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 idealization of a real pendulum using the following assumptions:

• The rod or cord on which the bob swings is massless, inextensible and always remains taut;
• Motion occurs only in two dimensions, i.e. the bob does not trace an ellipse but an arc.
• The motion does not lose energy to friction or air resistance.

The differential equation which represents the motion of a simple pendulum is

${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0 \quad\quad\quad\quad\quad(1)$

where g is acceleration due to gravity, $\ell$ is the length of the pendulum, and θ is the angular displacement.

This is known as Mathieu's equation.

"Force" derivation of (1) Figure 1. Force diagram of a simple gravity pendulum. Consider Figure 1, showing the forces acting on a simple pendulum. Note that the path of the pendulum sweeps out an arc of a circle. The angle θ 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 θ 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 θ 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: ${d^2\theta\over dt^2} + {g\over\ell}\sin\theta = 0$

"Energy" derivation of (1) 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 θ0, then y0, the vertical distance from the screw, is given by $y_0 = \ell\cos\theta_0\,$ similarly, for y1, 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 ${d\theta\over dt} = \sqrt{{2g\over \ell}\left(\cos\theta-\cos\theta_0\right)}\quad\quad\quad (2)$ This equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration ${d\over dt}{d\theta\over dt} = {d\over dt}\sqrt{{2g\over \ell}\left(\cos\theta-\cos\theta_0\right)}$ $\begin{align} {d^2\theta\over dt^2} & = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}{d\theta\over dt} \\ & = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}\sqrt{{2g\over \ell} \left(\cos\theta-\cos\theta_0\right)} = -{g\over \ell}\sin\theta \end{align}$ ${d^2\theta\over dt^2} + {g\over \ell}\sin\theta = 0,$ which is the same result as obtained through force analysis.

Small-angle approximation

The differential equation given above is not easily solved. However adding a restriction to the size of the oscillation's amplitude gives a form whose solution can be easily obtained. If it is assumed that the angle is much less than 1 radian, or

$\theta \ll 1\,$,

then substituting for sin θ into (1) using the small-angle approximation,

$\sin\theta\approx\theta\,$,

yields the equation for a harmonic oscillator

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

The error due to the approximation is proportional to θ ³ (from the Maclaurin series for sin θ).

Given the initial conditions θ(0) = θ₀ and dθ/dt(0) = 0, the solution becomes,

$\theta(t) = \theta_0\cos\left(\sqrt{g\over \ell\,}\,t\right) \quad\quad\quad\quad \theta_0 \ll 1.$

The motion is simple harmonic motion where θ₀ 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

$T_0 = 2\pi\sqrt{\frac{\ell}{g}} \quad\quad\quad\quad\quad \theta_0 \ll 1$

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 $\scriptstyle g\approx9.81$ m/s², and $\scriptstyle g/\pi^2\approx{1}$ (the exact figure is 0.994 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 inverting equation (1)

Figure 3. Deviation of the period from small-angle approximation.

Figure 4. Relative errors using the power series.

${dt\over d\theta} = {1\over\sqrt{2}}\sqrt{\ell\over g}{1\over\sqrt{\cos\theta-\cos\theta_0}}$

and integrating over one complete cycle,

$T = \theta_0\rightarrow0\rightarrow-\theta_0\rightarrow0\rightarrow\theta_0,$

or twice the half-cycle

$T = 2\left(\theta_0\rightarrow0\rightarrow-\theta_0\right),$

or 4 times the quarter-cycle

$T = 4\left(\theta_0\rightarrow0\right),$

which leads to

$T = 4\sqrt{\ell\over g}{1\over\sqrt{2}}\int^{\theta_0}_0 {1\over\sqrt{\cos\theta-\cos\theta_0}}\,d\theta.$

This integral can be re-written in the elliptic function of the first kind (also see Jacobi's elliptic functions), which gives little advantage since that form is also insoluble.

$T = 4\sqrt{\ell\over g}F\left(\csc{\theta_0\over2},{\theta_0\over 2}\right)\csc {\theta_0\over 2}$

or more concisely, using the $\theta\,$ to $u\,$ substitution, $\sin{u} = \frac{\sin{\theta\over 2}}{\sin{\theta_0\over 2}}$

$T = 4\sqrt{\ell\over g}F\left(\sin{\theta_0\over 2}, {\pi \over 2} \right)$

where F(k,φ) is Legendre's elliptic function of the first kind defined as

$F(k,\varphi) = \int^\varphi_0 {1\over\sqrt{1-k^2\sin^2{u}}}\,du\,.$

When ϕ = π / 2, F(k,φ) is referred to as the complete elliptic function (or integral) of the first kind.

Figure 3 shows the deviation of T from T₀, the period obtained from small-angle approximation.

Value for the complete elliptic function can be computed using the rapidly convergent, highly accurate arithmetic-geometric mean method (see elliptic integrals), or alternatively computed using the following series:

$\begin{alignat}{2} T & = 2\pi \sqrt{\ell\over g} \left( 1+ \left( \frac{1}{2} \right)^2 \sin^2\left(\frac{\theta_0}{2}\right) + \left( \frac{1 \cdot 3}{2 \cdot 4} \right)^2 \sin^4\left(\frac{\theta_0}{2}\right) + \left( \frac {1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right)^2 \sin^6\left(\frac{\theta_0}{2}\right) + \cdots \right) \\ & = 2\pi \sqrt{\ell\over g} \cdot \sum_{n=0}^\infty \left[ \left ( \frac{(2 n)!}{( 2^n \cdot n! )^2} \right )^2 \cdot \sin^{2 n}\left(\frac{\theta_0}{2}\right) \right]. \end{alignat}$

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.

For a swing of exactly 180° the bob is balanced over its pivot point and so T = ∞.

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.

For example, the period of a pendulum of length 1 m on Earth (g = 9.80665 m/s²) at initial angle 10 degrees is $4\sqrt{1\ \mathrm{m}\over g}F\left({\sin {10^\circ\over 2}},{\pi\over2}\right) \approx 2.0102\ \mathrm{s}$, where the linear approximation gives $2\pi \sqrt{1\ \mathrm{m}\over g} \approx 2.0064\ \mathrm{s}$.
The difference (less than 0.2%) is much less than that caused by the variation of g with geographical location.

By using the following Maclaurin series:

$F \left(x, {\pi \over 2} \right)={\pi \over 2} \left( 1 + \frac{1}{4}x^2 + \frac{9}{64}x^4 + \frac{25}{256}x^6 + \frac{1225}{16384}x^8 + \cdots\right)$

$\sin \left({\theta_0 \over 2}\right)=\left(\frac{1}{2}\theta_0 - \frac{1}{48}\theta_0^3 + \frac{1}{3840}\theta_0^5 - \frac{1}{645120}\theta_0^7 + \cdots\right)$

The equivalent power series is:

$\begin{alignat}{2} T & = 2\pi \sqrt{\ell\over g} \left( 1+ \frac{1}{16}\theta_0^2 + \frac{11}{3072}\theta_0^4 + \frac{173}{737280}\theta_0^6 + \frac{22931}{1321205760}\theta_0^8 + \frac{1319183}{951268147200}\theta_0^{10} + \frac{233526463}{2009078326886400}\theta_0^{12} + . . . \right) \end{alignat}$

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°

• 

  Pendulum with just barely enough energy for a full swing

• 

  Pendulum with enough energy for a full swing

• 

  Initial angle of 180°, unstable equilibrium.

Compound pendulum

A compound 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:

α is the angular acceleration.

τ 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:

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

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.

See also

• Blackburn pendulum
• Conical pendulum
• Double pendulum
• Inverted pendulum
• Spring pendulum
• Mathieu function
• Pendulum equations (software)

Further reading

• The Pendulum: A Physics Case Study, Gregory L. Baker and James A. Blackburn, Oxford University Press, 2005
• Kenneth L. Sala, “Transformations of the Jacobian Amplitude Function and its Calculation via the Arithmetic-Geometric Mean”, SIAM J. Math. Anal., vol. 20, no. 6, pp. 1514–1528, Nov. 1989.
• Karlheinz Ochs, “A comprehensive analytical solution of the nonlinear pendulum”, European Journal of Physics, vol. 32, no. 2, pp. 479–490, 2011.

External links

• Mathworld article on Mathieu Function