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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

Asked by Megha M 27th September 2012, 9:13 PM
Answered by Expert
Answer:
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%
Answered by Expert 28th September 2012, 10:37 AM
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