Solve the Pendulum

A pendulum is an object consisting of a mass suspended from a pivot so that it can swing freely. The mathematics of pendulums are governed by the differential equation

• <math>\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{l}\sin\theta = 0</math>

which is a nonlinear equation in <math>\theta.</math> Here, <math>g \approx 9.8\ \text{m}\,\text{s}^{-2}</math> is the gravitational acceleration, and <math>l</math> is the length of the pendulum. Simple pendulums can be used to measure the local gravitational acceleration to within 3 or 4 significant figures.

Steps

Small Angle Approximation

1. Make the small-angle approximation.
   • The governing differential equation for a simple pendulum is nonlinear because of the <math>\sin\theta</math> term. In general, nonlinear differential equations do not have solutions that can be written in terms of elementary functions, and this is no exception.
   • However, if we assume that the angle of oscillation is small, e.g. <math>\theta < 15^{\circ},</math> then it is reasonable to make the approximation that <math>\sin\theta \approx \theta.</math> We see that <math>\theta</math> is the first term in the Taylor series for <math>\sin\theta</math> about <math>\theta =

0,</math> so our error in this approximation is on the order of <math>O(\theta^{3}).</math>

1. • • <math>\sin\theta = \theta - \frac{\theta^{3}}{3!} + \frac{\theta^{5}}{5!} - \frac{\theta^{7}}{7!} + \cdots</math>
   • We then obtain the equation for a simple harmonic oscillator. This equation is linear and has a well-known solution.
     • <math>\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{l}\theta = 0</math>
2. Solve the differential equation using the small-angle approximation. Since this is a linear differential equation with constant coefficients, our solution must either be in the form of exponentials or trigonometric functions. For physical reasons, we expect that the equation of motion be oscillatory (trigonometric) in nature.
   • Obtain the characteristic equation and solve for the roots.
     • <math>r^{2} + \frac{g}{l} = 0</math>
     • <math>r = \pm \sqrt{\frac{g}{l}}i</math>
   • Since our roots are imaginary, our solution is indeed oscillatory, as expected. From the theory of differential equations, we obtain our solution below. We write the angular frequency <math>\omega = \sqrt{\frac{g}{l}}.</math>
     • <math>\theta(t) = c_{1}\cos\omega t + c_{2}\sin\omega t</math>
3. Write the equation of motion in terms of amplitude and phase factor. A more useful formulation of the solution involves making the following manipulation.
   • Multiply the solution by <math>\frac{\sqrt{c_{1}^{2} + c_{2}^{2}}}{\sqrt{c_{1}^{2} + c_{2}^{2}}}.</math>
     • <math>\sqrt{c_{1}^{2} + c_{2}^{2}}\left(\frac{c_{1}}{\sqrt{c_{1}^{2} + c_{2}^{2}}} \cos\omega t+ \frac{c_{2}}{\sqrt{c_{1}^{2} + c_{2}^{2}}}\sin\omega t\right)</math>
   • Draw a right triangle with angle <math>\phi,</math> hypotenuse length <math>\sqrt{c_{1}^{2} + c_{2}^{2}},</math> opposite side length <math>c_{1},</math> and adjacent side length <math>c_{2}.</math> Replace the constant <math>\sqrt{c_{1}^{2} + c_{2}^{2}}</math> with a new constant <math>\Theta,</math> denoting amplitude. Now we can simplify the quantities in parentheses. The result is that the second arbitrary constant has been replaced with an angle.
     • <math>\begin{align}\theta(t) &= \Theta(\sin\phi\cos\omega t + \cos\phi\sin\omega t) \\

&= \Theta\sin(\omega t + \phi)\end{align}</math>

1. • Because <math>\phi</math> is arbitrary, we can also use the cosine function as well. Mathematically, the two phase factors are different, but in terms of finding the equation of motion given initial conditions, only the form of the solution matters. Writing it in terms of cosine is slightly more common because it fits initial conditions well (imagine a pendulum being let go at some angle - the cosine function fits this situation naturally).
     • <math>\theta(t) = \Theta\cos(\omega t + \phi)</math>
2. Solve for initial conditions. Initial conditions are solved in the usual manner with regards to second-order differential equations given the general solution.
   • Assume initial conditions <math>\theta(0) = \theta_{0}</math> and <math>\frac{\mathrm{d}\theta}{\mathrm{d}t}(0) = 0.</math> This is equivalent to saying that we release a pendulum without any force at some angle <math>\theta_{0}</math> from equilibrium, provided that <math>\theta_{0}</math> is not too great.
   • Substitute these conditions into the general solution. Differentiate the general solution and substitute these conditions into that as well. We immediately obtain <math>\phi = 0</math> and <math>\Theta = \theta_{0}.</math>
     • <math>\theta(t) = \theta_{0}\cos\omega t</math>
   • If you are given numbers, then simply follow the above steps with the appropriate numbers substituted.
3. Find the period of a simple pendulum.
   • Physically, the angular frequency is the number of radians rotated per unit time. It is therefore related to the period via the relation <math>\omega = \frac{2\pi}{T}.</math> We can then solve for the period <math>T.</math>
     • <math>\omega = \sqrt{\frac{g}{l}}</math>
     • <math>T = 2\pi\sqrt{\frac{l}{g}}</math>
   • The order of <math>g</math> and <math>l</math> can get confusing. If it does, we go back to physical intuition. Intuitively, a longer pendulum should have a longer period than a shorter pendulum, so <math>l</math> should be on top.

Arbitrary Angle

1. Write the differential equation of a pendulum without the small-angle approximation. This equation is no longer linear and is not easily solved. It turns out that the period of such a pendulum can be written exactly in terms of elliptic integrals - integrals that historically were studied to find the arc length of ellipses, but naturally arise in the study of pendulums as well.
   • <math>\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{l}\sin\theta = 0</math>
   • To make things simple, we are given the same initial conditions as before: <math>\theta(0) = \theta_{0}</math> and <math>\frac{\mathrm{d}\theta}{\mathrm{d}t}(0) = 0.</math>
2. Multiply the equation by <math>2\frac{\mathrm{d}\theta}{\mathrm{d}t}</math>.
   • <math>2\frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{2g}{l}\frac{\mathrm{d}\theta}{\mathrm{d}t}\sin\theta = 0</math>
   • We can then make use of the chain rule for both terms.
     • <math>2\frac{\mathrm{d}\theta}{\mathrm{d}t}\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^{2}</math>
     • <math>\frac{\mathrm{d}}{\mathrm{d}t}\cos\theta = -\sin\theta\frac{\mathrm{d}\theta}{\mathrm{d}t}</math>
   • Then, we arrive at the following equation.
     • <math>\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^{2} = \frac{2g}{l}\frac{\mathrm{d}}{\mathrm{d}t}\cos\theta</math>
3. Integrate with respect to time. The integration introduces an integration constant. Physically, this constant represents the cosine of an initial angle. There are two solutions because the pendulum can move counterclockwise or clockwise.
   • <math>\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^{2} = \frac{2g}{l}(\cos\theta - \cos\theta_{0})</math>
   • <math>\frac{\mathrm{d}\theta}{\mathrm{d}t} = \pm\sqrt{\frac{2g}{l}}\sqrt{\cos\theta - \cos\theta_{0}}</math>
4. Set up the integral to find the period.
   • From our previous results, we found that <math>\Theta =

\theta_{0}</math> was the amplitude of oscillation. This suggests that half the period is the time taken for the pendulum to traverse from <math>-\theta_{0}</math> to <math>\theta_{0}.</math>

1. • • <math>\frac{T}{2} = \sqrt{\frac{l}{2g}}\int_{-\theta_{0}}^{\theta_{0}}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta - \cos\theta_{0}}}</math>
   • Because <math>\cos\theta</math> is even, we can factor out a 2.
     • <math>T = 2\sqrt{\frac{2l}{g}}\int_{0}^{\theta_{0}}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta - \cos\theta_{0}}}</math>
   • This integral is tough, and cannot be evaluated using elementary methods. However, it can be evaluated exactly in terms of the Beta function if we assume that <math>\theta_{0} = \pi/2,</math> i.e. the angle of oscillation is 90°. This is large enough to be outside the scope of the small-angle approximation. We do this calculation in the next step.
2. Solve for the period given an oscillation angle of 90°.
   • When <math>\theta_{0} = \pi/2,</math> <math>\cos\theta_{0} = 0</math> and we obtain the following integral.
     • <math>T = 2\sqrt{\frac{2l}{g}}\int_{0}^{\pi/2}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta}}</math>
   • This integral still does not have an antiderivative that can be written in terms of elementary functions, but it can be evaluated exactly in terms of the Integrate Using the Beta Function, itself written in terms of the Integrate Using the Gamma Function.
     • <math>\frac{\Gamma(\alpha)\Gamma(\beta)}{2\, \Gamma(\alpha + \beta)} =

\int_{0}^{\pi/2}\cos^{2\alpha - 1}\theta\sin^{2\beta - 1}\theta\mathrm{d}\theta</math>

1. • We see from direct comparison that <math>\alpha = 1/4</math> and <math>\beta = 1/2.</math> Given that <math>\Gamma(1/2) = \sqrt{\pi},</math> we arrive at the following answer.
     • <math>\int_{0}^{\pi/2}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta}} = \frac{\sqrt{\pi}}{2}\frac{\Gamma(1/4)}{\Gamma(3/4)}</math>
   • We now make use of Euler's reflection formula to simplify, since <math>\Gamma(3/4)</math> is related to <math>\Gamma(1/4).</math>
     • <math>\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}</math>
     • <math>\int_{0}^{\pi/2}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta}} = \frac{\Gamma^{2}(1/4)}{2\sqrt{2\pi}}</math>
   • Combining with our previous result, and setting the period of the pendulum with small-angle approximation <math>T_{0},</math> we arrive at the following result. Note that <math>\Gamma(1/4)</math> is transcendental.
     • <math>T = \sqrt{\frac{l}{g}}\frac{\Gamma^{2}(1/4)}{\sqrt{\pi}} = \frac{\Gamma^{2}(1/4)}{2\pi^{3/2}}T_{0} \approx 1.18T_{0}</math>
   • Thus, the period of a pendulum given an amplitude of 90° has a period about 18% longer than that given by the simple harmonic oscillator.
2. Rewrite the period in terms of elliptic integrals.
   • We first restate the integral to be evaluated.
     • <math>T = 2\sqrt{\frac{2l}{g}}\int_{0}^{\theta_{0}}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta - \cos\theta_{0}}}</math>
   • Make use of the following substitutions. The third line immediately follows from the second substitution.
     • <math>\cos\theta = 1 - 2\sin^{2}\frac{\theta}{2}</math>
     • <math>\sin\frac{\theta}{2} = \sin\frac{\theta_{0}}{2}\sin\phi</math>
     • <math>\frac{1}{2}\cos\frac{\pi}{2}\mathrm{d}\theta = 2k\cos\phi\mathrm{d}\phi</math>
   • For simplicity, let <math>k = \sin\frac{\theta_{0}}{2}.</math> Notice that when <math>\theta = 0,</math> <math>\phi = 0,</math> and when <math>\theta = \theta_{0},</math> <math>\phi = \pi/2.</math>
     • <math>\begin{align}T &= 2\sqrt{\frac{2l}{g}}\int_{0}^{\theta_{0}}\frac{\mathrm{d}\theta}{\sqrt{\cos\theta - \cos\theta_{0}}} \\

&= 2\sqrt{\frac{l}{g}}\int_{0}^{\theta_{0}}\frac{\mathrm{d}\theta}{\sqrt{\sin^{2}\frac{\theta_{0}}{2} - \sin^{2}\frac{\theta}{2}}} \\ &= 2\sqrt{\frac{l}{g}}\int_{0}^{\pi/2}\frac{1}{k}\frac{2k\mathrm{d}\phi}{\sqrt{1 - \sin^{2}\phi}}\frac{\sqrt{1 - \sin^{2}\phi}}{\sqrt{1 - \sin^{2}\frac{\theta}{2}}} \\ &= 4\sqrt{\frac{l}{g}}\int_{0}^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1 - k^{2}\sin^{2}\phi}}\end{align}</math>

1. • This integral is called the complete elliptic integral of the first kind, denoted by <math>K(k).</math> This integral does not have a solution expressible in terms of elementary functions, but it can be expressed as a series by way of the Beta function again.
   • The period can thus be written exactly as follows.
     • <math>T = 4\sqrt{\frac{l}{g}}K\left(\sin^{2}\frac{\theta_{0}}{2}\right)</math>
2. Evaluate the elliptic integral using the Beta function. A more detailed explanation of this evaluation can be found here.
   • We must make use of the binomial series.
     • <math>(1 - x)^{\alpha} = \sum_{m=0}^{\infty}{\alpha \choose m}(-1)^{m}x^{m}</math>
   • <math>\begin{align}K(k) &= \int_{0}^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1 - k^{2}\sin^{2}\phi}} \\

&= \int_{0}^{\pi/2}\mathrm{d}\phi\sum_{m=0}^{\infty}{-1/2 \choose m}(-1)^{m}k^{2m}\sin^{2m}\phi \\ &= \sum_{m=0}^{\infty}{-1/2 \choose m}(-1)^{m}k^{2m}\int_{0}^{\pi/2}\sin^{2m}\phi\mathrm{d}\phi \\ &= \sum_{m=0}^{\infty}\frac{(-1)^{m}\Gamma(1/2)}{m!\Gamma(1/2-m)}\frac{\Gamma(1/2)\Gamma(1/2+m)}{2\,m!}k^{2m} \\ &= \frac{\pi}{2}\sum_{m=0}^{\infty}\frac{(-1)^{m}\Gamma(1/2+m)}{(m!)^{2}\Gamma(1/2-m)}k^{2m} \\ &= \frac{\pi}{2}\sum_{m=0}^{\infty}\frac{(-1)^{m}\Gamma^{2}(1/2+m)}{\pi (m!)^{2}}\sin(\pi(m + 1/2))k^{2m} \\ &= \frac{\pi}{2}\sum_{m=0}^{\infty}\left[\frac{(2m-1)!!}{2^{m}m!}\right]^{2}k^{2m} \\ &= \frac{\pi}{2}\left[1 + \frac{1}{2^{2}}k^{2} + \left(\frac{3\cdot 1}{2\cdot 2}\right)^{2}\frac{1}{2!^{2}}k^{4} + \left(\frac{5\cdot 3\cdot 1}{2\cdot 2\cdot 2}\right)^{2}\frac{1}{3!^{2}}k^{6} + \cdots\right]\end{align}</math>

1. • In this derivation, we used the binomial series, the relation between the Gamma and the factorial functions <math>\Gamma(z) = (z - 1)!,</math> Euler's reflection formula to simplify the <math>\Gamma(1/2 + m)</math> and <math>\Gamma(1/2 - m)</math> terms, the fact that <math>(-1)^{m}\sin(\pi(m + 1/2)) = 1</math> for all integers <math>m,</math> and the double factorial identity relating it to the Gamma function, written below.
     • <math>(2m - 1)!! = \frac{2^{m}\Gamma(m + 1/2)}{\sqrt{\pi}}</math>
2. Examine the series. This is a very important series, and from this, we obtain the period of a true pendulum. Let <math>T_{0} = 2\pi\sqrt{\frac{l}{g}}</math> be the period of the pendulum using the small-angle approximation. The series clearly demonstrates the deviation from this approximation as <math>\theta_{0}</math> gets larger. Since the region of convergence is <math>|k| < 1,</math> we see that at 180°, the series diverges, corresponding to a pendulum at unstable equilibrium. Remember that <math>k = \sin\frac{\theta_{0}}{2}</math> in this relation.
   • <math>T = T_{0}\left[1 + \frac{1}{2^{2}}k^{2} + \left(\frac{3\cdot 1}{2\cdot 2}\right)^{2}\frac{1}{2!^{2}}k^{4} + \left(\frac{5\cdot 3\cdot 1}{2\cdot 2\cdot 2}\right)^{2}\frac{1}{3!^{2}}k^{6} + \cdots\right]</math>
   • The graph above shows the elliptic integral in red, along with its series expansions truncated to 2nd, 4th, and 10th order. We can clearly see the divergence here, as well as the series being progressively better approximations the more terms we keep.

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