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For a constant-amplitude force on $m\ddot{x}+b\dot{x}+kx=F_0\cos \Omega t$, the cycle-averaged absorbed power is $b{F}_0^2{\Omega }^2/[2((k-m{\Omega }^2{)}^2+b^2{\Omega }^2)]$. It peaks at ${\omega }_0=\sqrt{k/m}$. Its half-power frequencies are ${\Omega }_{\pm }=(\sqrt{b^2+4mk}\pm b)/(2m)$, hence their separation is exactly $b/m$ and $Q={\omega }_0/\Delta \Omega$. A displacement-amplitude resonance has different half-power frequencies away from weak damping; the measured observable must be specified.

An acoustic wavelength measurement makes an invisible spatial pattern readable with a ruler. The authored guide Interesting and inexpensive experiments for high school physics by Joe Wolfe places a loudspeaker near a pipe's open end, drives it with an oscillator, and slides a microphone attached to a metre rule through the pipe. An oscilloscope or computer displays the microphone signal. Its oscillation amplitude rises and falls with position; adjacent minima locate pressure nodes of a standing wave a distance $d=\lambda /2$ apart. A cap gives a pressure antinode at the closed end; an open end has a pressure node just outside it. These are pressure patterns, distinct from displacement patterns. The guide advises wavelengths longer than the pipe diameter, sufficient pipe length to locate both a node and an antinode, and frequency spectrum analysis to distinguish the intended tone from other sounds.^[ref]

The inference follows from superposition of two oppositely directed acoustic plane waves. In an ideal uniform, lossless pipe, taking equal pressure amplitudes $P$ and a rigid cap at $x=0$ givesThe time-varying pressure perturbation $p^{\prime }$ has local oscillation amplitude $2|P\cos (kx)|$. Its zeros are separated by $\pi /|k|=\lambda /2$. Unequal incident and reflected amplitudes leave nonzero minima, but their spacing remains $\lambda /2$ in this ideal model. A microphone detects that pressure variation through a changing electrical signal, rather than watching a wave crest move. Record the tone's frequency too, then infer phase speed $v_p=2fd$.

For proposed controls, keep the drive frequency and amplitude fixed during a scan, verify the frequency on the recorded signal, and repeat the scan in both directions. Fit several minimum positions $x_j=a+jd$ using linear regression to reduce dependence on one pair of positions. Check the ruler scale and use one consistent reference point on the microphone. Compare with the source switched off to locate background measurement noise; use frequency spectrum analysis to check for additional tones, and reduce drive if the trace ceases to resemble a sinusoid. Record air temperature and avoid strong air motion. These are suggested controls, not additional procedures or observations attributed to the guide.

The pipe must support approximately plane waves over the measured region: transverse structure at shorter wavelengths, the microphone's finite size, and disturbance by the ruler can compromise the interpretation. Weak reflection gives shallow minima, while damping and other reflections can shift the extrema of the simple two-wave model. Measure interior spacings rather than assuming the first minimum is exactly at the pipe lip. A short scan may not include enough minima to estimate $d$ reliably, and drift during sequential scanning can imitate spatial variation.

For an explicitly hypothetical calculation, $f=1000\mathrm{Hz}$ and $d=0.1715\mathrm{m}$ give $\lambda =0.3430\mathrm{m}$, $K\simeq 18.3183\mathrm{rad}{\mathrm{m}}^{-1}$, and $v_p=343.0\mathrm{m}{\mathrm{s}}^{-1}$. No such measurements are claimed here. If independent small measurement errors in $f$ and fitted $d$ have standard deviations ${\sigma }_f$ and ${\sigma }_d$, first-order error propagation gives ${\sigma }_{v_p}/v_p\simeq \sqrt{({\sigma }_f/f{)}^2+({\sigma }_d/d{)}^2}$. A systematic scale error or changing air conditions requires separate assessment.

An air track supports a cart on a cushion of air to reduce contact friction. It provides a nearly one-dimensional platform for a damped harmonic oscillator; residual drag remains part of the baseline damping.

A base-excited oscillator has a moving support rather than a directly applied periodic force. If its damping depends on velocity relative to the support, $z=x-y$ obeys $m\ddot{z}+b\dot{z}+kz=-m\ddot{y}$. A model with damping against absolute velocity instead obeys $m\ddot{x}+b\dot{x}+kx=ky$ after subtracting its static equilibrium. These conventions give different transfer functions.

An original reduced model makes the calibration explicit. Let the torsion pendulum have restoring torque $-\kappa \theta$, with torsion constant $\kappa$, and viscous damping torque $-b\dot{\theta }$, with $b>0$. Allow an unwanted axial torque ${\tau }_p(t)$:Here ${\omega }_0$ is the undamped natural angular frequency and $Q$ the quality factor. The forced harmonic oscillator response must be included before a twist can be interpreted as a torque.

Using the complex amplitude convention $m_z(t)=\mathrm{Re}(\tilde{m}{e}^{\mathrm{i}\omega t})$ and similarly for $\theta$ and ${\tau }_p$, differentiation multiplies by $\mathrm{i}\omega$. Substitution givesThis is the torque-to-twist transfer function. With ${\tilde{\tau }}_p=0$ and real positive $\tilde{m}=m_1$, the twist amplitude isFor $\gamma <0$, the twist phase relative to $m_z$ is $\pi /2-atan2(b\omega ,\kappa -I{\omega }^2)$. At $\omega ={\omega }_0$ it is zero, although the reaction torque leads $m_z$ by $\pi /2$. A phase measured at the mirror is not automatically the phase of the applied torque.

A fixed-amplitude reference torque ${\tau }_c\cos \omega t$ measures $|H_{\tau }|$ directly. The Einstein-de Haas effect instead contains an extra factor $\omega$ when $m_1$ stays fixed. Consequently its twist amplitude peaks exactly at ${\omega }_0$ in this model, whereas the fixed-torque peak is at ${\omega }_0\sqrt{1-1/(2Q^2)}$ when $Q>1/\sqrt{2}$. To verify the first statement, divide the squared denominator of $A_{\mathrm{EdH}}$ by ${\omega }^2$: the quantity to minimize is $(\kappa /\omega -I\omega {)}^2+b^2$, minimized at ${\omega }_0$. Both response curves tend to zero at high frequency; only the magnetomechanical curve also tends to zero at zero frequency.

For a fully stipulated example, choose $I=2.00\times 10^{-10}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =8.00\times 10^{-10}\mathrm{N}m/rad$, $b=8.00\times 10^{-11}\mathrm{N}\mathrm{m}s/rad$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and $m_1=1.76\times 10^{-4}\mathrm{A}{\mathrm{m}}^2$. These teaching inputs give ${\omega }_0=2.00rad/s$, $Q=5.00$, reaction torque amplitude $2.00\times 10^{-15}\mathrm{N}\mathrm{m}$ at ${\omega }_0$, andAn optical lever at distance $D=1.00\mathrm{m}$ gives small-displacement spot amplitude $x\simeq 2DA=25.0\mu \mathrm{m}$; its full peak-to-peak width is $4DA=50.0\mu \mathrm{m}$. For a fast reversal from $+m_1$ to $-m_1$, the same inputs give $\Delta L_{\mathrm{body}}=-2.00\times 10^{-15}\mathrm{kg}{\mathrm{m}}^2/s$ and initial $\Delta \dot{\theta }=-1.00\times 10^{-5}rad/s$, provided the external angular impulse is negligible.

One independent mechanical calibration adds known moment of inertia $\Delta I$, keeping $\kappa$ unchanged. If $T_a,T_b$ are the undamped oscillation periods before and after adding it, thenFor $\Delta I=I$ in the stipulated example, $T_a=\pi \mathrm{s}$ and $T_b=\sqrt{2}\pi \mathrm{s}$, recovering the chosen $I$ and $\kappa$. Real damped periods require a damping correction: a free decay has angular envelope $e^{-bt/(2I)}$ and oscillation angular frequency $\sqrt{\kappa /I-(b/(2I){)}^2}$. Measuring that decay determines $b/(2I)$. Calibrating the optical scale, mechanical response and change of magnetic dipole moment separately is what makes the inferred gyromagnetic ratio meaningful.

A Cavendish torsion balance is a torsion balance arranged to measure gravitational torque from nearby source masses. Its suspended horizontal rod rotates about a vertical fiber. In the linear elastic regime, a twist angle $\theta$, measured from the fiber's untwisted orientation, gives a restoring torque ${\tau }_{\mathrm{fiber}}=-\kappa \theta$, where $\kappa >0$ is the torsion constant. The minus sign is essential: the fiber resists the twist. Under an approximately constant gravitational torque, the rotational equation using moment of inertia $I$ and angular acceleration $\ddot{\theta }$ isAt equilibrium, the two torques balance. The angle is a displacement from the fiber's reference orientation, not the largest excursion during an oscillation. ^[ref]

For the toy parameters in the Cavendish experiment, the predicted equilibrium twist isHere $L|{\theta }_{\mathrm{eq}}|/r\simeq 0.0107$, consistent with using a small geometric displacement in the toy model, but not an estimate of its total measurement error. Reversing the source positions ideally takes the equilibrium angle from $+{\theta }_{\mathrm{eq}}$ to $-{\theta }_{\mathrm{eq}}$: the magnitude of the change is $2|{\theta }_{\mathrm{eq}}|\simeq 0.6119^{\circ }$. A stable zero offset then cancels in the difference. A real reading may track a reflected light spot; the law of reflection doubles a mirror's rotation in the outgoing beam, so the spot's angular change must not be confused with the rod's angle. ^[ref]

How can the fiber's torsion constant be found without already knowing the gravitational force? Measure the oscillation period and calculate the moment of inertia. For the same toy rod, neglecting its mass and treating both test masses as point masses, $I=m{L}^2+m{L}^2=2m{L}^2$. With the source torque absent or treated as constant, let $\varphi =\theta -{\theta }_{\mathrm{eq}}$ be a small departure from equilibrium. Neglect damping and any change of gravitational torque with angle. The rotational equation becomes simple harmonic motion:Here $A$ is the angular amplitude, $\delta$ specifies the initial phase, and $\omega$ is the angular frequency. A softer fiber gives a larger equilibrium twist but also a longer oscillation period: greater sensitivity costs observation time. An undamped oscillation does not settle at equilibrium; damping is needed for settling. ^[ref]

The toy moment of inertia and undamped oscillation period areTiming this oscillation period would recover the assumed torsion constant. The units check explains the result: torque divided by torsion constant is an angle in radians, and $I/\kappa$ has units of time squared when radians are treated as dimensionless in SI.

Combining the equilibrium twist with the period calibration, the two-near-pair model would infer the gravitational constant asThe test mass cancels in the second expression because it enters both the gravitational torque and the toy moment of inertia. If the measured quantity is the magnitude $|\Delta \theta |$ of the change between opposite source positions, use $|{\theta }_{\mathrm{eq}}|=|\Delta \theta |/2$. Substituting the toy twist and oscillation period returns the input $G$; it is a consistency check, not independent experimental evidence.

The sensitivity of a Cavendish torsion balance also explains its practical difficulty. Air motion can exert unwanted torque; unequal temperatures can drive that motion. Cavendish discusses this effect on p. 471 and describes observing from outside a closed room to avoid disturbing the instrument. ^[ref] Mechanical vibrations can excite the same slow oscillations used to calibrate the torsion constant, while changes in the fiber or source separations alter the inferred force. Precision therefore depends on resolving the equilibrium angle, calibrating the full suspended moment of inertia and torsion constant, and modeling the actual source geometry. The two-near-pair model omits opposite-source attraction, rod contributions, finite-size effects, damping, and the variation of gravitational torque as the rod turns. ^[ref]

Consider an original scalar response model for a small collective angle $u$. Stipulate effective inertia $I>0$, restoring curvature $K\ge 0$, damping rate $\Gamma >0$, and a drive torque ${\tau }_0\cos \omega t$:The undamped resonance scale is ${\omega }_0=\sqrt{{\omega }_L^2+{\Omega }^2}$, where ${\Omega }^2=K/I$. This illustrates the role of an additional collective restoring energy. It is not the full coupled spin dynamics of either helium phase, and its parameters are not fitted to historical observations.

The steady response and mean absorbed power follow directly:With constant damping, differentiating this last expression places its positive-frequency maximum exactly at ${\omega }_0$. Setting chosen $\Omega =1$ and ${\omega }_L=2$ moves the peak from $2$ to $\sqrt{5}\simeq 2.236$, about $11.8\%$. At ${\omega }_L=10$ the same curvature moves it only to $\sqrt{101}\simeq 10.050$, about $0.50\%$. At large ${\omega }_L$, the shift is ${\Omega }^2/(2{\omega }_L)$ to leading order. A smaller shift at larger field therefore need not mean that the restoring curvature vanished.

Critical damping is the repeated-root case of the viscous damping model: its two characteristic roots coincide at $r=-{\omega }_0$. The solution with arbitrary initial conditions isThe extra factor $t$ is essential: one exponential function alone cannot fit arbitrary initial displacement and velocity when the characteristic equation has a double polynomial root. Substitution, or taking the limit of the underdamped formula as its angular frequency tends to zero, gives the displayed expression.

The precise fastest-return statement is this: hold $m$, $k$, $x_0>0$ and $v_0=0$ fixed, and compare all $\zeta \ge 1$. Every response decreases positively toward equilibrium; the critically damped responseis smaller than every overdamped response at each $t>0$. It therefore reaches and stays within any specified band $|x|\le ϵx_0$, $0<ϵ<1$, sooner than those responses. Exact arrival at rest takes infinite time. This is a comparison among nonoscillatory regimes for these particular initial conditions, not a universal minimum of every possible settling-time criterion.

One short comparison argument uses $q=\partial X/\partial \zeta$. Its initial conditions are zero and $q^{\prime \prime }+2\zeta q^{\prime }+q=-2X^{\prime }$. For release from rest, $X^{\prime }<0$ at $\tau >0$; the causal Green function of a damped oscillator is positive at positive time for $\zeta \ge 1$. Its convolution with $-2X^{\prime }$ gives $q>0$: increasing damping beyond the critical value increases the remaining displacement.

There is no repeated oscillation at critical damping, but there can be overshoot once. For $x_0>0$, the factor in square brackets crosses zero precisely when $v_0<-{\omega }_0{x}_0$, atFor example, $v_0=-2{\omega }_0{x}_0$ gives $x=x_0(1-{\omega }_{0}t)e^{-{\omega }_{0}t}$, crossing equilibrium at $t=1/{\omega }_0$ as shown above. An overdamped response can also cross once for suitable initial conditions. Conversely, specially chosen initial conditions can remove its slow exponential decay entirely. Thus the generic slow-root argument must not be applied to a response whose slow-root coefficient is zero.

Adding a velocity-proportional force $-b\dot{y}$ produces a damped harmonic oscillator. The fastest non-oscillatory release from rest is critical damping, with the repeated-root solution in the title. Reaching fraction $\epsilon$ of the initial displacement requires $(1+s)e^{-s}=\epsilon$, $s={\omega }_{0}t$. At one percent, $s=6.638352\dots$, rather than $\log 100$. Exact arrival at equilibrium takes infinite time; the tolerance reference must be stated.

A damped torsional oscillation is rotation about a fiber's fixed reference orientation with a restoring torque and dissipative torque. In the ideal viscous model, damping contributes $-b\dot{\theta }$ with constant $b\ge 0$. This is a rotational damped harmonic oscillator. For known constant moment of inertia $I$ and torsion constant $\kappa$, define $\gamma =b/(2I)$. In the oscillatory regime ${\gamma }^2<\kappa /I$,$A$ and $\phi$ are set by the initial state. Successive positive peaks measured from the same equilibrium have magnitudes $p_n$ and $p_{n+1}$; their ratio gives $\delta =\ln (p_n/p_{n+1})=\gamma T_d$. Consequently a measured damped oscillation period and decay together recoverUsing only $4{\pi }^{2}I/T_d^2$ underestimates the torsion constant in this model. The toy curve in torsion constant sets $I=2.0\times 10^{-6}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, and $\gamma =0.020{\mathrm{s}}^{-1}$. It gives $T_0\simeq 88.86\mathrm{s}$, $T_d\simeq 92.64\mathrm{s}$, and $\delta \simeq 1.853$. Ignoring the decay recovers $9.2\times 10^{-9}\mathrm{N}m/rad$, an $8\%$ underestimate; including it recovers the stipulated torsion constant.

These equations assume a single rotational mode, a stationary zero, linear elasticity, and viscous damping. Internal fiber losses need not behave as a constant viscous coefficient: Fleischer and colleagues model their precision apparatus using a loss angle. Their response is evidence for specifying a damping model, not for applying the viscous correction universally. ^[ref]

A wave equation with a linear damping term and positive wave speed $c$ separates into spatial normal modes and damped harmonic oscillators. For a spatial eigenvalue $k^2$, the time factor obeys $T^{\prime \prime }+\alpha T^{\prime }+c^2{k}^{2}T=0$. Its characteristic roots are $-\alpha /2\pm \sqrt{{\alpha }^2/4-c^2{k}^2}$. Thus underdamping requires $|\alpha |<2c|k|$, critical damping gives a repeated root and the second solution $t{e}^{-\alpha t/2}$, and overdamping gives two real exponential rates. Negative $\alpha$ amplifies rather than damps the motion.

Damping dissipates the energy of an oscillation, reducing its amplitude. A linearly damped string applies this mechanism to each normal mode.

Why can stronger damping make a return to equilibrium slower? A weak brake lets a spring exchange potential energy with kinetic energy many times; a very strong brake makes the final motion creep. Critical damping marks the boundary between these behaviours. The distinction depends on the force law: the classification below concerns a single damped harmonic oscillator with viscous damping, rather than every mechanism that removes energy.

Let $x(t)$ be displacement from equilibrium, $m>0$ the moving mass, $k>0$ the spring constant, and $b\ge 0$ the constant coefficient of viscous damping. With no driving force, Newton's second law givesThe two initial conditions specify displacement and velocity. Define the undamped eigenfrequency as an angular frequency ${\omega }_0=\sqrt{k/m}$ and the damping ratio as $\zeta =b/(2\sqrt{mk})$. The ordinary differential equation becomes $\ddot{x}+2\zeta {\omega }_0\dot{x}+{\omega }_0^{2}x=0$. A constant weight in a vertical spring merely shifts equilibrium; it is already absorbed in the definition of $x$.^[ref]

For example, the ideal parameters $m=1\mathrm{kg}$ and $k=4\mathrm{N}{\mathrm{m}}^{-1}$ give ${\omega }_0=2{\mathrm{s}}^{-1}$ and critical damping at $b=4\mathrm{kg}{\mathrm{s}}^{-1}$. Values $b=1,4,8\mathrm{kg}{\mathrm{s}}^{-1}$ give damping ratios $0.25,1,2$, respectively. These are model choices, not measured values. For release from rest with $x_0>0$, set $X=x/x_0$ and $\tau ={\omega }_{0}t$; then $X^{\prime \prime }+2\zeta X^{\prime }+X=0$, where primes now denote derivatives with respect to $\tau$.

The left panel uses identical initial conditions $X(0)=1$, $X^{\prime }(0)=0$. Underdamping produces repeated crossings of equilibrium; critical damping and overdamping give positive decreasing responses. The right panel keeps critical damping but changes initial velocity: $X^{\prime }(0)=-2$ gives one crossing, although there is no oscillation. Both panels are exact model predictions.

A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

Use a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$. Here $\Gamma$ is a decay rate in inverse seconds and also the full width at half maximum of the natural linewidth expressed in angular frequency. The corresponding full width in hertz is $\Gamma /(2\pi )$; the angular half width is $\Gamma /2$. These conventions determine every factor of two and $2\pi$ below.

Let $s$ be the on-resonance saturation parameter of each beam, with total excitation weak enough that saturation and coherent coupling between beams can be neglected. The independent-beam scattering rates and net force areThe force is the photon momentum per absorption multiplied by the difference in absorption rates. With dimensionless laser detuning $d=\delta /\Gamma$ and dimensionless velocity $u=kv/\Gamma$, writeFor the example $d=-0.5$ and $u=0.25$, the positive and negative beam factors are respectively $4/13$ and $4/5$. Thus $F(0.25)=-32/65\simeq -0.4923$: the atom is moving right and the net force points left. All values in this example are dimensionless.

Subtracting the two fractions gives an especially useful sign check:For red detuning, the force opposes nonzero $v$ throughout this ideal model; its useful magnitude eventually falls at large $|u|$. Near zero, a Taylor series gives dampingAt $d=-0.5$, $F^{\prime }(0)=-2$ and $\alpha =\hslash k^{2}s$. For blue detuning the sign reverses and the small-velocity motion is amplified; at exact resonance the opposing mean forces cancel in this symmetric model. Its derivation assumes weak excitation, a closed two-level system, equal beam intensities and nonrelativistic velocities; real multilevel atoms need additional physics.

Damping removes random kinetic energy, but the discrete photon kicks cause diffusion in momentum. Their balance gives a nonzero temperature. Define the momentum diffusion coefficient $D_p$ by the noise contribution $d⟨p_x^2⟩/dt=2D_p$. With atom mass $m$ and linear damping coefficient $\alpha$,at stationary balance, where $k_B$ is the Boltzmann constant and mean momentum is zero.

For ideal three-dimensional optical molasses with three identical independent beam pairs and isotropic spontaneous emission, the $x$-pair absorption rate at rest is $R_x=\Gamma s/(1+4d^2)$. Absorption from that pair contributes $(\hslash k{)}^2{R}_x$ to the noise growth of $⟨p_x^2⟩$. Emission from all three pairs contributes the same amount: its total rate is $3R_x$ and each emitted photon has mean squared $x$-momentum $(\hslash k{)}^2/3$. Hence $D_p=(\hslash k{)}^2{R}_x$. Using the Doppler cooling force coefficient givesPut $a=-d>0$. Then $(1+4d^2)/(-d)=a^{-1}+4a\ge 4$, with equality at $a=1/2$. The ideal weak-field Doppler cooling limit is therefore $T_D=\hslash \Gamma /(2k_B)$ at $\delta =-\Gamma /2$. It is the minimum of this specified diffusion-and-damping model, rather than a universal lower bound for laser cooling. The diffusion approximation also assumes recoil energy much smaller than $\hslash \Gamma$. In a one-pair geometry, emission patterns and transverse recoil change the diffusion bookkeeping. ^[ref].

A linearly damped oscillator $m\ddot{x}+b\dot{x}+kx=0$ has $x=A{e}^{-\gamma t}\cos ({\omega }_{d}t+\varphi )$, with $\gamma =b/(2m)$ and ${\omega }_d^2=k/m-{\gamma }^2>0$. After one damped period $T=2\pi /{\omega }_d$, both displacement and velocity are multiplied by $e^{-\gamma T}$. Their quadratic mechanical energy is therefore multiplied exactly by $e^{-2\gamma T}$, regardless of the starting phase. In particular successive same-direction equilibrium crossings have this ratio even though potential energy is zero at each. A cosine phase maximum is not exactly a displacement turning point when damping is nonzero.

Michael Faraday's induction ring separates two questions: can one electrical circuit affect another without an electrical connection, and does that effect continue under steady conditions? The surviving apparatus uses insulated copper windings around an iron ring; the Royal Institution dates the successful experiment to 29 August 1831.^[ref]

Faraday connected one winding group to a battery, the other to a galvanometer. Making battery contact deflected the needle; maintaining contact let it return to rest; breaking contact reversed the response. Reversing the battery reversed the deflections. His separate cylinder comparison enhanced induction with iron, whereas replacing it with copper gave no enhancement over the windings alone.^[ref]

The contrast between a switching response and a quiet steady state makes mutual induction tangible. A steady primary electric current can maintain magnetic flux without sustaining a secondary electromotive force. Changes in magnetic flux linkage matter, as expressed by Faraday's law of induction. This is why a driven electrical transformer uses repeated changes in magnetic flux to supply secondary power.

The galvanometer readout of an induction pulse also separates the measured motion from the inferred electric current. Its moment of inertia and damping mean that a deflection is not an instantaneous trace of electric current. The following coupled RL induction model calculates currents in a stipulated circuit, rather than fitting Faraday's needle motion or reconstructing his switching times.

For the ideal uniform wave equation on a string, the wave speed is $c=\sqrt{\tau /\mu }$, where $\tau$ is constant tension and $\mu$ is linear mass density. Assume small transverse slopes, negligible bending stiffness and no damping. Fix both ends at zero displacement, giving the homogeneous Dirichlet boundary conditions $y(0,t)=y(L,t)=0$. Our standing wave already satisfies the first. For a nonzero amplitude, the second requires $\sin (kL)=0$, henceThe value $n=0$ gives zero displacement in this sine family, not a vibrating normal mode. Each positive $n$ fits $n$ half-wavelengths into the string: it has $n$ antinodes, $n-1$ interior nodes of a standing wave, and two endpoint nodes of a standing wave. For $L=1\mathrm{m}$ and $c=20\mathrm{m}{\mathrm{s}}^{-1}$, $f_n=10n\mathrm{Hz}$; the third normal mode has $f_3=30\mathrm{Hz}$ and wavelength $2/3\mathrm{m}$.

The equally spaced eigenfrequencies follow from this string's uniform wave speed and fixed-end boundary conditions. They are not a universal spectrum for every vibrating object. Changing an endpoint constraint changes the allowed mode shapes; changing tension changes wave speed. At fixed $L$ and $\mu$, quadrupling $\tau$ doubles every $f_n$. The more general construction of arbitrary initial motion belongs to normal mode and wave equation on a string.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

The galvanometer current sensitivity is the change in settled deflection angle per change in electric current. In the linear model its derivative is constant, $S_I=NAB/k$. The illustrative design has $S_I=3000rad/A$, or $0.1719^{\circ }$ per $\mu \mathrm{A}$. A chosen full-scale angle ${\theta }_{\max }=0.180\mathrm{rad}$ corresponds to $|I|=60.0\mu \mathrm{A}$; a straight-line extrapolation beyond that range does not establish real instrument behavior.

Finite sensitivity and finite readable angle set different limits. Under the stipulated $0.10^{\circ }$ tolerance, a reading indistinguishable from zero means $|I|\lesssim 0.582\mu \mathrm{A}$ only if the zero is stable and the model applies. More turns can increase galvanometer current sensitivity while also increasing winding electrical resistance; the response to an applied electric potential difference depends on both. Friction, zero drift, departures from the radial magnetic field, and spring nonlinearity can spoil the calibration. Damping sets how long one must wait for a settled reading. A rapidly reversing electric current can produce little mean pointer deflection even when its instantaneous magnitude is large.