1. The problem statement, all variables and given/known data A simple pendulum of length L is pulled aside to angle θm then released from rest. For arbitrary angle θ on the path of motion the acceleration along the path is -gSinθ. The speed at this point is V(θ)= √[2gL (cos θ - cos θm)]. Find the period T of the pendulum (do not assume simple harmonic motion). I see how to solve this in the standard way but I also tried to solve it by calculating the average speed of the bob and then just deviding the path length by this speed to get the period. I believe that the two approaches should give the same answer. They do not. 2. Relevant equations standard approach: ds = L dθ = √[2gL (cos θ - cos θm)] dt dt so ∫dt= √( L/2g) . ∫ dθ/√(cos θ - cos θm) which gives the solution T= 4√(L/g) ∫ dx/√(1-A [Sinx.Sinx]) where integration limits are 0, ∏/2 and √A= Sin(θm/2). As θm goes to 0 T approaches 2∏√L/g as expected for SHM 3. The attempt at a solution the average speed <V> = [∫V(θ)dθ]/[∫dθ] integration lims 0 and θm The path length = 4Lθm so the period T = 4Lθm/ <V> integrating gives T= (16/∏)√L/g as low angle limit! where have i gone wrong?