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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]

The torsion constant, or torsional spring constant, is the positive restoring torque per unit twist angle of a complete suspension in its linear elasticity range:Its SI unit is $\mathrm{N}m/rad$. The minus sign describes restoration toward the untwisted orientation. The derivative specifies a local stiffness if the response is not globally linear; using one constant over a finite range additionally requires a straight calibration curve. Return to the same zero after unloading tests recoverable elastic deformation; repeat loading in both directions to check whether the response depends on its previous loading. Staying below permanent deformation is necessary, but does not alone prove proportionality between torque and twist.

For an ideal uniform circular wire of length $\ell$, radius $r$, and shear modulus $G_s$, the filament torsional stiffness is $C=G_s{J}_p$, with polar second moment of area $J_p=\pi r^4/2$. Uniform twist gives $\theta /\ell$ as the twist per unit length, soHere $U$ is stored elastic energy. Doubling the wire's radius multiplies the ideal torsion constant by $16$; doubling its length halves it. This explains why a long thin fiber is useful for weak torque measurement. The formula assumes linear elasticity with uniform material and circular section; it does not calibrate arbitrary ribbons, clamps, or a compound suspension. Within this ideal model, the equipartition theorem predicts an equilibrium thermal mean-square angular displacement $⟨{\theta }^2⟩=k_B\mathcal{T}/\kappa$, where $k_B$ is the Boltzmann constant and $\mathcal{T}$ the uniform absolute temperature. A softer suspension magnifies both the signal and its thermal fluctuations.

Static calibration supplies known torques without assuming the unknown interaction being measured. Apply a known perpendicular force $F$ at lever arm $a$, record the change from the unloaded twist, and useSeveral positive and negative settings test proportionality; fit the slope of applied torque against twist and retain a possible intercept. A nonzero intercept must not be silently absorbed into the slope. Convert degrees to radians before reporting $\mathrm{N}m/rad$. PASCO's procedure turns the Coulomb balance on its side and adds small known masses to generate a gravitational weight at the sphere's lever arm, returning the rod to its index mark by turning the dial. Its tabulated coefficient is force per degree: multiply by the lever arm and $180/\pi$ to obtain our torsion constant. ^[ref]

The figure's ideal static calibration uses $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$: an applied torque of $2.0\times 10^{-10}\mathrm{N}\mathrm{m}$ gives $0.020\mathrm{rad}$. Both signs lie on the same straight line. The plotted points are calculated examples, not experimental observations; they cannot demonstrate a real fiber's linearity.

Dynamic calibration instead measures the oscillation period $T_0$ with known total suspended moment of inertia $I$. For negligible damping, small free oscillations about a fixed reference obey simple harmonic motion:Include the rod, mirrors, and any attachments in $I$ unless their contributions are demonstrably negligible. Keep the applied torque absent or independent of angle. If an external torque has derivative $g_{\tau }=d{\tau }_{\mathrm{ext}}/d\theta$ at equilibrium, linearization gives effective stiffness $\kappa -g_{\tau }$: timing alone then measures that combination, and stability requires $\kappa -g_{\tau }>0$. A constant applied torque merely shifts equilibrium. Non-negligible damping changes the measured period; damped torsional oscillation gives a correction under a specified damping model. Agreement between static and dynamic calibrations is a useful check on the assumed moment of inertia, reference stability, and restoring law, rather than a reason to ignore their assumptions.