(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Period of arbitrary amplitude pendulum using average speed

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