Question

The period of oscillation of a simple pendulum in the experiment is recorded as $2.63s$, $2.56s$, $2.42s$, $2.71s$ and $2.80s$ respectively. The average absolute error is

• A. $0.11s$
• B. $0.01s$
• C. $1.0s$
• D. $0.1s$

Answer 1

The correct option is A $0.11s$

Find true or mean value of time period of oscillation.
${\alpha }_{\text{true}}={\alpha }_{\text{mean}}={\alpha }_0$
$=\frac{{\alpha }_1+{\alpha }_2+{\alpha }_3+...+{\alpha }_n}{n}=\frac{1}{n}{\sum }_n^i=1a_i$
$=\frac{2.63+2.56+2.42+2.71+2.80}{5}$
$=2.62\text{sec}$
Find mean absolute error in time period of oscillation.
$\frac{\Delta {\alpha }_{\text{mean}}=\left|\Delta alph{a}_1\right|+\left|\Delta {\alpha }_2\right|+\left|\Delta {\alpha }_3\right|+..{.}_{\cdot }+\left|\Delta {\alpha }_n\right|}{n}$
${\Delta }_{{\alpha }_{\text{mean}}}$ is also represented as ${\overline{\Delta }}_{\alpha }\Rightarrow a$
$={\alpha }_0\pm \overline{\Delta \alpha }$
Average value
$\left|\Delta T_1\right|=2.63-2.62=0.01$
$\left|\Delta T_2\right|=2.62-2.56=0.06$
$\left|\Delta T_3\right|=2.62-2.42=0.20$
$\left|\Delta T_4\right|=2.71-2.62=0.09$
$\left|\Delta T_5\right|=2.80-2.62=0.18$
Mean obsolute error,
$\frac{\Delta T=\left|\Delta T_1\right|+\left|\Delta T_2\right|+\left|\Delta T_3\right|+\left|\Delta T_4\right|+\left|\Delta T_5\right|}{5}$
$=\frac{0.54}{5}$
$=0.108$
$=0.11\text{sec}$
Final answer : (b)