Determine g from least squares fit line?

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To determine the value of gravitational acceleration (g) from a least squares fit line, the slope (m) of the line represents g when plotting acceleration (y) against the ratio of masses (x). The relationship derived from the experiment is a = (m1/(m1+m2))g, which aligns with the linear equation y = mx + c. In this case, the gradient m is found to be 0.09651, indicating that g is approximately 0.09651 in the units used. Since the y-intercept is close to zero, it suggests minimal experimental error, confirming the accuracy of the slope as the value of g. Thus, the analysis is complete, and no further work is needed to validate that m equals g.
noname1
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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?
 
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Perhaps it would help to provide the variables that you are plotting in your graphical analysis?
 
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
 
Err...no not the values of the variables...I need to know what the variables in question are. ie what x and y represent.
 
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)
 
noname1 said:
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)
Air track? Hm...I guess I have to trouble you to explain the set-up a bit.
 
maybe this can help better
 

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From the worksheet, they already derived the relation:
a = \frac{m_{1}}{m_{1} + m_{2}}g​
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.
 
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 don't know if i am right or not, but i was going to do

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

thanks for trying to help me
 
  • #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.
 
  • #11
so there is no work i can show to demonstrate this correct?
 
  • #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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