Homework Help: Need verification on a spring force problem regarding a bungee cord.

1. Nov 18, 2009

Hunter Arcanu

1. The problem statement, all variables and given/known data

A bungee jumper of mass m=70 kg is riding a bungee cord with spring constant k=50 N/m. Its unstretched length is L=9.0 m. What is the amplitude of the jumper's oscillation?

m=70 kg
k=50 N/m
L=9 m

2. Relevant equations

mg(L+x)=(1/2)kx^2
x(t)=Bcos(omega(t)+alpha)
omega = (k/m)^(1/2)

3. The attempt at a solution

(70)(9.8)(9+x) = (1/2)(50)x^2
25x^2 - 686x - 6174 = 0
x = 34.58

(L+x)/2 = (9+34.58)/2 = 21.79 m

I'm not really sure how to use the equation for simple harmonic motion to determine the amplitude, but shouldn't the amplitude just be (L+x)/2 given that air friction is negligible and total energy is conserved?

2. Nov 18, 2009

Delphi51

What is the initial height of the jumper? I've never jumped myself, so no idea!
If the cord is hanging, unstretched, and the guy takes hold and drops, then the height should be just x, not 9+x. The 9+x means he is taking the end of the cord and climbing a ladder up to where the cord is tied before jumping.

Yes, I think you are supposed to assume no loss of energy since no measure of the loss is given.

3. Nov 18, 2009

Hunter Arcanu

The height of the jumper isn't given, so when I set up my potential energy equation, I assumed he was just high enough so that he barely touched the ground (the length of the cord plus the max displacement of the cord) so that I didn't have to deal with gravitational potential energy on both sides of the equation. I'm also assuming that he jumps from where the cord is attached so that he'll have fallen 9 m before the cord stretches.

4. Nov 18, 2009

Delphi51

Okay - sounds scary!
Your solutions looks great.