Error in t (ms): (0.005*t+0.05)
v0 = (1.800 ± 0.013) m/s
g = (9.49 ± 0.08) m/s2
= 0.128061 for 8 degrees of freedom. (
probability:
99.9999)
Note: This last plot illustrates a rather extreme assumption: the timer
is supposed to have a 0.5% accuracy error, and this is accounted for
in the same manner as we do it for multimeter measurements ( 0.5% + 1/2 digit )
- not quite the same expression, but the same idea.
Apart from an expected increase in the error for g
it is nothing special here :-) ... Unless you are amazed by the 99.99% in
probability; but this is also expected, since the errors are so big that any
other new draw of points, with the same big error, will also accommodate the
previously calculated line of best fit equally well.
Read more about the meaning of
on Dr. Harrison's
web-page,
and particularly pages 7 and 8 of the syllabus
on fitting principles posted on
his page.
Conclusion
There are a few interesting things one can learn from the above plots.
First, the mean value for g is constantly
less than expected. Typically it's 9.45 ... 9.48 m/s2 with errors
of 0.02 m/s2 or such.
Even for a "good fit"
like (5), g is far away from the expected
9.80 m/s2 value.
Observe that I say "far away" because the 9.80 m/s2
is neither in a 1-sigma, nor 2-sigma distance from the mean value of g
we got from the experiment.
Secondly, in all cases displayed above, the experimental points were lying
on a straight line pretty well. You cannot see any spread in the points location.
The residuals (in the rectangular frame on each plot) show you the region around
the best fit line, zoomed-in - so you see how points lay around the line of
best bit! But for a hand-made graph you shouldn't expect to see any deviation
from that line, given the scale you may have for the plot.
Third,
the error bars are too small to be displayed (in all cases -
except for plot (7) where the errors are overestimated).
This means that you cannot apply the swinging/shifting of the fit line through
the points in order to get an estimate of the errors in the slope from the graph. See Lab.
manual (pag. 136) for details on how this procedure is supposed to work.
The fit is a straight line as expected, but the value of g is
consistently smaller than the known value (9.80 m/s2 ).
This points out to the fact that the acceleration
of the plastic ball is smaller
due to some other effects, most probably because of the friction.
There is a catch here though,
since friction depends on the velocity, and therefore
the acceleration is not constant.
Hence, using the frictionless theory (which assumes constant acceleration)
to get the acceleration with friction
it's not appropriate in principle.
What can we do if we would like to get some estimate for the error in g ?
It's clear that even in the ideal case (of a nice
spread around the line, and not a perfect alignment as we got it) we
do not get by hand a value of error for g
so precise as the fit program gives it. Perhaps we get it bigger, say by
a factor of 2 or 3. This is a reasonable expectation.
As for our situation we
are still "out of luck", because we cannot do the swinging/shifting
required for error evaluation from the graph.
So how do we get an error, anyway?
I have some clues about what can be done ...
Hint: try to draw the graph by hand and
see how accurately you are able to:
a) put the points in their places,
Hint question:
are you able to plot the values of
t and s/t
given with 4 significant digits?
b) draw the "exact" line of best fit,
c) read the raise and run of the line right from the graph,
to calculate the
slope.
Well, I guess you already figured it out that the situation is far for being
desperate. There are plotting and
reading-from-the-graph errors in all the steps above!
You just have to account for them and get the error in g.
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Last updated: April 7, 2003
© Sorin Codoban, 2003.