How Do You Calculate Potential Energy in a 2D Oscillator System with Springs?

[forestmine]

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$^{2}$ with r$^{2}$ = x$^{2}$+ y$^{2}$.

Homework Equations

The Attempt at a Solution

I'm honestly at a loss with this problem. I know that the total force is F=-kr, where F$_{}x$ = -kx and F$_{}y$ = -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.

 

[fzero]

You've got things backwards: the force is the gradient of the potential, $\mathbf{F} = - \nabla U$. Also, the forces should involve the quantity $a-l_0$.

 

[forestmine]

Whoops, that was a bad mistake. Thanks for catching that.

Though I'm still not really sure what to do. My potential can then be written in terms of x and y components, right?

For instance,

U_x = 1/2 * k * rx^2, but I don't really see what r ought to be. I understand that there should be some dependence on l_0, but I honestly have no idea how. Unless we can essentially rewrite the equations for r_n, with (a-l_0)^2, (a+l_0)^2, etc.

Sorry, I'm so lost on this problem. I know that there are two methods, one of which involves a taylor expansion, and the other, a second derivative of the potential function, but again, I don't even know what my potential function looks like in this case.

 

[vela]

Set up the coordinate system so that the mass is sitting at the origin when it's in equilibrium and the springs lie along the axes. The end of the springs lying on the +x axis will be at the point (a,0). What's the potential energy in the spring when the mass is at (0,0)? What's the potential energy in the spring when the mass is at the point (x,y)?

 

[paranormal]

Hi there,

It looks like you are on the right track with your work so far. The potential energy for a two-dimensional oscillator can be written as U = 1/2 * k * r^2, where r^2 is the sum of the squared displacements in the x and y directions. In this case, r^2 = x^2 + y^2, as you correctly stated.

To find the potential energy for this specific system, we need to take into account the four springs that are attached to the puck. As you mentioned, the total force on the puck will be F = -kx(i) -ky(j), where i and j are unit vectors in the x and y directions, respectively. To find the potential energy, we need to take the negative gradient of this force, which will give us the potential energy as a function of position (x,y).

To do this, we can write the potential energy as U = -∂F/∂x * x - ∂F/∂y * y. Plugging in our expression for F, we get U = kx^2/2 + ky^2/2, which is equivalent to U = 1/2 * k * r^2, where r^2 = x^2 + y^2.

So, in summary, the potential energy for this system can be written as U = 1/2 * k * r^2, where r^2 = x^2 + y^2. This is the standard form for the potential energy of a two-dimensional oscillator, and it makes sense intuitively because the potential energy will increase as the displacement from the equilibrium position increases in either the x or y direction.

I hope this helps guide you in the right direction. Good luck with your studies!