The simple pendulum has been the object of much mathematical study. A device as simple as the pendulum may not seem to warrant much mathematical attention, but the study of its motion was begun by some of the greatest scientists of the 17th century. The reason was that the pendulum allows one to keep time with great accuracy. The development of pendulum clocks made sea travel safer during a time when new worlds in the Americas were just being developed and discovered. The scientist Christian Huygens studied the pendulum extensively from a geometric standpoint, and Issac Newton's theory of gravitation made it possible to study the pendulum from an analytic standpoint, which is what we will do in this project. The compound pendulum is still studied today because of its complicated "chaotic" behavior.

Figure 46.1: The Pendulum

A simple pendulum is a small mass suspended from a light rigid rod. Ideally the mass will be small enough so that it will not deform the rod, but the rod should be light so that we can neglect its mass. (We will consider what happens if the rod stretches later in the project.) If we displace the bob slightly and let it go, the pendulum will start swinging. We will assume at first that there is no friction. In this idealized situation, once the pendulum is started, it will never stop. The only force acting on the pendulum is the force of gravity.

**46.1 Derivation of the Pendulum Equation**

The force acting on the pendulum can be broken into two components, one in the direction of the rod and the other in the direction of the pendulum's motion.
If the rod does not stretch, the component directed along the rod plays no part in the pendulum's motion because it is counterbalanced by the force in the rod.
The component in the tangential direction causes the pendulum to move.
If we impose a coordinate system centered at the point of suspension of the pendulum and let
denote the vector giving the position of the pendulum bob, then Figure 46.1 illustrates the force of gravity on the pendulum and the resolution of this force into two components * F_{R}* and

We will use the vector form of Newton's law, * F=mA*, to determine the equation of motion of the pendulum.
We assume that the origin of our fixed inertial coordinate system is at the pivot point of the pendulum shown in the figure.
The axes have their usual orientation.

If the vector

Show that

Prove that

The next step in finding the equations of motion of the pendulum is to use some basic vector geometry.
The gravity vector is simple in *x*-*y* coordinates, but we need to express it in radial and tangential components.
You should be able to remember a simple entry in your geometric-algebraic vector lexicon of text Chapter 15 that lets you compute these components as perpendicular projections.

The magnitude of the gravitational force on the bob of mass *m* is *mg* and it acts down.
The constant *g* is the universal acceleration due to gravity (which we discussed in Galileo's law in Chapter 10.) *g=9.8* in mks units.
As a vector, the force due to gravity is

for the scalar quantities you compute.

The derivative of a general position vector with respect to *t* is the associated velocity vector.
The second derivative of the position vector with respect to *t* is the acceleration vector.

You can verify this by writing

Calculate the second derivative of the position vector
with respect to *t* and express the result in terms of the unit vectors * U_{R}* and

We can now use * F=mA* to determine the equation of motion of the pendulum.
We first need to note that the rod from which the bob is suspended provides a force that counteracts the radial component of gravity and prevents any acceleration in the radial direction.
Hence when we equate

Why doesn't the equation of motion for the pendulum depend on the mass? What does this mean physically?

**46.2 Numerical Solutions of the Pendulum Equation**

This system can then be solved by the computer program

Figure 46.7 illustrates the phase plane trajectories of solutions to the system of differential equations for various initial conditions. The displacement angle is plotted on the horizontal axis, and the angular velocity is plotted on the vertical axis.

Figure 46.2: Phase Diagram for the Pendulum Equation

canceling

Prove that is constant on solutions of the pendulum equations.

*
Make a contour plot of the function
.
(HINT: See the main text, Section 24.4.) *

**Problem 46.1**
Mathematical and Physical Experiments

Modify theFlow2Dprogram to build a flow solution of the pendulum equations. We suggest that you use initial conditions (L=9.8meter,g=9.8m/sec). =^{2}initxys = Table[{x,0},{x,-2 Pi,2 Pi,.6}]and =initxys = Table[{0,x},{x,-2 Pi,2 Pi,.6}]As you can see from the phase plane there are two types of motion that the pendulum displays, the motion corresponding to the closed flow trajectories and that corresponding to the oscillating flow curves. Explain physically what type of motion is occurring when the flow trajectories are closed curves. Explain the type of motion occurring when the flow trajectories oscillate without closing on themselves. Use the

AccDEsolnprogram to plot explicit time solutions associated with each kind of flow line.

We suggest that you also try some calculations with various lengths. A 9.8 m pendulum is several stories high and oscillates rather slowly. It would be difficult to build such a large pendulum and have it swing through a full rotation. How many seconds is a full oscillation (that is, what is the period)? How does the period depend on the length?

We suggest that you try some physical experiments as well, at least for small oscillations. With small variations in , a string will suffice for the rod. These experiments are easy to do and you should measure the period of oscillation for various lengths.

**Problem 46.2**
The Period of the Pendulum

Verify physically and by numerical experiments that the period depends on the length but not on the mass.

We would like to have a formula for the period, so we imagine the situation where we release the pendulum from rest at an angle . For the time period while is a decreasing function, say one-quarter oscillation, , we could invert the function , . (See Project 21 and note that , so .) Now we proceed with some more clever tricks.

Compute the derivative

Use the pendulum equation and integrate with respect to

Finally, separate variables and integrate to the bottom of the swing, one-quarter period,

This integrand is discontinuous at and may cause your computer trouble. We do some trig,

Now

with the change of variables and differentials , with when and when , using , we get finally,

This is an "elliptic integral of the first kind," or in

**46.3 Linear Approximation to the Pendulum Equation**

This is a second-order linear constant-coefficient differential equation and can be solved explicitly for given initial conditions using the methods of text Chapter 23.

**Problem 46.3**
The Linearized Pendulum's Period

Show that the period of the linearized pendulum is a constant .

The true period of the pendulum differs from this amount more and more as we increase the initial release angle.

Figure 46.3: The Ratio

As you demonstrated above the period of the linearized pendulum is a constant independent of the initial displacement (and mass). This should give you some idea of why pendula are such accurate timekeepers. If the pendulum is released at a certain angle to start and friction causes the amplitude of the swings to diminish, it still swings back and forth in the same amount of time. In a clock, friction is overcome by giving the pendulum a push when it begins to slow down. The push can come from weights descending, springs uncoiling, or a battery.

The comments about the invariance of period apply to the linearized model of the pendulum. That model is not accurate for large initial displacements.

Another way to compare the periods is in the **Flow2D** solutions.

with

*
Compare the flow of the linearized model to the flow of the "real" (rigid rod frictionless) pendulum.
Why do the dots of current state remain in line for the linear equation and not remain in line for the nonlinear flow? *

Nonlinear and Linear Pendula

The basis for our linear approximation to the equation of motion for the pendulum is the approximation
. where if . Use this equation with

Substitute into this equation with

Use this to show that . If a pendulum swings no more than , how large is the error between and ? What is the relative error compared to the maximum angle? How does this compare with the relative error of 1 minute per week? One minute per month?

with

*
How nearly do the dots remain in line for the nonlinear flow under only "" oscillations? *

How many minutes per month might you expect for a pendulum clock that has oscillations varying from 10 degrees fully wound to 8 degrees at the end of its regular winding period? Remember that the clock can be adjusted so that 9 degree oscillations would produce perfect time even though they do not have period .

**46.4 The Period of the "Real" Pendulum**
***

**46.5 Friction in the Pendulum (Optional)**

**46.6 The Spring Pendulum (Optional)**

Figure 46.4: The Spring Pendulum

In order to solve the system of second-order differential equations numerically, we must again introduce phase variables to reduce it to a system of first-order differential equations.
If we let *y _{1}=L*, , , and
, then the system becomes:

*
Modify the AccDEsoln program to solve this system numerically.
Use g=9.8, L_{0}=1.0, k=9.0, and m=1.0 for the acceleration due to gravity, the unstretched length of the spring, the spring constant, and the mass of the bob.
Start the pendulum by stretching the spring and giving it a small angular displacement.
For example, take y_{1}(0)=2.5 and y_{3}(0)=0.05. Use 0 as initial conditions for y_{2} and y_{4}. How does the amplitude of the swings change as time goes on? In particular, does the amplitude of the swings increase, decrease, or remain unchanged? *

Figure 46.5: A Lissajous