Pendulum Period
y = ∫₀^(π/2) 1/sqrt(1 − x² · sin(u)²) du
Open in the app The dials and keys named below are the app's.
What it draws
The curve is an integral, computed at every point: y = ∫₀^(π/2) du / √(1 − x²·sin²u). Read x as k = sin(θ₀/2), θ₀ the angle a pendulum is released from. The integral is then K(k), the complete elliptic integral of the first kind, and a pendulum of length L under gravity g swings with period T = 4·√(L/g)·K(k). The plot is that period against the amplitude, in units of 4·√(L/g).
At x = 0 the integrand is 1 and the integral is π/2 ≈ 1.571, which gives T = 2π·√(L/g), the small-angle period. The curve barely rises at first: at x = 0.5, a swing of 60°, it is 1.686, only 7% more. At x = 0.9, a swing of 128°, it is 2.28. At x = 1 the integrand blows up at u = π/2 and the curve goes vertical: a pendulum balanced upside down takes forever to fall. Beyond ±1 the square root is imaginary and there is nothing to draw.
Why the swing slows
The force pulling a pendulum back is proportional to sin θ, not to θ. Near the bottom the two agree and the period is fixed. Pulled further out, the bob lingers near the ends of its swing, where the pull back is weakest, and the period grows. Energy conservation gives the time as an integral, and the substitution sin(θ/2) = k·sin u puts a quarter period into the form on screen.
History
Galileo claimed around 1602 that a pendulum's period does not depend on its amplitude, nearly true for small swings. Christiaan Huygens showed in his Horologium Oscillatorium of 1673 that the circular pendulum is not isochronous, and that a bob swinging along a cycloid is. Carl Friedrich Gauss found in 1799 that the arithmetic–geometric mean computes it fast. Adrien-Marie Legendre named the three kinds of elliptic integral, this the first, in his Exercices de calcul intégral of 1811.
Try
- Divide by the small-angle value, 2/π·integral(1/sqrt(1 − x²·sin(u)²), u, 0, π/2): the curve is the period as a multiple of the small-angle one, 1.18 at x = 0.707, a swing of 90°.
- Plot against the angle itself, integral(1/sqrt(1 − sin(x/2)²·sin(u)²), u, 0, π/2), and raise Span of x to 4: x is now θ₀ in radians and the vertical stands at π. Past it the curve comes down, since sin(x/2) falls again.
- Cut the series short, π/2·(1 + x²/4 + 9x⁴/64), three terms of its series: it hugs the curve out to x ≈ 0.5, then falls behind, 2.03 at 0.9 against the integral's 2.28.
- Take the second kind, integral(sqrt(1 − x²·sin(u)²), u, 0, π/2): E(k) falls from π/2 to exactly 1 at x = 1. That is a quarter of the perimeter of an ellipse with semi-axes 1 and √(1 − x²).