Question

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

• A. $0.11\,s$
• B. $0.12\,s$
• C. $0.13\,s$
• D. $0.14\,s$

Answer 1

Answer: (A)

The mean period of oscillation of the pendulum is
$T_{mean}=\frac{\displaystyle \sum_{i=1}^n T_i}{n}$ ;
$T_{mean}=\frac{\left(2.63 + 2.56 + 2.42 + 2.71 + 2.80\right)}{5}\,s$
$=\frac{13.12}{5}\,s=2.624\,s=2.62\,s$
(Rounded off to two decimal places)
The absolute errors in the measurement are
$\Delta T_{1}=2.62\,s-2.63\,s=-0.01\,s$;
$\Delta T_{2}=2.62\,s-2.56\,s=0.06\,s$
$\Delta T_{3}=2.62\,s-2.42\,s=0.20\,s$;
$\Delta T_{4}=2.62\,s-2.71\,s=-0.09\,s$
$\Delta T_{5}=2.62\,s-2.80\,s=-0.18\,s$
Mean absolute error is
$\Delta T_{mean}=\frac{\displaystyle \sum_{i=1}^n\left|\Delta T_{i}\right|}{n}$
$\Delta T_{mean}=\frac{\left(0.01+0.06+0.20+0.09+0.18\right)}{5}s$
$=\frac{0.54}{5}s=0.11\,s$