The period of oscillation of a simple pendulum in the experiment is recorded as `2.63 s , 2.56 s , 2.42 s , 2.71 s , and 2.80 s`. Find the average absolute error.
A. `0.1 s`
B. `0.11 s`
C. `0.01 s`
D. `1.0 s`
A. `0.1 s`
B. `0.11 s`
C. `0.01 s`
D. `1.0 s`
Correct Answer – b
Average value `= (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/(5)`
`= 2.62sec`
Now `|Delta T_(1)| = 2.63 – 2.62 = 0.01`
`|Delta T_(2)| = 2.62 – 2.56 = 0.06`
`|Delta T_(3)| = 2.62 – 2.42 = 0.20`
`|Delta T_(4)| = 2.71 – 2.62 = 0.09`
`|Delta T_(5)| = 2.80 – 2.62 = 0.18`
Mean absoltude
`Delta T = (|Delta T_(1)| + |Delta T_(2)| + |Delta T_(3)| + |Delta T_(4)| + |Delta T_(5)|)/(5)`
`= (0.54)/(5) = 0.108 = 0.11sec`