Bungee Jumper/Oscillatory Motion

  • Thread starter jromega3
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Homework Statement


A person named Larry is found dead, hanging at the 7-story level of a 15 story (45m) building. Larry has a mass of 90kg. When the bungee cord is cut, it has a relaxed length of 5 stories. Detective say he hit the ground while bungee jumping, you KNOW he was murdered. What is your evidence?


Homework Equations


Fs=-Ks (X)
X(t)= Acos(wt + theta)
Fs=Fg when he's resting

The Attempt at a Solution


So at the 7 story level Fs=Fg if there's no acceleration. That's a change in X of 3 floors, or 9m. 90kg*9.8=9m*Ks
Ks=98N/M.

Know we know W= root (ks/mass), or 98/90, which gives me a w of 1.04

so X=A*cos(1.04T + theta)
I'd assume solving for A, or amplitude, from the resting point (7 floors up) would tell us if he hits the ground or not. If it's 21 or over, he hits it, if it's under, he clearly cannot hit the ground.
But, that's three unknowns.
So, I'm stuck, but just had an idea that I'm not sure is valid or not.
Using simple kinematics, if I were to choose 2 points, (say at 10m above and below the resting point), I could find the Avg. acceleration (Fg-Fspring)/mass, and knowing average acceleration and X position, I could solve for T.
Knowing T for X, I'd have two unknowns (A and theta) for two equations. I could then solve for A.

Would this be a correct line of thinking? Or am I erring somewhere? Thanks
 

Answers and Replies

  • #2
turin
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Did you think about using conservation of energy? What energies are involved in this problem?
 
  • #3
Redbelly98
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How about using energy conservation?

EDIT: turin beat me to it.
 
  • #4
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Ahh. The initial total Energy would be just
mgh + PEs, right?
Now what? Find what mgh + PEs works on the other side to equal the same total energy, and this would be the lowest point (where V=0). If so, that makes sense. Thanks...hopefully I'm on the right track ;)
Well, doing this, I got X to be 35m off the ground...hmm.
 
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  • #5
Redbelly98
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That doesn't seem right, since it puts Larry at about the 12th floor, above where he was found, and above the relaxed position of the bungee.

Perhaps you could say what value you got for the total initial energy, and also what equation did you set up to find the height? That would help us find where the error is.
 
  • #6
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That doesn't seem right, since it puts Larry at about the 12th floor, above where he was found, and above the relaxed position of the bungee.

Perhaps you could say what value you got for the total initial energy, and also what equation did you set up to find the height? That would help us find where the error is.
Yeah, I definitely erred somewhere along the road.

So the initial energy was mgh + 0.5Ks(delta x)^2
90*9.8*45 + 0.5*98*15^2 = 50,715 J

now to solve for x (lowest point, ie where all energy is potential)....50,715=90*9.8*X + 0.5*98*(x-15)^2
s0 50,715=882X + 49(-30x+225+X^2)
0=49X^2 + 588X -39690
X=-23.1, 35.1
35.1 is the only possible solution of the two...but that's obviously wrong.
 
  • #7
turin
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Ah. The bungee does not contribute to the potential energy at the top of the building, and it does not compress, so it only effectively stores energy one way (by stretching). It only starts to stretch at a particular floor, and that should be your zero point for the elastic potential energy. Otherwise, your approach looks good. (I did not read all of your algebra - this text based stuff is sure annoying.)
 
  • #8
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Ah. The bungee does not contribute to the potential energy at the top of the building, and it does not compress, so it only effectively stores energy one way (by stretching). It only starts to stretch at a particular floor, and that should be your zero point for the elastic potential energy. Otherwise, your approach looks good. (I did not read all of your algebra - this text based stuff is sure annoying.)

Yep. That would've been my problem. Without that he "lands" a few meters off the ground. Thanks for the help everyone
 

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