Question

The period of oscillation of a simple pendulam in the experiments is recorded as $2.63s,2.56s,2.42s,2.71s\text{and}2.80$ respectively. The average absolute error is

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

Answer 1

The correct option is D $0.11s$

Find the average value of period of oscillation.
Given:
$T_1=2.63s{T}_2=2.56s{T}_3=2.42s{T}_4=2.71s{T}_5=2.80s$
Therefore, the average value of period of oscillation is given by,
$\text{Average value}T=\frac{2.63+2.56+2.42+2.71+2.80}{5}\text{Average value}T=2.62sec$

Find the Absolute Error of period of oscillation.
Hence, the absolute error is given by,
$\left|\Delta T_1\right|=T_1-T=2.63-2.62=0.01\left|\Delta T_2\right|=T-T_2=2.62-2.56=0.06\left|\Delta T_3\right|=T-T_3=2.62-2.42=0.20\left|\Delta T_4\right|=T_4-T=2.71-2.62=0.09\left|\Delta T_5\right|=T_5-T=2.80-2.62=0.18$

Find the mean Absolute Error of period of oscillation.
Therefore,
Mean absolute error is given by,
$\Delta T=\frac{\left|\Delta T_1\right|+\left|\Delta T_2\right|+\left|\Delta T_3\right|+\left|\Delta T_4\right|+\left|\Delta T_5\right|}{5}\Delta T=\frac{0.54}{5}=0.108\approx 0.11sec$