Animating A Simple Pendulum Using Elliptic Functions 

A.C. Baroudy
acbarou@inco.com.lb 

Abstract 

The purpose of the present article is to built the equation of motion of a simple pendulum for large angles, then to use Maple commands: JacobiSN, to find the displacement angle for a given time.
A set of ordered pairs of coordinates for the bob is thus obtained and used for plotting.
The animation is somehow tricky since the set obtained doesn't allow for a straight forward animation and needs some careful manipulation of the data in the set.
The reward is double:
1- being able to put to good use Maple engine in finding Elliptic Functions (EF) with a simple command rather than resorting to tables for finding EF.
2- A good exercise in programming with Maple animation. 

Equation of Motion of a Simple Pendulum with Large Angles 

From Figure-1 below we can see that
1- the centripetal force just keeps the bob on a circular course.
   It is the result of the tension T & the radial component of mg,
2- the restoring force moves the bob along its circular path.
    It is the tangential component of mg.
The equation of motion is related to this force which is: 

                                            → r   
                                                                 → .                                       (1)
Note that the sign of the restoring force  is just opposite of the sign of the angular displacement θ:
- when θ is on the positive side, is to the left toward the equilibrium position and is negative,
- when θ is on the negative side, is to the right toward the equilibrium position and is positive . 

To solve the equation of motion (1) we use the following:
= = = ,
from which we get the differential equation for the motion of the pendulum:
                                                         = .                                             (2)
By integration we get : + constant
i.e.                                                    + constant.                                       (3)
If at A (Figure-1) the initial value of θ is α then
When θ = α , = 0 and the constant = ,
hence (3) becomes:                                       
                                                    .                                         (4)
Taking the square root of (4) we get :
                                               .
i.e.                                          .                                  (5) 

To make it simple we take r = g then by integrating (5) from θ = 0 to θ = α & t = 0 to t = we get the time from θ = α to θ = 0.
                                         = u.                                 (6)
For a complete period : from +α to -α and back, the time is 4 .
To show that this integral is an elliptic integral of the first kind we proceed as follows:
we use the identity :
& ,
and put :                                            x = .                                                      (7)
Differentiating (7):
                                                       ,
hence:
                                                        .                                                (8)
Equation (3) becomes:
= dθ
= =
replacing with its equivalent from (7) we get:

,
                                                      =                                           (9)  
where we put = . 

In case α = π/2 then
                                                   k = ,                                  (10)
and x being   while the integration at the very start was about θ and after integration we have to replace θ with α  then x = = 1 in  case α = π/2 .
The integral in x becomes:                         .                                  (11)
To put it in the Jacobi form we set x = sin Φ →  and (11) becomes:
,
                                                   =                                                  (12)
                                                   = F(sin Φ =1, k = ) = 1.854074677                              (13) 

Once we have the value of this integral i.e. the value of the time for descending from A to B ( or going upward from B to A) and which is equal to = 1.854074677 seconds, we divide it by 10 to get one small time increment : Δt. Then using Maple command JacobiSN(Δt,) we get sin (ΔΦ) corresponding to Δt.
From relation (7) we get θ which defines our pendulum position at Δt we finally get θ:
sin (ΔΦ) = x  = = →
                                                         → .                                      (14)
This is angle θ of the pendulum as counted from point B since ΔΦ is in fact from 0 to Φ = ΔΦ corresponding to time from t = 0 to t = Δt.
We repeat the same process for 2Δt to get Δ'Φ from which we get θ counted from point B to Δ'Φ corresponding to 2Δt and so on till we reach the last one i.e. 10Δt = = 1.854074677 corresponding to θ = always counted from point B.
                                                                               Note

For the geometrical relations between the angles α,θ & Φ see :
1- Figure-2 below,
2- Greenhill's "Elliptic Functions and Their Applications". 

Maple Program for plotting the pendulum bob 

As stated above we use Maple command JacobiSN to get sin (ΔΦ) from which we get angle θ. 

The following short program will give us just angle θ at successive time increments:
Δt,2Δt.......10Δt = 1.854074677. Then we get all x&y coordinates of the bob at these values of θ as a list.
To get the left half of the swing we just make the sign of all x-coordinates negative and set them in a 2d list.
Plotting is done for both lists with Maple command display. 

Maple Program for animating the pendulum 

Here we have to pay attention to the way we have to animate the pendulum. Care should be exercised when animating the pendulum over the upswings and downswing and whether it is on the right half or the left half.
We need to have the pendulum start from point A then down to B and again up into the left half. From there down to B and again up to A. The coordinates are arranged in the lists from B up to A so we need to reverse the order of plotting. Furthermore the pair [0,0] should be suppressed from the 2d list so that no hanging is observed when the bob is passing this point from either direction.
The following procedure will take care of all these critical points 

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