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Spring constant K question

  1. Aug 22, 2005 #1
    if you cut a spring with a constant "k" in half, does the new spring's "k" change?
  2. jcsd
  3. Aug 22, 2005 #2
    What do you think would happen and why?

  4. Aug 22, 2005 #3
    I'm not sure, I was thinking that the "k" would double, because by cutting the spring in half you are reducing the spring's displacement.
  5. Aug 22, 2005 #4
    When you cut it in two you are not changing the displacement at all. That is, if by the displacement you mean how much you are compressing it.
  6. Aug 22, 2005 #5
    okay, so now I'm really confused.
    My other idea about the question was that maybe the "k" doesn't change at all, because it is a constant, so it is not effected by the change in the size of the spring.
  7. Aug 22, 2005 #6


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    The units for the constant are in F/L where the L refers to the length of compression. Does it say anything about the spring's actual length?
  8. Aug 22, 2005 #7
    Hey, Fred Garvin, thanks for replying, but there's nothing about the actual length of the spring. I don't need an exact numerical value, I just need to know if the spring constant changes, and if so, how does it change.
  9. Aug 22, 2005 #8
    I think I found out the answer to my question. I think that k would be doubled, because the force of the spring is the same, and the length of compression is reduced. Can anybody tell me if that's right?
  10. Aug 22, 2005 #9

    The spring constant does not change.

    :rofl: :rofl:
  11. Aug 22, 2005 #10


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    What I was trying to get at was that the spring constant is a function of material and the way it is wound. It is not a function of the overall length of the spring. The only thing that cutting a spring in half does is halves the total distance you can compress it before you reach the spring's solid height.
  12. Aug 22, 2005 #11


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    The tacit assumption I believe you're making is:
    - This is in reference to a helically wound compression or extension spring.
    - The spring is literally cut in half so that the material, wire diameter and coil diameter stay the same. Only the number of active coils gets reduced by 1/2.

    If those are your assumptions, then you are correct. The k will double. The equation for spring constant for such a spring is:

    k = d^4 * G / (8 * D^3 * N)

    Where d = wire diameter
    G = Modulus of rigidity
    D = mean diameter of coil
    N = Number of active coils
  13. Aug 22, 2005 #12
    Thanks for that Q_Goest.

    I always had thought it was just a function of the material and the way it was wound.

    Sorry PhysicsStudent12 you were right all along :wink:
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