C.1 · IB Physics
Simple pendulum
Release a pendulum from rest and compare the full motion, integrated step by step, with the small-angle SHM model T = 2π√(L/g). At small amplitudes the two agree; at large amplitudes the real pendulum takes longer.
Predict first
Keep g = 9.81 N kg⁻¹ and a small amplitude. Set the length L to 0.50 m, then increase it to 2.00 m. What happens to the SHM period 2π√(L/g)?
T = 2π√(L/g), so T is proportional to √L. Multiplying L by 4 multiplies T by 2: from 1.42 s to 2.84 s.
Check it in the model: Move the Length L slider from 0.50 m to 2.00 m and compare the SHM period readout (1.42 s then 2.84 s).
real pendulum (RK4) small-angle SHM model kinetic energy
small angle: T₀ = 2π√(L ÷ g)
exact: T = T₀ ÷ AGM(1, cos(θ₀ ÷ 2))
E = mgL(1 − cosθ), v = L|θ′|
Try this
- Keep L = 1.00 m and change the mass from 0.50 kg to 5.00 kg. Does the period change? Then double L from 0.50 m to 2.00 m and compare the periods.
- Use the Small swing, then the Large swing (90°) preset and compare the gold curve with the dashed SHM curve on the θ–t graph. How much longer is the real period at 90°?
- Switch the graph to T against θ₀. At roughly what amplitude does the real period become 10 % longer than 2π√(L/g)?
Common error: saying the period never depends on amplitude; T = 2π√(L/g) holds only for small angles in radians, and at 90° the real period is about 18 % longer.