Stable points of a particle in a 2d potential field.

In summary, the conversation discusses a particle of mass m moving in 2d space in a potential V (x, y) = -1/2 kx2 + 1/2 λ0 x2y2 + 1/4λ1x, and how to determine the point (x0, y0) where the particle is in stable equilibrium. The equations ∂V/∂x=0 and ∂V/∂y=0 are used to find the solutions x=λ1/4 (k-y2 λ0) and λ0 x2 y =0, but it is discovered that these solutions do not satisfy the given conditions. It is suggested that this may be due to a typo
  • #1
sudipmaity
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Homework Statement



Let a particle of mass m moving in 2d space in a potential V (x, y) = -1/2 kx2 + 1/2 λ0 x2y2 + 1/4λ1x where k,λ01 > 0.At what point (x0, y0) is the particle in stable equilibrium? 2 marks

Homework Equations


∂V/∂x=0 ∂V/∂y=0; ∂2V/∂x2 > 0;∂2V/ ∂y2 > 0



The Attempt at a Solution


After using relevant equations I get x= λ1/4 (k-y2 λ0)
and λ0 x2 y =0.
If I take x or y=0 then the point become unstable.
Don't know what to do next.The problem however simplifies very much for a 2 mark question if the first term of the potential right side is positive.Does anybody think that it could be a typo error? Or is this problem solvable in the given form?
If the first term were to be positive then stable point would be (λ1 /4k , 0) .This would satisfy all the conditions.
 
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  • #2
That's funny: I get ## x = {\lambda_1/4 \over k-\lambda_0 y^2}## (but that's probably what you mean, right ?).
There is no taking x = 0 because it doesn't satisfy that equation. So all that's left is ##y = 0, x = {\lambda_1 \over 4k}## .
Indeed, at that point the second derivative wrt x is negative, so the answer is: nowhere.

I'm with you is suspecting this is a typo. Ask teacher !
 

1. What are stable points in a 2d potential field?

Stable points in a 2d potential field refer to points where a particle experiences a net force of zero, meaning it will remain at that point unless acted upon by an external force.

2. How are stable points identified in a 2d potential field?

Stable points are identified by looking at the gradient of the potential field. If the gradient is zero, it means there is no net force and the point is stable.

3. What is the significance of stable points in a 2d potential field?

Stable points are important because they represent points of equilibrium in the system. A particle at a stable point will remain there indefinitely unless acted upon by an external force.

4. Can stable points change in a 2d potential field?

Yes, stable points can change if the potential field itself changes. For example, if the shape of the potential field is altered, the location of stable points may also change.

5. How are stable points related to energy in a 2d potential field?

Stable points are related to energy in a 2d potential field because a particle at a stable point has minimum potential energy. This means that the particle is in a stable state and any movement away from the stable point would require an input of energy.

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