In a series of successive measurements in an experiment, the readings of the period of oscillation of a simple pendulum were found to be 2.63s, 2.56s, 2.42, 2.71s and 2.80 s`. Calculate
(i) the mean value of the period of oscillation
(ii) the absolute error in eah measurement
(iii) The men absolute error (iv) the relative error (v) the percentage error. Expresss the results in proper form.

Explanation

`t_(1) = 2.63s, t_(2) = 2.56s` ,
`t_(3) = 2.42s, t_(4) = 2.71s, t_(5) = 2.80s`
(i) `T_(m) = (t_(1) + t_(2) + t_(3) + t_(4) + t_(5))/(5) = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/(5)`
`T_(m) = (13.12)/(5)` 2.624s
`T_(m)` = 2.62s (Rounded off to `2^("nd")` decimal place)
(ii) Absolute error `|DeltaT| = |T_(m) - t|`
`|DeltaT_(1)| = |2.62 - 2.63|` = +0.01s
`|DeltaT_(2)| = |2.62 -2.56|` = +0.06s
`|DeltaT_(3)| = |2.62 - 2.42|` = +0.20s
`|DeltaT_(4)| = |2.62 - 2.71|` = +0.09s
`|DeltaT_(5)| = |2.62 - 2.80|` = +0.18s
(iii) mean absolute error = `(sum|DeltaT_(i)|)/(n)`
`DeltaT_(m) = (0.01+0.06+0.20+0.09+0.18)/(5)`
`DeltaT_(m) = (0.54)/(5) = 0.108`s = 0.11s (Rounded off to `2^("nd")` decimal place)
(iv) Relative error: `S_(T) = (DeltaT_(m))/(T_(m)) = (0.11)/(2.62) = 0.0419`
`S_(T) = 0.04`
(v) Percentage error in `T = 0.04 xx 100 4%`
(vi) Time period of simple pendulum = `T = (2.62 pm 4%)`