Simple pendulum

The simple pendulum is an approximate model of a pendulum, comprised of a zero-dimensional point mass hanging at the end of a mass-less wire attached with no Friction to a fixed point. The mass is only subject to gravity and ignores other forces like air drag. It is a specific case of the Foucault pendulum where the angle θ\theta is constant and angular speed ω\omega can be taken to be zero. This makes the pendulum swing on a plane and the system simplifies to a two-dimensional problem.

Starting from the Foucault pendulum equations of motion, equation (1)(1) becomes trivial when θ=const.\theta=\text{const.} and ω=0\omega=0. Equation (2)(2) becomes

ϕ¨+gLsin⁡ϕ=0(1)\ddot{\phi}+ \frac{g}{L}\sin\phi=0\tag{1}

which is a second order, non-linear differential equation in ϕ\phi.

Small swings

For small swings, where the angle ϕ\phi of oscillation is small, we can approximate sin⁡ϕ≃ϕ\sin\phi\simeq\phi. This leaves us with

ϕ¨+gLϕ=0\ddot{\phi}+ \frac{g}{L}\phi=0

which is the harmonic oscillator equation, for which we know the solution. gg is the gravitational acceleration and LL is the length of the wire. If we call ω0=g/L\omega_{0}=\sqrt{g/L} the angular frequency of oscillation, the general solution is

ϕ(t)=ϕ0cos⁡(ω0t+ψ0)\boxed{\phi(t)=\phi_{0}\cos(\omega_{0}t+\psi_{0})}

where ϕ0\phi_{0} is the amplitude of the oscillations. The period of oscillation is

T=2πω0=2πLgT=\frac{2\pi}{\omega_{0}}=2\pi\sqrt{\frac{L}{g}}

Since the frequency is constant, the period is also constant. This means that small oscillations are isochronal. It is also independent of the mass of the object.

General case

Let the initial conditions be ϕ(0)=ϕ0>0\phi(0)=\phi_{0}>0 and ϕ˙(0)=0\dot{\phi}(0)=0. These mean that the mass begins at some nonzero angle from the vertical and with no starting velocity. As there is no friction, motion must still be periodic, as in the small oscillation approximation. The period TT is however not so easily solved, as the non-linear differential is considerably more difficult to solve.

To procede, multiply (1)(1) by 2ϕ˙2\dot{\phi}:

0=2ϕ˙ϕ¨+2gLϕ˙sin⁡ϕ=ddt(ϕ˙2−2gLcos⁡ϕ)0=2\dot{\phi}\ddot{\phi}+ \frac{2g}{L}\dot{\phi}\sin\phi=\frac{d}{dt}\left(\dot{\phi}^{2}- \frac{2g}{L}\cos\phi\right)

Integrating gives us

ϕ˙2=2gL(cos⁡ϕ−cos⁡ϕ0)\dot{\phi}^{2}=\frac{2g}{L}(\cos\phi-\cos\phi_{0})

We can take the square root and pick the sign to be minus, so that the angle is decreasing (i.e. it's swinging towards the center). This way, we get

ϕ˙=dϕdt=−2gL(cos⁡ϕ−cos⁡ϕ0)\dot{\phi}=\frac{d\phi}{dt}=-\sqrt{\frac{2g}{L}(\cos\phi-\cos\phi_{0})}

Splitting the differentials, we can separate the variables

dϕcos⁡ϕ−cos⁡ϕ0=−2gLdt\frac{d\phi}{\sqrt{\cos\phi-\cos\phi_{0}}}=-\sqrt{\frac{2g}{L}}dt

We can now solve this by integration from ϕ0\phi_{0} to ϕ\phi on the left and 00 to tt on the right:

∫ϕ0ϕdψcos⁡ψ−cos⁡ϕ0=−2gLt\int_{\phi_{0}}^{\phi} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}=-\sqrt{\frac{2g}{L}}t

We can use the boundary condition ϕ(T)=−ϕ0\phi(T)=-\phi_{0} to get

∫ϕ0−ϕ0dψcos⁡ψ−cos⁡ϕ0=−2gLT\int_{\phi_{0}}^{-\phi_{0}}\frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}=-\sqrt{\frac{2g}{L}}T

If we extract TT we get

T=L2g∫−ϕ0ϕ0dψcos⁡ψ−cos⁡ϕ0=2Lg∫0ϕ0dψcos⁡ψ−cos⁡ϕ0T=\sqrt{\frac{L}{2g}}\int_{-\phi_{0}}^{\phi_{0}} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}=\sqrt{\frac{2L}{g}}\int_{0}^{\phi_{0}} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}

where we used symmetry between [−ϕ0,0][-\phi_0,0] and [0,ϕ0][0,\phi_{0}]. Unfortunately, this integral is not solvable in closed form, so we must resort to numerical methods.

Since the motion is periodic, we only need to solve for one period and then we can reuse the results for every subsequent swing. If solved numerically, intermediate positions can be interpolated from the sampled solutions. More specifically, we can solve for just the half period, which is the time difference between to angles where the pendulum has stopped (i.e. the time to go from one extreme of a swing to the other).

By looking at the integral, we can see that it is improper, as the integrand shoots up to infinity when ψ→ϕ0\psi \rightarrow \phi_{0}. For such an integral, we can use the Cauchy principal part:

∫0ϕ0−εdψcos⁡ψ−cos⁡ϕ0+∫ϕ0−εϕ0dψcos⁡ψ−cos⁡ϕ0\int_{0}^{\phi_{0}-\varepsilon} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}+\int_{\phi_{0}-\varepsilon}^{\phi_{0}} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}

(without stating the limit ε→0\varepsilon \rightarrow 0 as it doesn't make much sense on a computer. ε\varepsilon is nevertheless very small). The first term can be numerically solve as usual, as it has no discontinuities. In the second, we can expand the cosine in a Taylor series about ϕ0\phi_{0} like

cos⁡ψ=cos⁡ϕ0−sin⁡ϕ0(ψ−ϕ0)−12cos⁡ϕ0(ψ−ϕ0)2+O(ψ3)\cos\psi=\cos\phi_{0}-\sin\phi_{0}(\psi-\phi_{0})- \frac{1}{2}\cos\phi_{0}(\psi-\phi_{0})^{2}+\mathcal{O}(\psi^{3})

Changing variable to z=(ϕ0−ψ)z=(\phi_{0}-\psi) we get

∫ϕ0−εϕ0dψcos⁡ψ−cos⁡ϕ0=∫0εdzsin⁡ϕ0 z−cos⁡ϕ0 z22=…\int_{\phi_{0}-\varepsilon}^{\phi_{0}} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}=\int_{0}^{\varepsilon} \frac{dz}{\sqrt{\sin\phi_{0}\ z- \cos\phi_{0}\ \frac{z^{2}}{2}}}=\ldots

which can be solved analytically to get

…=2cos⁡ϕ0(π2−arcsin⁡(1−εtan⁡ϕ0))\ldots=\sqrt{\frac{2}{\cos\phi_{0}}}\left(\frac{\pi}{2}-\arcsin\left(1- \frac{\varepsilon}{\tan\phi_{0}}\right)\right)

As ε\varepsilon approaches zero, this approximation (and therefore the integral itself) approaches zero. This means that ε\varepsilon needs to be chosen with care to avoid floating point errors when the denominator approaches zero. In total, the approximation is

T≃∫0ϕ0−εdψcos⁡ψ−cos⁡ϕ0+2cos⁡ϕ0(π2−arcsin⁡(1−εtan⁡ϕ0))\boxed{T\simeq\int_{0}^{\phi_{0}-\varepsilon} \frac{d\psi}{\sqrt{\cos\psi-\cos\phi_{0}}}+\sqrt{\frac{2}{\cos\phi_{0}}}\left(\frac{\pi}{2}-\arcsin\left(1- \frac{\varepsilon}{\tan\phi_{0}}\right)\right)}