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thank you ... for example if we have like this ? wat should we do ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_52.pdf
it Q2 (b) - the table
thanks again
okay for the first value of t, which is 18.9 +/- 0.1
so the periodic time is 1.89 and if you square the periodic time, you obtain T^2 = 3.57
now here, firstly, you will need to consider the errors in periodic time, so:
if
(uncertainty in t) / t = (uncertainty in T) / T
We obtain a constant uncertainty in T for all the values in the table and this is 0.01
There after we calculate the uncertainty in T^2,
this will be: (uncertainty in T^2) / T^2 = ( uncertainty in T x 2)/ T
therefore, for eg if we are using the first value
(uncertainty in T^2) = [ (0.01 x 2) / 1.89 ] x 3.57
hence you'll obtain 0.04
Hope that helped!