The small angle approximation for a simple pendulum of length L gives a period T = 2*pi*[L/g]1/2. However, if one does not make the small angle approximation, the expression for the period T is
T = 4[L/g]1/2K(sin a/2)
where a is the initial amplitude of the pendulum measured relative to the vertical, and K is the complete elliptic integral of the first kind given by
K(k) = 0Ipi/2 dz/[ 1 - k2sin2z ]1/2
where the symbol 0Ipi/2 represents the integral from 0 to pi/2, and k = sin a/2. The table below provides some values of the integral K for various values of the argument k. Calculate the period of the pendulum using both the small angle approximation and the precise value up to initial amplitudes of 90o by evaluating the elliptic integral. Use whatever numerical integration scheme you feel appropriate to evaluate the integral. Select sufficient values in a that you can identify at what initial amplitude the two solutions depart by 0.1%, 1.0%, and 10%. Graph your results.
For this problem, I used a sixteen point Gaussian Quadrature integration scheme to integrate the elliptical integral. To find the inital angle which the two solutions depart I used Mullers method to solve the equation
(TLarge -T Small)/TLarge - difference=0
because Newton's method requires a derivative (yuck!) and the Muller's method tends to converge much faster. For all calculations, I chose the length of the pendulum to be one and g = 9.81 m/s 2. TSmall=2.006066680710647 in all further calculations.
| % Difference | Initial Amplitude (In Radians) |
| .1% | .12649636954594307 |
| 1% | .40016431970246170 |
| 10% | 1.26932970981731682 |
The dependence of the period on the initial amplitude is illustrated in the table and plot below.
| Initial Amplitude | Period (seconds) |
| 10 | 2.00989262729860052 |
| 20 | 2.02145125782457619 |
| 30 | 2.04098988951913023 |
| 40 | 2.06893785189358914 |
| 50 | 2.10593461928364287 |
| 60 | 2.15287466688051548 |
| 70 | 2.21097618054218626 |
| 80 | 2.28188592401963935 |
| 90 | 2.36784194757623713 |