# How can I approach the motion of a pendulum with large angle in the first seconds?

how can I approximate the angle $$\theta$$ of a pendulum $$0.01 \mathrm{s}$$ after releasing it with speed $$0$$ at an initial angle $$\theta_0<\pi$$? Recalling

$$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{g}{l}\sin\theta=0$$

And what in the case when I have two pendulms with different initial conditions? I just need to approximate the motion in the very first instants of time.

$$\sqrt{\frac{g}{l}}T=\int_0^{2\pi}\frac{du}{\sqrt{1-K^2\sin^2 u }}$$
$$\frac{1}{\sqrt{1-K^2\sin^2 u }}=1+\frac{1}{2}K^2\sin^2 u+\cdots$$
and thus $$T=2\pi\sqrt{\frac{l}{g}}\left( 1+\frac{1}{16}\theta_0^2+\cdots \right)$$ The fractional change in period due to finite amplitude $$\theta_0$$ is $$\frac{\Delta T}{T}=\frac{T(\theta_0)-T(\theta_0=0)}{T}=\frac{1}{16}\theta_0^2$$ For an amplitude of $$0.1$$ rad, about $$6^o$$, the period is increased by about 1 part in $$10^4$$, slowing a clock by rougly a minute a day. For larger amplitudes, higher order terms can be introduced.
If you want to know just for a small time you can approximate $$sin(\theta)$$ using Taylor series- $$\Rightarrow sin(\theta)\approx sin(\theta_0)+(\theta-\theta_0)cos(\theta_0)$$ $$\frac{d^2x}{dt^2}+\frac{gcos(\theta_0)x}{l}+\frac{gsin(\theta_0)}{l}=0$$ where $$x=\theta-\theta_0$$. Now if you assume $$x$$ as a power series in $$t$$, $$x=\Sigma_{i=0}^{\infty}a_ix^i$$ you can get an approximate answer for that which will work for small times, when the angle is not changed too much.