Two connected springs and potential energy as a function of x and y

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SUMMARY

The discussion focuses on the potential energy of two connected springs with natural length 'a' and spring constant 'C', as a function of two-dimensional displacement (x, y). The potential energy is derived using Hooke's Law, resulting in the equation U(x,y) = C/2((√((a+x)² + y²) - a)² + (√((a-x)² + y²) - a)²). The force vector is calculated using the gradient of the potential energy, expressed as \vec{F} = -\vec{\nabla} U. The solution presented is confirmed to be correct by participants in the discussion.

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Homework Statement


Two springs each of natural length a and spring constant C are connected at one end
(see figure). Consider a two dimensional displacement given by [itex](x, y)[/itex]
(a) Write the potential energy as a function of x and y.
(b) Find the force vector for a given [itex](x, y)[/itex] pair.
springs.jpg



Homework Equations


Hooke's Law. Potential Energy.


The Attempt at a Solution



a) The stretch lengths of A and B springs Da and Db are
[tex]D_A = \sqrt{(a+x)^2 + y^2} - a[/tex]
[tex]D_B = \sqrt{(a-x)^2 + y^2} - a[/tex]
Since potential energy of a spring is
[tex]U_{spring} = 1/2kx^2[/tex]
The total potential energy U can be written by
[tex]U(x,y) = U_A + U_B = C/2(D_A^2+D_B^2) = c/2((\sqrt{(a+x)^2 + y^2} - a)^2 + (\sqrt{(a-x)^2 + y^2} - a )^2)[/tex]
b) [tex]\vec{F} = -\vec{\nabla} U[/tex] and etc

Would you check my solution ? Is my answer correct ? Thanks for help in advance.
 
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