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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.

A deliberately restricted forced harmonic oscillator shows two effects of the baryon loading parameter without duplicating the full photon Boltzmann hierarchy. Use conformal time $d\eta =dt/a$ in seconds, a comoving wavenumber $k$ in inverse metres, constant $R=3{\rho }_b/(4{\rho }_{\gamma })$, and equal constant dimensionless Newtonian gauge gravitational potentials $\Phi =\Psi \ne 0$. The energy densities ${\rho }_b$ and ${\rho }_{\gamma }$ use the same units. Neglect expansion drag, Silk damping and evolution of the potentials. The photon-baryon sound speed is $c_s=c/\sqrt{3(1+R)}$. Writing ${\Theta }_0$ for the photon monopole temperature perturbation, the photon-baryon acoustic oscillator reduces toFor $y=({\Theta }_0+\Psi )/\Psi$ this gives the displayed forced equation. Constant loading and the neglected expansion drag make this a local teaching model, not a self-consistent expanding cosmology with $R$ fixed forever.

Choose $y(0)=1/3$ and $y^{\prime }(0)=0$, the static matter-era Sachs-Wolfe combination normalized by $\Psi$. With $q=k{c}_s\eta$, direct substitution givesAt $R=0$, successive extrema have equal magnitudes $1/3$. At the stipulated $R=0.6$, $y(\pi )=-23/15$ whereas $y(2\pi )=1/3$. The shifted equilibrium $y=-R$ makes alternating extrema unequal. This is an elementary illustration of how the baryon loading parameter changes acoustic amplitudes; these extrema are neither measured peak heights nor a complete Cosmic microwave background power spectrum. Comparing curves at fixed $q$ isolates the response; at fixed $k$ and $\eta$, increasing $R$ also reduces the photon-baryon sound speed and hence the phase.

For example, holding $c{\eta }_{\ast }=300\mathrm{Mpc}$ gives sound horizons $r_s=173.2\mathrm{Mpc}$ at $R=0$ and $136.9\mathrm{Mpc}$ at $R=0.6$. Their ratio is $1/\sqrt{1.6}$. Each term of the forced equation has units of inverse seconds squared because $y$ and $R$ are dimensionless and $k{c}_s$ is an inverse time. In the actual photon-baryon fluid, $R$ grows with the scale factor and the sound horizon is the integral of the evolving photon-baryon sound speed.

For a forced harmonic oscillator with constant viscous damping coefficient $m\gamma >0$, the average power divided by its maximum is ${\gamma }^2{\omega }^2/[({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2]$. It peaks at ${\omega }_0=\sqrt{k/m}$. Its two positive half-power angular frequencies are ${\omega }_{1,2}=(\sqrt{4{\omega }_0^2+{\gamma }^2}\mp \gamma )/2$. Hence their separation is exactly $\gamma$, and their product is ${\omega }_0^2$. The narrow-resonance approximation ${\omega }_{1,2}\simeq {\omega }_0\mp \gamma /2$ is needed for symmetry about the peak, not for the exact bandwidth in angular-frequency units.

For a lightly damped forced harmonic oscillator with known force amplitude $F_0$, measure the static displacement $x_s$, the resonance angular frequency ${\omega }_r$ and the peak displacement $x_p$. The response is $|x_0|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. To leading order at large quality factor,The exact displacement resonance is ${\omega }_r^2=k/m-b^2/(2m^2)$, whereas the undamped natural angular frequency is ${\omega }_0=\sqrt{k/m}$. A lightly damped half-power bandwidth obeys $\Delta \omega \simeq b/m$ and $Q\simeq {\omega }_0/\Delta \omega$. Half power corresponds to displacement $x_p/\sqrt{2}$, not $x_p/2$. Graph readings should be reported as estimates rather than extra significant digits from the model inversion.

The denominator of the travelling response vanishes atFor the positive speeds in the paper, a finite travelling response exists whenever $c\ne c^{\ast }$. At $c=c^{\ast }$, the dynamic condition would reduce to $p_1/\rho =0$, which is impossible. The finite-depth dispersion relation of a surface gravity wave with wavenumber $k$ is ${\omega }^2=gk\tanh (kh)$. At $k=1$, its phase speed $\omega /k$ equals $c^{\ast }$: the moving pressure pattern drives a natural surface gravity wave at its own speed, producing resonance.

The restriction concerns a bounded travelling response. To distinguish this from existence of a solution of an initial value problem, set $T=\tanh h$, $P=p_1/\rho$, and ${\omega }_{\ast }^2=gT$. Ifthe linearized free-surface boundary conditions reduce to the forced harmonic oscillatorFor example, if the forcing is switched on at $t=0$ from a flat surface with zero initial velocity, then, for $c\ne {\omega }_{\ast }$,This satisfies $H(0)=\dot{H}(0)=0$ and contains both free oscillations and the forced component. At $c={\omega }_{\ast }$ a solution of the initial value problem still exists, but its particular solution containswhich grows linearly with time. Thus the ideal inviscid flow model predicts resonant growth rather than a finite constant amplitude. Additional free oscillations can set the initial conditions but cannot remove that secular growth.

Put $D={\Omega }^2-{\omega }^2$, $E=2\gamma \omega$, and $N=D^2+E^2$. Substituting a sine-and-cosine trial solution for the forced harmonic oscillator givesThe complementary solution is $e^{-\gamma t}(C\cos \nu t+S\sin \nu t)$, where $\nu =\sqrt{{\Omega }^2-{\gamma }^2}$. Every nonzero such solution is unbounded as $t\to -\infty$, so boundedness on the whole negative time axis forces $C=S=0$. This fixes the phase as well as the amplitude. In particular,

The moving equilibrium produces a forced harmonic oscillator:A particular solution is $F{e}^{-\alpha t}$, where $F={\omega }_0^{2}d/({\alpha }^2-\gamma \alpha +{\omega }_0^2)$. The denominator equals $(\alpha -\gamma /2{)}^2+{\omega }_d^2>0$, so it cannot vanish for real $\alpha$ under the given underdamping condition. Imposing both initial conditions givesFor $\alpha =\gamma$, $F=d$, so the entire oscillatory term vanishes:The body is tracking an exponentially moving equilibrium without oscillation. There is no spring extension in this special motion: linear drag alone supplies $m\ddot{x}={\gamma }^{2}md{e}^{-\gamma t}$. An underdamped homogeneous equation does not require every forced solution to contain an oscillatory transient.

Divide the external force by $m$. Away from resonance a particular solution has the forcing frequency, so the general forced harmonic oscillator solution isThe two homogeneous terms cannot be discarded in an undamped system: they do not decay. If the same initial conditions are imposed, then $A=x_0-C\cos \varphi$, $B=-C(\omega /{\omega }_0)\sin \varphi$, where $C=F_0/[m({\omega }_0^2-{\omega }^2)]$.

For $\omega \gg {\omega }_0$, the forced displacement approaches $-F_0\cos (\omega t-\varphi )/(m{\omega }^2)$: its amplitude decreases quadratically and it is in antiphase with the force. This statement concerns the forced component; a freely excited component at ${\omega }_0$ can remain.

At resonance the preceding denominator is zero and the appropriate solution isIts amplitude grows linearly because the external force supplies energy coherently. For $x(0)=x_0$, $\dot{x}(0)=0$, choose $A=x_0$ and $B=F_0\sin \varphi /(2m{\omega }_0^2)$. Physical damping or nonlinear response ultimately limits the ideal resonant growth.

Take $m,k,\gamma ,F_0>0$ and $\omega >0$, and write $D_{\omega }=k-m{\omega }^2$ and $B_{\omega }=m\gamma \omega$. Substitution of $z=A\cos \omega t$ into the forced harmonic oscillator equation givesEquating the independent sine and cosine coefficients yieldsThus the required phase matching for a displacement referenced driven oscillator isFor a positive amplitude $A$, one must also choose the correct quadrant: $\cos {\varphi }_0=D_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$ and $\sin {\varphi }_0=B_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$. Squaring and adding the coefficient equations givesFor arbitrary forcing phase, the steady response is instead $z=A\cos (\omega t+{\varphi }_0-\delta )$, where $\delta$ has the sine and cosine just specified. A change of time origin can put this displacement into $A\cos \omega t$, but it also changes the stated forcing phase.

The source's requested parameter list for $A$ omits $\gamma$; in general it cannot be omitted. As a concrete counterexample, take $m=k=F_0=\omega =1$ and ${\varphi }_0=\pi /2$. The prescribed cosine response exists for every $\gamma >0$, but $A=1/\gamma$, so fixing all four listed parameters does not fix the amplitude. Likewise, with fixed ${\varphi }_0=0$, positive damping and nonzero $\omega$, the second coefficient equation prevents a nonzero pure-cosine response. The forcing phase is an essential part of the condition.

The homogeneous linear differential equation has roots $(-1\pm i\sqrt{3})/2$, so its complementary function decays:For a particular integral, set $f_p=B\cos 2t+C\sin 2t$. Substitution givesThe two coefficient equations are $-3B+2C=A$ and $-2B-3C=0$, so $B=-3A/13$ and $C=2A/13$. Consequently every solution obeysThe limiting behaviour is a periodic steady oscillation with amplitude $|A|/\sqrt{13}$, the long-time response of a forced harmonic oscillator. Unless $A=0$, there is no finite numerical limit: at $t=n\pi$ and $t=n\pi +\pi /2$, the two limiting values are $-3A/13$ and $3A/13$. For $A=0$, $f(t)\to 0$.

Let $\mathbf{E}=E\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{y}}$, and choose $\hat{\mathbf{z}}=\hat{\mathbf{x}}\times \hat{\mathbf{y}}$. Part (a) gives $y=0$. With the origin at ${\mathbf{x}}_0$, the remaining Cartesian components areThe initial conditions are $x=z=\dot{x}=\dot{z}=0$. Integrating the second equation gives $\dot{z}=\omega x$, soSolving this forced harmonic oscillator and then integrating $\dot{z}=\omega x$ yieldsThus the position vector is ${\mathbf{x}}_0+x(t)\hat{\mathbf{x}}+z(t)\hat{\mathbf{z}}$. The oscillation occurs at the signed cyclotron frequency, superposed on the usual E-cross-B drift.

The dynamic condition reduces toDifferentiate $H^{\prime }=kF$ and define the natural frequencyThen the surface amplitude obeys the forced harmonic oscillatorFor $\omega \ne \Omega$, its general solution and the corresponding potential amplitude are

If the fluid is initially quiescent, $H(0)=F(0)=0$, so

Let $D=1+{\omega }^2$, with $\omega >0$. A particular solution of the forced harmonic oscillator is $y_p=e^{-t}(at+b)$. Substitution gives $Da=2$ and $Db-2a=0$, hence $a=2/D$ and $b=4/D^2$. Adding $A\cos \omega t+B\sin \omega t$ and imposing the two initial conditions gives $A=-4/D^2$ and $B=2(1-{\omega }^2)/(\omega D^2)$. ThusThe first term tends to zero. The amplitude of the surviving sinusoid isAs an independent interpretation, the residual amplitude of an oscillator after an integrable force is ${\omega }^{-1}|{\int }_0^{\infty }F(t)e^{-i\omega t}dt|$. Here that integral is $2/(1+i\omega {)}^2$, giving the same result.

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 ${\omega }_0$.

For the steady response of a forced harmonic oscillator, write $x=\mathrm{Re}(A{e}^{i\omega t})$ and use the same complex amplitude convention for the driving force. Substitution givesThus the response magnitude and displacement phase relative to the force areThe phase goes from $0$ at low frequency, through $-\pi /2$ at ${\omega }_0$, towards $-\pi$ at high frequency. The static magnitude is $1/(m{\omega }_0^2)=1/k$, while the high-frequency magnitude is approximately $1/(m{\omega }^2)$.

Define the quality factor by $Q={\omega }_0/\gamma$. If $Q>1/\sqrt{2}$, the displacement peak occurs at ${\omega }_r=\sqrt{{\omega }_0^2-{\gamma }^2/2}$; for $Q\le 1/\sqrt{2}$ the magnitude decreases monotonically from its static value. For $Q\gg 1$ there is a tall narrow peak close to ${\omega }_0$, of height approximately $Q/k$ and width approximately $\gamma$. At very small $Q$ the resonant peak disappears and the response is broad. The plotted functions are normalized as $k|R|$, at fixed $m,k$.

The original diagram is a series RLC circuit, so the same electric current $i=\dot{q}$ passes through all three elements. Choose capacitor electric charge $q$ positive on the plate into which positive $i$ enters. Kirchhoff's voltage law givesDividing by inductance $L$ givesThe correspondence to the forced harmonic oscillator is $m\leftrightarrow L$, $b\leftrightarrow R$, $k\leftrightarrow 1/C$, and $x\leftrightarrow q$. The stored energy is $L{i}^2/2+q^2/(2C)$; its balance is $\dot{E}=Vi-R{i}^2$, confirming the damping sign.

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 the forced harmonic oscillator $m\ddot{z}+m\gamma \dot{z}+kz=F_0\cos (\omega t+{\varphi }_0)$, a response $z=A\cos \omega t$ requires $A(k-m{\omega }^2)=F_0\cos {\varphi }_0$ and $Am\gamma \omega =F_0\sin {\varphi }_0$. Squaring and adding gives the amplitude $A=F_0/\sqrt{(k-m{\omega }^2{)}^2+(m\gamma \omega {)}^2}$ for positive $A,F_0$. The two coefficient equations determine the full quadrant of the forcing phase, while their ratio alone can lose that information. In particular the amplitude cannot generally be independent of damping: at $m=k=F_0=\omega =1$ and ${\varphi }_0=\pi /2$, it is $A=1/\gamma$.