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coupled pendula

StarvingAutist

this is my working:
http://imgur.com/Z8aqSLv&yUVHKS0

i have no real idea what i'm doing

Reply 1

Gregorius

Original post by StarvingAutist

this is my working:
http://imgur.com/Z8aqSLv&yUVHKS0

i have no real idea what i'm doing

I've not worked through this, but I notice that you have immediately changed variables to

x-y

and

x+y

. Have you tried following literally the instructions of the question and working directly with

x, y

via

x = x_0 \cos(\omega t + \alpha)

and

y= y_0 \cos(\omega t + \beta)

?

A quick google suggested to me that this is the standard way to go for the coupled pendulum!

Reply 2

atsruser

Original post by StarvingAutist

this is my working:
http://imgur.com/Z8aqSLv&yUVHKS0

i have no real idea what i'm doing

This looks like a normal modes problem which is the kind of problem which arises when you couple oscillators together. It turns out that you can find a number of independent modes of oscillation of the system, and the general motion of the system is a linear combination of those modes.

A "mode" is a motion where, say, all of the masses are moving in the same direction at the same time, then in terms of your problem, the expression

x+y

is useful, or when 1/2 the masses are moving in the -ve direction and 1/2 in the +ve direction, then the expression

x-y

is useful.

These expressions are "useful" because they allow you to write DEs purely in terms of those expressions e.g. you get say:

\ddot{x}+\ddot{y} = k(x+y)

\ddot{x}-\ddot{y} = k(x-y)

and you can introduce new variables

p=x+y,q=x-y

and get the decoupled DEs

\ddot{p}=kp \\ \ddot{q}=kq

When you do this, the

p,q

(or their

x,y

equivs) are called the normal coordinates of the system. They are the combinations of the original coords that allow you to write solveable DEs.

To formalise this, you usually write things in matrix form so you turn your eqns into, say,

\begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} =\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}

(*)

and then take a trial solution of:

\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} A \\ B \end{pmatrix} e^{i \omega t}

When you plug this into (*), you get an eigenvalue problem, whose eigenvectors are the A,B column vector, and whose eigenvalues give you the fundamental frequencies of vibration of the system (i.e. you will get a set of possible values for

\omega

- those are the fundamental frequencies of the normal modes).

It turns out that the eigenvectors specify the combinations of

x,y

which give you your normal modes, so for this problem, you would expect to find an eigenvector:

\begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

You'll note that the process of solving for the eigenvectors only gives the ratio of A to B, rather than their specific values.

At the end, you form a general linear combination of the normal modes to give the general solution, and at that point you can combine the complex