Stable equilibium of a potential

In summary, the given problem involves a particle with a potential energy function U(x) = U_0[2(x/a)^2 - (x/a)^4], where U_0 and a are positive constants. The questions asked are (i) the angular frequency w of oscillations about the stable equilibrium point, (ii) the minimum speed required for the particle to escape to infinity, and (iii) the position of the particle at time t=0 with a positive velocity equal to the escape speed. The solution involves understanding the potential function and equating it to the kinetic energy at the maxima of the graph, as well as finding the angular frequency of small oscillations around x/a=0.
  • #1
rbwang1225
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0

Homework Statement


A particle of mass moving in one dimension has potential energy ##U(x)=U_0[2(x/a)^2-(x/a)^4]##, where #U_0# and ##a## are positive constants.
(i)What is the angular frequency ##w## of oscillations about the point of stable equilibrium?
(ii)What is the minimum speed the particle must have at the origin to escape to infinity?
(iii)At ##t=0## the particle is at the origin and its velocity is positive and equal in magnitude to the escape speed of part (d). Find ##x(t)##

The Attempt at a Solution


I have no idea about (i), though I know the stable equilibrium point is x=0.
My result of (ii) is ##\sqrt{\frac{-2U_0}{m}[2(x/a)^2-(x/a)^4]}##
But, here comes the problem. The speed at x=0 is 0, so it must stay at the origin forever.
I might get something wrong in obtaining the expression of (ii).
Any help would be appreciated.
 
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  • #2
Plot the potential function. From that, you can understand what energy the particle needs to escape from the potential well.
As for (i) you need to find the angular frequency of small oscillations around x/a=0. If it is small, the fourth power can be ignored with respect to (x/a)^2 and the potential energy function approximates that of a spring.
I do not understand qestion (iii) What is x(x)? was not that v(x) instead?

ehild
 
  • #3
The graph would be something like this ,
the potential at maxima (x=+-a) needs to be equated to KE
you will get velocity (min)
 

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Related to Stable equilibium of a potential

What is a stable equilibrium of a potential?

A stable equilibrium of a potential is a state in which a system is in balance and has minimal energy. This means that any slight disturbance to the system will cause it to return to its original state without any additional energy input.

How is a stable equilibrium different from an unstable equilibrium?

A stable equilibrium is characterized by a system returning to its original state after a disturbance, while an unstable equilibrium is a state in which the system will not return to its original state after a disturbance, but will instead move towards a new equilibrium point.

What factors affect the stability of a potential equilibrium?

The stability of a potential equilibrium is affected by the shape of the potential curve, the strength of the forces acting on the system, and the presence of any external factors such as friction or air resistance.

How does the concept of stable equilibrium apply to real-world systems?

The concept of stable equilibrium applies to many real-world systems, such as pendulums, atoms, and planets in orbit. In these systems, the equilibrium is maintained by the balance between the forces acting on the system.

What are some practical applications of studying stable equilibrium of a potential?

Studying stable equilibrium of a potential has many practical applications, such as in engineering, where it is used to design stable structures, and in physics, where it is used to understand the behavior of various systems. It also has applications in economics, biology, and other fields.

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