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Homework Help: Finding Max Velocity For Simple Harmonic Spring problem?

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data

    A 100kg bungee jumper attached to a bungee cord jumps off a bridge. The bungee cord stretches and the man reaches the lowest spot in his descent before beginning to rise. The force of the stretched bungee cord can be approximated using Hooke's law, where the value of the spring constant is replaced by an elasticity constant, in this case, 100kg/s^2. If the cord is stretched by 30m at the lowest spot of the man's descent, then what is the man's highest velocity as he goes back up to equilibrium of the spring?

    2. Relevant equations
    I know that total work is = kinetic energy final - kinetic energy inital. I also know that Potential energy = 1/2kA^2 or amplitude^2 or amount stretched x^2.
    Lastly, Potential Energy max = Kinetic Energy Max(since all the potential energy gets transfered)

    1. I was wondering if we were able to just use the potential energy max equation for a simple harmonic spring to find the max velocity?

    2. If they gave us the total distance from where the man was standing to the lowest point the man fell, would this potential energy (mgh) be different from the springs potential energy?
    A: I think it would, because this is now talking about mechanical energy in the system, vs the energy of the spring.


    3. The attempt at a solution
    Last edited: Aug 31, 2014
  2. jcsd
  3. Aug 31, 2014 #2
    I believe that you are correct. You can find the kinetic energy with the potential energy equation. The total mechanical energy is not related to the potential energy found in a spring.
  4. Aug 31, 2014 #3


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    Be careful assuming everything is SHM. A bungee cord cannot produce a force under compression.
    Consider the moment at which the cord first becomes taut. Does the jumper have KE then? Will the jumper return to that vertical position? If so, what will happen next?
  5. Sep 1, 2014 #4


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    How many forms of energy are involved ? The information you have connects two of them at the lowest point, which gives you what you wanted under 2.
    Then: what is the mathematical way to express some variable is at a maximum ? Where does that happen if you know weight and elasticity constant ?

    Oh, and: draw a few diagrams of the various energies, position, velocity and acceleration as a function of time. Quantitatively at first. Check and correct; make them quantitative.
  6. Sep 1, 2014 #5
    I thought since they gave us k and x, we know the amplitude of the string from it's equilibrium point. With that I think we can find the potential energy = 1/2KA^2. Therefore, the max Kinetic Energy = max potential energy. However, there is the force of gravity... I am really confused now.

    But just to be clear, total mechanic energy is unrelated to the potential energy you can find "in" the spring?
  7. Sep 1, 2014 #6


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    I would say no. Total mechanical energy is very much related to the energy in the cord (spring). Jumper gathers speed until the cord is taut, and a little bit longer. Namely until the cord and the graviational pull are equal. At that point the (downward) acceleration is zero, meaning the velocity is at an extreme (remember that for later on!). That's the mathematical tidbit I was hinting at. But since speed is non-zero, stretching of the cord continues until v = 0: the bottoming out. At that point, there is no kinetic energy. All potential energy from the gravity field has been converted to stretching energy. You have the right equation there, so why not show some work: ##{1\over 2}kx^2 = ??##. Pick a reference point for the potential energy due to gravity.

    Now think what happens after the bottoming out: cord pulls bloke (or lady) up with considerable force, while gravity is pulling the other way. Hence acceleration. Upward speed increases until acceleration = ?? And when (/where) is that, precisely ? That's all they want to know ! (well, you still have to convert the energy into speed, but that's easy).

    I'm giving away far too much. Sorry, PF keepers... Must admit I had great fun following my own instructions drawing the diagrams. Used an equation solver at work. Beautiful.
  8. Sep 1, 2014 #7
    1/2KA^2 =1500J. The upward speed will increase until acceleration is zero. Which is at the max height as the lady goes back up. And the velocity as the lady goes back up from the lowest point to the point of equilibrium for the chord is = Square root of ((2*1500)/100)) = 5.477m/s. And eventually this velocity goes to zero due to the force of gravity, and then the woman comes down with the force of gravity.

    Btw, I sort of changed this problem from my book, therefore, it might not have all the necessary variables to be able to figure this problem out. It normally asked for the acceleration at the lowest point. Which was Fspring+force of gravity = ma. Which was F=-kx (3000N) and Fg=mg.(1000N) The answer I got is, 20m/s^2.

    Perhaps I need to use total conservation of mechanical energy. Kei + Pei = kef + Pef. Kei at the top of the bridge is 0, and the potential energy is mg(some height that i need to add) = Pef is zero, and kinetic energy final is the velocity right before the woman reaches her lowest point.

    Thanks BvU for helping me out! I am just trying to understand this concept better!
  9. Sep 2, 2014 #8


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    Oops, oops, oops. It clearly says the cord stretches 30 m. K is 100. So a lot more Joules of spring energy on my prehistoric calculator.

    Then: acceleration zero marks the extremum of the speed. That's right. There is no law that says "which is at the max height ...". (On the contrary: a max height is where the speed is zero ! That's what you should pick up from all my blabla!). Fortunately friend Newton has left us a law that links Force and acceleration. One zero, the other zero. And where please, is the point where the forces cancel ? (Think long term, when the bouncing up and down is over and the victim is hanging upside down without any more motion).

    Next: Eventually... can be a bit misleading. Could well be you understand the whole thing now, just that the wording leaves me in doubt.

    Have you picked a reference point for the potential energy from gravity yet ? Apparently the bridge. Fine with me. I think we can even make do without working out the length of the cord (i.e. the total plunge height difference minus 30 m). That's for later post. Have to go to work...

    Sounds good if g = 10 m/s2. You do have all the info you need to solve all of this.
  10. Sep 5, 2014 #9
    The forces are in opposite direction as the string is stretched down, because the restoring force is pointing back up and the gravitational force is pointing down. However, the restoring force is stronger than the force of gravity. Therefore, we get an upward acceleration of 20m/s^2.

    OOOO, I think I am seeing it. There is a difference between the kinetic energy of the spring, and total kinetic energy of the system. However, in certain situations, the total kinetic energy can equal to just the spring when the only force that is acting on the spring is the restoring force of the spring.
    Yet, In our scenario, we have two forces acting on the spring. If we knew the height of the bridge and the height of the string before it stretched we could find the Kinetic Energy there. (That is where Force restore -Force of gravity = ma(zero). As the lady falls right before the spring stretches that is also where our other highest velocity is at. But as the string starts to stretch the velocity goes to zero. We can have two kinetic energies. One, using the total mechanic energy with mg(h=the height before it stretches)=kinetic energy, and the other, using the kinetic energy of the spring, the lady goes back up from the lowest stretching point to the point of spring equilibrium for the chord is = Square root of ((2*1500)/100)) = 5.477m/s. Therefore, we can have two speeds. Hmmm, I don't see how the kinetic energy of the spring, is not hampered as it goes up do to the force of gravity. I can see the spring reaching it's max potential energy based on the stretch, but as the lady goes back up, I do not see that system achieving the equal magnitude of the potential energy because of the force of gravity.

    I might be wrong about where the forces cancel out. It might be that none of the forces cancel each other out. Because at the bottom, we have restoring force, and at the top we have gravity force. Maybe somewhere in between.

    Dang it! I feel mentally strained! Thank you for bearing with me BvU! Physics is too cool to hate, but definitely freaking frustrating.
  11. Sep 5, 2014 #10


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    We're not there yet. If the spring stretches 30 m, the mechanical energy stored in the spring is ##{1\over 2}\,k\,x^2 = {1\over 2} \, 100 \, {\rm kg}/{\rm s}^2 \, \left ( 30 {\rm m} \right )^2##, which is not, repeat NOT 1500 kg m2/s2.

    There is only ONE force acting on the spring: the pulling force that the person (in the original post it is said that it is a man) executes on the spring. There are TWO forces acting on the man.

    Is correct, if you are talking about the forces on the man at the bottom point (and if you use g = 10 m/s2).

    Yes, there sure is. In fact the total kinetic energy in the system is the total kinetic energy of the man. (The cord is considered massless). The kinetic energy starts with zero and starts to increase when he jumps.

    It keeps increasing even if the distance jumped is equal to the length of the spring/coil. At that point the acceleration is still g, so considerable. That means the speed keeps increasing, hence the kinetic energy too.

    It is important to understand that the speed (and hence the kinetic energy) keeps increasing until the acceleration of the man is zero. That is (because F = m a) at the point where the force, exercised on the man by the stretched cord is equal in magnitude to the force, exercised (in the opposite direction) on the man by gravity. In fact, we can calculate how far the spring is stretched at that point. And we have to do that, because we will need that amount of stretch later on. Oblige me and do this calculation.

    This is definitely NOT correct. I can't even understand what you try to express here.
    The coil exercises a force k x on whatever is pulling on it at all times when the stretch x ##\ge## 0.

    Definitely not true. At that point force from the spring is zero.

    No, as explained above. And is was a man when he jumped :smile:

    Yes! It will in fact even change sign at the bottom point and continue to increase (if you take upwards as the positive direction).

    No we can not. There is only one object with mass in the entire exercise: the man.

    I don't have time to comment on all that.

    You are absolutely right ! Are you now ready to abandon the idea that the maximum upward speed is reached at the point where the cord is no longer stretched ? Because from the point where the speed is maximum to the point where k x = 0 , the force on the man from gravity has been bigger in magnitude than the upward pulling force from the cord, so the acceleration has been downward, meaning the upward speed has been going down!

    Are you familiar with acceleration a is the time derivative of velocity v : ##\ \vec a = {d\vec v\over dt}## ?

    Do you know how to find extrema of functions x(y) by requiring ##{dy\over dx}=0## ?

    Do you see a similarity with your exercise ?
  12. Sep 5, 2014 #11
    Ahh thank you! It makes much more sense.
  13. Sep 6, 2014 #12


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    Good. Just to be sure, could you post your solution now ?
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