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# Pendulum (mathematics)

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 Title: Pendulum (mathematics) Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### 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

Animation of a pendulum showing the velocity and acceleration vectors.

A so-called "simple pendulum" is an idealization of a "real pendulum" but in an isolated system using the following assumptions:

• The rod or cord on which the bob swings is massless, inextensible and always remains taut;
• The bob is a point mass;
• 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

(Eq. 1)

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

 "Force" derivation of (Eq. 1) 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
 "Torque" derivation of (Eq. 1) 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.