Pendulum
The period depends on length and gravity, not on the mass, and (almost) not on the amplitude.
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What's happening
Gravity provides a restoring torque proportional to sin(θ). For small angles sin(θ) ≈ θ, giving simple harmonic motion with period T = 2π√(L/g).
At large starting angles the measured period is slightly longer than the small-angle formula. Doubling the length does not double the period; it multiplies it by √2, because T grows with the square root of L.
- Period of a pendulumT = 2π √(L / g)
On the syllabus
- A-Level Physics AQA 3.6.2, OCR, Edexcelsimple harmonic motion and the simple pendulum
- GCSE Physicsrequired practical skills on timing oscillations
Challenge
Predict first: to double the period, how much longer must the string be? Test your answer by measuring at two lengths.
The string must be four times longer: T grows with √L, so doubling the period needs 4× the length.
Learn this properly
Lessons from GCSE Combined and Separate Science on this topic: worked steps, then exam-style questions with new numbers every time.
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FAQ
Does the mass of a pendulum change its period?
No. A heavier bob needs more force to accelerate, but gravity pulls it harder in exactly the same proportion, so the period T = 2π√(L/g) has no mass in it.
Why is the period longer at big angles?
The formula uses sin θ ≈ θ, which only holds for small angles. At larger angles the restoring force is weaker than the approximation assumes, so each swing takes a little longer: about 3% longer at 40°.
How long is a pendulum with a period of 2 seconds?
L = g(T/2π)², so on Earth L = 9.81 × (2/2π)² ≈ 0.994 m. That is the classic seconds pendulum.
How do I find g from a pendulum experiment?
Time many swings, divide to get one period, then use g = 4π²L/T². A 0.800 m pendulum with T = 1.795 s gives g = 4π² × 0.800 / 1.795² = 9.80 m/s². Pick Gravity g under Solve for to see each step, and keep the swing small so the formula holds.