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Computer Laboratory


Applications of Differential Equations

The Simple Pendulum

(continued from last page...)

Your output from the command I just gave you should have looked something like this:


Sine Series

The last term shown here is called the order term, and is read as "big oh of y to the eleventh." It simply means that all remaining terms in the infinite series are powers of y greater than or equal to 11.

Now, look at the size of the denominators of each term the further you go out in the series. Here's some initial evidence that the terms are shrinking in size as you move out into the series' higher degree terms.

But hold on! This affect could be spoiled by the exponents in the series. After all, big powers usually mean big numbers, and these exponents are growing in size. How do we know that they're not growing so quickly as to defeat the shrinking affect produced by the denominators.

Well, one way we can guarantee some size control is by restricting the size of y. Remember, when you put a large exponent on a number less than 1 it actually shrinks in size. In fact we can take this idea to an extreme if we say that all of the terms beyond the first are of negligible size. Our conclusion:

for small values of y, sin(y) is approximately equal to y

This is exactly the same conclusion that we reached when we took the graphical approach. It seems that there is adequate evidence at this point to allow ourselves to make the substitution of y in place of sin(y), provided that the value of y is small. In doing so, we say that we are linearizing the initial value problem.

Making the Substitution

Once we make the replacement mentioned above, our initial value problem goes from:

Actual Initial Value Problem

y'' + (g/L) sin(y) = 0
y(0) = yo
y'(0) = vo

to this:

Linearized Initial Value Problem

y'' + (g/L) y = 0
y(0) = yo
y'(0) = vo

which is only valid when y is small.

Solving the Linearized Problem

Now we have a homogeneous, linear differential equation with constant coefficients, therefore it's solvable.

The problem would be a little easier to think about if we were to use concrete values for our two constants, so let's set the values of g and L in the equation to be L=1 m, and g=9.8 m/s2. The initial value problem then becomes:

y'' + 9.8 y = 0
y(0) = yo
y'(0) = vo

Before we have Mathematica solve it, we're also going to specify actual values for the initial conditions. Let's say that the bob is initially hanging straight down, and it's given a clockwise velocity of 1 radian/second. Then we need to set the values of yo and vo by issuing the commands:

y0=0, and v0=1

Note the use of single, not double, equals signs here since we're making an assignment, not declaring an equation. Go to Mathematica and enter these commands on two separate lines.

Let's now discuss your result...

Compass If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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