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Calculating Spring Constant

  1. Jul 28, 2013 #1
    Suppose I hang a spring vertically from a table. I attach a mass on it. Am I correct in saying that the net force on the spring is 9.8*mass ?

    If I note the displacement, can I calculate the spring constant by using:

    K = -F/x = -g*mass / displacement ?

  2. jcsd
  3. Jul 28, 2013 #2

    Doc Al

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    Yes, but lose the minus sign. (The spring constant is unsigned.)
  4. Jul 28, 2013 #3

    Simon Bridge

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    ... though the displacement and the force will be signed, it is usually easier to work in magnitudes.
    The standard approach to these kinds of problems is to, 1st, draw a free body diagram.
    I'd encourage anyone to get into the habit.
  5. Jul 29, 2013 #4


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    I don't think you can mean "net force". If the system is in equilibrium (When the spring is supporting a stationary mass), the net force is zero. It is the force, due to the mass (weight) that is mg, which will also be the tension, in equilibrium. When it's bouncing, the force will have a sinusiodal variation (SHM), from zero, through mg, to 2mg and back.
  6. Jul 29, 2013 #5


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    In addition to what has been said, there is another factor that needs to be mentioned here, and this is a common mistake that I see students make in intro physics labs. How do you know that the relationship F = kx works for that spring? Because not knowing the context of the question (i.e. is this simply a textbook question, or you are actually doing this and testing it out, etc.?), it is hard to know what can assumed to be correct.

    Remember, F=kx explicitly implies that there is a linear relationship between F and x. While this may be true in "usual" case of springs being used in many lab courses, it is part of the practice to do this for a series of different masses to VERIFY that this relationship works and thus, F = kx can be used. It is not uncommon to find a spring that has been deformed slightly, and where F = kx no longer works! Thus, using that relationship is invalid.

    Incidentally, I gave a similar scenario in our last PF Trivia/Quiz in relation to Ohm's Law.

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