# Potential energy of mass spring system

1. Feb 22, 2010

### seanryan

Ok, so the system is a ring with four identical springs connected to a single mass. Each spring is in a different quadrant and at an angle alpha from the horizontal. L is the relaxed length of each spring, k' is the spring constant, and R is the radius of the ring. and ignore gravitational and frictional forces.
See diagram

The mass is oscillating in the x direction only.

I am trying to find and equation for the potential energy and the sum of the forces from each spring but I can never get it right.

The paper that this came from is "mechanical analogs to the landau zener model" by Shore et al. Init he has the potential energy as

$$U=k'(2(R-L)^{2} + 2(1-\frac{L}{R}sin^{2}(\alpha))*x^{2})$$

I've tried finding the displacement and putting it into $$U=\frac{1}{2} kx^{2}$$ and ive tried finding the sum of the forces and $$U=-\int {F \bullet dx}$$. Neither gave me anything close to the paper.

If anyone could help me out that would be great. Thanks.

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Last edited: Feb 22, 2010
2. Feb 22, 2010

### turin

What did they give you? Without telling us that, it's difficult to figure out what is confusing you, but I suspect that you are not interpreting those expressions properly in terms of x. Remember:
F=0 when the spring is relaxed.
There are four distinct springs to consider.
According to your drawing, "x" is not the displacement of a given spring from its relaxed state.

3. Feb 22, 2010

### seanryan

What I first did was imagine that the mass was displaced a distance d to the right. Then I found what the new length of the springs are. For the left 2 springs i got
$$r_{1}^{2}= R^{2}+d^{2}+2Rdcos(\alpha)$$

for the right
$$r_{2}^{2}=R^{2}+d^{2}-2Rdcos(\alpha)$$

I then broke these into vector components. The Y components cancel and leave me with
$$F= -2k(r_{1}-L)(\frac{R}{r_{1}}cos(\alpha)+\frac{d}{r_{1}})-2k(r_{2}-L)(\frac{R}{r_{2}}cos(\alpha)-\frac{d}{r_{2}})$$

I plugged in the r2 and r1 then I integrated this with respect to d and got

$$U=k[2L*(\sqrt{R^{2}+x^{2}+2Rxcos(\alpha)}+\sqrt{R^{2}+x^{2}-2Rxcos(\alpha)})-2x^{2}]$$

Last edited: Feb 23, 2010
4. Feb 23, 2010

### seanryan

I also tried taking the x component out of the force then squaring it and putting that into 1/2kx^2 but that didn't work either

5. Feb 23, 2010

### turin

Two problems: 1) you have a sign error in your expression for force, and 2) apparently that paper assumes that d << R (neglecting terms of order (d/R)2 and higher). Are you familiar with Taylor series?

6. Feb 23, 2010

### seanryan

Ok, so the sign error just makes the potential energy

$$U=k[-2L*(\sqrt{R^{2}+x^{2}+2Rxcos(\alpha)}+\sqrt{R^{2}+ x^{2}-2Rxcos(\alpha)})+2x^{2}]$$

If I expand this I would get

$$-4kLR-2k\frac{L}{R}sin^{2}(\alpha)x^{2}$$

I can account for the 1-L/R by not expanding the x^2 so I have

$$k[2(-2LR)+2(1-\frac{L}{R}sin^{2}(\alpha))x^{2}]$$

but now I have to get -2LR to be (R-L)^2

If I subtract a 2L^2 and a 2R^2 I get (R^2+L^2-2LR) which reduces to (R-L)^2

so now I have

$$k[2(R-L)^{2}-2L^{2}-2R^{2}+2(1-\frac{L}{R}sin^{2}(\alpha))x^{2}]$$

So how do I get rid of the extra terms or what did I do wrong to get the extra terms?

They only change the potential energy by a constant and do not change the effective spring constant but I still would like to match the paper

7. Feb 23, 2010

### turin

If the only difference in potential energy is a constant, then you have the same physical result.

Probably, in the paper they simply started with the known form of the elastic potential energy:

$$U=k^\prime\left(\left(r_1-L\right)^2+\left(r_2-L\right)^2\right)$$

whereas you approached the problem with a more physical motivation by considering the actual physical meaning of the potential energy as the work that is required against the force in order to achieve the given configuration. This means that, for you, the potential energy when the mass is in the exact center is zero, whereas, in the paper, the potential energy at the center is some constant positive value. Neither choice is incorrect, unless you want to make other considerations.

8. Feb 23, 2010

### seanryan

Oh, thats how they did it. Significantly less work than what I was doing. Thanks for all your help.