# Pendulum (mathematics)

Template:Dynamics Template:Seeintro The mathematics of pendulums are in general quite complicated. Simplifying assumptions can be made, which in the case of a simple pendulum allows the equations of motion to be solved analytically for small-angle oscillations.

## Simple gravity pendulum

A so-called "simple pendulum" is an idealization of a "real pendulum" but in an isolated system using the following assumptions:

• The rod or cord on which the bob swings is massless, inextensible and always remains taut;
• The bob is a point mass;
• Motion occurs only in two dimensions, i.e. the bob does not trace an ellipse but an arc.
• The motion does not lose energy to friction or air resistance.

The differential equation which represents the motion of a simple pendulum is

## Small-angle approximation

The differential equation given above is not easily solved, and there is no solution that can be written in terms of elementary functions. However adding a restriction to the size of the oscillation's amplitude gives a form whose solution can be easily obtained. If it is assumed that the angle is much less than 1 radian, or

$\theta \ll 1\,$ ,

then substituting for sin θ into Template:EquationNote using the small-angle approximation,

$\sin \theta \approx \theta \,$ ,

yields the equation for a harmonic oscillator,

${d^{2}\theta \over dt^{2}}+{g \over \ell }\theta =0.$ The error due to the approximation is of order θ 3 (from the Maclaurin series for sin θ).

Given the initial conditions θ(0) = θ0 and /dt(0) = 0, the solution becomes,

$\theta (t)=\theta _{0}\cos \left({\sqrt {g \over \ell \,}}\,t\right)\quad \quad \quad \quad \theta _{0}\ll 1.$ The motion is simple harmonic motion where θ0 is the semi-amplitude of the oscillation (that is, the maximum angle between the rod of the pendulum and the vertical). The period of the motion, the time for a complete oscillation (outward and return) is

$T_{0}=2\pi {\sqrt {\frac {\ell }{g}}}\quad \quad \quad \quad \quad \theta _{0}\ll 1$ which is known as Christiaan Huygens's law for the period. Note that under the small-angle approximation, the period is independent of the amplitude θ0; 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 $g\approx 9.81$ m/s2, 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,

$\ell \approx {\frac {T_{0}^{2}}{4}},$ $T_{0}\approx 2{\sqrt {\ell }}$ ## Arbitrary-amplitude 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 (Template:EquationNote), Figure 3. Deviation of the "true" period of a pendulum from the small-angle approximation of the period. "True" value was obtained using Matlab to numerically evaluate the elliptic integral.
File:Pendulum Rel Error90a.png
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,

$T=t(\theta _{0}\rightarrow 0\rightarrow -\theta _{0}\rightarrow 0\rightarrow \theta _{0}),$ or twice the half-cycle

$T=2t\left(\theta _{0}\rightarrow 0\rightarrow -\theta _{0}\right),$ or 4 times the quarter-cycle

$T=4t\left(\theta _{0}\rightarrow 0\right),$ $T=4{\sqrt {\ell \over 2g}}\int _{0}^{\theta _{0}}{1 \over {\sqrt {\cos \theta -\cos \theta _{0}}}}\,d\theta .$ 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 re-written in terms of elliptic integrals as

$T=4{\sqrt {\ell \over g}}F\left({\theta _{0}},\csc {\theta _{0} \over 2}\right)\csc {\theta _{0} \over 2}$ $F(\varphi ,k)=\int _{0}^{\varphi }{1 \over {\sqrt {1-k^{2}\sin ^{2}{u}}}}\,du\,.$ $K(k)=F\left({\pi \over 2},k\right)=\int _{0}^{\pi /2}{1 \over {\sqrt {1-k^{2}\sin ^{2}{u}}}}\,du\,.$ 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/s2) 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 Template:EquationNote 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{alignedat}{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 {(2n)!}{(2^{n}\cdot n!)^{2}}}\right)^{2}\cdot \sin ^{2n}\left({\frac {\theta _{0}}{2}}\right)\right].\end{alignedat}} Figure 4 shows the relative errors using the power series. T0 is the linear approximation, and T2 to T10 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 x-axis, 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:

{\begin{alignedat}{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{alignedat}}. ### Arithmetic-geometric mean solution for elliptic integral

Given Template:EquationNote and the Arithmetic-geometric mean solution of the elliptic integral:

$K(k)={\frac {\pi /2}{M(1-k,1+k)}},$ This yields an alternative and faster-converging formula for the period:

$T={\frac {2\pi }{M(1,\cos(\theta _{0}/2))}}{\sqrt {\frac {\ell }{g}}}.$ ## Examples

The animations below depict several different modes of oscillation given different initial conditions. The small graph above the pendulums are their phase portraits.

## 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:

$\tau =-mgL\sin \theta \,$ where:

L is the distance from the pivot to the center of mass of the pendulum
θ is the angle from the vertical
$\alpha \approx -{\frac {mgL\theta }{I}}$ This is of the same form as the conventional simple pendulum and this gives a period of:

$T=2\pi {\sqrt {\frac {I}{mgL}}}$ 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: if θ0 is the maximum angle of one pendulum and 180° − θ0 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.