Derive the pendulum approximation and test its boundary

1. Frame the question

A pendulum bob constrained to a circular arc has a tangential gravitational force −mg sinθ toward the bottom, with θ measured from downward vertical. Tension is radial and does not supply that tangential restoring component. The arc displacement is s=Lθ when θ is in radians; the support length, not the bob’s mass, sets the geometrical conversion.

2. Build the mechanism

Newton’s tangential equation is m aₜ=−mg sin(s/L). For small angles, sinθ≈θ, producing a≈−(g/L)s. Comparison with a=−ω²s gives T≈2π√(L/g). Mass cancels because gravitational force and inertial resistance are proportional to the same bob mass. This is the simple pendulum approximation, not the period of every distributed swinging object.

Lengthening the pendulum increases its approximate period; increasing gravitational acceleration decreases it. Changing bob mass alone does not change it in the ideal model. Air drag, support motion, finite bob size and large angular amplitude can invalidate assumptions. Length must be measured from the pivot to the point-mass location, not merely along a visible string segment.

The model compares sinθ/θ at the selected endpoint. This ratio approaches one near zero and falls as the angle grows. The fractional discrepancy in the restoring term is not numerically equal to the fractional period error: actual timing depends on motion throughout the cycle. We report only the small-angle period and this clearly separated diagnostic, without claiming an exact large-angle correction.

3. Work an example

With L=1 m and g=10 m/s², T≈2π/√10≈1.987 s. At 10°=0.1745 rad, sinθ/θ≈0.99493, a restoring-term discrepancy about 0.51%. At 30°, the discrepancy is about 4.51%; neither number is an exact period-error percentage.

4. Predict and investigate

What changes if the bob mass doubles but length and local gravity remain fixed?

Compare lengths 0.5 and 2 m at one amplitude, then compare 5° and 30° at fixed length. Explain why the displayed approximate period stays unchanged in the second comparison even though the approximation diagnostic worsens.

5. Transfer to a new case

Try a new practice case

A small-angle pendulum has period 2 s. Gravity becomes one-quarter as large at unchanged length. Find the new approximate period.

Your numerical result (s) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Explain the boundary

Why is replacing sin30° by the number 30 invalid when deriving the linear pendulum equation?

These are bounded virtual models and supplied numerical cases, not independent measurements or an observed physical investigation. Do not construct spring, pendulum or collision apparatus from these instructions. Practical evidence requires an appropriate supervised setting.

Sources

Original explanations, worked cases and numerical models. Framework checked 30 October 2026: AP Physics 1 Course and Exam Description, topics 7.1; 7.2. Mechanism reference: OpenStax University Physics 1, simple harmonic motion and pendulums. No official assessment items reproduced.