Audit a pendulum small-angle approximation

Resolve forces and torque about the pivot

The point bob experiences gravity downward and ideal tension along the suspension. Tension has zero torque about the pivot. With angle theta from downward vertical, gravity gives restoring torque −mgl sin(theta). The point-mass inertia is ml², so theta double-dot=−(g/l)sin(theta). The arc displacement is s=l theta; radians are required for that relation.

Linearize only for small angular amplitude

When |theta| is small in radians, sin(theta)≈theta. The resulting equation theta double-dot=−(g/l)theta is harmonic, with omega=sqrt(g/l), T=2pi sqrt(l/g) and f=1/T. Bob mass cancels because gravitational torque and inertia are both proportional to it. The cancellation assumes the point-mass/massless-suspension model.

Measure a force approximation, not a period correction

At selected angular amplitude theta0, the ratio of exact restoring torque to its linear approximation is sin(theta0)/theta0. The model reports one minus this ratio relative to the linear value. That number measures the restoring-torque approximation at one angle; it is not the finite-amplitude period error. Larger amplitudes make the exact differential equation nonlinear.

Independently compare bottom speeds

Energy from rest at theta0 gives mgl(1−cos(theta0))=mv_bottom²/2, so v_bottom=sqrt(2gl(1−cos(theta0))). The harmonic approximation predicts theta0 sqrt(gl). Comparing these speeds checks an independent consequence of the small-angle approximation. Neither computation establishes actual damping, string stretch, pivot friction or clock calibration.

Worked original case

For l=1 m,g=10 and theta0=10 degrees≈.174533 rad, linear T≈1.986918 s. The restoring ratio is about .994931. Exact bottom speed is about .551221 m/s, while the linear speed prediction is about .551922 m/s. These are synthetic model calculations, not measured apparatus accuracy.

Investigation

Compare amplitudes5,15,30 degrees at one length, then double the length. Convert to radians, calculate restoring ratios and both bottom-speed predictions. Explain why changing length affects the linear period while the restoring ratio at the same angle does not. Identify an independent timing observation required to study the actual finite-amplitude period.

Public numerical transfer practice

Try a new practice case

If the length of an ideal small-angle simple pendulum is multiplied by9, by what factor does its period change?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax harmonic motion and pendulums. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.