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A normal mode is a small-oscillation motion with every coordinate proportional to one common sinusoid, at a fixed amplitude ratio and relative phase. Expanding an arbitrary small motion in normal mode coordinates decouples its linear equations into independent oscillators. With mechanical mass matrix $\mathsf{M}$ and mechanical stiffness matrix $\mathsf{K}$, the generalized eigenvalue equation is $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$.

Horizontal gravitational restoring force on each bob is $-m_i(g/l)x_i$. The two springs add $k(x_2-x_1)$ on the first bob, $k(x_1-2x_2+x_3)$ on the middle bob, and $k(x_2-x_3)$ on the last. Dividing by the bob masses gives the stated equations withReflection symmetry separates an antisymmetric end motion from two symmetric motions. The three coupled pendulums with equal end masses have independent modesCall their positive frequencies ${\omega }_0,{\omega }_a,{\omega }_s$ respectively. The middle amplitude in the last mode ensures $m{x}_1+M{x}_2+m{x}_3=0$. The modes are orthogonal in the mass-weighted inner product, rather than generally in the unweighted coordinate dot product.

The initially stationary displacement decomposes asThere are no sine terms because every initial velocity is zero. ThereforeThese give $(0,0,d)$ at $t=0$ and solve all three coupled equations. The spring constant is assumed positive; for $k=0$ the frequencies become degenerate and the same superposition reduces to uncoupled pendulum motion.

The original example uses $M=2m$ and $k/m=g/l$ to display the mode vectors and resulting superposition. Those illustrative parameter ratios are not additional supplied data.

For a mass change in a viscously damped spring oscillator, use linear viscous damping $m\ddot{x}+b\dot{x}+kx=mg$, holding the spring constant $k$ and damping coefficient $b$ fixed. Measuring $x$ from $mg/k$ removes the constant force. The envelope is $e^{-\gamma t}$, whereDoubling $m$ halves both $k/m$ and $\gamma$. ThereforeNeglecting damping in the frequency gives $0.5/\sqrt{2}=0.35355\mathrm{Hz}$, indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

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.

The tension is supplied by Stokes drag on the free bead. With bead radius $b=0.5\mu \mathrm{m}$, the stipulated isolated-sphere approximation givesThe trap also balances drag on the trapped bead, but that bead's drag is not an additional force transmitted through the chain. The polymer contour length is $L=Na=6\mu \mathrm{m}$. Use a room-temperature estimate $T\simeq 300\mathrm{K}$, consistent with the water-viscosity approximation imported from part (a); the expressions below retain the dependence on $T$.

For a theta solvent, Gaussian chain elasticity gives $F_{\mathrm{el}}=3k_{B}T{R}^2/(2N{a}^2)$ and henceThis is the mean extension along the force, rather than the unforced end-to-end distance of a polymer chain $a\sqrt{N}=60\mathrm{nm}$. At $300\mathrm{K}$, $fa/(k_{B}T)=0.410$; the finite-link freely jointed chain result $R=L[\coth (fa/k_{B}T)-k_{B}T/(fa)]$ is $0.810\mu \mathrm{m}$, close to the Gaussian estimate.

The good-solvent force-extension scaling changes the elasticity. In the tensile blob regime, a blob has size ${\xi }_t=k_{B}T/f$, contains $g\sim ({\xi }_t/a{)}^{5/3}$ segments, and the aligned string of $N/g$ blobs has lengthOnly the scaling and order of magnitude of the good-solvent value are determined: solvent-dependent swelling amplitudes are not supplied. The nonlinear force law is the essential correction to extrapolating a weak-force coil spring.^[ref]

To expose the unrealistic extrapolation, the unforced good-solvent size estimate is $R_0\sim a{N}^{3/5}=151\mathrm{nm}$. Treating this coil as a linear spring with spring constant $3k_{B}T/R_0^2$ predicts $R\simeq 5.17\mu \mathrm{m}$ at $300\mathrm{K}$, although its required condition $f{R}_0\ll k_{B}T$ fails by a factor of about $103$. Dropping the order-one factor three, as in the rough estimate $k\sim k_{B}T/R_0^2$, even gives $R\sim 15.5\mu \mathrm{m}>L$. An extension beyond the polymer contour length is an artefact of extending the coil's linear response outside its range. The tensile blob estimate avoids that violation but here ${\xi }_t\simeq 1.46\mathrm{nm}$ and $g\sim 4.4$, so it is only marginally within a many-segment-blob regime. finite extensibility of a polymer chain, chain drag and flow disturbance by both large beads further limit quantitative realism. The specified neglect of chain drag is a model assumption, not evidence that it is physically negligible.

TODO: biomolecular force assays are user-deferred. The following notes cover the generic physical apparatus, calibrated dielectric handles and synthetic attachments only.

In optical tweezers, a tightly focused laser produces a restoring gradient force on a suitable transparent bead. Scattering also transfers momentum along propagation, so a stable three-dimensional trap needs axial as well as lateral restoration. An objective with high numerical aperture, correctly filled entrance pupil, controlled beam polarization and low-aberration sample chamber are practical design elements. Collect transmitted/scattered light with a condenser and measure bead displacement with calibrated imaging or back-focal-plane photodetection. A steerable beam or translation stage sets relative positions; the force signal comes from bead displacement, not commanded stage displacement alone.

Near a stable operating point, use a harmonic optical trap with $U(x)=k{x}^2/2$. For a slowly changing external force, $F_{\mathrm{ext}}=k⟨x⟩$ in that coordinate; the optical restoring force is $-kx$. Calibrate each axis and the range over which its response is linear. Detector sensitivity must also be known: write $x=\beta (s-s_0)$, with measured detector voltage $s$, offset $s_0$ and position sensitivity $\beta$ in $m/V$. A spring constant in arbitrary detector units is not a force calibration.

Three complementary physical calibrations are useful. At thermal equilibrium the equipartition theorem givesThis needs an independently calibrated position scale and a variance corrected for drift, detector noise and exposure averaging. In an overdamped Newtonian medium, the optical trap Langevin calibration obeysUsing a one-sided position power spectral density per hertz,The corner frequency and spectral amplitude test both relaxation and fluctuation size. If drag is known, $k=2\pi \zeta f_c$. The isolated-sphere value $\zeta =6\pi \eta a$ must be corrected near boundaries; otherwise measured height dependence can be mistaken for a changed trap stiffness. An independent steady optical trap drag calibration moves fluid past the trapped sphere: $k⟨x⟩=\zeta u$. Use small positive and negative speeds to verify slope, offset and force range.

One can also determine position sensitivity without assuming the bulk drag. In a low-frequency sinusoidal-stage optical trap calibration, let the known local fluid velocity amplitude be $u_0$ at drive angular frequency $\Omega$. The coherent bead displacement amplitude isIf the corresponding detector amplitude is $A_s$, thenAfter removing the coherent drive line from the thermal record, combine this sensitivity with equipartition theorem to recover $k$, and then infer $\zeta$ from $f_c$. A published calibration combines thermal spectra and stage-driven motion without taking bead drag as an input.^[ref]

Controls determine whether these formulas measure the apparatus rather than its noise. Record blocked-light offsets and independent fixed-bead noise; calibrate length with known motion; keep sampling and electronics bandwidth adequate around $f_c$; and include exposure filtering and aliasing in a spectral fit. motion blur reduces measured variance and therefore makes naive equipartition theorem overestimate stiffness; additive position noise has the opposite bias. Changing exposure time provides a direct diagnostic.^[ref] Check temperature and laser heating, wall separation, bead size and refractive index, alignment and power stability. Repeat after changing power or optical geometry and compare thermal, spectral and drag estimates. For a synthetic force handle, also measure the attachment's compliance and geometry so that bead motion is not automatically identified with its extension.

Every element of the uniformly distributed mass has speed squared ${\dot{r}}^2+r^2{\dot{\varphi }}^2$, so the total kinetic energy is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2$. The extension of the ring's full circumference is $2\pi (r-r_0)$, and the stated spring constant therefore givesThe magnetic coupling for each charge element is $dq\mathbf{A}\cdot \mathbf{v}$. With the symmetric gauge $\mathbf{A}=(Br/2)\hat{\mathbf{\varphi }}$ and tangential speed $r\dot{\varphi }$, integration over the ring gives $qB{r}^2\dot{\varphi }/2$. ThusThis idealization neglects charge self-energy, radiation and deformation away from a circular ring. The elastic constant refers to extension of the whole circumference; replacing it by a radial spring constant without the factor $4{\pi }^2$ changes the problem.

Potential energy assigns energy to a system's arrangement, so a change of arrangement accounts for the work done by a conservative force. For a time-independent potential energy $V(\mathbf{r})$,The gradient points toward increasing potential energy; the force points toward decreasing potential energy. Only differences matter: adding a constant to $V$ changes neither the force nor its work. This endpoint rule follows by substituting $-\nabla V$ into the line integral for work. ^[ref]

Near Earth's surface, stipulate constant gravitational acceleration $g=9.81m/{\mathrm{s}}^2$ and measure height $h$ upward from a chosen zero. Then gravitational potential energy is $V=mgh$. Raising a $2\mathrm{kg}$ mass by $0.5\mathrm{m}$ increases its gravitational potential energy by $9.81\mathrm{J}$, while gravity does $-9.81\mathrm{J}$ of work. These are computed model values, not measurements of a particular lift. An ideal spring obeying Hooke's law instead has $V=k{x}^2/2$, with displacement $x$ from its relaxed state and spring constant $k$: $k=100N/m$ and $x=0.2\mathrm{m}$ give $V=2\mathrm{J}$ relative to $x=0$. ^[ref]

A restoring force acts toward an equilibrium when a body is displaced from it. In one dimension, taking displacement $x=0$ at equilibrium, this means $xF(x)<0$ for sufficiently small nonzero $x$. An ideal linear spring has $F(x)=-kx$ with positive spring constant $k$, producing simple harmonic motion for a constant mass without damping. More general restoring forces need not be linear.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

A uniform rigid body disk of mass $m$ and radius $a$ moves in a frictionless plane. Let $N\ge 3$ identical springs attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and spring constant $k$. Write the small centre displacement as $\mathbf{q}$ and the rotation as $\theta$. For a spring with radial and tangential unit vectors $\mathbf{n},\mathbf{t}$, its length isThis follows by a second-order Taylor series for the distance between $\mathbf{q}+aR(\theta )\mathbf{n}$ and $(a+l)\mathbf{n}$. In the elastic potential energy $k(s-l_0{)}^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.

The equally spaced directions satisfy $\sum \mathbf{n}=\sum \mathbf{t}=0$ and $\sum \mathbf{n}{\mathbf{n}}^T=\sum \mathbf{t}{\mathbf{t}}^T=(N/2)I$. Summing the spring energies givesThe kinetic energy is $m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$, using the moment of inertia of a uniform disc. The Euler-Lagrange equations give the displayed squared angular frequencies. The translational normal modes are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.

A linear spring stores elastic energy $k{x}^2/2$ when displaced by $x$ from its relaxed position. The external force maintaining that displacement is $kx$; its restoring force on the moving attachment has the opposite sign. Its spring constant $k$ has units force per length.

The spring constant is the positive proportionality coefficient in the ideal spring restoring force $F=-kx$. Its SI unit is newtons per metre. With a constant attached mass $m$, it sets the angular frequency $\omega =\sqrt{k/m}$ of simple harmonic motion. Doubling $k$ doubles the restoring force at a fixed displacement, but increases that angular frequency only by a factor $\sqrt{2}$.

Consider two equal balls of mass $m$, four light rods each of length $L$, a fixed upper pivot and a light sliding lower collar. Each upper and lower rod makes angle $\theta$ with the spindle. A spring of spring constant $k$ is uncompressed at $\theta =0$ and is compressed by collar motion. Its compression is $\delta =2L(1-\cos \theta )$. Collar balance gives $2T_l\cos \theta =k\delta$, while vertical ball balance gives $(T_u-T_l)\cos \theta =mg$. Radial Newton's second law gives $(T_u+T_l)\sin \theta =m{\omega }^{2}L\sin \theta$. Eliminating the rod tensions proves the displayed relation for nonzero $\theta$.

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

Attach the table-end mass to a linear spring of spring constant k. Adding spring potential energy $k(e_0+x{)}^2/2$ gives restoring curvature $k-\rho g$ about equilibrium. The positive gravitational feedback from increasing hanging chain length subtracts from the spring stiffness. Stable oscillations require $k>\rho g$; equality is neutral in this ideal linear model, while $k<\rho g$ gives exponential departure. The frequency is independent of the preload, although the equilibrium position is not.

A pendulum bob and a horizontally moving support of equal mass $m$, with support spring spring constant $k$, have the quadratic potential energy $k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. With $\kappa =k/m$, $q=g/l$, the normal mode eigenvectors can be normalized to $(1,r_{\pm })$, where $r_{\pm }=(\kappa /2\mp \sqrt{{\kappa }^2/4+q^2})/q$. The lower mode is in phase and the upper mode in opposite phase. The positive eigenvalues of the symmetric stiffness-per-mass matrix give the displayed squared angular frequencies.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.

A three-dimensional ideal chain of $N$ independent segments of length $b$ has $⟨R^2⟩=N{b}^2=Lb$. The isotropic small-force susceptibility gives $⟨z⟩=F⟨R^2⟩/(3k_{B}T)$, so $k_s=3k_{B}T/(Lb)$. For a long worm-like chain, the statistical segment is $b=2l_p$. This stiffness describes entropic end-to-end extension, not axial stretching of the material backbone, whose spring constant is $EA/L$.

Two positive masses linked in series between fixed walls by three identical springs of spring constant $k$ have equations of motion $m_1{\ddot{x}}_1=-2k{x}_1+k{x}_2$ and $m_2{\ddot{x}}_2=k{x}_1-2k{x}_2$. A normal mode $x_j=A_j\cos \omega t$ exists when $(2k-m_1{\omega }^2)(2k-m_2{\omega }^2)-k^2=0$, giving the displayed angular frequencies. Its amplitude ratio is $A_2/A_1=(2k-m_1{\omega }^2)/k$: positive for the lower angular frequency, negative for the higher one. Equal masses give amplitude ratios $+1,-1$ and squared angular frequencies $k/m,3k/m$.