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Units ?

  1. Jun 19, 2007 #1
    units ??

    1. The problem statement, all variables and given/known data

    This should be straightforward but I'm not sure what to do. If I make a graph and the Y axis is in cm, and the X is in g, does cm/g make sense?

    I'm trying to find the spring constant 'k' by using the slope of my graph. When I do this I get a huge number so I'm playing with units to bring it down.


    2. Relevant equations
    k = g/slope (g = 980 cm/s^2)


    3. The attempt at a solution
    using the info. gathered my k = (980 cm/s^2) / 0.073 = 13520.55
    this sounds extremely high. My slope units are what's throwing me I think. Does 0.073 cm/g make sense? Then, would the units be 13520.55 g/s^2 ?

    I believe k is supposed to be in N m, so I'm not sure this is right ?!
     
  2. jcsd
  3. Jun 20, 2007 #2

    chroot

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    The spring constant is in units of force per unit displacement, N/cm. I don't know what you're plotting, but it sounds like you haven't done it right.

    You should give us the entire problem, exactly as it was given to you, and we can help you with it.

    - Warren
     
  4. Jun 20, 2007 #3

    malawi_glenn

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    No the spring constant is N/m

    just check it:

    Potential energy stored in a compressed spring = (1/2)*k*x^2 ; where k is the spring constant and x is the compression in meters.

    [J]=k[m^2]
    k=[J/m^2]=[Nm/m^2]=[N/m]
     
  5. Jun 20, 2007 #4
    I'm actually plotting two graphs and comparing the k values.

    The first graph is measuring displacement vs mass. So, in my case its cm vs grams.

    The second is measuring period squared (T^2) vs mass. In my case, its seconds^squared vs grams.

    I draw line of best fit and determine the slope. From here, K is determined using one of two equations. The first graph is k=g/slope, the second graph 4pi^2/slope.

    Drawing Max/Min lines for determining slope error, I get the error for slope. Maybe this is where I go wrong.

    I need to do the math for discrepancy which is the difference between the k values, divided by the square root of the sum of the squares of their errors (I used slope error here - giving me a value around +/- 0.003)

    My question about units comes here. Because my k is huge (14000) and (15000) for example, when I do the discrepancy equation I get a ridiculous answer like 333333.33 when the answer should be close to 2.

    I thought units may be the problem for such a large value, but I'm not sure. The K's I'm comparing will both have different values (graph 1: cm/g, graph 2: s^2/g) so how do I make them the same to compare?
     
  6. Jun 20, 2007 #5
    Got it! :smile:
     
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