Question

In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be $2.63 \,s , 2.56\, s , 2.42\, s , 2.71 \,s$ and $2.80\, s$. Calculate the mean absolute error.

• A. $0.11\, s$
• B. $0.42\, s$
• C. $0.92\, s$
• D. $0.10 \,s$

Answer 1

Answer: (A)

The mean period of oscillation of the pendulum,
$T_{\text {mean }} =\frac{2.63+2.56+2.42+2.71+2.80}{5} $
$=\frac{13.12}{5}=2.624=2.62\, s$
The absolute errors in the measurements are
$\Delta T_{1}=2.63 \,s -2.62 \,s =0.01 \,s$
$\Delta T_{2}=2.56\, s -2.62\, s =-0.06 \,s $
$\Delta T_{3}=2.42 \,s -2.62\, s =-0.20\, s$
$\Delta T_{4}=2.71\, s -2.62\, s =0.09\, s$
$\Delta T_{5}=2.80\, s -2.62\, s =0.18 \,s$
The arithmetic mean of all the absolute errors is
$\Delta T_{\text {mean }} =\displaystyle\sum_{i=1}^{5} \frac{\left|\Delta T_{i}\right|}{5} $
$=[(0.01+0.06+0.20+0.09+0.18)] / 5$
$=0.54 / 5=0.108=0.11 \,s$