A person of mass $m$ is attached to a spring that its loose length is $l_0$. He is standing on a bridge and jumps off it, so he travels down a distance $2 l_0$ when measured from the bridge up to the moment the spring pulls him up.

Find the spring constant.

So, If I divide the jump into 3 parts then the man is in free fall up to the moment he travels a distance $l_0$. His velocity at that moment is $v_1=\sqrt{2gl_0}$. Then he keeps moving down up to the equilibrium point $l_1$ and then moves a one more distance $l_2$ until he gets to the minus height $2l_0$. So that $l_2 +l_1 +l_0=2l_0$. I am confused with the second part.

How the motion could be described there? Or, what is the jumper's velocity at (minus) height $l_0 +l_1$?

  • $\begingroup$ +1 nice homework, i rememeber it the next time (or the 1st time) i go bungee jumping :) $\endgroup$
    – Nikos M.
    Commented Oct 29, 2014 at 13:11
  • $\begingroup$ There are a lot of lousy homework questions that pop up from time to time, but this is a nice one. $\endgroup$
    – Sean
    Commented Oct 29, 2014 at 13:31

2 Answers 2


From the moment that the spring (bungee) comes under tension, the motion of the jumper can be described as a combination of simple harmonic oscillator and a constant acceleration - that is, a sine wave with an offset (the offset is the "equilibrium point" you mentioned).

But you don't need to go there. If you say that "at the bottom of the drop" the entire gravitational energy (accumulated from start of fall to bottom of drop) is converted to stored elastic energy ($\frac12 k x^2$). You know $x=l_0$, and you know the total energy ($2mgl_0$). $k$ can now be trivially computed...

Unless the question explicitly asks you to compute the motion at every point, going straight for the answer will save you a lot of time. These kinds of questions are often most easily answered by looking at energy rather than motion.

If you really want to know the equation of motion, do this:

For displacement $x$ below the point where the bungee becomes taut (where I use "up=positive"), the force on the person is

$$F = -kx - mg$$

Thus the equation of motion is

$$m\ddot x = -kx - mg$$

If we now transform to a new coordinate system $y = x - \frac{mg}{k}$ (so the equilibrium position corresponds to $y=0$), we get

$$m\ddot y = -ky$$

Which is an equation for simple harmonic motion. This is easily integrated; then plug in the value for velocity at the moment that x=0 (free fall over $l_0$) and you get velocity and displacement as a function of time. Now you can add the constraint that velocity is zero when $x=l_0$. You will get to the same conclusion. It's just a lot more work.

  • $\begingroup$ I need to find the point at which the jumper had the maximum velocity. So that I do need using the equation you wrote. Now, if we take the "up" direction to be positive, then the particle (i.e jumper in this case) is subject to gravity in the negative direction, and the spring force in the positive direction. So that the equation of motion is actually $$F=+kx-mg$$ $\endgroup$
    – E Be
    Commented Nov 2, 2014 at 10:57
  • $\begingroup$ Another thing: if $y = x - \frac{mg}{k}$ then $x = y + \frac{mg}{k}$. Plugging this into the equation of motion gives $m\ddot y = -ky-2mg$. Am I missing something? $\endgroup$
    – E Be
    Commented Nov 2, 2014 at 11:05
  • $\begingroup$ Simple harmonic motion about equilibrium point - so that is the point of greatest velocity $\endgroup$
    – Floris
    Commented Nov 2, 2014 at 13:51
  • $\begingroup$ Regarding your expression - the vertical force is greater when the displacement is more negative. So if bungee more taut happens when x is negative, I'm pretty sure that you need a minus sign in front of $kx$ in which case the two factors of $\frac{mg}{k}$ cancel in the expression containing just $y$. If they don't, you should fix your signs until they do (because that was the whole point of the exercise). And if you get that right, max velocity is reached at $y=0$. $\endgroup$
    – Floris
    Commented Nov 2, 2014 at 22:42

While it may be interesting and instructive, you don't really need to know much about the motion of the jumper. You can answer the quesetion by using a simple energy balance.

At the bottom of the fall where the velocity of the jumper is 0, all the potential energy of the jumper from there to the starting point has been absorbed by the spring:

potential energy of m at height 2l0 = energy stored in spring elongated l0

You should be able to write the appropriate equation and then solve for the spring constant. Note that you don't need to know anything about the velocity of the jumper over time. This is assuming we are ignoring losses due to air resistance, losses in the spring, etc.

  • $\begingroup$ Isn't this just repeating the first two (or three) paragraphs of my answer? $\endgroup$
    – Floris
    Commented Oct 29, 2014 at 15:38
  • $\begingroup$ @Floris: I don't know, as that would require actually reading your answer. I saw a bunch of equations and "harmonic oscillator" mentioned in your first sentence, so thought I'd give a simpler answer. $\endgroup$ Commented Oct 29, 2014 at 15:43
  • $\begingroup$ If you'd seen the original answer, there were no equations - just the mention of "harmonic oscillator" to address the "what kind of motion is this" part of the question. I then expanded it with a "if you really wanted to know, you would do this". But paragraph 2 in particular addresses exactly what you are saying: "don't do this the hard way - there's an easy way". Maybe I buried the signal in the noise - it's a bad habit of mine. $\endgroup$
    – Floris
    Commented Oct 29, 2014 at 15:52

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