So I was given a problem about bungee jumping,and it's obviously important to know in advance how far the cord of unstretched length L will stretch for a given weight of person. Consider the cord to be a weak spring with constant k and the person is a point mass m. Air resistance and pendulum motion are ignored. Point x=0 would be the point where the person freely fell until the entire slack of the cord is extended to length L. After the person passes x=0, the cord is stretched x(t).
Simple enough right? Well I was having trouble:

find a model for x(t) defined only on the first interval of time 0<=t<=T for which x>=0. Solve for x(t) and then determine maximum elongation. I do know that I should use x(t)=A sin(ωt+ φ) but after than I'm stuck. Could anyone possibly help me?

  • $\begingroup$ Maybe the The Physics Of Bungee Jumping will help? See equation 13 and the substitution they make for a. Regards $\endgroup$
    – Amzoti
    May 2, 2013 at 3:31
  • $\begingroup$ are you taking into account gravity? $\endgroup$
    – Alex R.
    May 2, 2013 at 4:08

1 Answer 1


The general solution of harmonic oscillator equation is a combination of $\sin \omega t$ and $\cos \omega t$ where $\omega = \sqrt{k/m} $. Since $x(0)=0$, we only have the sine. So, $x(t)=A\sin\omega t$. To find $A$, you can use the relation $x'(0)=A\omega$, where $x'(0)$ can be found from energy consideration. Indeed, jumping from height $x=-h$, the person acquires kinetic energy $mgh $ by the time that $x=0$. Hence, $x'(0)=\sqrt{2gh}$.


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