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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.

Botond
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  • What is $\alpha$? Which angle does it represent? – Gert Nov 19 '17 at 17:51
  • @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. – Botond Nov 19 '17 at 18:00
  • For horizontal displacement the effective spring constant is just the sum of the two along constants. For vertical displacements the answer is given here – Floris Nov 19 '17 at 19:05
  • @Floris I don't think so. I think the vertical movement depends on the horizontal, and vice versa. – Botond Nov 19 '17 at 19:36
  • Vertical and horizontal motion are two normal modes. For small displacement the behavior is linear so there is no mixing. – Floris Nov 19 '17 at 20:08
  • @Floris so $\sqrt{x^2+y^2}=x$ and also $\sqrt{x^2+y^2}=y$? – Botond Nov 19 '17 at 20:11
  • That is not what I said. The net horizontal force is unchanged by a small vertical displacement and the net vertical force is unchanged by a small horizontal displacement given that the springs are already under tension in equilibrium (that is an important condition!) – Floris Nov 19 '17 at 20:14
  • @Floris This does not seem mathematically correct to me, based on my equations. Or my equations should be bad. – Botond Nov 19 '17 at 20:19
  • If you assume displacement is $dx, dy$ then the new length is $L_1\sqrt{(L+dx)^2+dy^2}$. When you expand this and ignore higher order terms you can see $L_1-L = L(1+\frac{2dx}{L} + \left(\frac{dx}{L}\right)^2+\left(\frac{dy}{L}\right)^2)^{1/2} - L \approx dx$ if you ignore higher order terms (which is ok when $dx \ll L$ and $dy \ll L$ - "small displacements"). Then you expand these forces in the horizontal and vertical directions and you find the force in $x$ does not depend on $y$ and vice versa. Did you look at the derivation in the linked duplicate? – Floris Nov 19 '17 at 22:35
  • @Floris yes, several times, and I've just seen your update. I will try to compute later based on your answers.The Tension T is the force in the spring when the mass is in the equilibrium, so $T=k*(L-L_0)$? You've left an additional dollar sign in your update, but I can't remove it. – Botond Nov 20 '17 at 08:00
  • @Botond thanks for pointing out the typo - fixed now. Yes the tension changes a little bit but as I showed the first order effects allow separation of the X and Y motion without coupling. – Floris Nov 20 '17 at 08:04
  • @Floris But what exactly is the tension? The force in the spring when the mass is in the equilibrium? – Botond Nov 20 '17 at 08:15
  • It is the actual force in the stretched spring. For the purpose of computing the y force, the change in tension cam be neglected. For the net force in X, it cannot - but only the X displacement matters (first order in dX, second order in dY) – Floris Nov 20 '17 at 13:57
  • @Floris Thank you very much, I think I've got it. Does the $F_y=2T \frac{dy}{L}$ equals with $2k*dy$? – Botond Nov 20 '17 at 16:53

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