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George Mason UniversityZakaria Tarik ZerhouniSource: Home > Project 3 > Step 4Math 447: Numerical Analysis
Computer Problem 6.3.7 asks to find the smallest value for which the damped pendulum with an oscillating pivot will destabilize when set to the stable equilibrium of the downward position. By setting \(y = y_1 = 0\) and the initial velocity \( v = y_2 = 1\), we were able to produce a slight movement in the perfect equilibrium of the downward position. What we discovered is that this still remains stable by remaining inside of the interval \( (\frac{-\pi}{2}, \frac{\pi}{2} )\) for any forcing parameter A within \( (0, 12.261989630558716) \). However, the downward position becomes unstable at just \(A = 12.261989630558718\). The code used to compute this value can be found here. The videos below demonstrate the difference inherent in the \( 0.000000000000002 \: (2 \times 10^{-15}) \) change of the parameter \(A\). Stable at A = 12.261989630558716Click here to see the code used to generate these animations. Click here to download the video. Unstable at A = 12.261989630558718Click here to see the code used to generate these animations. Click here to download the video.
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