Small oscillations around equilibrium point in polynomial potential

[roflol12]

Hi guys i am a bit confused about this problem,
a particle of mass, m, moves in potential a potential u(x)=k(x⁴ - 7 x² -4x)
I need to find the frequency of small oscillations about the equilibrium point.

I have worked out that x=2 corresponds to the equilibrium point as

- dU/dx = F = -k(4x³ - 14x -4)=0 at x=2

I tried to solve using the Lagrangian with
L=1/2 m v² -k(x⁴ - 7 x² -4x)

using Euler–Lagrange equation d/dt(dL/dv)=dL/dx
to get: ma= -k(4x³ - 14x -4)
I thought about dropping the x^3 term as it would be very small with small oscillations but was confused what that left me with.

Any help appreciated!

 

[CompuChip]

Sounds like a good approximation.
What confused you?

We are talking about oscillations, right? If the oscillation frequency is $\omega$, what is the stereotypical solution that you try first?

 

[espen180]

The potential you gave has two equilibrium points, one at x=2 and one at x=-1.7071 .

Whenever you have an equilibrium point, you can generally approximate locally by an x² potential.

My suggestion is therefore to construct approximate k'(x-a)² potentials at the equilibria.

How small "small" oscillations are, depends of course on your resolution, and the value you find for the frequency of the oscillations may depend on the resolution.

 

[Vanadium 50]

The key is you can work using generalized coordinates. As espen180 points out, (x-a) is small if a is an equilibrium point. You can use the approximation (x-a) is small: not the approximation that x is small.

 

[roflol12]

think maybe i have solved it now

i took the ma= -k(4x^3 - 14x -4)
and rewrote x=2+r for a small r to get
ma= -k(4(2+r)^3 - 14(2+r) -4)
expanded to get,
ma= -k(34 r + 24 r^2 + 4 r^3)
considering r is small, i ignored the r^3 and r^2 terms to get
a=-34k/m r
which is shm with angular frequency, w^2=34k/m

look alright?

 

[espen180]

Looks good.

There is still the other equilibrium point though.

 

[roflol12]

Only interested in the x=2 equilibrium point :)
thanks all!