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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.

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.

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.

An experimentally useful question is whether the disappearing oscillation follows viscous damping, and whether increasing the brake changes $b$ rather than $m$ or $k$. A published teaching protocol uses a cart on an air track, attached between two springs, with ceramic magnets providing eddy-current damping and an ultrasonic motion sensor recording position against time. Replace the magnets with equal masses in the baseline trial so the moving mass stays fixed. Obtain each spring constant from the slope of applied force versus extension, then use $k=k_1+k_2$. Time many baseline cycles to check ${\omega }_0\simeq \sqrt{k/m}$. Compare one-magnet and two-magnet free decays; the protocol calls for a fit of the full trace including its offset. ^[ref]

The following additional controls make the inference testable. Level the air track and hold its air supply fixed. Check the ultrasonic motion sensor at several stationary flag positions against a ruler, including the full planned travel range; record the offset and scale correction. Check elapsed acquisition time against an independent clock, and choose a sampling rate that resolves many samples within the shortest expected oscillation period. Record the resting equilibrium before and after each run. Release the cart from a small measured displacement with minimal initial velocity, with the driving apparatus inactive. Repeat at several release sizes and magnet settings while keeping the springs and total mass unchanged. These are proposed calibration and repeatability checks, not reported measurements from the source.

For underdamping, fit the recorded position to $y(t)=y_{\mathrm{eq}}+A{e}^{-\alpha t}\cos ({\omega }_{\mathrm{d}}t+\varphi )$ by nonlinear regression. The readout to inspect is a sequence of alternating extrema about $y_{\mathrm{eq}}$ with shrinking amplitudes, and the deviations of the fitted curve from the recorded points. The model predictsCompare the right-hand side, determined independently from mass and spring constants, with the fitted left-hand side. A good fit supports a constant-coefficient single normal mode over the observed range; it does not establish a universal microscopic force law. The published protocol denotes the decay rate by $b$ and the velocity-proportional coefficient by $c$; here they are $\alpha$ and $b$, respectively.

At critical damping or overdamping, fit the corresponding nonoscillatory response including its offset and both initial conditions, rather than forcing a damped cosine onto the data. Adjusting the brake can explore the transition, but the cited protocol does not establish that its one-magnet or two-magnet settings attain critical damping. A near-critical trace over a short interval can be hard to distinguish from neighbouring damping ratios.

Important limitations are finite sensor range, missed or reflected acoustic wave pulses, measurement noise near equilibrium, nonzero release velocity, and position-dependent magnetic braking. Changing the release size tests for quadratic drag or nonlinear spring response; inspecting deviations near turning points can expose kinetic friction. Slow drift of the offset can imitate a long overdamped tail. Report uncertainties in fitted rates and independently measured $m,k$; absence of a visible crossing only bounds the behaviour within the instrument's sensitivity and recording interval. No numerical experimental results are asserted here.

The damping ratio is the dimensionless ratio $b/b_{\mathrm{crit}}$. Substituting the exponential function $x=e^{rt}$ into the viscous damping model gives the characteristic equation and its characteristic roots:For $0<\zeta <1$, underdamping gives a pair of complex conjugate roots with negative real part. For $\zeta =1$, critical damping gives the repeated negative real root. For $\zeta >1$, overdamping gives two distinct negative real roots. At $\zeta =0$ the roots are purely imaginary and the harmonic oscillator is undamped; this boundary case has no energy loss. With $m,k>0$, negative $b$ would inject energy and lies outside the dissipative model.

The left panel locates $r/{\omega }_0$ in the complex plane; the imaginary part gives the damped angular frequency, while the negative real part gives the exponential decay rate. The right panel shows the positive rates $-\mathrm{Re}(r_{\pm })/{\omega }_0$. Beyond critical damping, one root moves toward zero, explaining slow late motion for generic initial conditions. The repeated-root solution also has a polynomial prefactor, so these rates alone do not describe its complete response.

Eddy-current damping is electromagnetic damping in which motion of an electrical conductor through a magnetic field induces circulating electric currents, producing an opposing force and Joule heating. In the cart experiment it approximates viscous damping over the fitted range; spatial variation of the magnetic field or changes in velocity can limit that approximation.

For the convention $x=\mathrm{Im}(RF{e}^{i\omega t})$, viscous damping $b$ gives $R=[m({\omega }_0^2-{\omega }^2)+ib\omega {]}^{-1}$. Put $\delta =\omega -{\omega }_0$ and $\gamma =b/(2m)$. If both are small relative to ${\omega }_0$, the displayed approximation follows. Its squared modulus has half-width $\gamma$, and the displacement lags the force by $\pi /2$ at $\omega ={\omega }_0$.

For $\zeta >1$, let $r_{+}>r_{-}$ be the real characteristic roots. The response with the specified initial conditions isThe generic late response is governed by $r_{+}$, the root nearer zero. For very large damping ratio, $r_{+}\sim -{\omega }_0/(2\zeta )=-k/b$: an increasingly strong brake delays the return to equilibrium. This approximation comes from balancing the spring and viscous damping terms after the fast transient. Choosing $v_0=r_{-}{x}_0$ makes $C_{+}=0$ and leaves only the faster root; that exception requires different initial velocities as $b$ varies.

Use the $e^{i\omega t}$ convention for the harmonic force and displacement complex amplitudes. Substitution in the forced harmonic oscillator equation givesIts magnitude and phase relative to the force areAt zero frequency, $|R|=1/k$ and $\varphi =0$. At ${\omega }_0=\sqrt{k/m}$, $\varphi =-\pi /2$; at high frequency the phase approaches $-\pi$. For light viscous damping, the quality factor $Q=m{\omega }_0/b$ is large and there is a tall narrow displacement peak close to ${\omega }_0$.

Substituting $x=A{e}^{pt}$ into the damped harmonic oscillator equation givesFor nonnegative viscous damping, the cases areThe PDF's request to express these conditions in terms of $\gamma /m$ is dimensionally inconsistent with its definition $\gamma =b/m$: $\gamma$ already has units ${\mathrm{s}}^{-1}$. With the unnormalised equation $m\ddot{x}+b\dot{x}+kx=0$, the equivalent comparisons are $b/m\gtrless 2{\omega }_0$, or $b^2\gtrless 4mk$. Also ${\omega }_0$ is the undamped natural angular frequency, not in general the driven displacement peak.

A mass $M$ on a spring of stiffness $K$, subject to linear viscous damping and a sinusoidal drive, obeys the forced harmonic oscillator equationHere $x$ is displacement, $2M\gamma$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, and $F_0$ is a force in newtons. A series RLC circuit is an electrical analogue, with $M\leftrightarrow L$, $2M\gamma \leftrightarrow R$, $K\leftrightarrow 1/C$ and $x\leftrightarrow q$.

For free motion set $F_0=0$ and solve the characteristic equation $s^2+2\gamma s+{\omega }_0^2=0$. The roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$ distinguish three cases:

The constants are set by initial position and velocity. The mechanical energy and its rate of change in free motion areFor weak damping, the cycle-averaged energy decays approximately as $e^{-2\gamma t}$. The quality factor is $Q={\omega }_0/(2\gamma )$; for $Q\gg 1$ it equals $2\pi$ times stored energy divided by the loss per cycle. Logarithmic decrement of successive free-decay peaks determines $\gamma$ experimentally.

For driven motion the complete answer is the decaying free transient plus a steady response at the driving angular frequency:At low frequency, $X\to F_0/K$ and the displacement is nearly in phase with the drive. At $\Omega ={\omega }_0$, the lag is $\pi /2$ and $X=F_0/(2M\gamma {\omega }_0)$. At high frequency, $X\simeq F_0/(M{\Omega }^2)$ and the lag tends to $\pi$. The displacement resonance maximum lies atonly when $\gamma <{\omega }_0/\sqrt{2}$. Thus an underdamped oscillator need not have a nonzero-frequency displacement resonance peak.

In the steady state the mean input power equals the mean dissipated power:This absorption peaks exactly at $\Omega ={\omega }_0$, rather than at ${\Omega }_r$. Its half-power points have bandwidth $\Delta \Omega =2\gamma$, so $Q={\omega }_0/\Delta \Omega$ for this linear model. For weak damping, a high quality factor therefore means a narrow strong resonance and a long free-decay time. Exactly zero damping at resonant driving has no finite steady response: a secular amplitude grows with time.

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.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

Viscous damping means a force proportional to, and opposite to, velocity: $F_{\mathrm{d}}=-b\dot{x}$ with $b\ge 0$. Its coefficient has units $\mathrm{kg}{\mathrm{s}}^{-1}$, equivalently $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The name describes a force law; eddy-current damping can approximate the same law without a viscous fluid. Constant $m$, $b$ and $k$ produce a linear ordinary differential equation, allowing superposition of responses. Kinetic friction with approximately constant magnitude and quadratic drag obey different laws, so their decay need not have the exponential decay or critical threshold derived here.