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SCIENCE · INVESTIGATE AND EXPLAIN

Test where the small-angle approximation is useful

Compare a linear formula with the full sine restoring law and choose a stated error tolerance.

Learning outcome: Quantify period underestimation, distinguish numerical from model error and explain why amplitude independence fails beyond the small-angle limit.

Starting recap

The replacement sin(theta) ≈ theta uses radians and is accurate near zero. A useful approximation needs a purpose and an error tolerance.

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Try

Two pendulums have the same length but release angles of 10° and 60°. Must their periods agree?

Open the experiment

Predict

At 1 m, predict whether the small-angle formula overestimates or underestimates the 60° period, and whether its error remains below 1%.

Experiment

Compare 5°, 10°, 20°, 30° and 60° at 1 m. Calculate discrepancy as a percentage of the nonlinear reference. Find the largest offered angle meeting a 1% period tolerance, then change length and see whether that percentage changes.

Test the small-angle pendulum approximation

Worked teaching model · synthetic values · Point bob of mass 1 kg on a massless inextensible string, fixed pivot, planar motion, uniform g = 9.81 m/s², released from rest. No drag, pivot friction, driving or finite-bob effects. Small-angle period uses sin(theta) ≈ theta in radians. Nonlinear reference period uses the full sine restoring law with a bounded 256-interval Simpson integral, not measured data. The geometry shows the release configuration, not a time animation. Changing controls prepares a separate release.

Reset keeps your prediction, captures and practice history.

Release configuration · filled bob: release · open bob: lowest point0 m0.5 m1 m1.5 m2 mAngle 10° · length 1 m · rise 0.0152 mLength is pivot to bob center; drawing scale stays fixed
Compare one full cycle · shared period scale00.511.522.533.5Small-angleNonlinear referencePeriod (s) · small differences may need the numerical ledger
Pendulum timing, geometry and approximation
QuantityValueRelationship
Release angle0.174533 rad10 × π/180
Small-angle period2.006067 s2π√(L/g)
Nonlinear reference period2.009893 sFull sine restoring law; numerical integral
Period underestimate0.190356%100 × (reference − small-angle) / reference
Release angular acceleration-1.703489 rad/s²−(g/L) sin(theta)
Linearized angular acceleration-1.712168 rad/s²−(g/L) theta
Height above bottom0.015192 mL(1 − cos(theta))
Release energy above bottom0.149036 Jmgh, fixed m = 1 kg
Bottom speed0.545960 m/s√(2gh), ideal energy conservation

The geometry is exact for the stated point-bob model. The reference period is a numerical evaluation of that model, not an exact decimal or a measured period. A useful tolerance depends on the task; the displayed discrepancy does not include drag, length uncertainty or timing errors.

Small-angle period 2.006067 s; nonlinear reference 2.009893 s.

The small-angle value underestimates the reference by 0.190% (reference period is the denominator). Release height 0.015192 m; ideal bottom speed 0.545960 m/s. Numerical integration error is separate from physical-model error.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Ask how accurate the approximation must be

The small-angle period formula predicts the same period for every release angle at a fixed length. The full sine restoring law does not. At L = 1 m, compare the 10° and 60° settings and decide whether that difference matters for a task requiring period prediction within 1%.

The simple estimate stays at 2.006067 s. The nonlinear reference is approximately 2.009893 s at 10° and 2.152875 s at 60°. The larger-angle motion takes longer. The approximation underestimates period because replacing sin(theta) with theta overstates restoring acceleration away from equilibrium.

Use radians in the restoring law

Ten degrees is about 0.174533 radians; sixty degrees is about 1.047198 radians. At 60°, sin(theta) ≈ 0.866025, noticeably smaller than theta itself. In the full model, release angular acceleration is −(g/L) sin(theta); the linearized model uses −(g/L)theta. Compare those two entries in the ledger.

A force discrepancy at the release point is not itself the period discrepancy: the period depends on the whole motion. The nonlinear reference accounts for changing restoring force across the swing. It does not merely adjust the starting acceleration and assume it remains constant.

State the denominator and tolerance

The lab reports 100 × (Treference − Tsmall)/Treference, the percentage by which the approximation is below the reference. With 10°, this is about 0.190%; with 30°, about 1.711%; with 60°, about 6.819%. A calculation dividing by Tsmall would produce a different percentage, so name the denominator.

For a teaching target of at most 1% period error, explore the offered 5° increments. The largest offered angle meeting that criterion is 20°, with about 0.761% discrepancy. At 25° it exceeds 1%. This is a result for the stated model and discrete control choices, not a universal definition of “small angle.” A timing task with a different tolerance would choose a different range.

Now change length while keeping angle fixed. Both periods scale with √L, so the percentage discrepancy stays the same. Absolute error in seconds grows with length. Decide whether a specification is relative or absolute before choosing an approximation.

What the reference calculation does

The reference keeps the point-bob, fixed-pivot, no-drag assumptions but removes the small-angle replacement. Its period follows from the full sine restoring law and conservation of energy. The browser evaluates a smooth transformed integral using 256 fixed Simpson intervals, a bounded numerical sum.

Optional: inspect the nonlinear period integral

Let theta0 be the release angle in radians and phi an integration variable from 0 to π/2. The period is 4√(L/g) times the integral of 1/√(1 − sin²(theta0/2) sin²(phi)) over that interval. This is a complete elliptic integral. As theta0 tends to zero, the integrand tends to 1, recovering 2π√(L/g). The transformed form avoids an endpoint singularity in a direct angle-time integral.

The NIST pendulum reference gives the elliptic-integral period relation. Numerical verification compares the implementation with an independent arithmetic-geometric-mean relation across the supported settings. You do not need elliptic integrals to compare the two period predictions.

A numerical reference is not a measurement and its printed decimal is not exact. Integration error, small-angle approximation error, and physical-model error are different. Real length uncertainty, bob size, pivot friction and air resistance are absent from both model predictions. More numerical precision cannot repair an incorrect physical model.

Transfer to a measurement decision

A timing instrument resolves 0.02 s. At a 1 m length and 10° release, the model difference between the two periods is about 0.003826 s. Can one rounded timing reading reliably distinguish the predictions? Would this observation prove that the small-angle model is exactly correct?

Check the approximation transfer

The difference is smaller than the instrument’s stated resolution, so one rounded reading does not reliably distinguish the predictions. Agreement at that resolution does not prove exact equality. Timing many cycles, defining consistent start/end events and assessing random and systematic uncertainty could improve the comparison, but repeated readings do not automatically remove systematic errors.

Apply

A small-angle calculation gives 2.00 s while a nonlinear reference gives 2.10 s. The underestimate as a percentage of the reference is approximately:

Public practice checks a specific claim. It does not establish independent mastery.

Explain why the full restoring law produces amplitude-dependent timing, state the chosen denominator and distinguish a small numerical integration error from physical-model limitations.

Continue

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