(FB/Math/BM – Question by Souradeep P.)
What would the equation of motion be for a real simple pendulum with frictionless pivot but non-negligible angular amplitude?
Let the angle from the direction downwards be φ(t) at time t, t = 0 at the lowest point and the amplitude φ₀. Since the pendulum is at rest at φ(T/4) = φ₀, where T denotes the period time, the energy law gives (pendul…um length L, mass m)
1/2 m (L dφ/dt)² + mgL(1 – cos φ) = mgL(1 – cos φ₀), from which
(L dφ/dt)² = 2gL(cos φ – cos φ₀)
dφ/dt = √(2g/L)·√(cos φ – cos φ₀)
∫ [0,φ(t)] dθ/√(cos θ – cos φ₀) = √(2g/L)·∫ [0,t] dτ
The integral is elliptic and cannot be expressed in elemntary functions but we get t as a function of φ(t),
t = √(L/(2g))·∫ [0,φ(t)] dθ/√(cos θ – cos φ₀), 0 ≤ t ≤ T/4.
Here the integral can be given ina standard form but to answer the question notice that the time asked for is T/4 corresponding to φ =φ₀ = π/2, i e
(*) T/4 = √(L/(2g))·∫ [0,π/2] dθ/√cos θ.
The substitution defined by 1 – x² = cos θ transforms the integral into
√2·K(1/√2), where K(k) ≡ ∫ [0,1] dx/√((1 – x²)(1 – k²x²)), 0 ≤ k < 1, is a standard form for a complete elliptic integral of the first kind. More generally, the period time for any amplitude φ₀ is given by
T = 4√(L/g)·K(k), where k = sin(φ₀/2). For small oscillations, φ₀ ≈ 0 we get the usual approximation T ≈ 4√(L/g)·π/2 = 2π√(L/(g).
In the particular case φ₀ = π/2 the integral in (*) also has a nice expression in terms of the gamma function,
(**) T/4 = √(L/g)·Γ(1/4)²/(4√π) ≈ 1.85·√(L/g).