Pendulums
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
A simple pendulum swinging through small angles is SHM, with period T = 2π√(L/g). Only the length and g matter. The bob’s mass and the size of the swing (as long as it’s small) don’t change the period.
1. The restoring force
2. Why mass doesn’t matter
A heavier bob feels a bigger restoring force, but it also has more inertia. The two cancel, just as all objects fall with the same acceleration.
L is measured from the pivot to the center of the bob.
“Small angles” means up to about 15°. For big swings the motion is no longer exactly SHM and the period gets slightly longer, but AP problems use the small-angle formula.
3. Example, then try it
Find the period of a 0.40 m pendulum. (g = 10 m/s²)
L/g = 0.40 ÷ 10 = 0.040; √0.040 = 0.20.
T = 2π(0.20) ≈ 1.3 s.
A clockmaker wants a pendulum with a period of 2.0 s on Earth (g = 10 m/s²). How long should it be? What would its period be on the Moon (g = 1.6 m/s²)?
Solve for L: T/2π = √(L/g), so L = g(T/2π)².
Plug in: L = 10 × (2.0 ÷ 6.28)² = 10 × 0.101 ≈ 1.0 m.
On the Moon: g is 10 ÷ 1.6 = 6.25 times smaller, so T is √6.25 = 2.5 times longer.
Moon period: 2.0 s × 2.5 = 5.0 s. The clock would run slow.
Answer: about 1.0 m long; 5.0 s on the Moon.
4. Common mistakes
You want to double the period of a pendulum. What should you do to its length?
T is proportional to √L. To make √L twice as big, L must be 4 times as big.
Ready to practice?
5 questions on Pendulums, each with a hint and an explanation for every wrong answer.