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.

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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.

T = 2π √(L / g)

A-Level Physics (AQA 3.6.2, OCR, Edexcel): simple harmonic motion and the simple pendulum. GCSE Physics: required 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.

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.