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Homework Help: Spring constant and stiffness constant

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A mass m hangs from a spring with stiffness constant k. The spring is cut in half and the same mass hung from it. WIll the new arrangement have a higher or a lower stiffness constant than the original spring?

    2. Relevant equations

    F= -kx

    3. The attempt at a solution

    I think that the spring will have the same spring stiffness. I dont feel that the spring constant is something that can be changed. If it was, then it shouldnt be called a constant.
  2. jcsd
  3. Jun 3, 2007 #2
    as far as you're concerned yes it doesn't change. also as far as i'm concerned it doesn't because i'm a noob but i do know that it does actually change because i know that cutting suspension springs changes their frequency of oscillation which is directly proportional to the root of [tex]\frac{K}{m}[/tex] which means the K does change. i just don't know how
  4. Jun 3, 2007 #3
    No, it is the spring constant. It does not change assuming it is the same spring. It will just half the amount of compression that the spring can take before you go past the limit.

    On a side note, I think my arrow keys are disable in the forum and when I hit the apostrophe key I get a quick search link. Is anyone else having the same problem?
    Last edited: Jun 3, 2007
  5. Jun 3, 2007 #4
    yup you're right, the cutting of the springs affecting springs rate is because it messes up the local annealing in the spring
  6. Jun 3, 2007 #5
    I don't really understand what you just said. I also don't understand what is meant by "rate".
  7. Jun 3, 2007 #6
    free frequency of oscillation is what i meant by spring rate,


    tempering i meant the tempering of the spring is affected
  8. Jun 3, 2007 #7
    If it isn't a car spring I highly doubt that you would use a acetylene torch to cut it. Say a mechanical pencil spring
  9. Jun 3, 2007 #8
    you were right, i was wrong, i was confused about why the K of the spring is affected.
  10. Jun 3, 2007 #9
    Whoah, I wasn't trying to be condescending, sorry if it came off like that.:frown::frown:
  11. Jun 3, 2007 #10
    i didn't think you came off as condescending:biggrin:
  12. Jun 3, 2007 #11
    thanks everyone for the help
  13. Jun 3, 2007 #12

    Doc Al

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    Staff: Mentor

    Think about the definition of spring constant. If a spring is cut in half, is it easier or harder to stretch it a given amount?

    Another way to look at it: Hang a weight from the original spring and it stretches the spring a distance X. Now imagine that the spring is really two springs attached together (each one half the size). How much does that weight stretch each half-spring?

    It's a constant for a given spring--as long as you don't stretch it too far. In Hooke's law (F = -kx), F and x change as you stretch the spring, but k remains constant. If you modify the spring by cutting it, you create a new spring with a new spring constant.
  14. Jun 3, 2007 #13
    what is the definition of the spring constant? is it not F/x
  15. Jun 3, 2007 #14

    Doc Al

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    Staff: Mentor

    That's correct.
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