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Physics of Bungee Jumping problem?

  1. May 25, 2015 #1
    You (77kgs) have volunteered to jump from a bridge while attached to a bungee cord. You fall 22 m from rest, beginning at time t = 0.0 s, before the cord begins to stretch. The cord stretches 14 m before you begin to move back upward. As you bounce up and down, it occurs to you that you are on an ideal bungee cord, and you will continue to oscillate until you or someone else figures out how to rescue you.

    What is the spring constant of the cord?

    I know I figured this out.
    TE = PE = 1/2 kdY^2 = 1/2 mV^2 = mgS = PE = TE; we solve for k = 2mgS/dY^2 = 2*77*9.8*22/14^2 = 169.4 N/M


    What is the time required for you to first reach (momentarily) zero velocity?
    A) 1.06 s
    B) 1.27 s
    C) 1.35 s
    D) 1.42 s
    E) 1.51 s

    Heres my work for this second part:

    Total time T = tf + tb; tf = sqrt(2S/g) = sqrt(44/9.8) = 2.1189sec.
    So it takes 2.1189 sec for the jumper to reach the point where the cord even begins to stretch. So how could any of the answers be correct???
     
    Last edited: May 25, 2015
  2. jcsd
  3. May 25, 2015 #2

    collinsmark

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    Don't forget about the potential energy associated with the 14 meter height difference. Yes, the gravitational potential energy of first 22m drop gets converted into the spring's potential energy, but so does the following 14 meters of drop too.

    Where does the "77" come from? Is your mass 77 kg? The mass is not specified in the problem statement. The mass is necessary to know in order to figure out the spring constant.
     
  4. May 25, 2015 #3
    Yes. The person jumping is 77 kgs. Sorry about that.
     
  5. May 25, 2015 #4

    collinsmark

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    Okay, then when finding the total change in gravitational potential energy, it's not just from the first 22 m, but it's the entire displacement of 22+14 meters.

    (Special bonus hint: there's a bit of a neat "trick" here involving conservation of energy in that you don't have to worry about kinetic energy at all if you don't want to, since you start and end at 0 m/s :wink:).
     
  6. May 25, 2015 #5
    So am I on the right track? I really have no idea how to finish off this hw, and will likely need a definitive answer.
     
  7. May 25, 2015 #6

    collinsmark

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    Yes, you are on the right track. But in your calculations when you set the spring's final potential energy equal to the other initial energies (which is ultimately the gravitational potential energy alone), you need to change your height (used in the mgh calculation) from 22 m to 22+14 m.

    I'm going to be away from my computer for the next several hours, but here is a hint on how to proceed:

    You won't be able to use your standard kinematics equations when the spring starts stretching because the acceleration is non-uniform.

    Have you been given a formula that gives the period of a spring-mass system? If so, use that in part to find the final time interval.

    If you haven't been such a formula you can find it yourself by solving an ordinary, second order, nonlinear differential equation. But that's the hard way. Use the formula if you have it.

    (Another hint: the time from when the rope starts stretching until your momentary speed is zero is not an entire period of a spring-mass system; it is only part of the period. I'll leave it in your hands to determine what fraction of a period it is.)
     
    Last edited: May 26, 2015
  8. May 25, 2015 #7
    Yes I am using T=2π√m/k and I get 4.2361 as the time for one period. divide it by 2 and i get 2.11. Well that is already bigger than the answer choices so I dont think adding more time to it would help...
     
  9. May 26, 2015 #8

    haruspex

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    Then you still have not made the correction collinsmark keeps pointing out. The correct value of k is a bit under 280N/m.
    On the other hand, I can't get any number as low as those offered either, even if I ignore the initial 2.12 seconds of free fall.

    Edit: Picking up on collinsmark's observation about halving the period (whoops!) if I ignore the initial 2.12 seconds I get a number lower than any offered.
     
    Last edited: May 26, 2015
  10. May 26, 2015 #9

    collinsmark

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    I'm back to my computer now.

    That's the right formula.

    Before we recalculate the spring constant, I must say:

    There does seem to be something wrong with the way the problem statement is worded. Like you said in the original post, if t = 0 at the time of jumping off the bridge, it takes longer to fall the 22 m than any of the given answers, which is impossible. ('Could be all the given answers are wrong, but whatever the case, something doesn't seem right [my interpretation is one of the possible errors, btw.].)

    Could you repeat the question, verbatim, word for word, because right now the details are not clear to me. Something does not seem right.

    When we do get to the bottom of this, are you sure you want to divide the period by 2? Think about what happens when a spring mass system oscillates. Starting at the point where the spring is not compressed, first the spring expands as the mass slows down, eventually coming to a halt. Then the mass shoots backwards where it came until the spring is no longer expanded again (and the mass's velocity is equal and opposite what it was at the start). But then what happens?

    [Edit: Btw, the way I'm presently interpreting the problem the exact solution is not quite as simple as dividing the period by 4 either. The way I'm interpreting the problem, this part gets tricky.]
     
    Last edited: May 26, 2015
  11. May 26, 2015 #10
    I suspect the cycle repeats itself in the same oscillating motion it was before.
     
  12. May 26, 2015 #11
    Here it is word for word from the hw.
    You (all brave 77 kg of you) have volunteered to jump from a bridge while attached to a bungee cord. You fall 22 m from rest, beginning at time t = 0.0 s, before the cord begins to stretch. The cord stretches 14 m before you begin to move back upward. As you bounce up and down, it occurs to you that you are on an ideal bungee cord, and you will continue to oscillate until you or someone else figures out how to rescue you. In the interim, you decide to calculate the following:

    Question 3 1 / 1 point
    a. The spring constant of the cord.

    Question options:
    A) 142 N / m
    B) 169 N / m--------->I chose this answer and it said that i am correct
    C) 182 N / m
    D) 191 N / m
    E) 211 N / m
    Question 4 0 / 1 point
    b. The time required for you to first reach (momentarily) zero velocity.

    Question options:
    A) 1.06 s
    B) 1.27 s
    C) 1.35 s
    D) 1.42 s
    E) 1.51 s
     
  13. May 26, 2015 #12

    collinsmark

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    I think this might be one of those situations where the truly correct answer does not correspond to the "given" correct answer (i.e., a mistake in the coursework).
     
  14. May 26, 2015 #13
    Ahh, that is what i suspected... If you had to take a guess what would you say the correct answer is given the answer choices?
     
  15. May 26, 2015 #14

    collinsmark

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    If I had to guess, I would try the following:

    Start with the given choice of the spring constant, which was accepted as correct. (169 N/m. This isn't truly correct though. 'Like I mentioned earlier, the actual spring constant calculation needs to be revised, given the other parameters in the problem statement. But use 169 N/m anyway for the rest of this.)

    In a truly ideal spring-mass system the mass would oscillate around a point somewhat below the point where the cord starts stretching. That's because when we account for gravity, a new equilibrium point is formed (the point at which kx = mg). We are going to ignore this difference and the time it takes for the jumper to first travel between the point where the cord starts stretching until crossing the new equilibrium point.

    Instead, we start the clock at the moment the jumper reaches the new equilibrium point. This is the point at which the jumper has maximum velocity. The clock ends when the jumper's velocity reaches 0 m/s.

    I'll leave it to you to find that time interval. It does give an answer that is listed as one of the "given" choices. (If you haven't realized by now, there are several "errors" with this method. But it does lead to one of the "given" answer choices.)
     
    Last edited: May 26, 2015
  16. May 26, 2015 #15

    collinsmark

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    Also, if I were you I would mention all of this to your instructor. At least that way it might get corrected for future students.
     
  17. May 26, 2015 #16
    x=Asin(wt)
    w=1.4832
    x=-7
    A=7
    m=77
    k=169.4

    -7=7sin(1.483t)
    -1=sin(1.483t)
    arcsin(-1)=1.483t
    t=1.059=1.06-->I think I have found the answer that my professor is looking for!!!

    Let me know if this is not correct, otherwise, Thank you all so much!!
     
  18. May 26, 2015 #17

    collinsmark

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    Well, I can't say that anything about it being really "correct." I suspect that the whole problem is fundamentally flawed.

    But if I had to guess what answer the instructor wants you to choose, it would be 1.06 s, yes.

    But again, I'd mention all of the flaws discussed in this thread to your instructor. I have a feeling that whoever created the question and answer choices made some mistakes along the way.
     
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