# Coupled oscilators equation of motion

There is something physically I do not understand, consider this situation: Lets assume there is no friction. if we defind $$x_1$$ and $$x_2$$ the displacement of mass $$M_1$$ and $$M_2$$ respectively, we get the equations of motion: $$\begin{array}{l} M_1 \ddot{x}_{1}=-k x_{1}+k\left(x_{2}-x_{1}\right)=-2 k x_{1}+k x_{2} \\ M_2 \ddot{x}_{2}=-k x_{2}+k\left(x_{1}-x_{2}\right)=-2 k x_{2}+k x_{1} \end{array}$$

But I do not understand why for example when we look on mass $$M_1$$, where is the contribution of the most right spring?, if $$x_1=x_2$$ we get $$F_1=-kx_1$$ but $$M_2$$ got change it's location by $$x_2$$ so there must be force from the most right spring that should impact $$M_1$$ but according to equation of motion the only force is from the most left spring.

• How can you tell which spring is acting when all of them have the same constant $k$. Redo the equations with different spring rates to see which contributes to what. – JAlex Oct 21 '20 at 16:44

Too long for a comment. It's not clear what is troubling you.

The rightmost spring does not communicate with $$M_1$$; it affects the displacement $$x_2$$ of $$M_2$$, which is already factored in the extension or contraction of the middle spring.

If you assume $$x_1=x_2$$, the middle spring has not changed its equilibrium length, so it does not exchange any force between the two masses, which are then only pulled/pushed by the respective walls. For example, this could be true for a special charmed moment of the motion.

Response to comment: The rightmost spring only affects $$x_2$$. If the middle spring is unextended, rigid, it won’t pull on the left or the right. You have essentially demanded that the two masses are not coupled--you might as well take away the middle spring if it does nothing. Adjust your intuition.

• But intuitively, if $x1_=x_2$ and lets assume we move it to right side, then there will be net force on $m_1$ to the left side the middle spring will stay in same lengh but the right most spring will get smaller, thus he applies net force to the left side as well, Why does it not affect $m_1$ – Sagigever Oct 21 '20 at 6:10

If I understand your question, it might make things more clear to show how $$x_{1}$$ and $$x_{2}$$ are defined. The below figure, has been modified to show this. In particular, $$x_{1}$$ and $$x_{2}$$ are measured relative to the equilibrium positions of each mass. In other words, if the masses are not moving, then $$x_{1}=0$$ and $$x_{2}=0$$. As shown in the figure, both masses are somewhat displaced from their equilibrium positions.

I hope this helps. In an effort to understand the effect of each spring, assign them different stiffness. Also pose the problem in matrix form where each contribution coefficient is seen on its own. What I got was the following system of equations

$$\mathbf{M} \ddot{\boldsymbol{X}} = - \mathbf{K} \boldsymbol{X}$$ $$\begin{bmatrix} M_1 & 0 \\ 0 & M_2 \end{bmatrix} \pmatrix{\ddot{X}_1 \\ \ddot{X}_2 } = - \begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2+k_3 \end{bmatrix} \pmatrix{X_1 \\ X_2}$$

So the first row is the acceleration of mass (1) and it involves springs $$k_1$$ and $$k_2$$, and the second row is the acceleration of mass (2) and it involves springs $$k_2$$ and $$k_3$$.

I don't see where the paradox is. It very logical to me, as those are the springs contacting the corresponding masses.