Two Compressed Springs -> Unstable Equilibrium

[brentd49]

I will make a crude visualization of this system:

|-------------O--------------|
<-----a------><------a------>

Identical springs: k1=k2=k
Natural Length: l > a

The problem is to prove that the system is unstable.

Obviously, a slight movement directed off the horizontal axis will cause the springs to unstretch to a natural position vertically above or below the current position. The setup is arranged on a frictionless horizontal table.
I know that the second derivative of the potential energy will tell me about the stability, so I am trying to write down the potential energy. My problem is how to write down the 'x' for the two springs, i.e.

$$U(x) = \frac{1}{2} k x^2_1 + \frac{1}{2} k x^2_2 , x_1=x_2$$
$$U(x) = k x^2$$

I suppose it is just a geometry question, but I'm not sure to find that compressed length $$x$$.

 

[brentd49]

^bump
.......

 

[NateTG]

Let's say that the 'midpoint' is displaced upwards by some distance $d$. Can you calculate the length the springs would have then?

 

[brentd49]

so, if I take my orgin at the far left, with $$l_o$$ the natural length.

$$x = l_f - l_o$$
$$x = \sqrt{a^2 + d^2} - l_o$$

so, now I need to replace d, right?

$$d = \sqrt{(l_o + x)^2 - a^2}$$

but that can't be right, because I would have $$x = x(x)$$. I must be missing something.

 

[brentd49]

anyone.............

 

[Astronuc]

Presumably if the net force is greater than zero, in either up or down then the system is unstable.

What is the net force if O is displaced upward by d?

If the springs were constrained in the horizontal, then one could establish an equation for SHM with one spring a+x(t) and the other a-x(t).