# Understanding the Terms in Coupled Springs Differential Equation

I am teaching differential equations and I got myself totally confused about the physics of a problem.

Consider a coupled spring system in series: there is a mass $m_1$ on a horizontal track which is connected to a wall by a spring (with natural length $L_1$ and spring constant $k_1$). Also attached to the first mass is a second mass $m_2$ on the same horizontal track and is connected to the first mass by a spring (with natural length $L_2$ and spring constant $k_2$).

The reference that I'm using goes through the standard derivation where $x_1$ is the displacement of the first mass and $x_2$ is the displacement of the second mass, deriving \begin{align*} m_1\frac{d^2x_1}{dt^2}&=-(k_1+k_2)x_1+k_2x_2\\ m_2\frac{d^2x_2}{dt^2}&=k_2x_1-k_2x_2. \end{align*}

I wanted to rewrite this system in terms of the stretch/compression of each spring. In particular, $y_1=x_1$ is the displacement of the first spring from its natural length and $y_2=x_2-x_1$ is the displacement of the second spring from its natural length. Substituting these into the differential equations, we get \begin{align*} m_1\frac{d^2y_1}{dt^2}&=-k_1y_1+k_2y_2\\ m_2\left(\frac{d^2y_1}{dt^2}+\frac{d^2y_2}{dt^2}\right)&=-k_2y_2. \end{align*} The first equation makes sense to me (in terms of the net force on the first mass), but I don't see where the force $$m_2\frac{d^2y_1}{dt^2}$$ is coming from. I tried to think of the second half of the system moving rigidly as $m_1$ moves, but this didn't lead to this differential equation.

TL;DR

What is the physical significance of the $$m_2\frac{d^2y_1}{dt^2}$$ term?

When you apply Newton's laws of motion you are assuming (usually without even thinking about it) that the frame(s) of reference are inertial frames.

In this case you can think of your coordinates $x_1, \, x_2$ and $y_1$ being measured in inertial frames of reference which are all fixed to the Earth.

However note that your coordinate $y_2$ has its zero referenced to a position $x_1$ which is accelerating relative to the Earth.

You therefore cannot directly apply Newton's laws of motion in that non-inertial frame of reference in which $y_2$ is measured.

In that non-inertial frame of reference you can measure a position coordinate $y_2$ of mass $m_2$ and you know the force applied by the spring on that mass $-k_2y_2$.

To use Newton's laws a pseudo force of magnitude $-m_2\ddot y_1 \,(=-m_2 \ddot x_2)$ must be introduced.

Applying Newton's second law in that non-inertial frame of reference one gets $$-k_2y_2-m_2\ddot y_1 =m_2 \ddot y_2$$ which is your fourth equation.

Just imagine the the second spring not being there and mass $m_2$ is connected by a rigid rod to mass $m_1$.

You are observing mass $m_2$ in the non-inertial frame in which you measure $y_2$.

In that non-inertial frame $y_2$ does not change but there is a force on mass $m_2$ due to the rod and that force is equal to $m_2\ddot y_1$.

If you now includes the pseudo force and apply newton's second law in that non-inertial frame $m_2\ddot y_1- m_2\ddot y_1 = m\ddot y_2 \Rightarrow \ddot y_2 =0$, which is what is observed.