Physics

In an experiment, the period of oscillation of a simple pendulum was observed to be $$2.63 s$$, $$2.56 s$$, $$2.42 s, 2.71 s$$, and $$2.80 s$$. The mean absolute error is:


ANSWER

$$0.11 s$$


SOLUTION
The mean period of oscillation of the pendulum is
$$\displaystyle T_{mean} = \frac {\sum_{i=1}^{n} T_i}{n}; T_{mean}\,=\ \frac {(2.63+2.56+2.42+2.71+2.80)}{5} s$$
$$\displaystyle \frac {13.12}{5} s = 2.624 s = 2.62 s$$
(Rounded off to two decimal places)
The absolute errors in the measurement are
$$\Delta T_1 =2.62s - 2.63s = 0.01 s;\Delta T_2 = 2.62s - 2.56s = 0.06s$$
$$\Delta T_3 = 2.62s - 2.42s = 0.20s;\Delta T_4 =  - 2.71s - 0.09s$$
$$\Delta T_5 =  2.62s - 2.80s = - 0.18s $$
Mean absolute error is
$$\displaystyle T_{mean} = \frac {\sum_{i=1}^{n} |\Delta T_i|}{n}$$
$$\displaystyle T_{mean} = \frac {0.01+0.06+0.20+0.09+0.18}{5} s$$
=$$\displaystyle \frac {0.54}{5}s = 0.11 s$$
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