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:

$0.11 s$

##### 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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Single Correct Medium Published on 18th 08, 2020
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