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Springs In Series

  1. Oct 8, 2003 #1
    Two springs are in series. You know the constants for each individual spring. How do you find the spring constant for the spring system?

    |-OOOOO---o---OOOOOOOOOO-| --->F
    In the middle of the spring system (where the "o" is) the net force is zero and we have
    k1x1=k2x2 disregarding sign
    F=k3x3
    x3 =x2+x1
    F=k3(x2+x1)
    F=k3(k1x1/k2+x1)
    F=k3*x1/k2*(k1+k2)
    but F=k1x1+k2x2
    F=2k1x1
    2k1x1=k3*x1/k2*(k1+k2)
    2k1k2=k3(k1+k2)
    k3=2k1k2/(k1+k2)

    Is this right?
     
  2. jcsd
  3. Oct 9, 2003 #2
    Looks OK.
    Now try to find a similar expression for N springs..:)
     
  4. Oct 9, 2003 #3

    Integral

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    Looks to me like you have an extra factor of 2 in there. Springs in series add as Resistors in parallel. So

    1/KT = 1/K1+ 1/K2

    OR


    KT= (K1+K2)/(K1K2)

    Also Springs in parallel add as resistors in series.

    KT=K1+K2

    I do not have time to do the derivation now. Should be able to post it later to day. A key is that in the series case the force seen by each spring is equal, while in the parallel case the extensions are equal.
     
  5. Oct 9, 2003 #4
    Ah, I see my mistake. I have F=k1x1+k2x2 but F=k2x1=k1x1. I need to be more careful.
     
  6. Oct 9, 2003 #5

    Integral

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    OK, that looks better!
    Here is my derivation. As long as I was at it I did the case for parallel springs also.

    BTW: it would be trivial to extend this to multiple springs.
     
  7. Oct 9, 2003 #6
    Thank you, Integral. I have put your work into my physics notebook (hope you don't mind).
     
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