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I need to find the equations of motion for the following system.

enter image description here

If $x_1$ is $m_1$'s extension and $x_2$ is $m_2$'s, then, I feel like for $m_1$ we just need to consider $x_1$ giving $$m_1 a_1 = x_1 K_C + x_1 K _B$$ because unless the extensions in both the left and right spring isn't the same, $m_1$ won't remain horizontal,

And for $m_2$, we just need to consider $x_2$ and $x_1$ giving $$m_2 a_2 = K_B (x_2 - x_1) + K_A x_2$$ because both the upper and lower springs exert forces on $m_2$.

But sadly, this isn't correct as this gives all 0 normal modes. Any help on where I went wrong would be really appreciated.

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  • $\begingroup$ In the first formula, the spring force of spring B is a function of $x_2$ as well. For example, the force in spring B would be larger if $x_2$ is 0, since the spring is stretched $\endgroup$
    – ROIMaison
    Sep 22, 2014 at 8:38
  • $\begingroup$ Does that mean it should be $m_1 a_1 = (K_C + K_B) x_1 + K_B x_2$? Is the second equation already correct? Thanks. $\endgroup$
    – Katie Adams
    Sep 22, 2014 at 8:42
  • $\begingroup$ It's the difference between $x_1$ and $x_2$ that's important for the spring force of $B$ (and for any spring that is ^^), as such, try writing it like $(x_1-x_2)$, or $(x_2-x_1)$, depending on your coordinate system, and see what equation comes out. $\endgroup$
    – ROIMaison
    Sep 22, 2014 at 8:50
  • $\begingroup$ Do you mean for all the springs above? Because that also gives all zero normal modes. $\endgroup$
    – Katie Adams
    Sep 22, 2014 at 9:05
  • $\begingroup$ The term in $x_1$ should come with a minus sign in the equation for $\ddot{x_1}=a_1$. The same for $x_2$ and $a_2$. Otherwise, your equation has no oscillating solution. Another point: the two masses are attached to the spring $B$. From Newton's laws, they must feel opposite forces from the spring. $\endgroup$
    – Tom-Tom
    Sep 22, 2014 at 10:19

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Too long for a comment, so in an answer:

Not all the springs are a function of $x_1$ and $x_2$, only spring $K_b$ is a function of both $x_1$ and $x_2$. Spring force is a function of how much a spring is stretched, e.g. how much difference er is between the beginning and the end of a spring, so $x_{begin}$-$x_{end}$ or $x_{top}$-$x_{bottom}$ for this case.

For spring A, the value of $x_{top}$ is fixed, so the spring force of A is a function of $x_{bottom}$ only, or in this case $x_2$. A similar reasoning can be applied to spring C, which only has a variable $x_{bottom}$, which is $x_1$. However, spring B has a moving $x_{top}$ and $x_{bottom}$, thus making this spring force a function of $x_2$ and $x_1$. You did recognize for your second equation, but failed to do so with the first equation. I have the feeling your making a mistake with positive and negative distances. Try to keep your coordinates consistent, and when in doubt, sanity check for physics. If, for example a mass moves, which direction will the spring force act? This is determined by the sign of the force in your equation.

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  • $\begingroup$ That means, $a_2$ should be a function of $(x_2 - x_1)$ and $x_2$ with $K_B$ and $K_A$ respectively. And $a_1$ should be a function of $x_1$ with $K_C$ and of $(x_1 - x_2)$ with $K_B$. Is this correct? Thanks! $\endgroup$
    – Katie Adams
    Sep 22, 2014 at 16:17
  • $\begingroup$ It all depends on your coordinate system. What I usually do is to first define a coordinate system, say e.g. down is postive for both forces and distances, and the fixation is at $x$ is zero . The distances (and their differences) are then fixed completely, e.g. $K_a \cdot (x_2), K_c \cdot (x_1), K_b \cdot (x_1-x_2)$. Subsequently you can determine the sign of the values of the $K$'s. For example, if you increase $x_2$ (thus down) and keep the rest constant, what will happen to the force of spring $A$, which way will it point? This should agree with the value of $(K_a \cdot (x_2))$. $\endgroup$
    – ROIMaison
    Sep 23, 2014 at 5:53

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