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

(N/A) The mean period of oscillation of the pendulum is:
$T = \frac{(2.63 + 2.56 + 2.42 + 2.71 + 2.80) \; s}{5} = \frac{13.12}{5} \; s = 2.624 \; s \approx 2.62 \; s$.
The absolute errors in the measurements are:
$|\Delta T_1| = |2.63 - 2.62| = 0.01 \; s$
$|\Delta T_2| = |2.56 - 2.62| = 0.06 \; s$
$|\Delta T_3| = |2.42 - 2.62| = 0.20 \; s$
$|\Delta T_4| = |2.71 - 2.62| = 0.09 \; s$
$|\Delta T_5| = |2.80 - 2.62| = 0.18 \; s$
The mean absolute error is:
$\Delta T_{\text{mean}} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} \; s = \frac{0.54}{5} \; s = 0.108 \; s \approx 0.11 \; s$.
The relative error is:
$\text{Relative error} = \frac{\Delta T_{\text{mean}}}{T} = \frac{0.11}{2.62} \approx 0.042$.
The percentage error is:
$\text{Percentage error} = \frac{\Delta T_{\text{mean}}}{T} \times 100\% = \frac{0.11}{2.62} \times 100\% \approx 4.2\% \approx 4\%$.