A pendulum has period 2s. What is its length?

T = 2π√(L/g) → L = g(T/2π)² = 9.8×(1/π)² ≈ 0.993 m ≈ 1 m.

How does the period change on Mars (g = 3.7 m/s²)?

T ∝ 1/√g → T_Mars = T_Earth × √(9.8/3.7) ≈ 1.63 × T_Earth.

Why does the small-angle formula fail for large swings?

True equation is d²θ/dt² = −(g/L)sin(θ); for large θ, sin(θ) < θ, giving longer period.

Pro 🔒~20 min

Pendulum Lab

Measure period and explore pendulum dynamics

How it works

A simple pendulum oscillates with period T = 2π√(L/g) for small angles (< 15°). The period depends only on length and gravity — not on mass or amplitude (for small angles). Large angles cause the true period to exceed this formula. On the Moon (g = 1.6 m/s²), the same pendulum would oscillate ~2.5× slower. Pendulums were historically used as precision timekeepers because of this mass-independence.

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Step-by-step

Drag the pendulum bob to an initial angle and release.
Use the stopwatch to measure 10 periods, then divide by 10 for accuracy.
Change length and repeat to plot T vs √L.
Try different masses to verify mass independence.

Key formulas

$T = 2\pi\sqrt{\frac{L}{g}}$ Pendulum period (small angle)
$f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$ Oscillation frequency

Frequently asked questions

A pendulum has period 2s. What is its length?: T = 2π√(L/g) → L = g(T/2π)² = 9.8×(1/π)² ≈ 0.993 m ≈ 1 m.
How does the period change on Mars (g = 3.7 m/s²)?: T ∝ 1/√g → T_Mars = T_Earth × √(9.8/3.7) ≈ 1.63 × T_Earth.
Why does the small-angle formula fail for large swings?: True equation is d²θ/dt² = −(g/L)sin(θ); for large θ, sin(θ) < θ, giving longer period.