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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.

The original apparatus has two small masses at distance $L/2$ from the suspension axis. Neglecting the rod's moment of inertia, their total moment of inertia is $I=2m(L/2{)}^2=m{L}^2/2$, consistent with the supplied torsion pendulum period. Treat $m$ and $L$ as exact because no uncertainties are specified. Rearranging the period relation givesThe torsion constant is a restoring torque per unit angle. In dimensional calculations the radian is dimensionless.

Using first-order propagation of uncertainty, with the quoted period uncertainty interpreted as a standard uncertainty,Consequently

Let $I$ be each sphere's moment of inertia about the fibre, and let $K_t$ be the common fibre filament torsional stiffness, so a length $\ell$ exerts restoring torque $-K_t\varphi /\ell$. The single-sphere torsion pendulum obeysFor the two coupled torsional pendulums, the upper fibre has twist ${\varphi }_1$ and the joining fibre has relative twist ${\varphi }_2-{\varphi }_1$. The joining fibre exerts equal and opposite torques on the two spheres:Using the measured ${\omega }_0$ givesThe fibre's own rotational inertia and damping are neglected, so its twists supply elastic torques without additional dynamic coordinates.

As $R\to 0$, the short joining fibre is very stiff and the lower normal mode has ${\varphi }_1\simeq {\varphi }_2$. The spheres therefore behave as one torsional load with combined moment of inertia $2I$, suspended by the upper fibre of torsion constant $K_t/{\ell }_1$. Its angular frequency satisfiesThe high-frequency opposite-twist mode costs increasingly large elastic energy and moves to arbitrarily high frequency in the massless-fibre model.

As $R\to \infty$, the joining fibre is very compliant. In the higher normal mode the lower sphere scarcely moves: the amplitude ratio tends to $A_2/A_1\simeq -1/R\to 0$. The upper sphere approaches the original single torsion pendulum, givingThe other mode is then a slow twist of the almost freely suspended lower sphere. These interpretations assume the formal limits do not invalidate the thin massless-fibre approximation.