To find the rotation of the plane of oscillation of Foucault's pendulum we consider a pendulum consisting of a massless string of length l with a mass at the end. Describe the position of the mass by a triple of Cartesian coordinates (x, y, z), where (0, 0, 0) is the equilibrium position of the pendulum, (1, 0, 0) and (0, 1, 0) are two orthogonal horizontal vectors and (0, 0, 1) is directed upwards.

As always when dealing with pendulums we assume that the maximum deflection is small. In this case it means that we ignore the z-component altogether, and that we approximate the restoring force of the pendulum by F = -k2(x, y), where k2 = g/l.

Normally we would be satisfied with this, but the whole point of the Foucault pendulum is to observe its behaviour over a longer time. We therefore also have to take into account the Coriolis effect. This produces a pseudo force C = 2w(sin L)(y', -x'), where L is the latitude of the pendulum and w is the angular velocity of Earth.

Now we release the pendulum from rest at (0, y0), and consider its motion during a half period, ie the time it takes it to get to roughly y = -y0.

Since the Corilois effect is relatively small, x will remain small during the half period, and hence the influence of the Coriolis effect on y is second order small. Therefore we can approximate the equation for y to y'' = -k2y, which as usual has the solution y = y0*cos kt. The period is T = 2π/k.

In finding the expression for x we must take the Coriolis effect into account. The equation of motion is

x'' = -k2x + 2w(sin L)y' = -k2x - 2wky0(sin L)(sin kt)

which has solution x = wy0(sin L)(t*cos kt - k-1sin kt).

Hence at t = T/2 the mass is at rest and (x, y) = (-wy0(sin L)T/2, -y0). The plane of oscillation of the pendulum has turned by an angle given by sin da = x/y. Since x/y is small this implies da = x/y, and thus da/dt = (x/y)/(T/2) = w*sin L. In 24 hours the angle turned by the plane is 2π*sin L, and the time taken to turn through 2π radians is 1/sin L.