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A uniform spring whose unstretched length .

  1. Mar 3, 2013 #1
    A uniform spring whose unstretched length.....

    A uniform spring whose unstretched length is L has a spring constant k. the spring is cut into two pieces of unstretched lengths L1 and L2, with L1 = nL2. What are the corresponding spring constants k1 and k2 in terms of n and k?

    The answer is k1 = (n+1)k/n and k2 = (n+1)k. I have no idea how they got that. I tried setting up the force equation F = kL and solving for k, k = F/L, but the substitution never worked out.
     
  2. jcsd
  3. Mar 3, 2013 #2

    tiny-tim

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    Welcome to PF!

    Hi Estrex! Welcome to PF! :smile:

    Instead of cutting the spring, just mark it where the cut should be, and remember that the reaction force across the mark will be the same in each direction, and will be the same as the force at the ends. (why? :wink:)
     
  4. Mar 3, 2013 #3
    Stop right there. The restoring force generated in a string is given by F = kΔx, where Δx is the length by which the string is compressed or elongated. F does not depend on L at all.
    Well now consider the original spring to be split up into two strings of spring constants k1 and k2, which are connected in series and consider what tiny-tim said. That's pretty much all you need to do here!
     
  5. Mar 3, 2013 #4
    Ok, I figured it out. I just forgot the formula for springs in series, 1/k = 1/k1 + 1/k2.
    Since L1 = nL2, k2 = nk1. 1/k = 1/k1 + 1/nk1. 1/k = n+1/nk1. k = nk1/n+1. k1 = k(n+1)/n. Solving for k2, 1/k = n/k2 + 1/k2. 1/k = n+1/k2. k = k2/n+1. k2 = k(n+1).
     
  6. Mar 4, 2013 #5
    good job... if u forget something, u can always derive it... works for me most of the time...
     
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