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Determine g from least squares fit line?

  1. Feb 22, 2010 #1
    I did a least squares fit project for physics and now i have to say the value of G and the slope.

    I know that slope is m from the equation

    y = mx+b

    but how do i determine G?
     
  2. jcsd
  3. Feb 22, 2010 #2
    Perhaps it would help to provide the variables that you are plotting in your graphical analysis?
     
  4. Feb 22, 2010 #3
    Sorry....

    The x values i am using are:
    0.03139
    0.05198
    0.09315
    0.1755

    The y values i am using are:
    0.00306
    0.00514
    0.00929
    0.01729
     
  5. Feb 22, 2010 #4
    Err...no not the values of the variables...I need to know what the variables in question are. ie what x and y represent.
     
  6. Feb 22, 2010 #5
    Y is acceleration (m/s²) and x is hard to explain.

    We measured these values using an air track where we had a mass hanging which is m1 and than we had a cart m2

    x values is (m1/m1+m2)
     
  7. Feb 22, 2010 #6
    Air track? Hm...I guess I have to trouble you to explain the set-up a bit.
     
  8. Feb 22, 2010 #7
    maybe this can help better
     

    Attached Files:

  9. Feb 22, 2010 #8
    From the worksheet, they already derived the relation:
    [tex]a = \frac{m_{1}}{m_{1} + m_{2}}g[/tex]​
    which is in the form y = mx + c.

    Hence, from the plot, we expect the gradient m to be g, and c to be zero. If c is not zero, then it probably indicates the existence of some experimental error. So, in fact, your gradient m is the value of g measured by the experiment.
     
  10. Feb 22, 2010 #9
    yes c is really close to 0, its 1.848x10^-4

    y = 0.09651x + 1.848x10^-4

    and sorry i didnt understand really well about the g part, could you explain in other words...

    I was thinking of this but dont know if i am right or not, but i was going to do

    g = (m1/(m1+m2))*a

    thanks for trying to help me
     
  11. Feb 22, 2010 #10
    I'll try. By comparing y = mx + c and a = g (m1/(m1+m2)) + 0, we can see that m in fact represents g - basically the gradient of the graph that you obtain when you plot acceleration against (m1/(m1+m2)) in this case is the value of the gravitational acceleration g.

    So, g would be 0.09651 when measured in the units that you are using.
     
  12. Feb 22, 2010 #11
    so there is no work i can show to demonstrate this correct?
     
  13. Feb 22, 2010 #12
    Well, the work of linearising the equation and choosing the variables to plot in fact already fixes the gradient m as g, so no, there is no further need to substantiate why m = g; it is clear from the variables plotted and the given equation from the beginning.

    The only task left for you was to determine the value of the gradient and hence g from your graphical analysis. So, essentially, you are done :)
     
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