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$D$ is the stiffness of the springs, and $l_0$ is their original length, ,with $l>l_0$. My try was this:

$$m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=-D(r+(l-l_0))\cos{\alpha}+D((l-l_0)-r)\cos{\alpha}=-2Dr\cos{\alpha}$$ $$m\frac{\mathrm{d}^2y}{\mathrm{d}t^2}=-Dr\sin{\alpha}$$

Where alpha is the angle between the spring and the horizontal axis, and $r^2=x^2+y^2$, but I can't solve it.

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  • $\begingroup$ What is $\alpha$? Which angle does it represent? $\endgroup$ – Gert Nov 19 '17 at 17:51
  • $\begingroup$ @Gert it's the angle between the horizontal axis and the spring, so it' 0 when the spring is fully horizontal, and pi/2 when vertical. $\endgroup$ – Botond Nov 19 '17 at 18:00
  • $\begingroup$ For horizontal displacement the effective spring constant is just the sum of the two along constants. For vertical displacements the answer is given here $\endgroup$ – Floris Nov 19 '17 at 19:05
  • $\begingroup$ @Floris I don't think so. I think the vertical movement depends on the horizontal, and vice versa. $\endgroup$ – Botond Nov 19 '17 at 19:36
  • $\begingroup$ Vertical and horizontal motion are two normal modes. For small displacement the behavior is linear so there is no mixing. $\endgroup$ – Floris Nov 19 '17 at 20:08