# Period of arbitrary amplitude pendulum using average speed

1. Dec 5, 2011

### nixed

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?

Last edited: Dec 5, 2011
2. Dec 6, 2011

### nixed

I solved the problem. The time for the motion is not given by the distance/ average of the velocity at each point on the path but it is given by the distance x the average of the reciprocal of the velocity at each point on the path. For this motion 1/<Vx> is not equal to <1/Vx>. In general

T= X/<Vt> = X . <1/Vx>

where <Vt> is the time average velocity and <1/Vx> is the space average velocity. Bit of elementary calculus of function of a function. Dooogh!

3. Dec 7, 2011

### nixed

For simple harmonic motion (small angle approximation of the pendulum) it is fairly easy to show that for the space average velocity

1/<V> = [8/ pi^2] . < 1/V>

so that the time averaged velocity <Vt> (= <1/V> ) and space averaged velocity <V> are related:

1/<Vt> = [pi^2/8] . 1/<V>

So the period calculated as dist/<V> = [16/pi] SQRT {L/g} should be multiplied by
(pi^2)/8 to give the true period = 2pi. SQRT{L/g}