Surprising implications of Newton's 3rd law when applied to a spring falling downwards with a weight

Question

I know that Hooke's law says that the force with which a spring tries to pull itself together is proportional to the amount you have extended the spring by. So, if you had a stationary spring (I will call it massless for the sake of simplicity) that has been extended due to a mass hanging off it, I'm pretty confident that Newton's 3rd law works beautifully in that the force the mass exerts downwards on the spring (which is just the mass's weight) is equal and opposite to the force that the spring exerts upwards on the mass. The reason that the spring doesn't accelerate downwards and that the mass doesn't accelerate upwards is because these 3rd law partner forces acting on different objects are also internal to the objects themselves, which isn't normally the case. (For example, as well as the mass exerting its weight force on the spring, it is also exerting its weight force on itself, meaning this weight force basically applies to two objects at once, so the mass can have balanced forces of its weight force downwards with the spring force upwards.)

Now, when you hang a mass on an initially stationary spring, and then let go, the spring and mass fall downwards together as the spring stretches, and the pair then oscillate up and down for a while before coming to rest at the stationary, or equilibrium, position that I just described in the above paragraph. When I tried to apply Newton’s 3rd law to situation 1, when the spring and mass were above the equilibrium position, falling towards it and situation 2, when the spring and mass were below the equilibrium position, moving back up to it, I drew some pretty surprising conclusions that I am not certain I got right, so I would love to know for sure if I am correct or not.

Conclusions drawn in situation 1: In this situation, the extension of the spring is less than when the spring is at its equilibrium position. Therefore, the upwards force exerted by the spring on the mass (say, 5N) must also be less than that force is when the spring is in its equilibrium position (say, 10N). When the spring is in its equilibrium position, this 10N upwards force exerted by the spring on the mass must equal the downward force exerted by the mass on the spring (so this must also be 10N). In its equilibrium position, the downward force exerted by the mass on the spring is equal to the weight of the mass, so the weight of the mass must also be 10N. However, due Newton’s 3rd law, if the spring is only exerting 5N on the mass in this situation 1, the mass can only be exerting 5N on the spring, even though the weight force of the mass is 10N. This does kind of make sense, because it feels intuitive that the mass wouldn’t be pulling so hard on the spring when the spring is doing what the mass (metaphorically) wants by falling downwards with it, as when the spring is stubbornly resisting the pull of the mass in the equilibrium position. It also makes sense in that it explains why the mass and spring would be moving downwards at all, since if the spring is pulling back on the mass with 5N, but gravity is pulling down on the mass with 10N, then there would be 5N resultant force on the mass at this point, just as there would be 5N on the spring from the mass, meaning both the mass and spring would be accelerating at the same rate at this point, if they both had the same mass. I suppose it’s just a question of separating one pair of Newton’s partner forces (force by spring on mass and force by mass on spring) from another pair (force by planet Earth on mass and force by mass on planet Earth) and therefore recognising that the force by mass on spring does not have to equal the force by planet Earth on mass, even though both involve the weight of the mass?

Conclusions drawn in situation 2: Here, the spring is more extended than in its equilibrium position, so it could be pulling the mass with a force of, say, 15N. Therefore, I think the mass must be pulling the spring with an equal force of 15N, even though the weight force of the mass is only 10N? This again makes some sense intuitively, since the spring is now doing the opposite of what the mass “wants” by pulling the mass up when it wants to go down, so the mass is resisting even more than when both have reached a compromise at the equilibrium position. Likewise, it again makes sense of why both objects are moving upwards, since the 15N upwards on the mass is greater than the 10N downwards of its weight force. So am I correct that in both these scenarios, the mass is not pulling the spring with a force equal to its weight force?

Answer 1

Yes, you are broadly correct. An object suspended from a light unextended spring does not initially impose a load on the spring equal to the object's weight. As it falls it imposes an increasing load in line with the increasing resistance as the spring is stretched. The load will exceed the weight of the object as it falls beyond the equilibrium point, causing the object to decelerate and eventually stop falling and start to rise. It is only at the equilibrium point that the force on the spring equals the weight of the suspended object.

What I have described above is somewhat idealised- in reality there would be effects arising from the mass of the spring etc.

Answer 2

A massless spring is what is called a force element. A thing that applies forces on its ends. This spring force is a function of extension $x$ and care must be taken to account for the sign convention. For a linear spring we have $F = -k x$ where positive $F$ denotes pushing the ends apart, and negative $F$ denotes pulling the ends together.

This extension is not the position of the mass, but a function of it. Let us call the position of the mass $y$ (+ direction is upwards) and $y_0$ the zero force height (where the spring is free).

$$ y = y_0 - x $$

The force $F = -k x = -k (y_0- y)$ is applied in equal and opposite measure on the ends of the spring. Since a positive force means pushing the ends apart, it is applied downwards on the mass, and upwards on the ceiling.

The equation of motion for the mass is thus

$$ \begin{aligned} m \ddot{y} & = -m g - F \\ & = - m\, g + k\, (y_0 -y) \\ & =(k\, y_0 - m\, g) -k\, y \end{aligned} $$

From the above you find the equilibrium height, when $\ddot{y} = 0$ as $$y_{\infty} = y_0 - \frac{m g}{k}$$ and change the equation of motion to $$ m \ddot{y} = -k (y-y_\infty) $$

Now the RHS of the equation above is the sum of forces acting on the mass, and when the position $y$ is above the equilibrium the force is downwards, and when it is above the equilibrium the force is upwards.