- Pendulum (mathematics)
The mathematics of

pendulum s 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 anellipse .

* The motion does not lose energy to friction.The

differential equation which represents the motion of the pendulum is:$\{d^2\; hetaover\; dt^2\}+\{gover\; ell\}\; sin\; heta=0\; quadquadquadquadquad(1)$

^{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 acquiredkinetic 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

:$\{d\; hetaover\; dt\}\; =\; 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 (θ

_{0}) 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 aharmonic oscillator ::$\{d^2\; hetaover\; dt^2\}+\{gover\; ell\}\; heta=0.$

Under the initial conditions $heta(0)=\; heta\_0$ and $\{d\; hetaover\; dt\}(0)=0$, the solution is

:$heta(t)\; =\; heta\_0cosleft(sqrt\{gover\; ell,\},t\; ight)\; quadquadquadquad\; heta\_0\; ll\; 1.$

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:$T\_0\; =\; 2pisqrt\{frac\{ell\}\{g$ 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 thatGalileo discovered.**Rule of thumb for pendulum length**:$T\_0\; =\; 2pisqrt\{frac\{ell\}\{g$ can be expressed as $ell\; =\; \{frac\{g\}\{pi^2\; imes\{frac\{T\_0^2\}\{4.$

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 $\{frac\{g\}\{pi^2approx\{1\}$ (the exact figure is 0.994 to 3 decimal places).

Therefore $ellapprox\{frac\{T\_0^2\}\{4$, 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)

:$\{dtover\; d\; heta\}\; =\; \{1oversqrt\{2sqrt\{ellover\; g\}\{1oversqrt\{cos\; heta-cos\; heta\_0$

and integrating over one complete cycle,

:$T\; =\; heta\_0\; ightarrow0\; ightarrow-\; heta\_0\; ightarrow0\; ightarrow\; heta\_0,$

or twice the half-cycle

:$T\; =\; 2left(\; heta\_0\; ightarrow0\; ightarrow-\; heta\_0\; ight),$

or 4 times the quarter-cycle

:$T\; =\; 4left(\; heta\_0\; ightarrow0\; ight),$

which leads to

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

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.:$T\; =\; 4sqrt\{ellover\; g\}Fleft(\{\; heta\_0over\; 2\},csc^2\{\; heta\_0over2\}\; ight)csc\; \{\; heta\_0over\; 2\}$

or more concisely,

:$T\; =\; 4sqrt\{ellover\; g\}Fleft(sin\{\; heta\_0over\; 2\},\; \{pi\; over\; 2\}\; ight)$

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

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

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:

:$egin\{alignat\}\{2\}T\; =\; 2pi\; sqrt\{ellover\; g\}\; left(\; 1+\; left(\; frac\{1\}\{2\}\; ight)^2\; sin^2left(frac\{\; heta\_0\}\{2\}\; ight)\; +\; left(\; frac\{1\; cdot\; 3\}\{2\; cdot\; 4\}\; ight)^2\; sin^4left(frac\{\; heta\_0\}\{2\}\; ight)\; +\; left(\; frac\; \{1\; cdot\; 3\; cdot\; 5\}\{2\; cdot\; 4\; cdot\; 6\}\; ight)^2\; sin^6left(frac\{\; heta\_0\}\{2\}\; ight)\; +\; cdots\; ight)\; \backslash \; =\; 2pi\; sqrt\{ellover\; g\}\; cdot\; sum\_\{n=0\}^infty\; left\; [\; left\; (\; frac\{(2\; n)!\}\{(\; 2^n\; cdot\; n!\; )^2\}\; ight\; )^2\; cdot\; sin^\{2\; n\}left(frac\{\; heta\_0\}\{2\}\; ight)\; ight]\; .end\{alignat\}$

Figure 5 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.

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 $4sqrt\{1over\; g\}Fleft(\{sin\; 10over\; 2\},\{piover2\}\; ight)\; =\; 2.0102$ seconds, whereas the linear approximation gives $2pi\; sqrt\{1over\; g\}\; =\; 2.0064$.

**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:

:$T\; =\; -\; m\; g\; L\; sin(\; heta)$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 $sin\; heta\; approx\; heta$,

:$a\; approx\; frac\{mgL\; heta\}\; \{I\}$

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

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

[

*[*]*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 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.**ee also***

Mathieu function

*Double pendulum **External links*** [

*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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