# Pendulum (mathematics)

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Pendulum (mathematics)

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 simple pendulum is an idealisation, working on the assumption that:
* The rod or cord on which the bob swings is massless, inextensible and always remains taut;
* Motion occurs in a 2-dimensional plane, i.e. the bob does not trace an ellipse.
* The motion does not lose energy to friction.

The differential equation which represents the motion of the pendulum is

:$\left\{d^2 hetaover dt^2\right\}+\left\{gover ell\right\} sin heta=0 quadquadquadquadquad\left(1\right)$ see derivation

This is known as Mathieu's equation. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance $h$ would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy.

The "first integral of motion" found by integrating (1) is

:$\left\{d hetaover dt\right\} = sqrt$2gover ell}left(cos heta-cos heta_0 ight)} quadquad(2) see derivation

It gives the velocity in terms of the angle and includes the initial displacement (θ0) as an integration constant.

Small-angle approximation

It is not possible to integrate analytically the full equations of a simple pendulum. A further assumption, that the pendulum attains only a small amplitude, that is

:$heta ll 1$

is sufficient to allow the system to be solved easily. Making the assumption of small angle allows the approximation

:$sin hetaapprox heta$

to be made. To first order, the error in this approximation is proportional to $heta^3$ (from the Maclaurin series for $sin heta$). Substituting this approximation into (1) yields the equation for a harmonic oscillator:

:$\left\{d^2 hetaover dt^2\right\}+\left\{gover ell\right\} heta=0.$

Under the initial conditions $heta\left(0\right)= heta_0$ and $\left\{d hetaover dt\right\}\left(0\right)=0$, the solution is

:$heta\left(t\right) = heta_0cosleft\left(sqrt\left\{gover ell,\right\},t ight\right) quadquadquadquad heta_0 ll 1.$

The motion is simple harmonic motion where $heta_0$ 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

:$T_0 = 2pisqrt\left\{frac\left\{ell\right\}\left\{g$ quadquadquadquadquad heta_0 ll 1

which is Christiaan Huygens's law for the period.Note that under the small-angle approximation, the period is independent of the amplitude $heta_0$; this is the property of isochronism that Galileo discovered.

Rule of thumb for pendulum length

:$T_0 = 2pisqrt\left\{frac\left\{ell\right\}\left\{g$ can be expressed as $ell = \left\{frac\left\{g\right\}\left\{pi^2 imes\left\{frac\left\{T_0^2\right\}\left\{4.$

If SI units are used (i.e. measure in metres and seconds), and an assumption is made the measurement is taking place on the earth's surface, then "g" = 9.80665 m/s², and $\left\{frac\left\{g\right\}\left\{pi^2approx\left\{1\right\}$ (the exact figure is 0.994 to 3 decimal places).

Therefore $ellapprox\left\{frac\left\{T_0^2\right\}\left\{4$, or in words:

On the surface of the earth, the length of a pendulum (in metres) is approximately one quarter of the time period (in seconds) squared.

Arbitrary-amplitude period

For amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)

:$\left\{dtover d heta\right\} = \left\{1oversqrt\left\{2sqrt\left\{ellover g\right\}\left\{1oversqrt\left\{cos heta-cos heta_0$

and integrating over one complete cycle,

:$T = heta_0 ightarrow0 ightarrow- heta_0 ightarrow0 ightarrow heta_0,$

or twice the half-cycle

:$T = 2left\left( heta_0 ightarrow0 ightarrow- heta_0 ight\right),$

or 4 times the quarter-cycle

:$T = 4left\left( heta_0 ightarrow0 ight\right),$

:$T = 4\left\{1oversqrt\left\{2sqrt\left\{ellover g\right\}int^\left\{ heta_0\right\}_0 \left\{1oversqrt\left\{cos heta-cos heta_0,d heta.$

This integral cannot be evaluated in terms of elementary functions. It can be re-written in the form of the elliptic function of the first kind (also see Jacobi's elliptic functions), which gives little advantage since that form is also insoluble.

:$T = 4sqrt\left\{ellover g\right\}Fleft\left(\left\{ heta_0over 2\right\},csc^2\left\{ heta_0over2\right\} ight\right)csc \left\{ heta_0over 2\right\}$

or more concisely,

:$T = 4sqrt\left\{ellover g\right\}Fleft\left(sin\left\{ heta_0over 2\right\}, \left\{pi over 2\right\} ight\right)$

where $F\left(k,phi\right)$ is Legendre's elliptic function of the first kind

$F\left(k,phi\right) = int^\left\{phi\right\}_0 \left\{1oversqrt\left\{1-k^2sin^2\left\{ heta\right\},d heta.$

Figure 4 shows the deviation of $T,$ from $T_0,$, the period obtained from small-angle approximation.

The value of the elliptic function can be also computed using the following series:

:

Figure 5 shows the relative errors using the power series. $T_0,$ is the linear approximation, and $T_2$ to $T_\left\{10\right\}$ include respectively the terms up to the 2nd to the 10th powers.

For a swing of exactly $180^circ$ the bob is balanced over its pivot point and so $T=infty$.

For example, the period of a 1m pendulum on Earth ("g" = 9.80665 m/s²) at initial angle 10 degrees is $4sqrt\left\{1over g\right\}Fleft\left(\left\{sin 10over 2\right\},\left\{piover2\right\} ight\right) = 2.0102$ seconds, whereas the linear approximation gives $2pi sqrt\left\{1over g\right\} = 2.0064$.

Physical pendulums

A physical pendulum is one where the rod is not massless, and the mass may have extended size; in this case the pendulum and rod have a moment of inertia $I$ around the pivot point.

The equation of torque gives:

:$T = I a$

where::$a$ is the angular acceleration.:$T$ is the torque

The torque is generated by gravity so:

:$T = - m g L sin\left( heta\right)$where::$L$ is the distance from the pivot to the center of mass of the pendulum:$heta$ is the angle from the vertical

Hence, under the small-angle approximation $sin heta approx heta$,

:$a approx frac\left\{mgL heta\right\} \left\{I\right\}$

This is of the same form as the conventional simple pendulum and this gives a period of:

:$T = 2 pi sqrt\left\{frac\left\{I\right\} \left\{mgL.$

[ [http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html Physical Pendulum ] ]

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.

ee also

*Mathieu function
*Double pendulum

* [http://mathworld.wolfram.com/MathieuFunction.html Mathworld article on Mathieu Function]

References

* 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.
* [http://www.ulb.tu-darmstadt.de/tocs/129360481.pdf "The Pendulum: A Physics Case Study", Gregory L. Baker and James A. Blackburn, Oxford University Press, 2005]

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