Pendulum (mathematics)

Figure 1. Force diagram of a simple gravity pendulum.

Please take the time to consider Figure 1 on the right, showing the forces acting on a simple pendulum. Note that the path of the pendulum sweeps out an arc of a circle. The angle \theta is measured in radians, and this is crucial for this formula. The blue arrow is the gravitational force acting on the bob, and the violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion. The direction of the bob's instantaneous velocity always points along the red axis, which is considered the tangential axis because its direction is always tangent to the circle. Consider Newton's second law,

F=ma\,

where F is the sum of forces on the object, m is mass, and a is the acceleration. Because we are only concerned with changes in speed, and because the bob is forced to stay in a circular path, we apply Newton's equation to the tangential axis only. The short violet arrow represents the component of the gravitational force in the tangential axis, and trigonometry can be used to determine its magnitude. Thus,

F = -mg\sin\theta = ma\,

a = -g \sin\theta\,

where

g is the acceleration due to gravity near the surface of the earth. The negative sign on the right hand side implies that \theta and a always point in opposite directions. This makes sense because when a pendulum swings further to the left, we would expect it to accelerate back toward the right.

This linear acceleration a along the red axis can be related to the change in angle \theta by the arc length formulas; s is arc length:

s = \ell\theta\,

v = {ds\over dt} = \ell{d\theta\over dt}

a = {d^2s\over dt^2} = \ell{d^2\theta\over dt^2}

thus:

\ell{d^2\theta\over dt^2} = -g \sin\theta

{d^2\theta\over dt^2} + {g\over\ell}\sin\theta = 0

Equation (1) can be obtained using two definitions for torque.

\mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \over {dt}}

First start by defining the torque on the pendulum bob using the force due to gravity.

\mathbf{ \tau } = \mathbf{ l \times F_g },

where \mathbf{l} is the length vector of the pendulum and \mathbf{F_g} is the force due to gravity.

For now just consider the magnitude of the torque on the pendulum.

\mathbf{ |\tau| } = -mg l \sin\theta,

where m is the mass of the pendulum, g is the acceleration due to gravity, l is the length of the pendulum and \theta is the angle between the length vector and the force due to gravity.

Next rewrite the angular momentum.

\mathbf{ L } = \mathbf{ r \times p } = m\mathbf{ r \times (\omega \times r) }.

Again just consider the magnitude of the angular momentum.

\mathbf{ |L| } = mr^2 \omega = m l^2 {d\theta \over dt} .

and its time derivative

{d \over dt}\mathbf{|L|} = m l^2 {d^2\theta \over dt^2} ,

According to { \mathbf{ \tau } = {d \mathbf{ L} \over dt} }, we can get by comparing the magnitudes

-mgl \sin\theta = m l^2 {d^2\theta \over dt^2} ,

thus:

{d^2\theta\over dt^2} + {g\over l}\sin\theta = 0,

which is the same result as obtained through force analysis.

Figure 2. Trigonometry of a simple gravity pendulum.

It can also be obtained via the conservation of mechanical energy principle: any object falling a vertical distance h would acquire kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by

\Delta U = mgh\,

change in kinetic energy (body started from rest) is given by

\Delta K = {1\over2}mv^2

Since no energy is lost, those two must be equal

{1\over2}mv^2 = mgh

v = \sqrt{2gh}\,

Using the arc length formula above, this equation can be rewritten in favor of {d\theta\over dt}

v = {\ell}{d\theta\over dt} = \sqrt{2gh}

{d\theta\over dt} = {1\over \ell}\sqrt{2gh}

h is the vertical distance the pendulum fell. Look at Figure 2, which presents the trigonometry of a simple pendulum. If the pendulum starts its swing from some initial angle \theta_0, then y_0, the vertical distance from the screw, is given by

y_0 = \ell\cos\theta_0\,

similarly, for y_1, we have

y_1 = \ell\cos\theta\,

then h is the difference of the two

h = \ell\left(\cos\theta-\cos\theta_0\right)

in terms of {d\theta\over dt} gives

{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 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\approx9.81 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 (Eq. 2), 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. 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 half-cycle T = 2 t\left(\theta_0\rightarrow0\rightarrow-\theta_0\right), or 4 times the quarter-cycle 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 re-written 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{1-k^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{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 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. 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:[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}. Arithmetic-geometric mean solution for elliptic integral Given Eq. 3 and the Arithmetic-geometric mean solution of the elliptic integral: K(k) = \frac {\pi /2}{M(1-k,1+k)}, where M(x,y) is the arithmetic-geometric mean of x and y. This yields an alternative and faster-converging 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 45° Initial angle of 90° Initial angle of 135° Initial angle of 170° 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 small-angle 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 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.[6] See also Blackburn pendulum Conical pendulum Double pendulum Inverted pendulum Kapitza's pendulum Spring pendulum Mathieu function Pendulum equations (software) 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 arithmetic-geometric 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, Saint-Petersburg, Russia, pp. 13-18. Further reading Baker, Gregory L.; Blackburn, James A. (2005). The Pendulum: A Physics Case Study. Oxford University Press.  Ochs, Karlheinz (2011). "A comprehensive analytical solution of the nonlinear pendulum".   Sala, Kenneth L. (1989). "Transformations of the Jacobian Amplitude Function and its Calculation via the Arithmetic-Geometric Mean". SIAM J. Math. Anal. 20 (6): 1514–1528.   External links Mathworld article on Mathieu Function Categories Articles with separate introductions Horology Mathematical physics Pendulums This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. 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