# Is there any tension in a massless spring that connects two free falling bodies in different horizontal planes?

Two bodies A and B of same mass $m$ are attached with a massless spring and are hanging from a ceiling with a massless rope. They are in same vertical plane but not in same horizontal plane. Now the string that connected A with the ceiling is cut and the system is experiencing free fall.

1. Is there any tension in the spring?

My attempt:

Now the whole system should descend with the acceleration $g$ and the body B (and also A) experiences a gravitational pull $mg$. Let the tension in the spring be T.

Therefore, from the free body diagram of B, $mg - T = mg$,ie. $T=0$.

1. But A also moves downwards, so puts a force on B, how to take account of that? Will there be an relative acceleration between A and B? I am confused about the free body diagrams of A and B.

2. Will the tension change if the mass of A and B are different?

There is tension in the spring. It it extended and hence there is tension! It is the centre of mass that falls with acceleration $g$ rather then each individual mass. So the equation $$mg-T=mg$$ is invalid. As the two masses fall they will oscillate (getting closer and further away) and the tension will cycle.

Let us call the distance fallen by mass $A$, $x_A$ and that fallen by mass B $x_B$ the equation of motion for each mass is given by: $$m \ddot x_A=mg+T$$ $$m \ddot x_B=mg-T$$ $T$ is a function of $x_A$ and $x_B$, ($T=k(x_B-x_A-L)$ where $k$ is the spring constant, and $L$ is the natural length) and we cannot assume that $\ddot x_A=g$ or $\ddot x_B=g$. These sorts of equations are called coupled differential equations and can be solved a number of ways.

The answer is that it depends on how your initial spring loaded mass is moving. But, the fascinating (but not too fascinating once you phrase it like this) part is that until the compression wave from the top interacts with bottom out on the slinky the dynamics of the bottom half won't change.

If we assume it was at rest, essentially the top mass will move fast enough that the center of mass will accelerate at $9.8 m/s^2$. When it reaches the true equilibrium length of the string without gravity it will start accelerating the bottom mass. At this point if you looked from the COM frame the spring would appear to be oscillating like it normally does. This is because the oscillations occurring here is called an eigenfrequency. The other eigenfrequency (since this problem has 2 independent variables) is the motion of the COM. With the motion of the COM and the motion of both masses about the COM you have all the information needed to reconstruct the dynamics of your masses.

A great demonstration of this is the slinky in this video, which is like a spring with an equilibrium length of zero: CC BY-SA 3.0 image by AllenMcC, see Wikipedia Commons