Period of arbitrary amplitude pendulum using average speed

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SUMMARY

The discussion focuses on calculating the period T of an arbitrary amplitude pendulum using average speed, specifically addressing the discrepancy between two methods of calculation. The standard approach yields T = 4√(L/g) ∫ dx/√(1 - A[Sinx.Sinx]) with limits from 0 to π/2, while the average speed method incorrectly suggests T = (16/π)√(L/g). The key conclusion is that the period is determined by the distance multiplied by the average of the reciprocal of the velocity, not simply the distance divided by the average velocity. This distinction is crucial for accurate calculations in pendulum motion.

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Homework Statement



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.

Homework 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



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:
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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!
 
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}
 

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