Bungee jumping - find spring constant 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 $l_0 +l_1$?
 A: 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.
A: 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.
