# Matlab "Linear Model" of a Pendulum via Euler's Method

Tags:
1. Nov 16, 2016

### Morgan Chafe

In my problem the linear modal is defined as the first term in the series expansion of $\sin(x)$ so:

$$\sin(x) = x - \frac{x^{3}}{3!}+\dots$$

$\sin(x) = x$ is the linear modal.

So with this, I then have to write $\frac{d^{2}x}{dt^{2}} = -\sin(x)$ as a system of $x^{\prime}$ and $y^{\prime}$, so:
$$x^{\prime} = y$$
$$y^{\prime} = \sin(x)$$

I tried the linear modal in Euler's method, with initial conditions X(1) = 1 and V(1)=0 :

Code (Text):
for i = 1:1000
V(i+1) = V(i)-(1.*s) ;
X(i+1) = V(i);
end
Where s is the step size. But apparently I'm supposed to get a circle when I plot V with respect to X which makes sense, but all I get is a straight line.

If I change it to:

Code (Text):
for i = 1:1000
V(i+1) = V(i)-(X(i).*s) ;
X(i+1) = V(i);
end
With s=0.8 I get a spiral, which looks like a development but I'm no closer to the circular shape that I am expecting. I think I just need a fresh pair of eyes to see where perhaps an obvious error lies.

2. Nov 18, 2016

### Staff: Mentor

In your example the sin(x) = x approximation has the understanding that x is in radians and that its true for small angles <10 degrees or so.

https://en.wikipedia.org/wiki/Small-angle_approximation

Also I'd try to hand calculate things or better yet have the computer print out x(i) and v(i) so you can see where it goes wrong.

EDIT:

I looked at your example, and it does produce a line by hand calculation the V goes 0.0 to -0.1 to -0.2 ... while the X values do the same but more abruptly 1, 0.0, -0.1, -0.2 ...

To get a circle, wouldn't V have to be something like $V(i+1) = sqrt(1.0 - x(i)*x(i))$
and then it wouldn't work when x(i) > 10 degrees?

Last edited: Nov 18, 2016