An exact solution to the equations of motion for a pendulum requires the use of elliptic integrals and elliptic functions, and the pendulum problem is one of the most basic illustrations of the usefulness of elliptic integrals and functions in physics and applied mathematics.
As everyone is no doubt aware, a pendulum may easily be idealized as simply a mass m suspended from a rigid rod (of negligible mass) of constant length L:
| \ L
mg sin θ |\
mg mg cos θ
Summing forces, we see that the weight component mg cos θ (where g is the acceleration due to gravity of 9.8 m/s2 or thereabouts) cancels out due to the restraint of the rod. The only weight component that contributes to the overall motion of the pendulum is mg sin θ, that is tangent to the path of motion. If we take s as the arc length of the path, then Newton's Second Law of motion F = ma gives us:
m ___ = -mg sin θ
The arc length s of a circle of radius L is in terms of the central angle θ obviously s = Lθ (where θ is in radians of course), so in terms of the angular position θ the equation of motion for the pendulum becomes:
___ + __ sin θ = 0
Note that the mass of the pendulum bob has cancelled out (so it's still true that the weight of the hanged man doesn't influence his motion, as Umberto Eco would put it). This differential equation is obviously nonlinear, as sin θ is a nonlinear function of θ, but we can integrate the equation to yield an equation in terms of the energy of the pendulum:
1 dθ g2
_ (__)2 - __ cos θ = C
2 dt L2
where C is a constant related to the initial potential energy of the pendulum, which we immediately see is given by g2/L2 cos θM, where θM is the maximum angular displacement of the pendulum (what Webster 1913 calls the arc of vibration). By solving for dθ/dt in that equation, we obtain:
__ = ±(__)1/2 sqrt(cos θ - cos θm)
We take t to be 0 when θ = 0, and dθ/dt > 0 (so the negative solution above is considered extraneous). integrating the equation from 0 to θM we get:
∫ (cos θ - cos θM)-1/2 dθ = sqrt(2g/L) t
This is a quarter of a cycle, so the time t we get once we evaluate the left hand integral is a quarter of the pendulum's period T. Since θ is always less than or equal to θM by definition, we can make the half-angle substitution sin(θ/2) = sin(θM/2) sin φ, so the integral above reduces to:
T = 4 sqrt(L/g) ∫ ___________________
0 sqrt(1 - k2 sin2 φ)
where k = sin(θM/2). This integral we recognize as one of Legendre's elliptic integral standard forms, specifically the (complete) elliptic integral of the first kind, so we finally get:
T = 4 sqrt(L/g) F(sin(θM/2)|π/2)
As an example, say we had a pendulum with a rod which is 0.600 m long, and we started the pendulum at a maximum angular displacement of 80.0 degrees. T = 0.990 F(sin 80.0/2 | π/2) = 0.990*1.78677 = 1.77 seconds (from tables in Abramowitz and Stegun). The approximation in piq's wu on the other hand yields 1.55 seconds, which is quite inaccurate. However, that approximation, which was first derived by Christiaan Huygens (1629-1695) in the 17th century (by linearizing the nonlinear equation of motion above to d2θ/dt2 = -(g/L)2θ), is pretty good when displacements are small, as the smaller θM is the closer F(k|π/2) approaches π/2. Using π/2 as an approximation for the elliptic integral produces 2π sqrt(L/g).
On the other hand, in order to solve for the angle of motion as a function of time, for instance to make a simulation of the movement of the pendulum, it becomes necessary to look into inverse functions of the elliptic integrals, the Jacobi elliptic functions.
Abramowitz, Milton and Irene A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, 1964.
Andrews, Larry C., Special Functions of Mathematics for Engineers, McGraw-Hill, 1992.
Arfken, George, Mathematical Methods for Physicists, Academic Press, 1985
Eco, Umberto. Foucault's Pendulum, Ballantine Books, 1990.
Greenhill, Alfred George, The Applications of Elliptic Functions, Macmillan and Co., 1892.