Are Oscillation Periods in a Quartic Potential Always Equal?

In summary: From there, you can start to plot these values as a function of x_1 and see if there is a correlation between the two values.
  • #1
PhyConnected
13
0

Homework Statement


Consider a quartic potential,
i.e. [tex]V(x) \equiv ax^4 + bx^3 + cx^2 + dx + e[/tex]
s.t. there are two local minimums for the potential.
233px-Polynomialdeg4.svg.png

For a given particle with energy E, prove that the period of oscillation around the two minimums are the same.

Homework Equations


[tex]dt \equiv \frac{dx}{\sqrt{(\frac{2} {m}) E-V(x)}}[/tex]
I suppose?

The Attempt at a Solution


No clue at all, seems impossible to evaluate the integral above directly?

P.S. This is not a homework/coursework question but rather "challenge" type question.
Thanks :smile:
 
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  • #2
PhyConnected said:

Homework Statement


Consider a quartic potential,
i.e. [tex]V(x) \equiv ax^4 + bx^3 + cx^2 + dx + e[/tex]
s.t. there are two local minimums for the potential.
233px-Polynomialdeg4.svg.png

For a given particle with energy E, prove that the period of oscillation around the two minimums are the same.

Homework Equations


[tex]dt \equiv \frac{dx}{\sqrt{(\frac{2} {m}) E-V(x)}}[/tex]
I suppose?

The Attempt at a Solution


No clue at all, seems impossible to evaluate the integral above directly?

P.S. This is not a homework/coursework question but rather "challenge" type question.
Thanks :smile:

I think the easiest way to do this would be to use the Lagrangian (have you learned Lagrangian mechanics yet?) to get the equations of motion. Then since you want to look at (small) oscillations around the local minima of the potential, assume one of your two local minima is at [itex]x=x_1[/itex] and see what your equations of motion look like for [itex]x(t)=x_1+\epsilon(t)[/itex].
 
  • #3
gabbagabbahey said:
I think the easiest way to do this would be to use the Lagrangian (have you learned Lagrangian mechanics yet?) to get the equations of motion. Then since you want to look at (small) oscillations around the local minima of the potential, assume one of your two local minima is at [itex]x=x_1[/itex] and see what your equations of motion look like for [itex]x(t)=x_1+\epsilon(t)[/itex].
Hi, thanks for the quick respond!
No, unfortunately I haven't learned Lagrangian mechanics.
I've tried using linear approximation by Taylor's Theorem to find a expression for SHM, but that doesn't seem to help since there's no simple expression to the root for a general case.
More importantly, the question does not say that it has to be a small oscillation and the actual figure that comes with the question (not this one) indicates that the energy level is well above the potential minimum.
I suppose the only constrain is that the energy doesn't exceed the local maximum so that there is a turning point.

Thanks again!
 
  • #4
PhyConnected said:
No, unfortunately I haven't learned Lagrangian mechanics.

I don't think that will be a problem. Newtonian mechanics should be fine here. What is the force on the particle in terms of the potential?

I've tried using linear approximation by Taylor's Theorem to find a expression for SHM, but that doesn't seem to help since there's no simple expression to the root for a general case.

Taylor expansion will be useful here.

More importantly, the question does not say that it has to be a small oscillation and the actual figure that comes with the question (not this one) indicates that the energy level is well above the potential minimum.
I suppose the only constrain is that the energy doesn't exceed the local maximum so that there is a turning point.

Certainly, the oscillations must not take the particle past the local maximum in between the two minima or the particle will no longer be oscillating around the same minima, so they can't be too large.
 
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  • #5
gabbagabbahey said:
I don't think that will be a problem. Newtonian mechanics should be fine here. What is the force on the particle in terms of the potential?
Taylor expansion will be useful here.
Certainly, the oscillations must not take the particle past the local maximum in between the two minima or the particle will no longer be oscillating around the same minima, so they can't be too large.
So [tex]F(x) \equiv 4ax^3 + 3bx^2 + 2cx + d[/tex]
and so around for small x around the equilibrium,
By Taylor Expansion,
[tex]F(x) \equiv F'(eq)x + H.O.T.[/tex]
How should I continue from here?
It doesn't seem to be quite true that the two roots of the F(x) is symmetric with respect to the local/global minimum of the F'(x).
(s.t. F'(x) is the same at for the two equilibrium points.)
Thanks.
 
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  • #6
PhyConnected said:
So F(x) is simply 4x^3 + 3bx^2 + 2cx + d

I think you are off by a negative sign, and I would rewrite this as

[tex]m \ddot{x} = -\frac{d V}{ d x}[/tex]

and so around for small x around the equilibrium,
F(x) = F'(x_equilibrium) * x
How should I continue from here?

I would actually start by writing out the full Taylor expansion of the right hand side of the above equation around one of your local minima [itex]x(t)=x_1+\epsilon(t)[/itex] (your potential is a quartic, so you only need to go up to the [itex]\frac{d^5 V}{ d x^5}[/itex] term). If [itex]x_1[/itex] is a local minimum, what can you say about [itex]\left. \frac{d V}{ d x} \right|_{x = x_1}[/itex] and [itex]\left. \frac{d^2 V}{ d x^2} \right|_{x = x_1}[/itex]?
 
  • #7
gabbagabbahey said:
I think you are off by a negative sign, and I would rewrite this as

[tex]m \ddot{x} = -\frac{d V}{ d x}[/tex]
I would actually start by writing out the full Taylor expansion of the right hand side of the above equation around one of your local minima [itex]x(t)=x_1+\epsilon(t)[/itex] (your potential is a quartic, so you only need to go up to the [itex]\frac{d^5 V}{ d x^5}[/itex] term). If [itex]x_1[/itex] is a local minimum, what can you say about [itex]\left. \frac{d V}{ d x} \right|_{x = x_1}[/itex] and [itex]\left. \frac{d^2 V}{ d x^2} \right|_{x = x_1}[/itex]?
I'm aware how to find the period for a S.H.M., but how would I find the period for a non-linear differential equation after the full Taylor Expansion?
So obviously [tex] \left. \frac{d V}{ d x} \right|_{x = x_1} [/tex] is zero at x= x1
but I don't really see any significance of [tex]\left. \frac{d^2 V}{ d x^2} \right|_{x = x_1}[/tex]
 
  • #8
PhyConnected said:
I'm aware how to find the period for a S.H.M., but how would I find the period for a non-linear differential equation after the full Taylor Expansion?
So obviously [tex] \left. \frac{d V}{ d x} \right|_{x = x_1} [/tex] is zero at x= x1
but I don't really see any significance of [tex]\left. \frac{d^2 V}{ d x^2} \right|_{x = x_1}[/tex]

After thinking about his problem a little more, I don't see why it would be true in general that the oscillations about the 2 minima must have the same period. Is there more to the problem than what you've posted? (If it is taken from a textbook, please give the problem number and name of the text).
 
  • #9
gabbagabbahey said:
After thinking about his problem a little more, I don't see why it would be true in general that the oscillations about the 2 minima must have the same period. Is there more to the problem than what you've posted? (If it is taken from a textbook, please give the problem number and name of the text).
It also confuses me since this result certainly seems non-trivial. (somewhat counter-intuitive by considering extreme cases)
I found the question from some local university physics exam.
"www.cap.ca/" + "sites/cap.ca/files/UPrize/cap_2008.pdf"
Thanks:smile:
 

FAQ: Are Oscillation Periods in a Quartic Potential Always Equal?

What is oscillation in a quartic potential?

Oscillation in a quartic potential is a type of periodic motion in which a system moves back and forth between two points in a quartic potential energy function.

How is a quartic potential related to oscillation?

In a quartic potential, the potential energy is proportional to the fourth power of the displacement from equilibrium. This results in a restoring force that is proportional to the displacement, making the system oscillate back and forth.

What is the equation for the motion of a particle in a quartic potential?

The equation for the motion of a particle in a quartic potential is given by Newton's second law: F = ma = -kx - cx3, where k is the spring constant and c is a constant related to the shape of the potential.

What are some real-life examples of oscillation in a quartic potential?

One example is a mass-spring system with a quartic potential energy function, in which the mass oscillates back and forth between two points. Another example is a simple pendulum with a quartic potential energy function, in which the pendulum bob oscillates back and forth.

How does the amplitude of oscillation change in a quartic potential?

The amplitude of oscillation in a quartic potential depends on the initial conditions and the energy of the system. As the energy increases, the amplitude of oscillation also increases. However, if the energy is too low, the system may not oscillate at all.

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