Question

This is a pendulum problem using Newton's Second Law and Simpson's Rule. I have put in the comments the full question and then how I've started it. I have Simpson's Rule set up, I just don't know how to get my f(x) to use in the rule.

Answer 1

Here's the full question:

The figure shows a pendulum with length L that makes a maximum angle θ0 with the vertical. Using Newton's Second Law, it can be shown that the period T (the time for one complete swing) is given by
\[T=4\sqrt{\frac{ L }{ g }} \int\limits_{0}^{\pi/2}\frac{ dx }{ \sqrt{1-k^2\sin ^2x} }\]
where \[k=\sin (\frac{ 1 }{ 2 }\theta _{0})\] and g is the acceleration due to gravity. If L=4 and \[\theta _{0}=44 degrees\], use Simpsons rule with n=10 to find the period. (use g=9.8 m/s^2. Round your answer to five decimal places.

Answer 2

I know that Simpson's rule will start like this:

\[4\sqrt{\frac{ 4 }{ 9.8 }}\ (\frac{ \pi/2 }{ 30 })\left[ f(0)+4f(\pi/20)+2f(2\pi/20)...+4f(9\pi/20)+f(\pi/2) \right]\]
I just don't know how to get the function, f(x), to plug in to the formula.

Answer 3

That's just the integrand, is it not?

Answer 4

I think so but I don't know how to find that integral I guess...

Answer 5

no, integrand. not integral. The integrand is the thing you're integrating - in this case, it is
\[ f(x) = \frac{1}{1-k^2\sin^2(x)} \]

Answer 6

OK - I tried that just now and didn't come up with a correct answer. I ended up with 3.04566 after calculating the integrand as the f(x)

Answer 7

Oops, sorry, that should be a square root in there - did you put that in?

Answer 8

no, I wondered about that....I'll add it and try again

Answer 9

Got it. Thank you! :)