- #1

- 203

- 0

## Homework Statement

A puck with mass m sits on a horizontal, frictionless table attached to four identical springs (constant k and unstreched length l_0). The initial lengths of the spring a are not equal to the unstretched lengths. Find the potential for small displacements x,y and show that it has the form 1/2 * k' * r[itex]^{2}[/itex] with r[itex]^{2}[/itex] = x[itex]^{2}[/itex]+ y[itex]^{2}[/itex].

## Homework Equations

## The Attempt at a Solution

I'm honestly at a loss with this problem. I know that the total force is

**F**=-k

**r**, where F[itex]_{}x[/itex] = -kx and F[itex]_{}y[/itex] = -ky.

I also know that my potential is minus the gradient of the force. If I were to take the gradient of F, where F = -kx(i) -ky(j), I get F= -k(i) -k(j). Not really sure where to go from here, or if I'm on the right track for that matter.

I'm obviously not looking for the answer, just some help in the right direction. I don't think I fully understand it conceptually to be able to work it analytically. Any tips would be greatly appreciated!

EDIT// Here's some work I've done since posting. Still unsure of how to continue.

To account for all possible positions of the spring,

r1^2=x^2 + (a-y)^2

r2^2=x^2 + (a+y)^2

r3^2=(x+a)^2 + y^2

r4^2=(x-a)^2 + y^2

Now, r will be some variation of the above, so I summed the above 4 equations. I would then think to use U=1/2 * k * r^2, where r^2 is the sum of the above equations. I feel like I'm missing something, though.

EDIT 2// In fact, I definitely don't feel as though that's suitable, because it says that the potential is not at all dependent on l, which of course is false.

Last edited: