Torsional Oscillator (Disk on a Shaft)

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Clamp one end of a shaft, hang a disk on the other, give it a twist, and let go: it rings. The disk winds the shaft up, the shaft winds it back, and the two trade energy back and forth at one fixed rate — the system’s natural frequency. This is the torsional cousin of a mass bobbing on a spring, and it is the single most important idea in machine dynamics, because every real driveline, crankshaft, and torsion-bar is one of these hiding in plain sight.

$\omega_n = \sqrt{\frac{k_t}{J_d}}, \qquad k_t = \frac{G J_p}{L}, \qquad f = \frac{\omega_n}{2\pi}, \qquad T = \frac{1}{f}$

The shaft is a torsion spring of stiffness $k_t = G J_p/L$ — literally the inverse of the twist the shaft-in-torsion page computes, $\theta = TL/GJ_p$. The disk is the inertia $J_d = \tfrac12 m_d R^2$ — literally what the flywheel page computes. Bolt the two together and you get an oscillator. Three things to notice:

• The pitch ignores how hard you ring it. There is no amplitude $\Theta$ anywhere in $\omega_n$. A violent swing and a gentle one oscillate at exactly the same frequency — the isochronism that Galileo spotted in a swinging lamp and that made the pendulum clock possible. What the amplitude does set is the stress: the peak surface shear $\tau_{max} = G d\,\Theta/(2L)$ is directly proportional to $\Theta$. Ring it twice as hard and the note holds while the stress doubles. Drive the amplitude knob up and watch the frequency sit still while the safety factor falls.
• The material enters only as $G/\rho$. Expand everything and the frequency is $\omega_n^2 = (G/\rho)\,d^4/(16 L R^4 t_d)$ — the material shows up only through the ratio of shear modulus to density, which is the square of the shear-wave speed. That ratio barely changes across metals, so a steel disk-shaft and an aluminium one of the same size ring at nearly the same pitch (swap the material in the widget and watch the frequency hardly move). Reach for a polymer, with a much lower $G/\rho$, and the pitch drops hard. Strength ($\sigma_y$) decides what amplitude the shaft survives, never what note it plays.
• Stiffness up, inertia down. A fatter (the $d^4$ is brutal) or shorter shaft is a stiffer spring and rings higher; a bigger or heavier disk carries more inertia and rings lower. This is why engine flywheels are sized as much for the torsional frequencies they create as for the energy they store.

One material, one simplification. Here a single material sets both the shaft and the disk, so you cannot make only the disk denser in the widget — increasing density raises $J_d$ (lowering the pitch) and changes $G$ at the same time. The counter-intuitive “a heavier disk rings lower” story still lives in the readouts: watch $J_d$ climb with density. Independent disk-and-shaft materials are a Phase 3 capability (multi-material binding slots); this page states the shared-material assumption up front rather than fake it.

The chain teaser (coming in Phase 4). This page is built to be chained. Its two governing inputs are exactly two other pages’ outputs: the shaft stiffness that torsion-shaft embodies and the disk inertia that flywheel-disk produces. The peak restoring torque $T_{dyn}$ it reports back is an honest torque you could feed into the shaft page to size it. Wire motor → shaft → flywheel and you have simulated a real driveline’s natural frequency, every number carrying its citation. The chain-builder that lets you draw those wires arrives in Phase 4; for now the ports are here, waiting.

A note on units: the natural frequency $f$ is measured in hertz and the spin-like $\omega_n$ in radians per second, and although both are “per second” the site treats them as different quantity kinds — you cannot silently wire a 35 Hz ring into a port expecting a 35 rad/s spin. The factor of $2\pi$ between them is always explicit. That distinction is exactly what the dimensional type system exists to protect.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget reports one clean natural frequency and the shear stress at a given amplitude. Real torsional systems fail in ways that clean single number quietly hides:

• Resonance is the whole danger. A natural frequency is harmless until something drives the system at it — and engines drive their crankshafts at every firing pulse, whose frequency sweeps with rpm. When a firing harmonic crosses $\omega_n$, the amplitude no longer stays where you put it: it builds, cycle on cycle, until damping (small, in a bare shaft) or a fractured shaft stops it. Marine and aircraft drivelines have torsional critical speeds they must pass through quickly or avoid entirely, and crankshafts carry tuned dampers built for exactly this. This page computes where the resonance sits; it says nothing about how hard reality will push on it.
• Fatigue lives at the amplitude, not the frequency. The safety factor here is against first shear yield in a single swing. But a shaft that rings passes through its peak stress twice per cycle, tens of millions of times a day at running speed. Torsional fatigue cracks start at the surface where $\tau_{max}$ lives — and far below the yield line — nucleating at the usual suspects: keyway corners, fillet steps, spline roots, press-fit edges. The clean $16T/\pi d^3$ field this page assumes is exactly what a stress concentration destroys.
• The lumped model throws away the shaft’s own inertia. Treating the shaft as a massless spring is only good while its rotational inertia is small next to the disk’s — the warn banner fires past $J_{shaft}/J_d > 0.1$. A heavier shaft stores kinetic energy too, so the true frequency is lower than $\sqrt{k_t/J_d}$ (a first correction lumps about one-third of the shaft’s inertia into the disk). A long uniform shaft is really a continuous system with an infinite ladder of higher torsional modes this one-disk model cannot see at all.
• Two disks, or ten, change the question. Add a second rotor and the system has two natural frequencies and an internal node where the shaft twists hardest; a full driveline (engine, clutch, gearbox, propeller) is a chain of inertias and torsional springs with a whole spectrum of modes, solved by the Holzer method or eigenvalues, not by one square root. This page is the single degree of freedom you build all of that from.
• Big amplitudes leave linear elasticity behind. Everything here assumes the shaft stays linear-elastic so that $k_t$ is a constant. Ring it hard enough to yield (the shear-stress warn) and the spring rate softens, the motion stops being a clean sinusoid, and the frequency itself drifts with amplitude — the isochronism that defined the page quietly breaks.

None of these are edge cases; they are the reason torsional-vibration analysis is its own discipline. The natural frequency is where that analysis starts, not where it ends.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: the torsion formula τ = 16T/πd³, the angle of twist θ = TL/GJ, and hence the torsional stiffness k_t = GJ/L).
• Hibbeler, R. C., Engineering Mechanics: Dynamics, 14th ed., Pearson, 2016 — §17.1 (mass moments of inertia; the uniform solid disk I = ½ m R² about its spin axis) and ch. 22 (undamped free vibration: a single mass on a spring oscillates at ω_n = √(k/m), the translational analogue of the torsional system here).
• Timoshenko, S. P., Young, D. H., & Weaver, W., Vibration Problems in Engineering, 4th ed., Wiley, 1974 — the torsional pendulum / disk-on-shaft system: a rotor of mass moment J on a shaft of torsional stiffness k_t oscillates in simple harmonic motion at ω_n = √(k_t/J).
• Juvinall, R. C., & Marshek, K. M., Fundamentals of Machine Component Design, 5th ed., Wiley, 2011 — shaft design and torsional vibration: when the shaft's own inertia is not small compared with the attached rotor, a lumped one-disk model over-predicts the natural frequency, and roughly one-third of the shaft inertia should be added to the disk.
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 7 (shafts and torsional systems), §3-12 (power transmission), and §5-4 (maximum-shear-stress theory, shear yield at S_sy = 0.5 S_y).