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Homework Help: Find the effective force constant ke

  1. Jul 31, 2009 #1
    This isn't really a homework problem, but I've been going over these Yale problem sets for physics 200 (http://oyc.yale.edu/physics/fundamentals-of-physics/content/resources/problem_set_3.pdf" [Broken]), and I'm having trouble with one of the questions.

    1. The problem statement, all variables and given/known data
    Two springs have the same unstretched length but different spring constants k1 and
    k2 . Find the effective force constant ke if they are connected in series and in parallel.
    (To ¯nd ke, imagine stretching the combination by an amount x and keeping track
    of the force needed.)

    2. Relevant equations
    f = ma
    fs = -kx
    Pupil = awesome

    3. The attempt at a solution
    So I figured the parallel spring constant out fairly easily, but I'm having trouble with the series one. The way I imagined it is we tie spring with constant [tex]k_1[/tex] to the wall, and tie spring with constant [tex]k_2[/tex] to spring with constant [tex]k_1[/tex], and finally pull [tex]k_1[/tex] by some amount A. Spring with [tex]k_1[/tex] will stretch by some amount A1 and spring with [tex]k_2[/tex] will stretch by some amount [tex]A_2[/tex]. Summing up, we have [tex]k_2A_2 + k_1A_1 = F[/tex]. The problem is I don't know what [tex]A_1[/tex] and [tex]A_2[/tex] are. I know the springs won't stretch the same amount, so I'm stuck with this ugly expression. Help me get unstuck (don't tempt me to look at the answers)! Thanks!

    EDIT: The title was supposed to be Spring Constant in Series...Don't know what happened.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 31, 2009 #2


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    Re: Spring

    k_2A_2 + k_1A_1 = F

    isn't right. Think about the point where the two springs connect. The first spring applies k1A1 to the second, and the second applies k2A2 to the first, so F=k1A1=k2A2. The problem should be easy from there.
  4. Jul 31, 2009 #3


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    Re: Spring


    That's not quite right. Springs in series experience the same force, so

    k1A1 = k2A2 = F

    However, the displacements add up, so we can also say

    Anet = A1 + A2


    EDIT: Hey ideasrule, great minds think alike :biggrin:
    Last edited by a moderator: May 4, 2017
  5. Jul 31, 2009 #4


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    Re: Spring

    x1 is the extension produced by the force F in spring 1, similar for x2

    The force on the first spring=force on the second spring
    so we have


    Now the effective spring constant relates to F by this

    F=kex (where x=x1+x2)

    i.e. we have [itex]F=k_e(x_1+x_2)[/itex]

    looks a bit unsettling right? BUT we know that [itex]F=k_1x_1=k_2x_2[/itex]

    you can put either 'kx' equal (replace F) to the formula above, and replace the 'other' x using the relation in the last formula. You will understand what I mean when you do it.

    For parallel:

    Each spring will stretch by the same amount, so finding ke is easier here.

    EDIT: seems ideasrule and Redbelly98 can type faster than me
  6. Jul 31, 2009 #5
    Re: Spring

    Holy crap! The force of the second on the first is equal to the force of the first on the second. I forgot Newton's third law! *Face palm* Thanks guys. You rock!
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