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Homework Help: Springs in Parallel

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data
    How do we calculate the effective spring constant of springs in parallel, where the springs have different spring constants and different stretch distances (because the original lengths of the springs are different).

    2. Relevant equations

    Hooke's Law --> Fx = kx

    3. The attempt at a solution

    If we assume the stretch distance of both springs are equal (x), and each spring constant is k1 and k2 respectively. Then the effective spring constant of the springs in parallel is:
    F = k1x + k2x = (k1+k2)x
    From above, one can see that the effective spring constant is k1+k2. However, I assumed the distance would be equal, therefore I was able to common factor it.

    So, now my problem is, what happens if the stretch distances are different. Then the above equation becomes:
    F=k1x1 + k2x2
    As one can see, I cannot common factor anything.

    EDIT: I'm ultimately looking to find the effective spring constant and in this case, I cannot common factor, so I cannot find the effective spring constant.
    Last edited: Nov 20, 2009
  2. jcsd
  3. Nov 20, 2009 #2


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    Of course if the springs stretch by a different distance, you get F=k1x1 + k2x2. Why do you need to factor this? It's a perfectly good equation giving the force applied by the spring.
  4. Nov 20, 2009 #3
    So how do I find the effective spring constant then?
  5. Nov 20, 2009 #4


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    There's no such thing as an effective spring constant because the force applied depends not only on position, but also on orientation (of whatever the springs are attached to).
  6. Nov 21, 2009 #5


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    Welcome to PF!

    Hi Morass! Welcome to PF! :wink:
    I agree with ideasrule :smile:

    If the springs are different lengths, then whatever's attached to them will have to start rotating, so how do you define a spring constant?

    Is this a question from a book or test (if so, what's the full question), or have you just made it up out of interest?
  7. Nov 21, 2009 #6

    This question is from a lab that I am doing in school. For the lab, I had to use 2 different springs of different length and spring constant. Then I attached various masses and recorded the stretch when the 2 springs are put in parallel. Then teacher asked me to find the effective spring constant using this data. So what I did was I measured the stretch of each of the springs (and they are different because they were originally of different length) and then I took the average of those two stretches and sort of assumed that as the mutual stretch of the two springs so that I could use the original equation to find the spring constant of parallel springs: F = k1x+k2x = (k1+k2)x. - where x is the average of the two stretches of the springs.

    I'm not sure if I can just average the two stretches but it seems reasonable :confused:
  8. Nov 21, 2009 #7


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    oh i see … the two springs remain the same length as each other because of some constraint on whatever is pushing or pulling …

    eg two plates constrained to remain parallel, joined by two springs of different original lengths and spring constants.

    Well, just calculate the two forces at each distance, y, and add them (I'm pretty sure it will depend on y and the "natural" lengths as well as on k1 and k2)
    erm :redface: … if you're not sure, then it isn't reasonable! :wink:
  9. Nov 21, 2009 #8
    Hmm the thing is, I know you can add up the forces to find the total spring force on the object but how would I find the effective spring constant in that case? I think I'm averaging the two stretches in order to mimic the ideal case where the stretches are the same.

    So i guess the final question is, how do I find the spring constant if I only know:
    The total force of the two springs (Fg because the system is at equilibrium)
    The stretch distance of each spring individually
  10. Nov 21, 2009 #9


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    Stop worrying about the future …

    get the basic equation first, and then decide how to make it neat.
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