# Pendulum (derivations): Wikis

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# Encyclopedia

Derivations from Pendulum (mathematics).

## Analysis of a simple gravity pendulum

Figure 1. Force diagram of a simple gravity pendulum.

To begin, we shall make three assumptions about the simple pendulum:

• The rod/string/cable on which the bob is swinging is massless, does not stretch, and always remains taut.
• The bob is a point mass.
• Motion occurs in a 2-dimensional plane, i.e. pendulum does not swing into and out of the page.

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 instantaneous 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:

$\ell{d^2\theta\over dt^2} = - g \sin\theta$
${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0 \quad\quad\quad (1)$
Figure 2. Trigonometry of a simple gravity pendulum.

This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h would have acquired 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 ${d\theta\over dt}$

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

h is the vertical distance the pendulum fell. Consider Figure 2, which shows 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

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

substituting this into the equation for ${d\theta\over dt}$ 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}
${d^2\theta\over dt^2} = -{g\over \ell}\sin\theta,$

which is the same result as obtained through force analysis.