Question

The period of oscillation of a simple pendulum in the experiment is recorded as \[2.63\;{\rm{s}},2.56\;{\rm{s}},2.42\;{\rm{s}},2.71\;{\rm{s}}\] and . Find the average absolute error.

Answer 1

Hint: This question is based on the period of oscillation. First, we have to know that the period is the least time interval in which the oscillation returns to its initial position. We have to find the average value of the given time period. And then use mean absolute error to get the result.

Complete step by step answer:
The time period recorded in the experiment is $2.63s,2.56s,2.42s,2.71s$ and \[2.80\;{\rm{s}}\].
We know that the period of oscillation is the ratio of the total time of oscillation to that of the number of oscillation. Oscillation is the complete cycle of swinging one way and returning to its mean position.

So, we have to find the mean period of oscillation of the pendulum is given as,
${T_{{\rm{mean}}}} = \dfrac{{\sum\limits_{i = 1}^n {{T_i}} }}{n}$
Now, we have to put the time period of an experiment we get,
$
{T_{{\rm{mean}}}} = \dfrac{{\left( {2.63\;{\rm{s}} + 2.56\;{\rm{s}} + 2.42\;{\rm{s}} + 2.71\;{\rm{s}}} \right)}}{5}\\
\implies {T_{{\rm{mean}}}} = 2.62\;{\rm{s}}
$

So, there are some errors in the time period with mean value.
Therefore, the absolute errors in the measurement for first time period is,
$\Delta {T_1} = \left( {2.63 - 2.62} \right){\rm{s}} = 0.01\;{\rm{s}}$
Absolute errors in the measurement for second time period is,
$\Delta {T_2} = \left( {2.62 - 2.56} \right){\rm{s}} = 0.06\;{\rm{s}}$

Absolute errors in the measurement for third time period is,
$\Delta {T_3} = \left( {2.62 - 2.42} \right){\rm{s}} = 0.20\;{\rm{s}}$

Absolute errors in the measurement for fourth time period is,
$\Delta {T_4} = \left( {2.71 - 2.62} \right){\rm{s}} = 0.09\;{\rm{s}}$

Absolute errors in the measurement for fifth time period is,
$\Delta {T_5} = \left( {2.80 - 2.62} \right){\rm{s}} = 0.18\;{\rm{s}}$

After finding the absolute errors, now we have to find the average absolute error we get,
$
\Delta T = \dfrac{{\left( {0.01 + 0.06 + 0.20 + 0.09 + 0.18} \right)\;{\rm{s}}}}{5}\\
\therefore \Delta T = 0.11\;{\rm{s}}
$

Therefore, the average absolute error is $0.11\;{\rm{s}}$.

Note:
In this question, students must have knowledge of the period of oscillation of a simple pendulum. The average time period is the addition of all the time periods and divided by the number of time periods of oscillation