The question is shared to me by my physics teacher, we have never solved it:

Imagine a spring standing vertically with one end pressed on the table, now you press down the spring from the upper end by x meter, then you remove your hand. Now if the compression is enough, the spring could raise its bottom off the table. It asks: what is the max height the top of the spring will reach.

Motion of the spring

There are point masses M1, M2 attached at bottom and top of the massless spring respectively. You are given that spring has natural length L, spring constant k, initially compressed by x meter, gravitational acceleration g.

I don't see a clear way to solve this, my best attempt was to produce a simulation as in the gif, but it didn't clear up much things. My best hope is a expression in terms of those parameters that expresses this max height.

Link to simulation: https://www.desmos.com/calculator/7mvofzasjn

Edit: I have seen people posting answers, but what they didn't realize is that the spring can be very stretched or compressed when the top of the spring is at its highest! This means there is EPE leftover in the spring. But how much? That would depend on how stretched it is then, which one probably would determine when one really solved the problem.


2 Answers 2


This is an interesting problem, and it turns out that you can solve for it exactly using the technique of normal coordinates. I'll try to give you the general outline of how it works in this case without violating this forum's "no complete answers" policy.

Let $x_1$ be the position of the lower mass and $x_2$ be the position of the upper mass. The force on the lower mass is $F_1 = - m g - k (x_1 - x_2 - L)$; the force on the upper mass is $F_2 = - m g + k (x_1 - x_2 - L)$. So the two equations of motion for the masses are \begin{align} \ddot{x}_1 &= - g - \frac{k}{m} (x_1 - x_2 - L), \tag{1}\\ \ddot{x}_2 &= - g + \frac{k}{m} (x_1 - x_2 - L). \tag{2} \end{align}

These two equations are coupled, which means that both quantities $x_1$ & $x_2$ appear in both equations and influence each other (which is what makes this a hard problem.) But let's try defining two new coordinates to describe the system: $X = (x_1 + x_2)/2$, which is the position of the center of mass of the system; and $y = x_1 - x_2 - L$, which is the amount of extension of the spring. You can hopefully see that if I tell you $X$ and $y$ at any given time, it describes the system just as well as if I told you $x_1$ and $x_2$; in fact, we have $$ x_1 = X + \frac{1}{2} (y + L) \qquad x_2 = X - \frac{1}{2} (y + L). \tag{3} $$ And if you take the derivatives of $X$ and $y$, you can also see that $$ \ddot{X} = \frac{1}{2} (\ddot{x}_1 + \ddot{x}_2) \qquad \ddot{y} = \ddot{x}_1 - \ddot{x}_2. $$ Plugging in our values for the accelerations from (1) and (2) and simplifying, we get $$ \ddot{X} = - g \qquad \ddot{y} = - \frac{2k}{m} y $$ These two equations are decoupled, in that neither equation contains both $X$ and $y$. What's more, the form of these equations should be familiar to you; what this says is that $X$ (the position of the CM) just moves in parabolic motion, while $y$ oscillates in simple harmonic motion. In particular, you can try to figure out what the solutions $X(t)$ and $y(t)$ are; and then you can figure out what $x_1(t)$ and $x_2(t)$ are using the relations in Equation (3); and then you can find the value of $t$ at which $x_2$ is maximized.

  • 1
    $\begingroup$ Thank you for this great answer. I've worked a lot on your thread, and so far I have got the exact equation for X. For the other function, Asin(wt-k) would work (w=sqrt(2k/m)), and I figured out you can calculate k if you know A, the amplitude. But then I became completely lost in trying to find A. When the string is fully stretched by A, each ball in fact will has velocity which are undetermined, and such an approach with energy soon fails. Can you please offer me some more hints, because I believe if I know A, I can pull the rest together with using your other equations. $\endgroup$
    – Simon
    Jan 5 at 4:40

By observation and for maximum height, we can say that the extension/compression of the spring at the topmost point will be zero. By assuming that and using conservation of mechanical energy,

#At initial compressed position, the bottom is at ground level. But the top is at (L-x) meters above the ground. #At the topmost point, bottom is at height h and since there is no compression, top is at height h+L So, $$\frac{1}{2}kx^2+M_2g(L-x)=M_1gh+M_2g(h+L)$$ By rearranging the terms we get, $$h_b=\frac{kx^2-2M_2gx}{2g(M_1+M_2)}$$ This is the height that the bottom of the spring reaches, so the top reaches a maximum height of $$h_t=\frac{kx^2-2M_2gx}{2g(M_1+M_2)}+L$$

  • $\begingroup$ But your observation is not correct! The spring can be very extended/compressed at the top point. This is what makes me believe there isn't a simple solution with energy conservation, one might need some equation to keep track of the phase for the vibration as well. $\endgroup$
    – Simon
    Jan 4 at 6:38
  • $\begingroup$ @Simon For attaining maximum height under ideal conditions, we can say that the total elastic potential energy is converted into gravitational potential energy right? $\endgroup$ Jan 4 at 6:50
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    $\begingroup$ But at highest point it could be that the spring is stretched so there is EPE left. Also, I think when the top point is at the highest doesn't even qualify as when GPE is highest for the system because the bottom ball also contribute to the GPE, so you could have a case where the top ball is falling while the bottom one is being pulled up really fast to compensate and to yet increase the GPE. $\endgroup$
    – Simon
    Jan 4 at 7:17

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