Why does large initial displacements/angles on a pendulum affect the period and the accuracy of oscillation? I know at small angles it does not effect the period however large angles (maybe above 30$^o$) do..
Thank you
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Intuitively, imagine if you gave a pendulum enough momentum that it would flip almost all the way over. Then it would hang for a very long time nearly balanced upside-down before swinging back. By giving it closer to that perfect amount of momentum so that it gets to the top and rests perfectly upside-down, the period can be made arbitrarily long. So the period must be amplitude-dependent. As Ross's answer shows, the harmonic approximation (which implies that the oscillation frequency doesn't depend on initial conditions) breaks down when the amplitude gets too large. – spaceisdarkgreen Aug 19 '17 at 05:01
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If you follow the derivation of the equation of motion, there is an approximation made that $\sin \theta = \theta$. Before the approximation the equation is $\ddot \theta + \frac gl \sin \theta =0$. Once you make the approximation you have a simple harmonic oscillator. As the Taylor series of $\sin \theta$ around zero is $\sin \theta \approx \theta - \frac 16 \theta^2 $ we are ignoring the cubic (and higher) term(s). This is valid when the angle is small.
Ross Millikan
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Ah okay cheers! so sin θ ≈ θ up to about 0.244 radians (14°) due to the small angle approximation with the taylor series. https://en.wikipedia.org/wiki/Small-angle_approximation Thanks for the understanding – Jr. Mathematician Aug 19 '17 at 04:58
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There is no specific value for which the approximation breaks down. It is a function of how accurate you want to be. – Ross Millikan Aug 19 '17 at 05:07