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In the approximation $$-(g/ \ell) \sin \theta \approx -(g/ \ell) \theta $$ we make an error $R$ which is $O(\theta ^3)$. If i did well my calculations it is estimated by $$R\leq|(g / \ell)(\theta^3/3!)|$$ however this isn't essential to my two questions.

Q#1: We do the estimation of the error for $\ddot \theta$. How does the error reflect on the solution? A first rough idea that comes to my mind is the following: we are negliging a factor $R$ in the second derivative of the angle, so approximating $R$ as constant, the error on $\theta$ must be something like: $$ \int _0 ^t \int _{0} ^{t'} R dt' dt'' = 0.5 R t^2 .$$

Q#2 After getting to the solution: $$\theta(t) \approx \theta _0 \cos (\sqrt \frac {g}{\ell} t) $$ having set $\dot \theta (0) =0$, our teacher did the following. He reconsiders the equation: $$\ddot \theta =-(g / \ell) \sin \theta \approx -(g / \ell) \theta (1-\theta ^2 /3!)\approx -(g / \ell) \theta (1-\frac{<\theta ^2 >}{3!}),$$ where: $$<\theta ^2>=\dfrac{1}{T} \int _0 ^T \theta ^2 \, dt .$$ Now, to evaluate the integral, he uses the first solution: $$\dfrac{1}{T} \int _0 ^T \theta ^2 \, dt =\dfrac {1}{T}\int _0 ^T \theta _0 ^2 \cos ^2 (\sqrt {.} t)dt=\frac{\theta _0^2}{2}$$ Hence: $$\ddot \theta \approx (-g/\ell)(1-\dfrac{\theta _0 ^2}{12})\theta ,$$ which should be better from the first solution. Now, apart from the lot of approximations that maybe will make you feel uncomfortable like me, second question is: is it logical to use the first rough approximation to calculate $<\theta ^2>$, which we need for the better approximation? Shouldn't the second solution be independent from the first to improve it?

Thanks to anyone who wants to help.

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