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

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SUMMARY

The discussion centers on determining the stable equilibrium point (x0, y0) for a particle of mass m in a 2D potential field defined by V(x, y) = -1/2 kx² + 1/2 λ0 x²y² + 1/4 λ1x, where k, λ0, and λ1 are positive constants. The necessary conditions for equilibrium, ∂V/∂x=0 and ∂V/∂y=0, lead to the equations x = λ1/4(k - y²λ0) and λ0x²y = 0. The participants conclude that if the first term of the potential were positive, the stable point would be (λ1/4k, 0), but the current form suggests instability, leading to the assertion that the problem may contain a typographical error.

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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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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 !
 

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