# Curve fitting for Gravity/Conservation of Energy Lab

• Dietrichw
In summary, the problem has to do with sig figs going down to 1. I've checked them multiple times by hand and with sigfig calculators but it is all the same. With 1 sig fig my standard deviation ends up being 0 which I am not sure that is acceptable.

## Homework Statement

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The problem has to do with sig figs going down to 1. I've checked them multiple times by hand and with sigfig calculators but it is all the same. With 1 sig fig my standard deviation ends up being 0 which I am not sure that is acceptable.
It makes sense as the points are super consistent for measuring G but when it comes to filling out the tables having 0 for most parts bothers me.

V^2=g(2H)
Y=mx+b

## The Attempt at a Solution

I have an excel document with all my data and calculations

#### Attachments

• measuring g.xlsx
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It seems that you calculated the h (or Δh) values incorrectly since you are using the same angle, θ or L2, for each run? Or did you keep the angle the same and released the glider at progressively lower points along the track?

Last edited:
andrevdh said:
It seems that you calculated the h (or Δh) values incorrectly since you are using the same angle, θ or L2, for each run? Or did you keep the angle the same and released the glider at progressively lower points along the track?
We kept the angle the same and changed the release point for each set of runs

0 is not the same as an error in one significant figure.
One significant figure means you take the first nonzero digit, so your ##\sigma_g## is 0.06

(In case the first nonzero digit is a 1, we often take the first two digits - because the step from 1 to 2 is so big - and still speak of one significant digit. But it depends: with the relative error in the sigma approximately ##1/\sqrt N## and 5 observations you understand that ##\sigma## itself isn't very accurate at all)​

So as a result you have ##g = 9.56 \pm 0.06 ## m/s2 for the light glider - and have to say something reasonable about the intercept ##0.055 \pm 0.024## (here I show two digits -- the intercept is 2##\sigma## from zero, which may or may not give you a reason to fit y = m x instead of y = mx + b) .

I forgot something -- how accurate is your theta ? This is a systematic error (all h are affected the same way by a deviation from the actual theta), so it has to be folded into the error you calculated for g (in fact you have determined ##g/\sin\theta## and report ##g## -- a 1% error in ##\theta## already dominates the statistical error ! You could be very sophisticated and report ## g = 9.56 \pm 0.06\, (\text stat) \pm 0.1 \,(\text syst) ## m/s2 . But I wonder if teacher will appreciate it...