How to derive duffing equation for a particular system?

1. Jan 31, 2013

great_sushi

Say I have a mass m on a non linear spring k with some damping b.

I start with the restoring force of the spring F=-kx+x^3........ the x^3 is the non linearity.

Set that equal to newtons second law F=mx'' = -kx+x^3

Add in the damping which is dependent of velocity bx'

............................mx'' = -kx + x^3 + bx'
............................= x'' + k/m*x + x^3 + bx'
............................= x'' + bx' + w0*x + x^3............ Now because my system is driven I add in a periodic force dependent on t.

Fcos(wt) = x'' + bx' + w0*x + x^3

Is this the correct method? I may have gotten mixed up with my signs I tend to do that :(

2. Jan 31, 2013

TSny

Your method is basically correct. But, yes, you do need to be careful with the signs.

Generally, there would be a constant factor in front of the x3 term: βx3

I would recommend going ahead and putting in the driving force at the beginning.

The damping force is usually written -bx' rather than +bx' since the force opposes the velocity.

So, mx'' = -kx +βx3 -bx' + Fcos(ωt)

Then you can easily rearrange the terms to get Fcos(ωt) alone on one side.

You could just as well write the nonlinear part as -βx3 so that

mx'' = -kx -βx3 -bx' + Fcos(ωt)

Then you will get all positive terms when you solve for Fcos(ωt).

[Edit: Be careful when dividing through by the mass m. All terms get divided by m. You should get the Duffing equation .]

Last edited: Jan 31, 2013
3. Jan 31, 2013

great_sushi

Oh thank you very much!
Apologies for asking the question twice, i didn't know if it was advanced or introductory..

So if I integrate the duffing equation with respect to x will I get the potential energy? Because the work = force* distance and so the force = work / distance.

4. Jan 31, 2013

TSny

Potential energy is only defined for forces that depend on position alone. (When you have motion in more than one dimension, there is an additional restriction.) Since you have forces that depend on velocity (damping force) and on time (driving force), there will not be a potential energy for the total force. However, there is a potential energy associated with the force -kx - βx3 which you can find by integrating with respect to x.

5. Feb 1, 2013

great_sushi

Ahhh I see that makes sense. So the integral of -kx + βx^3 = -1/2kx^2 + 1/4βx^4 = V(x) and in this case, that gives a double well potential.

6. Feb 1, 2013

TSny

The potential energy is the negative of the integral of the force. So, the signs in V(x) would be opposite. The shape will depend on whether β is positive or negative.

7. Feb 2, 2013

great_sushi

Oh great! I was wondering why my signs were the wrong way round. I didnt realise it was a negative integral.
Thank you very much, you have been extremely helpful :)

8. Feb 5, 2013

great_sushi

Can you tell me where the cubic term comes from? I know that it an odd power series but why?
Thanks