We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63s, 2.56s, 2.42s , 2.71s and 2.80s. Calculate the absolute errors, relative error or percentage error.

Explanation

Mean period,
`barT=(2.63+2.56+2.42+2.71+2.80)/(5)`
`=(13.12)/(5)`
`=2.624`
`barT~~2.62 s`
The erros in the measurements are,
`2.63-2.62=DeltaT_(1)=0.01`
`2.56-2.62=DeltaT_(2)=-0.06`
`2.42-2.62=DeltaT_(3)=-0.20`
`22.71-2.62=DeltaT_(4)=0.09`
`2.80-2.62=DeltaT_(5)=0.18`
Mean error,
`DeltaT=(|DeltaT_(1)|+|DeltaT_(2)|+|DeltaT_(3)|+|DeltaT_(4)|+|DeltaT_(5)|)/(5)`
`=(0.01+0.06+0.20+0.09+0.18)/(5)`
`=(0.54)/(5)`
Average absolute error `Delta barT~~0.11s`
We will write periodic time error by,
`T=(2.62+0.11)s`
`:.` Relative error `=(Delta barT)/(barT)=(0.11)/(0.62)=0.04198`
`~~0.042`
`:.` Percentage error `=(DeltabarT)/(barT)xx100%`
`=0.042xx100`
`=4.2%`