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I am a student and hope you can explain in simple terms the formula for the pendulum:

wikipedia says that the period of a simple pendulum for angles of small amplitude is $$T \approx 2\pi \sqrt\frac{L}{g} \qquad 0\lt\theta\lt 12^{\circ} ,$$

because the sine of an angle can be approximated to the angle $$ {d^2\theta\over dt^2}+{g\over \ell} \theta=0\qquad $$ in the general formula $${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0\\$$

wikipedia gives some solutions to that DE, and further to that I learned in another question that the exact general solution is given by the so-called Jacobi elliptic function

Those articles are too difficult for me, can you tell me what is the best/simplest formula to find the period of a pendulum? If you choose any angle between 45° and 90° and make a concrete example, that would be great

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  • $\begingroup$ It should be $0 \lt \theta \lt 12$ $\endgroup$ – Shailesh Sep 14 '16 at 7:38
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There is no simple solution as the period cannot be expressed with the ordinary functions.

The true expression is

$$T=4\sqrt{\frac Lg}K\left(\sin(\theta/2)\right)$$

where $K$ denotes a special function kwnon as the "complete elliptic integral of the first kind", defined as

$$K(k)=\int_0^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}.$$

(If you never heard of integrals, this will be meaningless to you.)

This function is tabulated or can be computed numerically with specific algorithms. Below, a plot of the function:

enter image description here

The starting value is $\pi/2$, showing that for small angles, the period is indeed $2\pi\sqrt{L/g}$. And given the "flatness" of the curve, the validity domain of the approximation isn't too bad.

Notice that for angles reaching $180°$, the period becomes infinite. Indeed an upside down pendulum will remain in equilibrium forever (at least in theory).

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After you get to $\dfrac{d^2\theta}{dt^2}+\dfrac g{\ell} \theta=0$, which is derived at Wikipedia:pendulum in the boxes at the end of the paragraph, this becomes a fairly simple differential equation in $\theta$, with a solution of $\theta=\cos(\sqrt{\dfrac g{\ell}t})$, because $\dfrac{d^2\theta}{dt^2}=-\dfrac g{\ell}\cos(\sqrt{\dfrac g{\ell}t})$.

$\sqrt{\dfrac g{\ell}}$ is $2\pi$ times the frequency of the pendulum swing, and as a pendulum is a simple harmonic oscillator, we can say that $T=\dfrac1f=2\pi\sqrt{\dfrac{\ell}{g}}$.

The full solution is more complicated, involving the arithmetic-geometric mean. The Wikipedia link is here.

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