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.

Drag the bob and compare large swings with a small-angle reference. Measure the longer period, then try damping or full rotations.

Laboratory bead speed $0\,\mathrm{m/s}$

Recorded time $0\,\mathrm{s}$

Principal angle $0\,\mathrm{rad}$

Unwrapped angle $0\,\mathrm{rad}$

Angular velocity $0\,\mathrm{rad\,s^{-1}}$

Relative kinetic energy $0\,J$

Gravitational energy $0\,J$

Mechanical energy $0\,J$

Viscous dissipation $0\,J$

Recorded energy-balance defect $0\,J$

Step-halving state difference $0$

Actual integration step $0\,\mathrm{ms}$

Initial-energy undamped period $0\,s$

Small-angle period $0\,s$

Latest acquired crossing interval $\text{Collecting}$

Measured vs initial-energy reference $0\,\%$

Initial conservative orbit $\text{Libration}$

Rod force toward pivot $0\,N$

Does a bigger swing change a pendulum's period?

Show small-angle referenceA gray ghost starts with the same angle and angular velocity but has no damping.Bob mass

m

1.00\,\mathrm{kg}

Rod length

L

1.00\,\mathrm{m}

Gravity

g

9.81\,\mathrm{m\,s^{-2}}

Release angle from the bottom

\theta_0

75.00\,{}^\circ

Initial angular velocity

\dot\theta_0

0.00\,\mathrm{rad\,s^{-1}}

Viscous damping ratio

\zeta

0.00\,

Maximum numerical step

h_{\max}

10.00\,\mathrm{ms}

Inspect recorded time

t/t_{\mathrm{end}}

0.00\,\%

Drag the purple bob to set the release angle. The faint motion curve is the small-angle solution for the same starting state. Period dots use only crossings recorded before the cursor; the period curve is a prediction for a release from rest. A negative rod force means the rod is in compression. Everything you can drag also has a keyboard slider.

The model is a point bob on a rigid rod with a fixed pivot in uniform gravity. The rod can carry tension or compression, so unlike a string it never goes slack. A viscous torque at the pivot is optional. The elliptic-integral period describes the undamped orbit at the starting energy; with damping, the interval between crossings changes as energy is lost. RK4 integration of the power, step halving and the signed rod force show the limits of the model and of the numerics. The diagnostic readings separate kinetic energy, gravitational energy and accumulated heat. Scene lengths are rescaled to fit, and the bead size, supports and face markings are schematic. The gray bob follows the undamped small-angle solution even where it does not apply, so it may be inaccurate for large swings. The gray reference is offset slightly in depth for visibility.

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

\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 velocity is tangent to the path. The gray bob is a separate small-angle reference.

Nonlinear equation of motion

\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=\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)=\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=\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

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

01
Calibrate near the bottom
Acquire the full record for Small-angle clock, then open Release and numerical settings to change the bob mass.
What to observe: The period stays the same, while energy and force grow in proportion to the mass.
02
Increase the swing
Compare Large swing and Near the upright barrier with the gray reference bob. Orbit the apparatus or use Swing plane.
What to observe: The harmonic reference pulls ahead, and the measured interval matches the longer period from the elliptic integral.
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.
04
Separate physics from numerical error
Open Recording and diagnostics, acquire a damped record, then halve the maximum numerical step under Release and numerical settings.
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.