Question
Thu September 27, 2012 By: Megha M
 

We measure the period of oscillation of a simple pendulam. In successive measurements, the reading is turn out to be 2.63, 2.56, 2.42, 2.71 and 2.80 seconds. calculate the absolute errors. Relative error or percentage error

Expert Reply
Fri September 28, 2012
a1=2.63
a2=2.56
a3=2.42
a4=2.71
a5=2.80
avg a=2.624=2.62
delta a=absolute error in the measurements are=(2.63-2.62)=0.01
                                                                      (2.56-2.62)=-0.06
                                                                       (2.42-2.62)=-0.20
                                                                       (2.71-2.62)=0.09
                                                                        (2.80-2.62)=0.18
mean absolute errors=(0.01+0.06+0.09+0.20+0.18)/5=0.11
relative error=±0.11/2.62=±0.041
percentage error=4.1%
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