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Homework Help: General relationship for springs connected in series

  1. Jul 8, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the effective force constant of 2 springs which are connected in series? What is the general relationship for n springs connected in series?

    2. Relevant equations
    im not sure if i should use

    or is the formula for a series circuit also relevant?

    3. The attempt at a solution
  2. jcsd
  3. Jul 8, 2012 #2
    Hi Vandetah! :smile:

    When massless springs are connected in series, the forces exerted by each is the same. Can you derive a relation using this fact?
  4. Jul 8, 2012 #3
    so if theres two springs then it would be
    k1 = k2?
  5. Jul 8, 2012 #4
    Not quite. Even though their forces are same, their extension will be different. For two strings, you have,

    [tex]x_1 = \frac{F}{k_1}[/tex]

    [tex]x_2 = \frac{F}{k_2}[/tex]

    And, [itex]x= x_1 + x_2[/itex]

    Do you see a relation between the spring constants now?
  6. Jul 8, 2012 #5
    u might laugh at this but, does it mean that two constant springs connected to each other create a signle force constant and therefore act as if it were a single spring?
  7. Jul 8, 2012 #6
    No, they produce a single force. Not the same force constant. That should be clear from the above equations :smile:
  8. Jul 8, 2012 #7
    so that is the general relationship, but what about the effective force constant? Can i use the equation i mentioned in the op?
  9. Jul 8, 2012 #8
    Noo...Use the equations I wrote in my earlier posts to derive the relation for the effective force constant!
  10. Jul 8, 2012 #9
    is there a difference between just a force constant and an effective one?
  11. Jul 8, 2012 #10
    Well, force constant is for a simple, single spring. The effective constant is the effective spring you replace a given setup with, so that you get the same force exerted by that single spring which the whole setup would exert.
  12. Jul 8, 2012 #11
    so if i derive the equations u mentioned i will get

    k1 = [itex]\frac{F}{x1}[/itex]


    k2 = [itex]\frac{F}{x2}[/itex]

    so k = k1+k2??
  13. Jul 8, 2012 #12
    Noo :redface:

    Use x = x1 + x2.
  14. Jul 8, 2012 #13
    could it be?

    x = [itex]\frac{F K_{2}+ F K_{1}}{K_{1}K_{2}}[/itex]
    Last edited: Jul 8, 2012
  15. Jul 8, 2012 #14
    Yes, but you need to find the effective constant, and not the final extension :wink:

    Take x = F/k.
  16. Jul 8, 2012 #15


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    Homework Helper

    Imagine that you measure the extension of both springs applying the same force to each. You get ΔL1 with the first spring and ΔL2 with the other spring. If you connect the springs, the same force acts on each - you pull the second one with F, it stretches by ΔL2, the second spring pulls the first one with the same force, it extends by ΔL1, so the connected springs stretch by the sum ΔL1+ΔL2.

    You put the connected springs in a black box to hide the point where they are connected. You say it is a spring inside and ask somebody to measure the spring constant, applying force F. The person measures the change of length ΔL, and calculates the spring constant k=F/ΔL. That is the effective spring constant of the device inside the black box.

    You know that ΔL=ΔL1+ΔL2, and also that ΔL1=F/k1, ΔL2=F/k2, so you can express the effective spring constant with the individual ones.


    Attached Files:

  17. Jul 8, 2012 #16
    lol that post above got me more confused :(
  18. Jul 9, 2012 #17
    Ehild put all I was saying in one post.

    Just use x=F/k in your last equation, where k is the effective constant. :smile:
  19. Jul 9, 2012 #18
    1. Let 2 bodies connected with inextensible string.
    If 1st body is pulled by a force and remained at rest, EQUAL force also exerted on 2nd object.

    Thus spring#1 and spring#2 extended with equal force, x1 and x2

    F=-k1x1=-k2x2 .....(1)

    2. Total extension is x1+x2 due to F.

    Subt. (1) in (2)

    Keff =k1k2/k1+k2 like resistors in parallel
  20. Jul 9, 2012 #19


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    Homework Helper

    You mean that Keff=k1k2/k1+k2, that is Keff=2k2? It is wrong.
    Or you just forgot the parentheses...

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