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In the diamond cubic structure of silicon, the longitudinal acoustic phonon branch and longitudinal optical phonon branch meet at $X=(2\pi /a,0,0)$. The two atomic basis sites are identical, and the associated crystal symmetry protects their boundary degeneracy of normal modes. In aluminium phosphide, the two sites contain different chemical atoms. The corresponding zinc blende structure lacks the site-exchange symmetry, and these two longitudinal branches generally split at $X$.

A simple way to see the splitting is a diatomic-chain dispersion relation with equal springs, representing successive $[100]$ atomic planes by alternating masses $M_1,M_2$ at spacing $d=a/4$. With effective spring constant $K$, its angular frequencies obeyAt $X$, $kd=\pi /2$, so the inter-sublattice coupling cancels and the two squared angular frequencies are $2K/M_1$ and $2K/M_2$. Equal masses give a contact; unequal masses giveIn aluminium phosphide, the heavier phosphorus plane moves in the lower-frequency edge mode, and the lighter aluminium plane in the higher-frequency mode in this model. Their similar atomic masses explain why the splitting is relatively small. Different electronic environments can also change the force constant matrix. The chain calculation illustrates the mechanism; a quantitative three-dimensional phonon spectrum requires its full dynamical matrix. It does not assert that all transverse acoustic and optical branches meet in silicon.

The orientation-dependent results in the official notes show that the measured spring constant is not a Lorentz scalar. A Lorentz transformation changes both the simultaneous extension and the measured three-force; their ratio depends on the orientation. This does not make the spring intrinsically anisotropic in its own rest frame, or require a preferred inertial frame. It is a quasistatic effective stiffness comparison between frames; an extended real spring also has finite internal signal speeds.

Let $a$ be a stable equilibrium separation: $V^{\prime }(a)=0$, $K=V^{\prime \prime }(a)>0$. With bond extension $\delta =x-a$, the Taylor expansion isThe harmonic approximation applies when the typical thermal and zero-point bond fluctuations satisfy $|\delta |\ll a$ and the higher terms are negligible, for example $|V^{\prime \prime \prime }(a)\delta |\ll 3K$ and $|V^{\prime \prime \prime \prime }(a){\delta }^2|\ll 12K$. The relevant small quantity is the relative displacement of neighbouring atoms. Small bond excursions about a stable minimum give an amplitude-independent spring constant $K=V^{\prime \prime }(a)$. Large thermal excursions, bond breaking or appreciable anharmonicity invalidate the approximation.

For the Lennard-Jones potential,Its minimum and harmonic spring constant areUsing the monatomic-chain dispersion relation,The maximum is at $|k|=\pi /a$. Taking the argon atomic mass as $m\simeq 39.95\mathrm{u}=6.63\times 10^{-26}\mathrm{kg}$ and $ϵ=1.602\times 10^{-21}\mathrm{J}$ givesThese are the stipulated one-dimensional nearest-neighbour model estimates; a three-dimensional argon crystal has additional neighbours and phonon polarizations.

Assume $k>0$, $m>0$ and $\gamma \ge 0$. The spring constant $k$ measures restoring force per displacement from the equilibrium position $x_0$. The term $-\gamma mv$ is linear drag, with damping rate $\gamma$ in inverse-time units. Writing ${\omega }_0=\sqrt{k/m}$, the equation isFor small positive damping it gives underdamped motion: oscillations about $x_0$ with a decaying amplitude. If $\gamma =0$, the motion is undamped simple harmonic motion.

The characteristic equation of the damped harmonic oscillator is $r^2+\gamma r+{\omega }_0^2=0$. In the underdamped case its roots are $-\gamma /2\pm i{\omega }_d$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}$. Thus the exact real solution is $y=A{e}^{-\gamma t/2}\cos ({\omega }_{d}t+\alpha )$. For $\gamma \ll {\omega }_0$, ${\omega }_d={\omega }_0+O({\gamma }^2/{\omega }_0)$, giving the requested leading approximationFor mass $m$ and spring constant $m{\omega }_0^2$, the mechanical energy is $E=\frac{1}{2}m({\dot{y}}^2+{\omega }_0^2{y}^2)$. Differentiating and using the differential equation gives $\dot{E}=-m\gamma {\dot{y}}^2$. In particular, energy decay of a viscously damped harmonic oscillator is not an exact pointwise exponential decay: $\dot{E}=0$ at a turning point. To leading order in $\gamma /{\omega }_0$, the period average isThe quality factor is $Q=2\pi \overline{E}/(\text{energy lost in one cycle})$. Using the period $T\simeq 2\pi /{\omega }_0$ and the small fractional loss $\gamma T$ gives $\boxed{Q\simeq {\omega }_0/\gamma }$. Equivalently, the energy-decay definition $Q={\omega }_0\overline{E}/{\overline{P}}_{\mathrm{loss}}$ yields the same leading result. Replacing ${\omega }_d$ by ${\omega }_0$ in the oscillatory phase is not uniform over arbitrarily long times; its phase error is $O({\gamma }^{2}t/{\omega }_0)$.

Assume a physical mass $m>0$, spring constant $k>0$, and nonnegative viscous damping parameter $\gamma$. The homogeneous damped harmonic oscillator has characteristic equationIts roots are $r=-\gamma /2\pm \sqrt{{\gamma }^2/4-{\omega }_0^2}$. They have nonzero imaginary parts precisely when ${\gamma }^2<4k/m$. Defining $\Omega =\sqrt{k/m-{\gamma }^2/4}$, the real oscillatory solutions areFor strictly decaying oscillations also require $\gamma >0$. The limiting $\gamma =0$ case is an undamped harmonic oscillator. Equality gives critical damping with $z_h=(C+Dt)e^{-\gamma t/2}$, and a larger positive $\gamma$ gives two real decaying exponentials rather than oscillations.

Put $k_1=k_3=k$ and $m_1=m_2=m$. The mass-normalized stiffness matrix isIts symmetric eigenvector $(1,1)$ has eigenvalue $k/m$, while its antisymmetric eigenvector $(1,-1)$ has eigenvalue $(k+2k_2)/m$. ThusThe symmetric frequency does not depend on the middle spring constant because that spring has no extension in an in-phase normal mode. The antisymmetric motion stretches the middle spring and has the higher frequency for $k_2>0$.

Let $x_1,x_2$ be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law giveFor a normal mode $x_j=a_j\cos \Omega t$, the determinant condition isSolving the quadratic for ${\Omega }^2$ givesThe angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form $k_1{x}_1^2+k_2(x_2-x_1{)}^2+k_3{x}_2^2$ positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.

Each spring force is its extension multiplied by the spring constant, acting to restore the equilibrium. The equations of motion areNo end is attached to a wall. The spring extensions are $x_2-x_1$ and $x_3-x_2$, giving potential energy $k[(x_2-x_1{)}^2+(x_3-x_2{)}^2]/2$.

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.

Define a common displacement coordinate $X=(x_1+q{x}_2)/(1+q)$ and a relative displacement $\delta =x_1-x_2$. Adding and subtracting the equations of motion givesThe general real motions of the two normal modes are thereforeHere $C,D,A,B$ are independent real constants. The zero mode includes uniform drift as well as a static common displacement; writing only a constant times $\cos 0t$ would miss the independent linear-in-time solution. Their superposition isIn the oscillatory normal mode, $m_1{x}_1+m_2{x}_2=0$ and the two amplitudes have ratio $1/q$, with opposite phases. Equivalently the oscillatory motion has zero total angular momentum about the hoop center.

As $q\to \infty$ with $m_1,k$ fixed, $\omega \to \sqrt{2k/m_1}$ and the heavier particle's oscillatory amplitude tends to zero for fixed lighter-particle amplitude. The lighter particle then oscillates against two effectively anchored springs with combined spring constant $2k$. A common uniform rotation remains an allowed zero mode; the heavy particle is stationary in the limit only if the initial common drift is zero or tends to zero, as it does for bounded total angular momentum.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

The diagram has a connection between every pair of indices $0\le i<j\le n$, including the fixed beam $0$. Two parallel strings, each with spring constant $k/2$, together have energy $k(X_i-X_j{)}^2/2$. Thus the printed $E_{ij}$ is the energy of a connecting pair; calling it the energy of one string would introduce a factor-of-two discrepancy.

Take vertical coordinates positive upwards, with $X_0$ fixed. Each bar has gravitational potential energy $m_{i}g{X}_i$ and kinetic energy $m_i{\dot{X}}_i^2/2$. The Lagrangian isDifferentiating gives the Euler-Lagrange equationsEquivalently, with $S={\sum }_{j=1}^n{X}_j$, the spring term is $k[(n+1)X_i-S-X_0]$. Gravity changes the equilibrium but not the stiffness matrix. Subtracting a static equilibrium and writing ${\eta }_i=X_i-X_i^{\mathrm{eq}}$ gives the normal mode systemThis is the spring system underlying normal modes of an anchored complete spring graph.

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

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.

Let $x$ be the downward displacement of $M$ from its initial position just beyond the table edge. The smaller mass moves towards the edge by the same distance, and the hanging chain length is $x$. All the chain and both endpoint masses move at speed $\dot{x}$, so their total moving mass is $\mathcal{M}=M+m+\rho L$. The hanging chain's gravitational potential energy, relative to the initial configuration, is $-\rho g{x}^2/2$. Thus the Lagrangian for the chain sliding over a table edge isThe Euler-Lagrange equation gives $\mathcal{M}\ddot{x}=g(M+\rho x)$. Put $\Omega =\sqrt{\rho g/\mathcal{M}}$. With $x(0)=\dot{x}(0)=0$, the solution isThese are also the smaller mass's displacement and speed towards the edge. The formula applies until $M$ reaches the floor, at $t_f={\Omega }^{-1}\mathrm{arcosh}(1+\rho y/M)$. Since $y<L$, the smaller mass has not yet reached the edge during this interval. As $\rho \to 0$, the result tends to $x=Mg{t}^2/[2(M+m)]$, as expected for a massless connector.

Let the spring's spring constant be $k$, and write its extension as $e_0+x$, allowing for an initial extension $e_0$. The new Lagrangian isThe specified equilibrium is $x_e=y-y_1$, with $k(e_0+x_e)=g(M+\rho x_e)$. Put $\xi =x-x_e$; the equation becomesThe spring-stabilized chain over a table edge oscillates only when $k>\rho g$. More hanging chain increases its own gravitational pull, reducing the effective restoring stiffness. The equilibrium height alone does not determine $k$ without information about the spring's natural length or preload.

The four supporting springs act in parallel, so their total spring constant is $k=4.00\times 10^4\mathrm{N}{\mathrm{m}}^{-1}$. The driving angular frequency is $\omega =2\pi (10)=62.83{\mathrm{s}}^{-1}$. The force-driven vibration isolation estimate givesAt this mass, $f_0=\sqrt{k/m}/(2\pi )=0.894\mathrm{Hz}\ll 10\mathrm{Hz}$, and $Q=\sqrt{mk}/b\simeq 14.2$, so the assumptions are consistent. The exact forced amplitude isIt is indeed close to $1\mu \mathrm{m}$, but it is slightly above the strict requested bound. Setting the exact expression to $10^{-6}\mathrm{m}$ gives the high-mass thresholdChoose a mass strictly greater than this threshold for a strict inequality; $1.28\times 10^3\mathrm{kg}$ is a suitable rounded choice. Simply rounding the high-frequency estimate down would not guarantee the bound.

Reading near the origin, an acoustic point is approximately $(q,\omega )=(1.9\times 10^9{\mathrm{m}}^{-1},0.9\times 10^{13}{\mathrm{s}}^{-1})$. The initial slope therefore estimatesCombining the slope with the inferred spring constant givesThis $a$ is the model's adjacent Na-plane to Cl-plane spacing; its cell length is $2a$. The slope is a graphical estimate, so values around $4.5$–$5.0\mathrm{km}{\mathrm{s}}^{-1}$ and spacings near $0.2\mathrm{nm}$ are consistent with the reading precision. A real ionic crystal has interactions beyond the scalar nearest-neighbour model; an intercept-and-slope fit need not reproduce the entire measured dispersion or independently measured structural spacing exactly.

Take nearest-neighbour separation $d$, a common spring constant $K$, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with $\omega \simeq d\sqrt{K/m}|k|$ near the origin. The first Brillouin zone ends at $\boxed{k=\pm \pi /d}$, where $\omega =2\sqrt{K/m}$ and the group velocity vanishes.

For alternating masses $m,M$, choose a two-atom cell of length $2d$. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at $\boxed{k=\pm \pi /(2d)}$. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at $k=0$ the atoms move oppositely with a stationary centre of mass and ${\omega }_{+}^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.

If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin (kd/2)|$ and $2\sqrt{K/m}|\cos (kd/2)|$, touching at $\sqrt{2K/m}$. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.