This is the scenario: We have a spring resting on a surface with no mass on it. Let the $y$ position of the top of the spring be $l_0$. When we put an object with some mass on top of the spring, the spring compresses for some length $l_{1,0}$ so the top of the spring is at position $l_1=l_0-l_{1,0}$. The spring now has some potential energy. If we now compress a spring even more with some external force it will compress even more now to length $l_{2,0}$ thus making the y position of top of the spring $l_2=l_1-l_{2,0}$. If we now remove the external force, the spring will oscillate between $l_0$ and $l_2$ but at the end it will settle at position $l_1$ if no external force is acting on it. Now suppose we apply some external force on the spring so that when we remove the force the object gets shot in the air above the position $l_0$.

What confuses me is this: when we exerted some external force on a spring so that object oscillates between $l_0$ and $l_2$, the potential energy of the spring between $l_0$ and $l_1$ is preserved, but when we apply external force large enough to eject the object into the air, that potential energy is converted into kinetic energy. I can't wrap my head around this. I would appreciate if you can describe this scenario your way, maybe I will have better understanding of this system.

  • $\begingroup$ If you say the spring will "settle" at position $l_1$ after the external force is removed, then the spring would not be an ideal spring because damping has occurred (e.g. due to friction in the spring). Is that what you meant? Or do I misunderstand something? Why doesn't it continue to oscillate? $\endgroup$
    – Bob D
    Mar 1, 2019 at 20:48
  • $\begingroup$ That's true, the spring would continue to oscillate if there was no friction. What bothers me is the exchange between forms of energy. I am playing with this scenario for a while and just can't get correct calculations. $\endgroup$ Mar 1, 2019 at 20:55
  • $\begingroup$ In both cases there is a conversion of potential energy into kinetic energy. The potential energy of the spring is not constant even for the case of small oscillations, when the object does not leave the top of the spring. And in both cases you need to consider the gravitational potential energy too. $\endgroup$
    – nasu
    Aug 10, 2023 at 1:32

1 Answer 1


There are three positions that are important:

  • $l_0$: the initial position of the spring.
  • $l_1$: the position of the spring with a mass on top.
  • $l_2$: position of spring with mass on top and us pressing down on it further.

When the spring is compressed and stationary, there is only one type of energy: potential energy. When the spring is let go (after being held down or pulled up) the spring and mass begin to oscillate, in what is called simple harmonic motion. The moment that the spring begins to oscillate, its potential energy turns to kinetic, and gradually to potential again and so on. Let us see how to describe this mathematically.

Set the initial position of the spring to zero, $l_0=0$ for simplicity. Also assume that the mass is placed on the spring such that the spring and mass are stationary (so basically it is not dropped on the spring). How do we determine the length that the spring compressed? The force on the mass is $mg$, and the force exerted by the spring is $-k(y-l_0)=-ky$. Setting these equal (why?) one sees that the spring compressed to $y=-\frac{mg}{k}$, which we have called $l_1$, that is, $l_1=-\frac{mg}{k}$.

Now we will apply another constant force $F$, say by pushing down with our hand, until the spring pushes back hard enough to stop us. The new length of the spring is $l_2=-\frac{F+mg}{k}$. See if you can show this yourself.

And what if the spring is released now? There are two cases to consider.

  1. If $|l_2-l_1|<|l_0-l_1|$: Then we placed a mass on the spring which compressed it by some length $l_0-l_1$ (which we call $l_{1,0}$). And after that we compressed it more by a smaller length $l_1-l_2$ which we call $l_{2,1}$. The spring oscillations are centered at $l_1$, and have an amplitude (peak to peak) $A=2\times l_{2,1} = 2(l_1-l_2)$.
  2. $l_{2,1}>l_{1,0}$: Now we have an interesting situation. If the mass is glued to the spring, there is practically no difference to above, the system just has larger oscillations. But if they are not glued, and if we suppose a (unrealistic) mass-less spring, we get the following: The spring reaches its initial position $l_0$ with the mass on top travelling at some velocity. The mass shoots upwards, while the spring remains perfectly still, since it has no mass and no momentum of its own. It waits there until the mass comes back and causes it to compress again.

Here you will find more information that might help you calculate the actual height that the mass rises to the air, or its velocity upon leaving the spring. you will need to use energy conservation for both of these.

  • $\begingroup$ Thank you for taking your time to explain this little bit more in-depth! One more question: If we wanted to calculate the height that object reaches after detaching from spring, can we convert the energy we gave to the spring by pushing it further with external force to the energy of the object when at position l0? By that I mean: If this is a closed system, then only energy that is added to the system is the one using external force. Could we then add that energy to the potential energy of the object (since it gets launched into air and gains some additional height)? $\endgroup$ Mar 1, 2019 at 20:49
  • $\begingroup$ No worries, welcome to stackexchange :) That is correct, you can calculate the energy at the compressed state, and this is the energy that the mass will have once the mass and spring reach $l_0$; no energy is left for the spring so all the energy must be the kinetic energy of the mass (assuming you measure the kinetic energy of the mass from $l_0$). This is also the energy the mass will have at the apogee of its motion. $\endgroup$
    – myorbs
    Mar 1, 2019 at 20:54
  • $\begingroup$ Ok, I'll try to play around with the scenario a little bit more. Thank you for your time! $\endgroup$ Mar 1, 2019 at 21:03

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