's ratio $\sigma =-e_2/e_1$ measures transverse contraction relative to longitudinal extension when the transverse stresses are zero. It is dimensionless; it is not the ratio of transverse stress to longitudinal stress.

Radial and hoop strains. Let a material point originally at radius $r$ move to $r+R(r)$, with small axisymmetric displacement. Two neighbouring radii separated by $dr$ move apart by $dr+dR$, so the radial fractional extension is $dR/dr$. A circle's circumference changes from $2\pi r$ to $2\pi (r+R)$, so its fractional extension is $R/r$. ConsequentlyThe absence of longitudinal stress is a plane stress condition, not a claim of zero longitudinal strain.

Equilibrium of an annular wedge. Consider a slice of unit axial length, radial thickness $dr$ and small angular width $d\varphi$. The difference of the outward and inward radial face forces is $(r{\tau }_r{)}^{\prime }drd\varphi$ to leading order. The two tangential face tensions have inward radial resultant ${\tau }_{\varphi }drd\varphi$: each contributes ${\tau }_{\varphi }dr\sin (d\varphi /2)$. With no body force, their balance gives the radial equilibrium of a thick elastic cylinder,This hoop contribution is why the radial stress need not be spatially constant in a thick wall.

Displacement equation. With ${\tau }_z=0$, the two Hooke's law equations readInvert them:Substitute in ${\tau }_{\varphi }=(r{\tau }_r{)}^{\prime }$. After cancelling the common factor, the right-hand side is $r{R}^{\prime \prime }+(1+\sigma )R^{\prime }$, soThe cancellation explains why this displacement equation does not depend on the value of Poisson's ratio.

Power laws and wall pressure. Try $R\propto r^n$. The indicial equation is $n(n-1)+n-1=n^2-1=0$, soThe resulting Lamé stress distribution in a pressurized cylindrical wall can be writtenThe pressure boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. Solving them yieldsHence the positive compressive radial pressure in the wall isIt equals $p_1$ at $a$ and $p_0$ at $b$. This directional radial pressure is not an isotropic fluid pressure: radial and tangential stresses differ. The associated displacement is $R=[(1-\sigma )Pr+(1+\sigma )Q/r]/E$. Also $e_z=-\sigma ({\tau }_r+{\tau }_{\varphi })/E=-2\sigma P/E$, generally nonzero even though ${\tau }_z=0$.

Maximum absolute tangential stress. For $p_1>p_0$, $Q>0$ and ${\tau }_{\varphi }^{\prime }=-2Q/r^3<0$. Therefore the maximum of its absolute value must be at one of the two surfaces. Keeping the specified external pressure gives the fully general answerFor the usual high internal pressure with negligible external pressure, the largest magnitude is the tensile stress on the inner wall:For finite $p_0$, the inner-wall expression is $[p_1(a^2+b^2)-2p_0{b}^2]/(b^2-a^2)$, but its absolute value is not always the larger endpoint value. For example, nearly equal internal and external pressures can leave the outer surface more compressive. With $p_1=p_0=p$, both stresses correctly reduce to $-p$ everywhere. These results use a homogeneous linear elastic wall without axial loading; a closed-ended pipe would require a different longitudinal-stress condition.

Apply equal compressive pressure $p$ in all directions; here $\sigma$ denotes Poisson's ratio. The three principal normal stresses are $-p$. For an isotropic linearly elastic material, Hooke's law givesand similarly for $y,z$. At small strain, the fractional volume change is the sum of these normal strains:The bulk modulus is $B=-p/(\Delta V/V)$, giving the isotropic elastic modulus relation,The result assumes linear elasticity and fixed isotropic moduli. In the incompressible limit $\sigma \to 1/2$ with finite Young's modulus, the bulk modulus tends to infinity.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

Configuration count. Choose the end-to-end direction as positive and let $N_R,N_L$ count rightward and leftward links. ThenA configuration is determined by choosing the positions of its leftward links, soThe counts must be nonnegative integers; in particular, allowed signed extensions occur in steps of $2a$. If distance means $|L|$ with neither direction selected, there are two corresponding sets for $L>0$, adding a factor of two. That constant does not affect the tension. There is no excluded-volume penalty in this one-dimensional idealized count.

Entropy from the multiplicity. Put $x=L/(Na)$, so $|x|<1$. Applying the leading Stirling's approximation to all three large factorials givesTerms subleading in $N$, including logarithmic factorial prefactors, have been omitted consistently. Expanding near zero extension givesStretching reduces the number of available configurations and hence the entropy.

Tension and the range of Hooke's law. The model assigns no extension-dependent internal energy. At fixed temperature, the externally applied tension is $\mathcal{F}=(\partial F/\partial L{)}_T=-T(\partial S/\partial L{)}_T$. Differentiating the full leading entropy givesIt is the external force needed to oppose the chain's restoring entropic force. Only when $|L|\ll Na$ can one put $\mathrm{artanh}x\simeq x$, yieldingThe paper's displayed linear law therefore requires this small-extension approximation, which is not explicitly stated in its premise $L<L_0$. It is not the exact force law up to full extension. The entropic spring constant grows with temperature because stronger thermal disorder makes a given reduction in configurational entropy cost more free energy. This is the one-dimensional entropic chain force-extension relation.

A sparse number of backward links. Write $q=N_L\ll N$, so $N_R=N-q$ and $L=a(N-2q)$. Exactly,For fixed $q$ and large $N$, each numerator factor approaches $N$, givingMore precisely, the logarithm of the product is $-q(q-1)/(2N)+O(q^3/N^2)$. A small relative error in the multiplicity requires $q^2/N\ll 1$; merely $q/N\ll 1$ need not make the omitted product close to one. The weaker sparse-link condition does suffice for the leading entropy density and leading high-extension tension. These are sparse backward links near full chain extension.

Tension in that limit. In the overlap regime $1\ll q\ll N$, use Stirling also for $q!$:ThusThe full leading entropy would give $(k_{B}T/(2a))\ln [(N-N_L)/N_L]$, which has the same dominant logarithm when $N_L/N\to 0$.

Approaching full extension. The continuum force law diverges logarithmically as $N_L\to 0$, or $L\to Na$. One must not literally apply Stirling to a zero factorial: at the discrete endpoint there is exactly one directed configuration, $\Omega =1$, and its entropy is zero. A continuous fixed-extension derivative is not defined at that endpoint.

An independent force-ensemble calculation makes the limiting physical statement precise. For applied tension $\mathcal{F}$, a link has partition function $z=2\cosh (\beta \mathcal{F}a)$, giving the exact mean extensionIt approaches $Na$ only as $\mathcal{F}\to \infty$ at positive temperature, and inversion reproduces the logarithmic force law for mean extension. In a real chain, bond extension or failure eventually replaces the ideal inextensible-link model.

Under a uniform compressive pressure $p$, each normal stress is $-p$. The isotropic Hooke's law gives each longitudinal strain asThus $\Delta V/V=-3p(1-2\sigma )/E$. By the definition of bulk modulus, $B=-p/(\Delta V/V)$, so the isotropic elastic modulus relation isPositive finite Young's modulus and bulk modulus requireThe incompressible limit is $\sigma \to 1/2$ with $B\to \infty$ at finite $E$. If positivity of the shear modulus $G_s=E/[2(1+\sigma )]$ is also imposed, the full stable isotropic range is $-1<\sigma <1/2$. Negative Poisson's ratio is not ruled out; the displayed bulk-modulus relation alone supplies the upper bound.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.