Coupled differential equations problem

This is about a system with two pendulums of mass m hanging from strings of length l connected by a spring with spring constant k such that the spring is neutral when the pendulums are vertical. The displacement from equilibrium is x₁ and x₂.

The first part of the question asks me to derive the coupled differential equations governing this system in matrix form in the form d²x/dt²=Ax which I have done:

But the second part asks:

Substitute a solution in the form

x = \psi \cos (\omega t + \phi)

to arrive at an eigenvalue-eigenvector problem for the special "spatial configuration"

\psi

and eigenfrequencies

\omega

.

I genuinely do not understand how I'm supposed to go about substituting this solution into here. Could anybody give me some advice? Thanks.

(edited 8 years ago)

Reply 1

Gregorius

Original post by Plagioclase

But the second part asks:

Substitute a solution in the form

Unparseable latex formula:

x = \psi \cost(\omega t + \phi)

to arrive at an eigenvalue-eigenvector problem for the special "spatial configuration"

\psi

and eigenfrequencies

\omega

.

I genuinely do not understand how I'm supposed to go about substituting this solution into here. Could anybody give me some advice? Thanks.

Well that's a bit vague, isn't it! Is it simply asking you to guess a form of the function

\psi

to use as it says "a solution in the form...". If so, then just choose

x_1 = \cos(\omega_1 t + \psi_1)

and

x_2 = \cos(\omega_2 t + \psi_2)

Reply 2

atsruser

Original post by Plagioclase

But the second part asks:

Substitute a solution in the form

Unparseable latex formula:

x = \psi \cost(\omega t + \phi)

to arrive at an eigenvalue-eigenvector problem for the special "spatial configuration"

\psi

and eigenfrequencies

\omega

.

I genuinely do not understand how I'm supposed to go about substituting this solution into here. Could anybody give me some advice? Thanks.

This is incomprehensible as written. You have a normal mode problem, and the usual approach is to use a trial solution of the form

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} Ae^{i\omega t} \\ Be^{i \omega t} \end{bmatrix}

to create an eigenvalue problem - you will need a zero determinant for non-trivial solutions for A and B. If they're not talking about that, then I have no idea what they mean.

In this case, you should end up with two normal modes, one with both oscillators moving in-phase, and the other with them moving 180 out of phase. The general solution for the motion is an arbitrary linear combination of those modes.

Reply 3

ServantOfMorgoth

Is this further maths?

Reply 4

Plagioclase

Original post by Gregorius

Well that's a bit vague, isn't it! Is it simply asking you to guess a form of the function

\psi

to use as it says "a solution in the form...". If so, then just choose

x_1 = \cos(\omega_1 t + \psi_1)

and

x_2 = \cos(\omega_2 t + \psi_2)

Okay so I've done this...

\vec{x}'' = A\vec{x}

(\vec{\psi} \cos (\omega t + \phi))'' = A\cos (\omega t + \phi)

-\omega ^2 \vec{\psi} \cos (\omega t + \phi) = A \vec{\psi} \cos (\omega t + \phi)

- \omega ^2 \vec{\psi} = A \vec{\psi}

(A+\omega ^2 I)\vec{\psi} = 0

Is that right?

(edited 8 years ago)

Reply 5

atsruser

Original post by Plagioclase

Okay so I've done this...

\vec{x}'' = A\vec{x}

(\vec{\psi} \cos (\omega t + \phi))'' = A\cos (\omega t + \phi)

-\omega ^2 \vec{\psi} \cos (\omega t + \phi) = A \vec{\psi} \cos (\omega t + \phi)

- \omega ^2 \vec{\psi} = A \vec{\psi}

(A+\omega ^2)\vec{\psi} = 0

Is that right?

See here for some more info:

http://www.thestudentroom.co.uk/showpost.php?p=62655331&postcount=3

Reply 6

atsruser

Original post by ServantOfMorgoth

Is this further maths?

It's university level physics/applied maths.

Reply 7

ServantOfMorgoth

Original post by atsruser

It's university level physics/applied maths.

Yeah cause I did this last week and I'm doing mechanical engineering lol

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Reply 8

Plagioclase

Well I did it, turned out it wasn't as hard as I thought, I just wasn't thinking sensibly about it. Thanks for everyone's help, I'll just post my solutions here in case anyone else has the same confusion...

Reply 9

atsruser

Original post by Plagioclase

Cool - note that the eigenvectors indicate the two modes that I mentioned earlier - in-phase or out-of-phase motion - by way of the signs.

(BTW, you edited your earlier post a bit - it didn't make sense to me before, but I see what you are doing now. And did you ever get an answer to your entropy problem?)

Reply 10

Plagioclase

Original post by atsruser

Thanks. And no, I haven't yet. However, I have arranged a thermodynamics tutorial when I get back to uni so hopefully I'll get it cleared up then, thanks for asking!