Suppose you have two massless springs, both of length $d$ and spring constant $k$, in the following arrangement:

As indicated by the gray springs, the two inital springs both get streched by a gravitational force $F_\text{G}$ of a mass $m$, by a vertical distance of $d$ (which is the same as the springs' initial length). Then, the system is stationary.

The question is what the spring constant $k$ has to be in order for this to happen.

There are two approaches to this problem:

1. Forces

In order for the mass to stay in place, the spring's forces have to add to $F_\text{G}$, where the horizontal component of each spring force is $k\Delta x\cdot\cos(45°)$, which adds up to the condition $F_\text{G}=2k\Delta x\cos(45°)$. This results in $k=\frac{mg}{2\Delta x\cos(45°)}$ or $$k=\frac{mg}{(\sqrt2d-d)\sqrt2}$$ because $\cos(45°)=1/\sqrt2$ and the distance each spring gets streched is $\Delta x=\sqrt2 d-d$.

(This should be the correct result.)

However, if we choose the following second approach, we get this:

2. Conservation of Energy

The mass $m$ has a decrease of potential energy of $mgd$. This has to be balanced by an increase of the two spring's energies, which are $\frac12 k(\Delta x)^2$ for each spring, so $k(\Delta x)^2$ in total, where again $\Delta x=\sqrt2d-d$. This results in the condition $mgd=k(\sqrt2d-d)^2$, which results in $$k=\frac{mgd}{(\sqrt2 d-d)^2},$$ which is of course a different result than above.

Question: Why do we get two different results for what seem to be two valid approaches to this problem?

  • $\begingroup$ The correct energy approach would be to find the potential energy as a function of the height variation and force $d$ to be a minimum of the function. Thus, you'll find the equilibrium position. $\endgroup$ Mar 1, 2019 at 22:32
  • $\begingroup$ There's no reason to believe that the potential energy would remain zero in both cases. Realistically, this system would enter a SHM if no external force did work to keep the mass stationary. $\endgroup$ Mar 1, 2019 at 22:33
  • $\begingroup$ Perhaps the total energy of the system is not the same with the body in either position. Otherwise it could be moved from one position to the other without any expenditure of work. $\endgroup$
    – Jim
    Mar 1, 2019 at 22:41

5 Answers 5


The approach that gave you the right answer was the force approach. Let's talk about the energy approach.

This is a conservative system, so we can write the potential energy $U$ as a function of the resultant force on the mass $m$, like this:

$$\vec{F} = -\nabla U$$

Or in terms of the resultant force's modulus:

$$F = -\frac{dU}{dx}$$

Where $x$ is the distance (positive below the initial position) of the mass $m$ from its initial position (where both springs had lenght $d$).

The equilibrium position is found when $F = 0$, therefore $\frac{dU}{dx} = 0$ so we just have to find $x$ such that the $U$ function has a mininum or a maximum (it's gonna be a minimum in this case).

If we set $U_{gravitational} = 0$ in the initial position, we have $U_{gravitational} = -mgx$.

We can then use the $U_{elastic} = k(\sqrt{x^2 + d^2} - d^2)$ and therefore

$U = -mgx + k(\sqrt{x^2 + d^2} - d^2)$

So all you have to do is take the derivative $\frac{dU}{dx}$ and this will give some equation with a solution $x_{equilibrium}$. But the problem told you already that this $x_{equilibrium} = d$, so all you have to do is solve the equation you found for $k$.

Why did your energy approach not work?

It's simple: the potential energy isn't the same in both situations at all. There's no reason to believe that. That's why if you actually release the mass $m$ in that initial condition, it will reach the equilibrium position with some kinetic energy given by the difference between the potential energies $U_{equilibrium}$ and $U_{initial}$. This is just conservation of energy. The particle will oscillate if released from the initial condition.

  • 1
    $\begingroup$ The particle will oscillate if it is released from the initial condition, but for a large displacement problem like this it will not perform simple harmonic motion. The vertical restoring force is not proportional to the displacement from the equilibrium position. Note, apart from that detail, the rest of the answer is correct. $\endgroup$
    – alephzero
    Mar 1, 2019 at 23:34
  • $\begingroup$ You're right; the displacement might be too big for the SHM approximation to work here. Thank you, I have edited the answer. $\endgroup$ Mar 1, 2019 at 23:38

You have made an incorrect assumption in the second case. If you drop the mass from the unstretched position it will actually overshoot the equilibrium position and oscillations will occur. Energy actually isn't conserved if you only consider gravity and the spring forces. Something else needs to do work to take energy out of the system so that the mass is at rest at equilibrium. Therefore, your second attempt is incorrect.

  • $\begingroup$ Wow, so many answers posted while I typed this up haha. $\endgroup$ Mar 1, 2019 at 23:00
  • $\begingroup$ I was thinking the same thing. A lot of answers posted in a very short amount of time heh $\endgroup$ Mar 1, 2019 at 23:01
  • 2
    $\begingroup$ ... and they even all agree! $\endgroup$
    – flaudemus
    Mar 1, 2019 at 23:03

Why do we get two different results for what seem to be two valid approches to this problem?

The primary reason is that the conservation of energy is not a valid method for finding the equilibrium position. Energy is conserved in all states of the system, not just at the equilibrium. So the conservation of energy does not uniquely specify any position.

Furthermore, for most initial states a strict conservation of energy will lead to oscillation rather than equilibrium. Transitioning to equilibrium typically involves a loss of mechanical energy from the system.


If the total energy of both states were the same, you could move from one to the other with no work done. But you have to put work in and lift the mass to go between the two states. They do not have the same energy.

Going from the neutral position to the equilibrium position (by releasing the weight) would find the mass still moving and therefore with KE. That KE is lost over time to friction.


I agree with the comment of João Vítor G. Lima in saying that you did not use the conservation of energy argument correctly. Conservation of energy applies to dynamical systems. Suppose you start your mass at rest with the two springs being horizontal. Gravitation will pull the mass downwards and an oscillatory motion around the equilibrium position will start. Whenever the mass moves through the equilibrium position, it will have kinetic energy. The sum of kinetic and potential energy will be constant in time.

Suppose the oscillation is subject to friction, such that it comes to rest at the equilibrium position eventually. Then it will have a lower energy than when it started, because energy has been dissipated by friction. This means, the potential energy of the system in the state with horizontal springs and lifted mass is necessarily higher than the potential energy in the equilibrium position.

The system behaves essentially like a single spring with a mass attached. There, the same applies: the potential energy of relaxed spring with lifted mass is higher than that of the mass at the equilibrium position, and the spring stretched.


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