{d\theta\over dt} = \sqrt{d\theta\over dt} \\ & = {1\over 2}{(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta\cos\theta_0\right)}}\sqrt \quad\quad\quad\quad\quad \theta_0 \ll 1
which is known as Christiaan Huygens's law for the period. Note that under the smallangle approximation, the period is independent of the amplitude θ_{0}; this is the property of isochronism that Galileo discovered.
Rule of thumb for pendulum length

T_0 = 2\pi\sqrt{\frac{\ell}{g}} can be expressed as \ell = {\frac{g}{\pi^2}}\times{\frac{T_0^2}{4}}.
If SI units are used (i.e. measure in metres and seconds), and assuming the measurement is taking place on the Earth's surface, then g\approx9.81 m/s^{2}, and g/\pi^2\approx{1} (0.994 is the approximation to 3 decimal places).
Therefore a relatively reasonable approximation for the length and period are,

\ell\approx{\frac{T_0^2}{4}},

T_0 \approx 2 \sqrt{\ell}
Arbitraryamplitude period
For amplitudes beyond the small angle approximation, one can compute the exact period by first inverting the equation for the angular velocity obtained from the energy method (Eq. 2),
Figure 3. Deviation of the "true" period of a pendulum from the smallangle approximation of the period. "True" value was obtained using Matlab to numerically evaluate the elliptic integral.
Figure 4. Relative errors using the power series.

{dt\over d\theta} = \sqrt{\ell\over 2g}{1\over\sqrt{\cos\theta\cos\theta_0}}
and then integrating over one complete cycle,

T = t(\theta_0\rightarrow0\rightarrow\theta_0\rightarrow0\rightarrow\theta_0),
or twice the halfcycle

T = 2 t\left(\theta_0\rightarrow0\rightarrow\theta_0\right),
or 4 times the quartercycle

T = 4 t\left(\theta_0\rightarrow0\right),
which leads to

T = 4\sqrt{\ell\over 2g}\int^{\theta_0}_0 {1\over\sqrt{\cos\theta\cos\theta_0}}\,d\theta.
Note that this integral diverges as \theta_0 approaches the vertical

\lim_{\theta_0 \rightarrow \pi} T = \infty ,
so that a pendulum with just the right energy to go vertical will never actually get there. (Conversely, a pendulum close to its maximum can take an arbitrarily long time to fall down.)
This integral can be rewritten in terms of elliptic integrals as

T = 4\sqrt{\ell\over g}F\left( {\theta_0}, \csc{\theta_0\over2}\right)\csc {\theta_0\over 2}
where F is the incomplete elliptic integral of the first kind defined by

F(\varphi , k) = \int_0^\varphi {1\over\sqrt{1k^2\sin^2{u}}}\,du\,.
Or more concisely by the substitution \sin{u} = \frac{\sin{\theta\over 2}}{\sin{\theta_0\over 2}} expressing \theta in terms of u,

T = 4\sqrt{\ell\over g}\,K\left( \sin{\theta_0\over 2} \right)


(Eq. 3)

where K is the complete elliptic integral of the first kind defined by

K(k) = F \left( {\pi\over 2}, k \right) = \int_0^{\pi/2} {1\over\sqrt{1k^2\sin^2{u}}}\,du\,.
For comparison of the approximation to the full solution, consider the period of a pendulum of length 1 m on Earth (g = 9.80665 m/s^{2}) at initial angle 10 degrees is 4\sqrt{1\ \mathrm{m}\over g}K\left( {\sin {10^\circ\over 2}} \right) \approx 2.0102\ \mathrm{s}. The linear approximation gives 2\pi \sqrt{1\ \mathrm{m}\over g} \approx 2.0064\ \mathrm{s}. The difference between the two values, less than 0.2%, is much less than that caused by the variation of g with geographical location.
From here there are many ways to proceed to calculate the elliptic integral:
Legendre polynomial solution for the elliptic integral
Given Eq. 3 and the Legendre polynomial solution for the elliptic integral:

K(k) = \frac{\pi}{2}\left\{1 + \left(\frac{1}{2}\right)^2 k^{2} + \left(\frac{1 \cdot 3}{2 \cdot 4}\right)^2 k^{4} + \cdots + \left[\frac{\left(2n  1\right)!!}{\left(2n\right)!!}\right]^2 k^{2n} + \cdots \right\},
where n!! denotes the double factorial, an exact solution to the period of a pendulum is:

\begin{alignat}{2} T & = 2\pi \sqrt{\ell\over g} \left( 1+ \left( \frac{1}{2} \right)^2 \sin^2\left(\frac{\theta_0}{2}\right) + \left( \frac{1 \cdot 3}{2 \cdot 4} \right)^2 \sin^4\left(\frac{\theta_0}{2}\right) + \left( \frac {1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right)^2 \sin^6\left(\frac{\theta_0}{2}\right) + \cdots \right) \\ & = 2\pi \sqrt{\ell\over g} \cdot \sum_{n=0}^\infty \left[ \left ( \frac{(2 n)!}{( 2^n \cdot n! )^2} \right )^2 \cdot \sin^{2 n}\left(\frac{\theta_0}{2}\right) \right]. \end{alignat}
Figure 4 shows the relative errors using the power series. T_{0} is the linear approximation, and T_{2} to T_{10} include respectively the terms up to the 2nd to the 10th powers.
Figure 5. Potential energy and phase portrait of a simple pendulum. Note that the xaxis, being angle, wraps onto itself after every 2π radians.
Power series solution for the elliptic integral
Another formulation of the above solution can be found if the following Maclaurin series:

\sin {\theta_0 \over 2}=\frac{1}{2}\theta_0  \frac{1}{48}\theta_0^3 + \frac{1}{3840}\theta_0^5  \frac{1}{645120}\theta_0^7 + \cdots.
is used in the Legendre polynomial solution above. The resulting power series is:^{[1]}

\begin{alignat}{2} T & = 2\pi \sqrt{\ell\over g} \left( 1+ \frac{1}{16}\theta_0^2 + \frac{11}{3072}\theta_0^4 + \frac{173}{737280}\theta_0^6 + \frac{22931}{1321205760}\theta_0^8 + \frac{1319183}{951268147200}\theta_0^{10} + \frac{233526463}{2009078326886400}\theta_0^{12} + . . . \right) \end{alignat}.
Arithmeticgeometric mean solution for elliptic integral
Given Eq. 3 and the Arithmeticgeometric mean solution of the elliptic integral:

K(k) = \frac {\pi /2}{M(1k,1+k)},
where M(x,y) is the arithmeticgeometric mean of x and y.
This yields an alternative and fasterconverging formula for the period:^{[2]}^{[3]}

T = \frac{2\pi}{M(1, \cos(\theta_0/2))} \sqrt\frac{\ell}{g}.
Examples
The animations below depict several different modes of oscillation given different initial conditions. The small graph above the pendulums are their phase portraits.

Initial angle of 0°, a stable equilibrium.





Initial angle of 180°, unstable equilibrium.

Pendulum with just barely enough energy for a full swing

Pendulum with enough energy for a full swing
Compound pendulum
A compound pendulum (or physical pendulum) is one where the rod is not massless, and may have extended size; that is, an arbitrarily shaped rigid body swinging by a pivot. In this case the pendulum's period depends on its moment of inertia I around the pivot point.
The equation of torque gives:

\tau = I \alpha\,
where:

\alpha is the angular acceleration.

\tau is the torque
The torque is generated by gravity so:

\tau =  m g L \sin\theta\,
where:

L is the distance from the pivot to the center of mass of the pendulum

θ is the angle from the vertical
Hence, under the smallangle approximation \scriptstyle \sin \theta \approx \theta\,,

\alpha \approx \frac{mgL \theta} {I}
This is of the same form as the conventional simple pendulum and this gives a period of:

T = 2 \pi \sqrt{\frac{I} {mgL}}
^{[4]}
And a frequency of:

f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{mgL}{I}}
Physical interpretation of the imaginary period
The Jacobian elliptic function that expresses the position of a pendulum as a function of time is a doubly periodic function with a real period and an imaginary period. The real period is of course the time it takes the pendulum to go through one full cycle. Paul Appell pointed out a physical interpretation of the imaginary period:^{[5]} if θ_{0} is the maximum angle of one pendulum and 180° − θ_{0} is the maximum angle of another, then the real period of each is the magnitude of the imaginary period of the other. This interpretation, involving dual forces in opposite directions, might be further clarified and generalized to other classical problems in mechanics with dual solutions.^{[6]}
See also
References

^ Nelson, Robert; M. G. Olsson (February 1986). "The pendulum — Rich physics from a simple system". American Journal of Physics 54 (2): pp. 112–121.

^ Carvalhaes, Claudio G.; Suppes, Patrick (December 2008), "Approximations for the period of the simple pendulum based on the arithmeticgeometric mean",

^ Adlaj, Semjon (September 2012), "An eloquent formula for the perimeter of an ellipse",

^ Physical Pendulum

^ Paul Appell, "Sur une interprétation des valeurs imaginaires du temps en Mécanique", Comptes Rendus Hebdomadaires des Scéances de l'Académie des Sciences, volume 87, number 1, July, 1878

^ Adlaj, S. Mechanical interpretation of negative and imaginary tension of a tether in a linear parallel force field , Selected papers of the International Scientific Conference on Mechanics "SIXTH POLYAKHOV READINGS", January 31  February 3, 2012, SaintPetersburg, Russia, pp. 1318.
Further reading
External links

Mathworld article on Mathieu Function
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