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.
 A: 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.
A: To first order, a pendulum simple harmonic motion and it's pendulum does not depend on the amplitude of its swing. However, the motion of a pendulum is not exactly simple harmonics  motion.
To calculate the time period, you need to solve what is called elliptic integral. However you can approximate it, In the following case if you work out it is turn out that :
$$\sqrt{\frac{g}{l}}T=\int_0^{2\pi}\frac{du}{\sqrt{1-K^2\sin^2 u }}$$
$$\frac{1}{\sqrt{1-K^2\sin^2 u }}=1+\frac{1}{2}K^2\sin^2 u+\cdots$$
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.
