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Help with spring constant units

  1. Sep 28, 2013 #1
    Part 1 (static method): We measured the displacement of a spring (in cm) after adding and subtracting masses (in g). The spring was placed in a vertical position. I am supposed to find the spring constant ks.

    Part 2 (dynamic method): We did something similar to the above, but we lifted the spring a bit so it would bounce. We then counted the number of oscillations of the spring for 20 seconds each time we changed the masses put on the spring. I am then supposed to find kd.

    2. Relevant equations
    Hooke's law: F = kx
    units of k = N/m (kg/s2)
    ks = Sg

    3. The attempt at a solution
    I created a graph and found S + ΔS = 25.3 ± 0.10 g/cm

    However, it's the equation ks = Sg that's messing me up. My instructions don't tell me what I'm supposed to do with g. I have no idea why it's stuck in there. According to this equation, ks ± Δks = 25.3 ± 0.10 g2/cm, but that doesn't seem to make any sense. How am I supposed to end up with the units of kg/s2 when I won't have any of those units until part 2? And if these are not the units I'm supposed to use, how do I know if I do have the correct units?
  2. jcsd
  3. Sep 29, 2013 #2

    Simon Bridge

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    Don't you have an equation that tells you what to do with it?

    But you have all those units right there ... what are the units of g?

    S is the slope of the graph for part 1?
    It looks like you plotted mass on the vertical (y-axis) and extension on the horizontal (x-axis).
    Is that correct?

    Why not convert S to SI units - then use the standard value for g in SI units?
  4. Sep 29, 2013 #3
    Ahh...it didn't mention in the instructions, but I just realized that g must mean the acceleration due to gravity. Gah! I knew it was something simple! With multiplying all of this by gravity, I end up with a much nicer 24.9 ± 0.10 kg/s2. Thanks!

    Yes, you are correct.


    Perhaps you can help me in calculating an error for part 2? Or do I have to make a new thread?


    My equation is kd = (4π2)/S. (This is a new slope with s2 on the y-axis and g on the x-axis.)
    From the first equation, I found kd = 24.1 kg/s2.

    I also have an equation for a function f(x) = Cxn, then... Δf/f = Cn(Δx/x). Δf is the error, C is a constant, and Δx/x is the fractional error in x.

    Therefore, Δf must equal fCn(Δx/x). Now, it seems to me that this means Δkd = kd(4π2)(-1)(ΔS/S). However, this would cause Δkd to be -0.00190973753, which doesn't make sense to me. Shouldn't it be a positive number?

    So it seems then that kd ± Δkd = 24.1 ± 0.0 kg/s2.
  5. Sep 29, 2013 #4

    Simon Bridge

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    It sounds like you were following instructions without understanding the physics behind them.

    When you hang a mass off the end of the spring, the the spring pulls on the mass with F=kg, and the mass pulls down with F=mg, so that mg=kx.

    If you plot m vs x, you get a line with equation: m=(k/g)x
    You could have plotted mg vs x, which would make the slope k.

    if you have a measurement ##x\pm\Delta x##, and ##z=1/x##, what is ##\Delta z##?
    Clearly it is not zero! That your problem?

    It sounds like you are using equations without understanding the physics behind them.

    well: xz=1 so $$\frac{\Delta x}{x}+\frac{\Delta z}{z}=\frac{\Delta 1}{1}\\
    \Rightarrow \frac{\Delta x}{x}=\frac{\Delta1}{1}-\frac{\Delta z}{z}\\
    \Rightarrow\frac{\Delta x}{x}=-\frac{\Delta z}{z}$$ ... since there is no uncertainty on a constant: Δ1=0.
    What this does is tell you that you are right - it's not just you using the wrong equation.
    It is important that you understand the equations that you use.

    So - using your understanding of errors, how would you handle it?
    What is the minus sign telling you in this case?

    Hint: The error is always written out as ##\pm## - why?

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