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Spring with constant k cut in two

  1. Jan 12, 2005 #1
    Been thinking for a long time and can't come up with a conclusion!

    If you have a spring with a constant k value and you cut the spring in two(not exactly in half), does the k value change in the springs or stay the same?

    I'm thinking that the K value does not change because the length of the new springs doesn't have anything to do with the k value. I have been asked to come up with mathematical answer to this question but can't seem to find one but F = -kx
  2. jcsd
  3. Jan 12, 2005 #2


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    Is this homework ? Because if it is, I can't give you an explicit answer.

    But I can give you a hint. Think of two springs that are put in series (one connected to the end of the other). What is the spring constant of the combined spring in terms of the spring constants of both smaller springs ? The same force acts through both but the total extension is the sum of the extensions of each smaller spring. The situation obtained by cutting a spring in two is the reverse of this.
  4. Jan 12, 2005 #3
    ok...i think i come up with a conclusion...but I'm not sure of how I did it exactly...was reading up on Young's modulus and "found" out that k is inversely affected by length...So if you cut the spring in 2, the k constant will change from k to k(initial length/final length) for one spring. I'm not sure if i am right...can anyone confirm?
  5. Jan 12, 2005 #4
    yep u r rite
  6. Jan 12, 2005 #5


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    Well, you don't have to bring the Young Modulus into the picture at all. The formula for the final spring constant [itex]k[/itex] of two springs with constants [itex]k_1[/itex] and [itex]k_2[/itex] which are in series is like the formula for adding resistances in parallel.

    [tex]\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}[/tex]

    Using either this, or considering the contribution of each part-spring to the total extension, you should be able to figure it out.
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