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The joule is the SI unit of energy, work, and heat. One newton of force acting along one metre of displacement does one joule of work. The relation $1\mathrm{J}=1\mathrm{N}\mathrm{m}$ makes mechanical and electrical transfers comparable in the same energy unit. ^[ref]

PHYWE's implementation rubs a $1.60\mathrm{m}$ rod supported $0.40\mathrm{m}$ from each end. Its end disc drives a dusty glass tube; a movable tuning slide selects a clear pattern. A ruler measures repeated positions and a thermometer records gas temperature. The two supports select a rod wavelength equal to its length.^[ref]

The University of Iowa also documents an electronic version: a loudspeaker drives the tube and a microphone on a plunger supplies an oscilloscope readout. This permits a spatial scan of sound pressure without relying on the dust pattern.^[ref]

For a proposed pressure-scan measurement, keep the drive frequency fixed and move a small microphone through the tube. Record its position and fitted sinusoidal amplitude after each move; repeat the scan in reverse to reveal drift. Locate several consecutive maxima, rather than estimating a wavelength from a single pair. Reducing the drive checks whether the inferred spacing is independent of amplitude in the linear acoustics regime. A drive-off record estimates electrical background. These are controls for the proposed measurement, not reported observations from the cited demonstrations.

A microphone and its access opening can perturb the field. End compliance, losses and transverse modes also challenge the one-dimensional description. Dust traces identify a repeating structure, but the grains are not fluid parcels: the equations below describe sound pressure and acoustic particle velocity, rather than a detailed grain-transport model. The temperature and gas composition must remain sufficiently uniform to interpret a single fitted speed of sound.

An ideal single-round-trip Michelson interferometer converts differential length into phase difference $\delta \varphi =4\pi \Delta L/\lambda$. If its dark-port power scale is $P_0$ and its static phase offset is ${\varphi }_0$, thenExactly at a dark fringe, ${\varphi }_0=0$, power is quadratic in small displacement and loses its sign. A deliberate offset supplies an optical reference field: interference with the small signal field produces a linear term. Equivalently, $|E_{\mathrm{ref}}+E_{\mathrm{sig}}{|}^2$ contains $2\mathrm{Re}(E_{\mathrm{ref}}^{\ast }{E}_{\mathrm{sig}})$. The photodetector measures power, not a signed displacement directly.

Advanced LIGO uses this DC homodyne principle with a small differential arm offset and an output mode cleaner. Its feedback control maintains optical resonance and alignment, using actuation distributed over the suspension stages.^[ref] The simple formula above omits resonant storage: arm and recycling cavities give a frequency-dependent optical gain and delay, so no constant pass-count multiplier describes the entire observing band.

To see why a locked detector can still measure a disturbance, use a linear feedback control model. Let $x_{\mathrm{ext}}$ be the external differential displacement, $d_{\mathrm{err}}$ the residual readout, $d_{\mathrm{ctrl}}$ the recorded control command, $C(f)$ the sensing response and $A(f)$ the actuation response. With the convention $x_{\mathrm{res}}=x_{\mathrm{ext}}-A{d}_{\mathrm{ctrl}}$, reconstruction givesHere tildes denote Fourier transforms; $C$ has readout units per metre and $A$ metres per command unit. Feedback suppresses residual motion but records the compensating command. Omitting that command would suppress the reconstructed low-frequency signal too. In the actual GW150914 experiment, a modulated auxiliary laser exerted calibrated radiation pressure on a mirror to measure the response; periodic excitations monitored calibration.^[ref]

The antiplane field $w=b\arg (x+iy)/(2\pi )$ has a smooth local gradient on the punctured plane and satisfies the differential strain compatibility condition there. Its circulation around the defect axis is nevertheless $b\ne 0$, excluding a globally single-valued displacement on that domain. A radial cut permits a branch with an axial jump $b$ between the two cut faces; joining faces with that mismatch is the Volterra construction of a screw dislocation with Burgers vector parallel to the axis. The stress has a singular core where linear elasticity requires microscopic or plastic regularization. This example separates a local differential integrability test from global closure around a hole.

For underdamping, let $x_n,x_{n+1}$ be successive same-sign extrema of displacement relative to equilibrium, one full oscillation period apart. The logarithmic decrement isComparing extrema $N$ full cycles apart instead gives $\delta =N^{-1}\ln (|x_n|/|x_{n+N}|)$. Adjacent extrema of opposite sign are only half a cycle apart and give half this decrement. This distinction matters when inferring a damping ratio. Near critical damping, a full second same-sign extremum may be too small to resolve above measurement noise; a full-trace fit with uncertainty is then more informative than a peak ratio.

A longitudinal wave has an oscillating displacement parallel to the direction of propagation. An acoustic wave in a fluid supplies a familiar example.

Machine compliance is the extra load-path displacement per applied force attributed to the apparatus in a specified measurement model. Its units are $\mathrm{m}{\mathrm{N}}^{-1}$, the reciprocal units of a spring constant. Here $C_m$ includes the frame, load cell, adapters and grips insofar as their deformation contributes to the recorded crosshead displacement $x$. It is a property of that configuration and readout, rather than a universal constant of the testing machine.

Assume a uniform specimen with initial length $L$, constant $A_0$ and Young's modulus $E$, small axial elastic deformation, quasistatic loading, no slip, and a linear, length-independent apparatus response over the chosen force window. Equilibrium makes the same force act through specimen and apparatus, while their extensions add. The two springs in series therefore giveThe offset $x_0$ describes the displacement origin after seating; the slope $q$ describes total compliance. Fitting an intercept avoids forcing an arbitrary zero through the origin, but cannot repair slip or a curved response.

If all of $x-x_0$ is incorrectly attributed to specimen extension, the inferred modulus isFor positive $C_m$, this is too small. The fractional bias is most serious for a short, thick, stiff specimen because its own compliance $L/(E{A}_0)$ is small. A local extensometer instead has $d(\Delta L_g)/dF=L_g/(E{A}_0)$ in this ideal model, excluding deformation outside its contacts.

A shaped specimen requires more care. With varying initial cross-sectional area $A(s)$ and uniform $E$, its axial compliance is $\int ds/[EA(s)]$ in the one-dimensional approximation. Shoulder and end deformation must be included explicitly or remain identical between tests and be absorbed into an effective intercept. Changing only the extensometer gauge length on an unchanged specimen does not change its total compliance and cannot perform a two-length machine compliance separation.

For two incompressible streams of equal density with a common tangential magnetic field parallel to their velocities, decaying normal modes and interface displacement/total-pressure continuity give $\rho (\sigma +k_x{U}_1{)}^2+\rho (\sigma +k_x{U}_2{)}^2=2k_x^2{B}^2$ in units ${\mu }_0=1$. The displayed dispersion relation follows, with $v_A^2=B^2/\rho$. Thus $|\Delta U|<2v_A$ excludes exponential Kelvin-Helmholtz instability; equality is marginal. Above threshold the growth rate is $|k_x|\sqrt{(\Delta U{)}^2/4-v_A^2}$. The transverse horizontal wavenumber affects the decay length through $k=(k_x^2+k_y^2{)}^{1/2}$ but not this growth threshold. Magnetic tension supplies the stabilizing restoring force.

For a static smooth ideal magnetohydrodynamics equilibrium with conservative external forces and surface terms eliminated by the boundary conditions, the linear displacement equation is $\rho \ddot{\mathbf{\xi }}=\mathcal{F}\mathbf{\xi }$, where the force operator is a self-adjoint operator in the appropriate integral pairing. Its restoring quadratic form $\delta W=-\frac{1}{2}\int {\mathbf{\xi }}^{\ast }\cdot \mathcal{F}\mathbf{\xi }dV$ is real. Multiplying the normal mode equation by the complex conjugate of the displacement and integrating gives ${\omega }^2\int \rho |\mathbf{\xi }{|}^{2}dV=2\delta W$. Hence squared frequencies are real, and a negative restoring form gives an unstable mode through the variational Rayleigh quotient, subject to the operator domain and spectral hypotheses. Background fluid flow adds terms to this argument; the static hypothesis is essential.

Infer mantle relaxation from dated relative sea levels and shorelines, geodetic displacements, gravity changes and a reconstructed ice/water load history. The inferred viscosity trades off with ice history, lithospheric thickness, radial structure and transient rheology, so independent constraints and uncertainty propagation are essential.

The mean displacement is the expected value of a particle's displacement over a time interval. It describes systematic transport, while the mean squared displacement also measures fluctuations. For one component with finite second moment,For constant drift speed $u$ and ordinary diffusion of coefficient $D$, the mean displacement is $ut$ and its centered variance is $2Dt$, so the uncentered mean squared displacement is $u^2{t}^2+2Dt$. With fixed initial velocity $v_0$ and free inertial Langevin dynamics, the mean displacement is $v_0{\tau }_v(1-e^{-t/{\tau }_v})$, where ${\tau }_v$ is the velocity relaxation time. Its fluctuation variance is given by the integrated Ornstein-Uhlenbeck displacement.

Three consecutive masses $m,M,m$ joined by four identical springs of constant $k$ between fixed walls have an antisymmetric normal mode with zero central displacement. The two adjacent spring forces cancel on the central mass, while each outer mass experiences restoring force $-2kx$. This proves that the displayed angular frequency is independent of $M$ for every positive central mass. The two symmetric modes satisfy $(2k-m{\omega }^2)(2k-M{\omega }^2)-2k^2=0$ and generally do depend on $M$. For $M=m$, all three mode shapes are $(1,\sqrt{2},1)$, $(1,0,-1)$ and $(1,-\sqrt{2},1)$, with squared frequencies $(2-\sqrt{2})k/m$, $2k/m$, $(2+\sqrt{2})k/m$.

A constraint that depends explicitly on time can exert a normal force that does actual work, even when its force does no work on an allowed instantaneous virtual displacement. The surface motion gives the constrained particle a normal component of actual velocity.

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

A Newton step for Newton's method in optimization is the displacement $s$ solving the displayed linear system, where $H$ is the Hessian matrix and $\nabla f$ the gradient. The next iterate is $p+s$. This makes the first order Taylor expansion of the gradient at the new point vanish. An invertible Hessian matrix determines a unique step; it need not give a descent direction when the Hessian matrix is indefinite.

A node of an ideal standing wave is a position whose oscillation amplitude is zero: its displacement stays zero for all time. For a nontrivial normal mode $u(x,t)=\varphi (x)q(t)$, this means $\varphi (x)=0$. A momentary zero caused by $q(t)=0$ does not make every position a node. In more dimensions the zeros of a mode shape can form curves or surfaces.

For the $n$th normal mode of a uniform fixed-end wave equation on a string, the nodes of a standing wave areThere are $n+1$ nodes of a standing wave including the two endpoints, and $n-1$ interior ones. For $n=3$ and $L=1\mathrm{m}$, they lie at $0$, $1/3$, $2/3$, and $1\mathrm{m}$. Each remains stationary through every snapshot in the normal mode diagram; an antinode instead marks a maximum oscillation amplitude. ^[ref]

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

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

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

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

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

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

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

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

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