l d2ψ= − g sin ψ.
dt2

(15)

If we multiply by 2dψ/dt and integrate, we obtain

( )2 = 2gcos ψ + const.,
dt l

(16)

which is seen to be equivalent to (14). If the pendulum oscillate between the limits ψ = ±α, we have

( δψ)2 = 2g(cos ψ − cos α) = 4g(sin2 1⁄2α − sin2 1⁄2ψ);
dt ll

(17)

and, putting sin 1⁄2ψ = sin 1⁄2α. sin φ, we find for the period (τ) of a complete oscillation

τ = 4 ∫1⁄2π0 dtdφ = 4√ l· ∫1⁄2π0
g√(1 − sin2 1⁄2α · sin2 φ)