How this calculator works
The amplitude correction comes from the nonlinear pendulum equation. A complete elliptic integral gives the period; its arithmetic-geometric mean form makes the numerical calculation fast. The table compares release angles for your solved length and gravity. The model assumes a point bob on a rigid massless rod with no friction or air resistance. With a string, releases above 90 degrees can make the string slack, so those rows describe a rigid rod instead.
Formula
Let A = AGM(1, cos(theta0/2)). f0 = sqrt(g/L)/(2 pi); f = A f0; T = 1/f; L = g[A/(2 pi f)]²; g = L(2 pi f/A)². Equivalently T = 4 sqrt(L/g) K(sin(theta0/2)).
Worked example
For L = 1 m and g = 9.869604401 m/s², the small-angle frequency is 0.5 Hz. A 90 degree release lowers it to 0.4236065424 Hz, with a period of 2.360681198 s. At the same amplitude and gravity, a 0.5 Hz design needs a length of about 0.717770011 m.
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Frequently asked questions
What is one complete oscillation?
It is a trip from one turning point to the other and back. Passing through the bottom twice counts as one cycle.
Why can I not enter 180 degrees?
At the upright position, release from exact rest does not produce a finite repeating swing. The limiting period grows without bound as the release approaches that position.
Does zero amplitude mean the bob is moving?
No. At exactly zero amplitude it stays at rest. The displayed value is the frequency approached by arbitrarily small nonzero swings.
Can I use this to measure gravity?
Use the frequency averaged over several complete cycles and measure pivot-to-centre length. Keep the swing amplitude steady enough to justify one amplitude value; friction and a heavy rod can bias the result.