When you are using a graph to analyse experimental data you generally want to graph your raw data, or very close to the raw data. That is, you are graphing the things that you measured.
In this case we measured the length 𝓁 and we measured the time T, so those are the things you graph.
There are two reasons for this.
Firstly you are often wanting to check that your data obeys the equation you think it does. In this case our theory says that T² ∝ 𝓁, and we test that by graphing T² against 𝓁 and checking that we get a straight line (which we do).
Secondly the graph can do a lot of calculation for you, and in a way that minimises errors.
@JohnRennie We could have also made a graph with r and T^2 and found k by substituting the average of the lengths, right? That is, using r instead of \ell?
When you have a lot of data points on a graph the best fit straight line combines all the points so their errors tend to cancel, and this improves the overall error. As a rough guide if you have N points on the graph the errors are reduced by a factor of √N due to the averaging out of the errors.
@Bml We could do, yes. It's just that in this experiment we kept 𝑟 constant and varied 𝓁.
We could have done a different experiment where we kept 𝓁 constant and varied 𝑟.
Varying 𝓁 is easy of course, since we can easily change the length of the wire we use. We cannot change 𝑟 except by finding a different wire with the different radius.
We know there is some error in our measurements of 𝓁 and T, but the advantage of a graph is that the error in the gradient can be estimated from the scatter in the points.
So we don't need to put in the errors in T and 𝓁 by hand because they are estimated from the graph.
So we find there is about a 5% error in the gradient obtained using linear regression just from the graph.
I realized that one goal of the experiment is to calculate the error on each individual measurement... And report it on the graph! How can this be done?
The errors in T and 𝓁 come from your measurement, and you have to judge how big they are. e.g. I would guess timings are only accurate to about ¹⁄₂ a second and lengths to about a millimetre.
@JohnRennie My professor said that the error on each individual measurement should be calculated either by partial derivatives, or by some error propagation calculation, including the sensibility of the instrument...
@JohnRennie The problem is that I did not understand what he means.... He says that the errors on each of the 40 measurements should be different!
I can tell you how a working experimental physicist would do the analysis. If your prof wants something different then I guess you have to do what he says.
@JohnRennie And after that, represent with a bar in the graph the error, that is, how far the point really deviates from the line...
@JohnRennie My professor told me about partial derivatives today, but I am not sure if that is what I knew with respect to the error of the mean. I knew another method.
@JohnRennie No, I mean: in addition to dividing the single measurement by \sqrt{5}, he wants us to include the error of the single measurement (0.01) in the error of the mean.
@JohnRennie It means that if we put 3.6/√5, we calculate the error in the mean without taking into account the error in the single measurement (0.01), while instead the latter should be included. Or not?
Furthermore, he referred to a more complicated error propagation expression in which it would not be very easy to enter 0.1 as error to calculate the error in the mean.
@JohnRennie Sorry, I am very confused... You said earlier that the error in the mean was calculated with that equation you presented earlier, right? Now, why the standard deviation?
I said there are two ways of calculating the error in the mean:
> So you can calculate the standard deviation of the five values and use that as an estimate of the error. Or you can estimate the error from how accurately you think you can do the measurement.
The first method does not use your estimate of what you think the error is. It just assumes that the error can be estimated from the scatter in the points.
That's what calculating the standard deviation does. It is derived from how different all the values are from each other.
