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Sandbox Physics

M062 · Nonlinear oscillations / measuring the period

Finite-Amplitude Pendulum

Does a bigger swing change a pendulum's period? It does. A bigger swing takes longer than the small-angle formula predicts, by an amount given by an elliptic integral. At a 90-degree amplitude the period is about 18 percent longer. As the swing approaches the very top, the period grows sharply.

Turn a three-dimensional rigid-rod pendulum around and drag its starting state in the coordinate view. Compare its nonlinear motion with a harmonic reference, time successive passes through the bottom, and check the measured period against an independent prediction from the elliptic integral. Try large swings, the slowdown near the upright barrier, damping and complete rotations.

Interactive modelFinite-Amplitude Pendulum
Laboratory bead speed0 m/s0\,\mathrm{m/s}
Recorded time0 s0\,\mathrm{s}
Principal angle0 rad0\,\mathrm{rad}
Unwrapped angle0 rad0\,\mathrm{rad}
Angular velocity0 rad s−10\,\mathrm{rad\,s^{-1}}
Relative kinetic energy0 J0\,J
Gravitational energy0 J0\,J
Mechanical energy0 J0\,J
Viscous dissipation0 J0\,J
Recorded energy-balance defect0 J0\,J
Step-halving state difference00
Actual integration step0 ms0\,\mathrm{ms}
Initial-energy undamped period0 s0\,s
Small-angle period0 s0\,s
Latest acquired crossing intervalCollecting\text{Collecting}
Measured vs initial-energy reference0 %0\,\%
Initial conservative orbitLibration\text{Libration}
Rod force toward pivot0 N0\,N

Physics tutorial

When amplitude becomes part of the clock

BackgroundThe restoring torque is proportional to the sine of the angle. Near the bottom that is almost linear, but on large swings the bob lingers near its turning points.

Why it mattersOpenStax section 15.4 gives the pendulum equation and the limit of the small-angle approximation. The prediction for finite swings comes from the energy integral, and the complete elliptic integral in it is evaluated independently with NIST DLMF 19.8.5.

Start with the essentials

Focus question
How much timing error does the small-angle approximation create?
One-sentence intuition
Keep two things apart: the period predicted for the undamped orbit, and the period actually measured between two successive crossings.

Core mathematical model

Position and velocity in the lab frame

r=L(sin⁡θ,0,−cos⁡θ),v=Lθ˙(cos⁡θ,0,sin⁡θ)\mathbf r=L(\sin\theta,0,-\cos\theta),\quad \mathbf v=L\dot\theta(\cos\theta,0,\sin\theta)

The vertical axis points up. The blue arrow is tangent to the path, and the gray bob is a separate small-angle reference.

Nonlinear equation of motion

θ¨+2ζω0θ˙+ω02sin⁡θ=0,ω0=g/L\ddot\theta+2\zeta\omega_0\dot\theta+\omega_0^2\sin\theta=0,\quad \omega_0=\sqrt{g/L}

The viscous damping torque is proportional to the moment of inertia and to the angular velocity.

Period of an undamped swing

T=4ω0K(k),k=sin⁡θmax⁡2T=\frac{4}{\omega_0}K(k),\quad k=\sin\frac{\theta_{\max}}2

The turning angle comes from the energy, and the release may include an initial velocity.

Elliptic integral from a separate algorithm

K(k)=π2AGM⁡(1,1−k2)K(k)=\frac{\pi}{2\operatorname{AGM}(1,\sqrt{1-k^2})}

The prediction uses its own numerical algorithm, not the integrator that computes the motion.

Energy and dissipated heat

E=12mL2θ˙2+mgL(1−cos⁡θ),E+Q=E0E=\tfrac12mL^2\dot\theta^2+mgL(1-\cos\theta),\quad E+Q=E_0

Heat is integrated from the viscous power, so any mismatch left in the balance is still an independent check.

Rod force, with its sign

Frod=m(gcos⁡θ+Lθ˙2)F_{\mathrm{rod}}=m(g\cos\theta+L\dot\theta^2)

Positive force pulls toward the pivot; negative force pushes away. A rigid rod supports both.

Common difficulties

A prediction is not a measurement

Typical misconceptionThe period shown at release should already be a measurement.

Better mental modelThe prediction is available at once. A measurement has to wait until the bob passes the bottom twice in the same direction.

Damping changes orbit energy

Typical misconceptionOne elliptic-integral period describes every damped cycle.

Better mental modelThe fixed reference uses the starting energy. As the swing shrinks, later crossing intervals change.

A rod can push

Typical misconceptionAny release from a high angle must make the support go slack.

Better mental modelThat is true for a string, which can only pull. This apparatus uses a rigid rod on purpose.

Run the experiment

  1. 01

    Calibrate near the bottom

    Acquire the full record for Small-angle clock, then change the bob mass.

    What to observe: The period stays the same, while energy and force grow in proportion to the mass.
  2. 02

    Increase the swing

    Compare Large swing and Near the upright barrier with the gray reference bob. Orbit the apparatus or use the side view.

    What to observe: The harmonic reference pulls ahead, and the measured interval matches the longer period from the elliptic integral.
  3. 03

    Switch from swinging to rotating

    Choose Complete rotations, then reverse the initial angular velocity.

    What to observe: The angle shown within one turn jumps back, but the unwrapped angle stays continuous, and so does the timing of full turns in the same direction.
  4. 04

    Separate physics from numerical error

    Switch to the energy view, acquire a damped record, then halve the requested time step.

    What to observe: The damping loss remains, while the independent energy mismatch and the difference between the two runs shrink as the step gets smaller.
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