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The real idea behind -kx

  1. Jan 6, 2006 #1

    What is the real idea behind -kx?

    is X really the displacement of "relaxed" state and "compressed" state?

    or is something else!?
  2. jcsd
  3. Jan 6, 2006 #2
    I assume you're talking about the force exerted by a spring.

    Imagine taking a spring and attaching it to something solid, say a wall, and you set it up in such a way that the spring is resting on a table (so that it doesn't sag down or anything). Let the spring sit for a bit and it will settle at some length, this is the equilibrium position of the spring. If you then carefully marked out some distance measurements on the table you could see that as you pushed the spring in it would push back with a force equal to some constant, k, times the distance you had pushed it back. Simmilarly it would also pull back if you pulled it towards you. The distance which you pulled it from that equilibrium position (where you defined the zero of your length scale when you were marking up the table) is the x in the equation you gave.
  4. Jan 6, 2006 #3


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    Another way of looking at it is that the force acting on a mass (e.g. due to a spring) is generally a function of the displacement, F = F(x). Generally, F(x) can be a complicated function. If the force has an equilibrium point, say x = 0, then for small displacements about x = 0 the force can be approximated by

    [tex]F(x) = F'(0) x + F''(0) x^2/2 + \cdot \cdot \cdot[/tex]

    which is just a Taylor series expansion (remember, F(0) = 0 since x = 0 is an equilibrium point). If we restrict ourselves to sufficiently small values of x then [itex]F(x)[/itex] is well-approximated by keeping only the leading term so we write [itex]F(x) = F'(0) x[/itex]. Again, since we're dealing with motion about an equilibrium then F'(0) must be a negative number (i.e. the force is a restoring force).

    It is convenient to represent F'(0) with a number like -k where k is positive and is often simply called the "spring constant." The process I've just described is called "linearization" and is generally applicable when dealing with small oscillations. Of course, there are many situations where the linear approximation does not hold such as when a spring is stretched or compressed in such a way that it becomes permanently deformed (i.e. the mass will not return to it's starting point!)

    Since the linear equations do apply to many situations and provide considerable insight into how things work the time it takes to learn and use them is time well spent.
  5. Jan 6, 2006 #4
    The better to understand is that that is how k is defined. It is further proved that the work done on the spring is directly proportional to the deformation, x given in length.
    If you are talking about the negative sign, the -kx determines the work done by the string. If someone pulled that spring, the work done by him is kx. Moreover understand that this is not because ork is a vector, it is a scalar. It is because some work has been done on it. The workdone is positive if some work is done by a body. Anyway what others told clearly explains the remaining part in your doubt that how is it derived and when does the formula hold.
  6. Jan 10, 2006 #5
    I have a question about the meaning of the way spring constants are expressed.
    Most sites say the spring constant is a measure of a spring's "stiffness" or "stretchability". That being the case, what are we saying when we express this as N/m? A given spring might have a spring constant of 33N/m or 24N/m or 17N/m. What is the "N/m" part saying about its "stiffness" or "stretchability"?
  7. Jan 10, 2006 #6

    Doc Al

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    I'm probably missing the intent of your question, but what's unclear about calling a spring that takes 1000 N to stretch it one meter (k = 1000 N/m) stiffer than one that only takes 10 N to stretch one meter (k = 10 N/m)? More force required = harder to stretch = "stiffer".
  8. Jan 10, 2006 #7


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    In automobile suspensions, stiffer means a higher spring constant, since it will take more force to move a stiffer spring the same distance as a weaker spring. There will be less body reaction to roll and pitch due to accleration forces with stiffer springs.
  9. Jan 10, 2006 #8
    You've answered my question. It wasn't clear to me that N/m meant "the amount of newtons it would take to stretch this spring to a length of one meter.'
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