# Determine g from least squares fit line?

1. Feb 22, 2010

### noname1

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. Feb 22, 2010

### Fightfish

Perhaps it would help to provide the variables that you are plotting in your graphical analysis?

3. Feb 22, 2010

### noname1

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

4. Feb 22, 2010

### Fightfish

Err...no not the values of the variables...I need to know what the variables in question are. ie what x and y represent.

5. Feb 22, 2010

### noname1

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)

6. Feb 22, 2010

### Fightfish

Air track? Hm...I guess I have to trouble you to explain the set-up a bit.

7. Feb 22, 2010

### noname1

maybe this can help better

#### Attached Files:

• ###### scan0001.jpg
File size:
60.1 KB
Views:
52
8. Feb 22, 2010

### Fightfish

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.

9. Feb 22, 2010

### noname1

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

10. Feb 22, 2010

### Fightfish

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. Feb 22, 2010

### noname1

so there is no work i can show to demonstrate this correct?

12. Feb 22, 2010

### Fightfish

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 :)