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In 5 successive measurements, the values of period of oscillation of a simple pendulum are obtained as T_(1)=2.63,T_(2)=2.56,T_(3)=2.42,T_(4)=2.71 and T_(5)=2.80 (in s). Demonstarte the different errors and range of a measurement. |
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Answer» Solution :`T_("mean")=underset(f)(SUM)T_(i)//5=(2.63+2.56+2.42+2.71+2.80)//5` `=(13.12)/(5)=2.624` s [mean value of T] `|DeltaT_(i)|=|2.63,2.62|,|2.56-2.62|,|2.42-2.62|,|2.71-2.62| and |2.80-2.62|` `=0.01,0.06,0.20,0.09 and 0.18` (s) [ABSOLUTE errors] `DeltaT_("mean")=underset(i)(sum)|DeltaT_(i)|//5=(0.01+0.06+0.20+0.09+0.18)//5=0.54//5` `=0.11`s [mean absolute error] RANGE of a measurement `T_("mean")+-DeltaT_("mean")=2.62+-0.11s` Relative error`=DeltaT_("mean")//T_("mean")=(0.11)/(2.62)xx100=4%` |
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