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Homework Help: Spring Constant in Trampolines

  1. May 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Hey I'm doing a project at school on trampolines trying to calculate how high you will go based on how much force is exerted by the jumper (driving force). The trampoline has 96 springs, not sure if this is relevant or not. To find the spring constant I need to use hooke's law k= mg/x but do I do this for an individual spring, or on the trampoline as a whole?

    2. Relevant equations
    k = mg/x

    3. The attempt at a solution
  2. jcsd
  3. May 15, 2008 #2
    It would of course be more accurate if you did each spring, and then the springy-ness of the fabric as-well. It should work fine to just talk about the "effective spring" that is the sum of all of the smaller aspects - so the trampoline as a whole.
  4. May 15, 2008 #3


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    I dont think the number of springs or the spring constant is relevant. The height only depends on the way in which you jump, and how much energy you can use etc. The difference between jumping on the ground and jumping on a trampoline is that you dont lose your energy when you fall back on a trampoline. So, when you fall back after the first jump, you can jump again with twice as much energy and reach twice the height (if you do it with the right timing). Ofcourse, its not so perfect, so maybe you can measure how much energy you lose per jump or something.
    Last edited: May 15, 2008
  5. May 15, 2008 #4
    In order to find the spring constant if needed eventually, I removed a spring and suspended 60 pounds then measured how much the weight caused it to stretch. After converting the units to metric here's what I came up with.

    k = (27.2155 kg)(9.81 m/sec^2) / (0.05715 m)
    k = 4671.63 N/m

    That seems like a ridiculously high number. Do you guys have any good ways of going about how to solve for how high a person will be elevated based on 'x' force being applied to the trampoline?
  6. May 15, 2008 #5
    That constant sounds about right.
    Figure out how high you can jump without a trampoline; how much energy would be required to get you to that height?
    Add that to the potential energy that would be stored in the trampoline if you were standing on it. How high will that energy get you?

    I'm really not sure how accurate this will be, i wouldn't be surprised if it was quite accurate... or if it wasn't at all.
    Alot about how trampolines get you to jump so high is due to resonance effects - in which case you have to think of it not as just a simple harmonic oscillator, but a driven one.
  7. May 15, 2008 #6
    Alright dude, I'm in my high school Honors Physics course.. Sorry for being for the most part helpless, but how would I go about doing that?
  8. May 16, 2008 #7
    No worries man.

    Lets say you can jump 0.8 meters off the ground. Then your potential energy at the top of your jump will be Ug=mgh so like 70kg*9.8m/s^2*0.8m = something like 560 Joules of energy.
    So thats about how much energy a 70 kg person can put into jumping. Now, on top of that, there is potential energy stored in the stretched springs of the trampoline (just before you jump) that will also contribute to your maximum height.
    The equation for potential energy of a spring is Us=0.5*k*x^2 where k is the spring constant and x is the displacement of the spring from its equilibrium position (how far down it sinks with you standing on it).

    Now, if you add those 2 energies together, that makes a good estimate of how much energy the system (you and the trampoline) have stored, and ready for jumping. Assuming all of that energy gets perfectly converted to height (in potential energy again, but with a different height), how high up will you go?
  9. May 16, 2008 #8
    I don't think that will be a very accurate calculation, but it's still a nice way to do it...

    It won't be very accurate since the technique you use for jumping on solid ground is completely different from jumping on a trampoline. If you use the same technique on a trampoline you won't go very high, I think the most energy comes from the springs, and you pushing your weight into the springs.
    (Think about it, if you land on a trampoline and you keep your legs stretched, you will only go a little less higher than your previous jump. If you contract your legs quickly on landing you can even manage to stay on the trampoline alltogether)
  10. May 16, 2008 #9
    Nick is totally right, the key to trampolines' is in more complex physics - when you land from your first small jump, you compress the spring farther than before; then by key timing in jumping a second time - you jump higher. When you land the second time you compress the spring even farther, letting you jump higher, and repeating the process until your jumping significantly higher than you would be able to just on the ground.
    This is a driven harmonic oscillator effect that the above technique doesn't explore at all. There's only one way to see how effective the above simplification is.

    I hope you'll share your results!
  11. May 16, 2008 #10
    Thank you guys for all the ideas, hopefully I'll get some testing done this weekend and let you guys know how it goes and if I need further help. :smile:
  12. May 18, 2008 #11
    All right, so earlier tonight my friend and I did a little testing. He weighs 70.4 kg and has a vertical of 0.4699 m (18 1/2 inches). And the spring is displaced a half inch (0.0127 m) when he was standing in the center of the trampoline. So here are the calculations...

    U(s) = (1/2)(k)(x^2)
    U(s) = (.5)(4681.364 N/m)(.0127 m)^2
    U(s) = .3775 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(s+g) = 324.571 Joules
    324.571 Joules = (70.4 kg)(9.80 m/s^2)(h)
    h = .47044 m = 18.5212 inches

    So we didn't really solve a whole lot.. the jump hardly increased which makes sense because your first jump isn't very high at all, but is there a way we could solve for jump two or three?
  13. May 18, 2008 #12
    Trying measure how far he compresses the spring when he lands from jump 1 --> use that as your stored spring energy --> see how high it gets you... you can repeat for a third jump.
  14. May 18, 2008 #13
    good thinking lzkelley, you've been huge help on this project! i'll let you know how it goes tomorrow :rofl:
  15. May 18, 2008 #14
    Cool, good luck. NP man, go physics.
  16. May 19, 2008 #15
    U(jump1) = (1/2)(k)(x^2)
    U(jump1) = (.5)(4681.364 N/m)(.022225 m)^2
    U(jump1) = 1.1562 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(jump1+g) = 325.3492 Joules
    325.3492 Joules = (70.4 kg)(9.80 m/s^2)(h)
    height of 2nd jump = .47158 m = 18.5661 inches

    U(jump2) = (1/2)(k)(x^2)
    U(jump2) = (.5)(4681.364 N/m)(0.03175 m)^2
    U(jump2) = 2.3596 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(jump2+g) = 326.553 Joules
    326.553 Joules = (70.4 kg)(9.80 m/s^2)(h)
    height of 2nd jump = .47332 m = 18.6346 inches

    Is there any other quick way (due wednesday) to get these jumps to show a bigger increase per jump. I know that the jumps increase by more than a tenth of an inch each time :confused:
  17. May 19, 2008 #16
    According to your equations, standing on the trampoline it was compressed by 0.02 meters (less than an inch) the first time, and only 0.03 the second time? i thought it would have been 5 times that...
  18. May 20, 2008 #17
    1/2 inch standing, 7/8 of an inch first jump, and 5/4 of an inch when landing 2nd jump

    does the number of springs have anything to do with that, or the fact that it is a large trampoline.. not a small exercise one or whatever. also, i suppose we could make up numbers to make our project work, but i'd really prefer not to lol
    Last edited: May 20, 2008
  19. May 20, 2008 #18
    making up numbers is never the answer...
    How are you measuring those displacements?
  20. May 20, 2008 #19
    Yeah, how are you measuring them? I can't imagine any easy and accurate way of measuring it when someone is jumping up and down. The displacement is only kept for a very little period of time...

    Maybe you can try videotaping it? If you make sure the camera is as perpendicular as you can get it to the center of the height of the trampoline, you can quite accurately measure it (as long as you use enough frames per second so you don't miss the largest displacement)... You can accurately measure the height of the trampoline (in real-life, not on the video) from which you can then measure the displacement of a single jump (on the video, on your pc from example).

    5/4 of an inch... I'm not good with inches, but I believe that is about 0.03 m or 3 cm? Isn't that very very VERY small?!
    I own a trampoline myself (round, 4.2m in diameter), it's about a meter high and if you really try you can get as deep as 20 cm off the ground (so about 80cm displacement)... In a regular jump it's definately about the 0.5m range...
    Last edited: May 20, 2008
  21. May 20, 2008 #20
    Wait a second lol.. When I am measuring the displacement in U(s) am I supposed to be measuring the displacement of the spring or the displacement of the mat?
  22. May 20, 2008 #21
    Hm sorry I thought you meant the displacement of the mat... I think you can do both...

    However, the question remains... How are you measuring the displacement of the spring if it happens so very quickly?
  23. May 20, 2008 #22
    no no! Its very important you measure the displacement of the mat!
    The spring expansion at such an oblique angle with make the force much less linear - and much much less accurate!
    Measure the vertical displacement of the center of the trampoline if you can.
  24. May 21, 2008 #23
    Alright guys, the final product is finally complete! Here are the equations after measuring the displacement of the MAT and not the spring haha.

    U(t) = (1/2)(k)(x^2)
    U(t) = (.5)(4681.364 N/m)(.3048 m)2
    U(t) = 217.455 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(t+g) = 541.648 Joules
    541.648 Joules = (70.4 kg)(9.80 m/s^2)(h)
    height of 1st jump = .7851 m = 30.909 inches
    U(jump1) = (1/2)(k)(x^2)
    U(jump1) = (.5)(4681.364 N/m)(0.5334 m)2
    U(jump1) = 665.96 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(jump1+g) = 990.153 Joules
    990.153 Joules = (70.4 kg)(9.80 m/s^2)(h)
    height of 2nd jump = 1.435 m = 56.496 inches
    U(jump2) = (1/2)(k)(x^2)
    U(jump2) = (.5)(4681.364 N/m)( 0.6096 m)2
    U(jump2) = 869.826 Joules

    U(g) = (m)(g)(h)
    U(g) = (70.4 kg)(9.80 m/s^2)(.4699 m)
    U(g) = 324.193 Joules

    U(jump2+g) = 1194.019 Joules
    1194.019 Joules = (70.4 kg)(9.80 m/s^2)(h)
    height of 3rd jump = 1.731 m = 68.136 inches

    These numbers make so much more sense than the previous attempts. Thanks a ton for the help guys. My friends and I took a video just for fun of us doing "tricks" if you'd like to watch. I can't post URL's since I've only made 12 total posts so search Youtube for Trampoline Bueno and it'll be the top video on the list!
    Last edited: May 21, 2008
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