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A Hopf bifurcation creates or destroys a small periodic orbit near an equilibrium point as a pair of complex conjugate eigenvalues crosses the imaginary axis. It explains how a resting state can give way to sustained oscillations: the linearization of a dynamical system supplies rotation and growth, while a nonlinear term can saturate that growth at a nonzero amplitude.

The simplest supercritical Hopf bifurcation is the exact Hopf normal formHere $z$ is a complex number, $\mu$ is the bifurcation parameter, and $\omega$ is the angular frequency. In polar coordinates $z=r{e}^{i\theta }$, the equations become $\dot{r}=\mu r-r^3$ and $\dot{\theta }=\omega$ for $r>0$. At $\mu <0$, each turn shrinks towards the equilibrium point. At $\mu >0$, small radii grow and large radii shrink towards $r=\sqrt{\mu }$, producing a limit cycle. For example, $\mu =0.04$ and $\omega =2$ give radius $0.2$, period $\pi$, and frequency $1/\pi$ cycles per unit time. The limiting angular frequency stays nonzero while the amplitude tends to zero: a tiny newborn oscillation can still repeat at a finite rate.

For a precise generic statement, consider a smooth function $f$ defining an autonomous differential equation $\dot{x}=f(x,\alpha )$ on ${\mathbb{R}}^n$, with a smooth branch of equilibrium points $x_{\ast }(\alpha )$ and critical parameter $\alpha =0$. Require:Then a locally unique branch of small periodic orbits, up to phase, emerges on the side $\beta a<0$. On the two-dimensional centre manifold, the branch attracts for $a<0$ and repels for $a>0$. Attraction in the full phase space additionally requires all remaining eigenvalues to have negative real part; any transverse unstable direction makes the periodic orbit unstable there.^[ref]

An eigenvalue-crossing bifurcation test alone cannot establish these nonlinear conclusions. For example, the purely linear equation $\dot{z}=(\mu +i\omega )z$ has the same crossing but no isolated periodic orbit: at $\mu =0$ every nonzero radius is constant, and at $\mu \ne 0$ none is. If the cubic coefficient vanishes, higher-order terms must be examined; a generalized Hopf bifurcation is one possible degeneracy. The usefulness of the generic theorem is that a large system can have its small oscillatory onset understood through one Hopf normal form, subject to these hypotheses and local limits.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

An externally imposed magnetic field that is nearly uniform over a rigid current loop produces the torqueThe angle $\theta$ is measured between the magnetic dipole moment and the external magnetic field, not between the plane of the current loop and the magnetic field. The magnetic dipole potential energy uses a dot product, whereas the torque uses a cross product. The torque is largest when the magnetic dipole moment is perpendicular to the magnetic field; parallel alignment minimizes the potential energy.^[ref]

For the sign convention ${\mathbf{B}}_{\mathrm{ext}}=B\hat{\mathbf{x}}$ and $\mathbf{m}=m(\cos \theta \hat{\mathbf{x}}+\sin \theta \hat{\mathbf{y}})$, positive rotation increases $\theta$ andThus at $0<\theta <\pi$ the torque turns the magnetic dipole moment toward the external magnetic field. Both parallel and antiparallel alignment have zero torque, but they are respectively a stable equilibrium and an unstable equilibrium. With no damping, a freely released current loop can undergo oscillation around the stable equilibrium; settling requires damping.

In the twelve-turn example, take $B=0.020\mathrm{T}$ and $\theta =60°$. The predicted torque is ${\tau }_z=-2.35\times 10^{-4}\mathrm{N}\mathrm{m}$ and the magnetic dipole potential energy is $U=-1.36\times 10^{-4}\mathrm{J}$. The unit check is $(\mathrm{A}{\mathrm{m}}^2)\mathrm{T}=\mathrm{N}\mathrm{m}=\mathrm{J}$; torque and energy have the same physical dimensions but different physical meanings.

A short derivation starts with the magnetic force on a current-carrying wire, $d\mathbf{F}=Id\mathbf{r}\times {\mathbf{B}}_{\mathrm{ext}}$. In an exactly uniform magnetic field, the net force is $I({\oint }_{C}d\mathbf{r})\times {\mathbf{B}}_{\mathrm{ext}}=0$, but its torque need not vanish: opposite sides produce a turning couple. Integrating $\mathbf{r}\times d\mathbf{F}$ gives $\mathbf{m}\times {\mathbf{B}}_{\mathrm{ext}}$. Integrating $dU=-{\tau }_{z}d\theta$ at fixed $m$ and $B$ gives the orientation potential energy. This is the mechanical orientation potential energy for a maintained electric current, with the external source held fixed; the energy accounting for field and power supply is a separate problem. A varying external magnetic field requires corrections across a finite current loop, and can also give a net force.

For incident light with linear polarization, an ideal linear polarizer transmitswhere $I$ is light intensity and $\theta$ is the angle between the incident electric field's line of oscillation and the polarizer's pass axis. The input and output are compared in the same medium. The classical explanation is an orthogonal projection: for a unit vector $\mathbf{u}$ along the pass axis,The dot product reduces the transmitted wave amplitude by $\cos \theta$. For a plane electromagnetic wave in the same medium, light intensity is proportional to squared electric field wave amplitude, giving Malus's law. For example, $\theta =60^{\circ }$ transmits one quarter of the incident light intensity, although the transmitted wave amplitude is one half. ^[ref].

Start with light already in linear polarization along the horizontal $0^{\circ }$ axis and let its light intensity be $I_0$. Two ideal absorbing linear polarizers with pass axes $0^{\circ }$ and $90^{\circ }$ give $I_0{\cos }^2{0}^{\circ }{\cos }^{2}90^{\circ }=0$. Insert a $45^{\circ }$ polarizer between them: the first leaves $I_0$ unchanged, the middle transmits $I_0/2$ with linear polarization along $45^{\circ }$, and the last transmits $I_0/4$ with linear polarization along $90^{\circ }$. Each angle in Malus's law is relative to the preceding output, so the final step uses $90^{\circ }-45^{\circ }=45^{\circ }$:

For these ideal absorbing polarizers, the rejected electric field component transfers energy to the filter. In the three-filter arrangement the middle absorbs the fraction $I_0/2$, and the last absorbs $I_0/4$, leaving $I_0/4$ transmitted. Conservation of energy holds at each step. Inserting the middle polarizer changes the linear polarization arriving at the last one; it does not create light. The effect illustrates why successive orthogonal projections cannot generally be replaced by a single comparison of the first and last axes.

The initially horizontal assumption matters. For unpolarized light of light intensity $I_0$, the first ideal linear polarizer transmits $I_0/2$: averaging ${\cos }^2\theta$ over uniformly distributed incident orientations gives $1/2$. The same $0^{\circ },45^{\circ },90^{\circ }$ sequence then transmits $I_0/8$. This extra first-stage factor does not apply to the already horizontally polarized input in the figure.

Why do methane, ammonia, and water have different shapes when each central chemical atom is surrounded by eight valence electrons? The answer separates the positions of atomic nuclei from the electron domains used in a VSEPR model.

Molecular geometry describes the three-dimensional arrangement of the atomic nuclei in a molecule, including bond lengths and bond angles. For a larger molecule, a shape name often describes just the neighbours of one central chemical atom. A Lewis structure records connections and electron bookkeeping; its flat layout does not fix molecular geometry.

In the VSEPR model, electron pair geometry includes both bonding electron domains and lone pairs, whereas molecular geometry counts only the chemical atoms. In methane, four bond pairs and no lone pairs give tetrahedral molecular geometry. In ammonia, three bond pairs and one lone pair give trigonal pyramidal molecular geometry. In water, two bond pairs and two lone pairs give bent molecular geometry. All three use the same ideal tetrahedral electron pair geometry; this does not make their actual bond angles equal. See Flowers, Theopold, Langley and Robinson, Chemistry 2e, §7.6: electron-pair and molecular geometries.

A static drawing also needs a physical interpretation. The atomic nuclei undergo oscillations; an arrangement at a minimum of potential energy and an experimentally averaged arrangement need not have identical bond lengths or bond angles. This distinction matters when comparing measurements with quantum mechanics calculations; see the IUPAC Gold Book: experimental bond-length conventions. Gas electron diffraction gives a concrete route from a laboratory signal to molecular geometry rather than inferring shape from a drawing alone.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

Negative feedback closes a causal loop whose net effect opposes an initiating change. In signal transduction, activation followed by a downstream inhibition of the earlier activating stage is one example of negative feedback. Loop sign alone does not establish asymptotic stability or complete signaling adaptation; see the worked signaling feedback example.

A useful biological example is chemotaxis feedback regulation: an Escherichia coli biological cell adjusts its response to a sustained attractant instead of remaining at its immediate response. Increased CheA activity promotes CheB phosphorylation; CheB reverses receptor methylation, opposing their activation. After attractant addition, reduced activity instead favors receptor methylation by CheR, restoring activity. This is a return from receptor activity to the machinery regulating that activity. The response need not return exactly at every dose. ^[ref]

A motivating measurement is the chemotaxis FRET assay. CheY and CheZ carrying yellow and cyan fluorescent tags were observed in immobilized Escherichia coli biological cells on a coverslip in a flow chamber. Adding and removing attractant changed the yellow-to-cyan emission ratio, reporting the association of phosphorylated CheY with CheZ. This measures a molecular readout upstream of motor behavior. Comparisons with biological cells lacking CheR, CheB, or both tested the dependence on receptor modification; these perturbations also change starting activity and sensitivity, so they are not pure changes of one feedback gain. ^[ref]

The later chemotaxis feedback regulation experiment used only the Tar receptor and nonmetabolizable methylaspartate, supplied and removed by flow. Long recordings separated slow recovery from incomplete signaling adaptation; correction for bleaching addressed drift in the optical readout. Activity recovered incompletely after sufficiently large additions, with detectable imprecision already at $10\mu \mathrm{M}$ in that preparation. Restricting the receptor type and preventing consumption of the stimulus help interpret the perturbation, but limit transfer to other receptors and conditions. The readout infers CheA activity through CheY-CheZ association, rather than directly measuring every molecular step. ^[ref]

Negative feedback can also overshoot or sustain oscillations if its return arrives too late. The worked signaling feedback example gives the contrasting instantaneous case; the delayed negative feedback model below isolates the effect of timing. Loop sign describes causation, while asymptotic stability describes what happens to displacements over time.

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.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.

The difference of two wave phases at the same place and time, considered modulo $2\pi$. It determines whether two same-frequency oscillations reinforce, cancel, or are in phase quadrature.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]