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Cutting a spring in half and then in 1/4 and 3/4

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    A helical spring of stiffness k is cut into two halves and a mass m is connected to the two halves. The natural time period of this system is found to be 0.5 sec. If an identical spring is cut so that one part is one-fourth and the other part is three-fourths of the original length, and the mass m is connected to the two parts, what would be the natural period of the system?

    2. Relevant equations


    3. The attempt at a solution
    Attached is the solution. Why is k2 = 4k? I understand the solution up until that point.

    QVGb9pA.png
     
  2. jcsd
  3. Oct 20, 2014 #2

    BiGyElLoWhAt

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  4. Oct 20, 2014 #3
  5. Oct 20, 2014 #4

    BiGyElLoWhAt

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    Oh wait, I think I misunderstood the question.

    F=kx => k =F/x and if we halve x we double k.

    So basically if you take a spring, with some number N coils and stretch it by some length x, each winding or coil stretches by a factor of x/N. Now halve N (you're cutting the spring in half). Stretch the spring by the same length, each coil now stretches by a factor of 2x/N, effectively doubling the spring constant. (It takes 2x as much force to stretch or compress the spring by the same amount)
     
  6. Oct 20, 2014 #5

    BiGyElLoWhAt

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    Yea, sorry about that.
     
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