# Homework Help: Simple question about elasticity

1. Oct 18, 2006

### student85

Imagine a bungee jumper dropping from a high cliff. When he reaches the point where the rope is just about to be stretched we can say he has reached his max. velocity. Then the rope will be stretched and when it has reached its maximum deformation, the jumper will "bounce" up and again make the rope stretch, this time a smaller distance.
My question is: if I want to calculate the constant, k, of the rope, I think I should use the Hooke's equation, where F=kx, where F is going to be the jumper's mass times g, and x is the maximum deformation of the rope. However...when the jumper stretches the rope for a second time, "x" is going to be smaller, but of course the jumper's weight is the same. So it is obvious that something extra needs to be put into the equation to account for the speed the jumper had when he first dropped and stretched the rope. How do you correctly calculate k??

2. Oct 19, 2006

### tim_lou

if you know the length of the rope and the maximum extension of the rope during that bungee jump, then just use energy:
$$mgh=.5kx^2$$
pick the reference point at the maximum extension point. x is simply maximum extension-length.

now, if you do not know the maximum... let's say you know how long it takes for the jumper to jump from top to maximum extension, then it gets a little tougher. First, express the equilibrium position (denote it xeq) in terms of k and other constants, then use d=.5gt^2 from 0 to the the height of -L. so now you have the time going from 0 to -L.

calculcating the time from -L to -L-xeq is trickier... since the rope is beginning to get stretched, the force is no longer the simple mg, (it is dependent on x). you need some kind of integral. Integrate using dt=dx/v, express v (velocity) using energy and integrate dx/v over -L to -L-xeq.

now, the third involves going from -L-xeq to -L-xmax, how do we know the time it takes if we do not even know what xmax is?
well...the third part involves calculating the period, since from -L-xeq to -L-xmax, the jumper undergoes simple harmonic motion, the period is only dependent on k and m (xmax is not needed). calculate 1/4 of the period.

then add up all 3 parts and equate that to the observed time. it should be an expression of k. well... solve it and you'll get k. (well the calculation will be UGLY):yuck:

Last edited: Oct 19, 2006
3. Oct 19, 2006

### quinn

In reality I would imagine that a bungee cord is not very well described by Hooke's law.

4. Oct 19, 2006

### Ja4Coltrane

"When he reaches the point where the rope is just about to be stretched we can say he has reached his max. velocity."

This is untrue. When the rope has just begun to stretch, there is almost no upward force. F=-kx and if x is almost zero, force will be too. Because of this, the persons weight mg is still greater than kx. At some value of x, kx will equal mg and therefore, any aditional stretch will cause a net upward force and this will cause the person to slow. This x value is x=mg/k

5. Oct 19, 2006

### tim_lou

i guess a bungee jump rope isn't really a good example of the hooke's law...

well, let's say you have a spring with some unknown constant k that obeys the hooke's law, and there is fiction during the oscillation so that the period decays.

the thing is, even if there is damping, the period is still independant of the energy of the system (as long as the potential function involves only the second order term). just calculate the time it takes for one complete cycle and use whatever period formula that you needs.

well, in general, if you have a potential function, $$U=kx^n$$
then, the period is proportional to:
$$T\sim{}E^{\frac{1}{n}-\frac{1}{2}}$$
so that if n=2, period is independent of the energy of the system (this is true for all harmonic oscillator).

Last edited: Oct 19, 2006