# What are the equations modelling a vertical spring system with two masses?

Modeling a vertical spring system with one mass is a pretty common problem. I looked around online and found some horizontal spring systems with two masses, but no examples of a vertical one.

I'm curious, how would you set up equations modeling a vertical spring system like this: $$----\\ \wedge \\ \vee \\ \wedge \\ \vee\\ (m_1)\\ \wedge\\ \vee\\ \wedge\\ \vee\\ (m_2)$$

Where the first spring has constant $k_1$, and the second $k_2$. I'll let $y_1(t)$ be the position of the top mass away from its equilibrium, and $y_2(t)$ the position of the bottom mass away from its equilibrium. I choose down to be the positive direction.

For the bottom mass, there is an upward force of $k_2y_2$, and a downward gravitational force of $m_2g$. So one equation should be $$m_2y_2''=-k_2y_2+m_2g$$ For the top mass, the first spring pulls up with force $-k_1y_1$ and a downward gravitational force of $m_1g$. I'm not sure how to account for the forces of the second spring and second mass acting of the first mass. Is the equation something like $$m_1y_1''=-k_1y_1+m_1g+\text{ something?}$$

I'm just curious how you would correctly set up the equations for this system. Thanks.

Let $k_i$ denote the spring constant for the spring above mass $m_i$, $i=1,2$. Then by Newton's Second Law $F=ma$ and Hooke's Law which says that the restorative force of the $y$ units beyond its equilibrium position, i.e. $F_\text{restorative}=-ky$, we get \begin{align} m_1y''_1&=\overbrace{-k_1y_1}^{\text{Hooke's Law from above}}\underbrace{-k_2(y_1-y_2)}_{\text{Hooke's Law from below}},\\ m_2y''_2&=\underbrace{-k_2(y_2-y_1)}_{\text{Hooke's Law from above}} \end{align} which is a linear coupled system in $y_1$ and $y_2$.
To see how Hooke's Law leads to these terms, first displace mass $m_1$ by $y_1$ units (taking "down" as the positive direction). Then Hooke's Law says spring 1 has a restoring force of $-k_1y_1$. Now pull mass $m_2$ down $y_2$ units. Since the total displacement of mass $m_2$ from its equilibrium position is $y_2-y_1$, this explains the Hooke's Law term $-k_2(y_2-y_1)$ in the second equation. Finally, to see where the second Hooke's Law term in the first equation comes from, just note that that will be the same force we just found for the second mass, $-k_2(y_2-y_1)$, but when it acts on the first mass, it will be in the opposite direction, so it will be $-k_2(y_1-y_2)$ which we can write as $+k_2(y_2-y_1)$ if you prefer.
• Thanks JohnD, can you explain how Hooke's law gives the terms $-k_2(y_1-y_2)$ and $-k_2(y_2-y_1)$? Are $y_1-y_2$ and $y_2-y_1$ displacements of something? (Sorry, it's been about 9 years since I took physics.) – Selena Tay Nov 22 '14 at 20:23
• Yes, $y_1$ and $y_2$ are the position of mass $m_1$ and m_2\$, respectively. I edited to add that explanation. – JohnD Nov 22 '14 at 20:41