Question

We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63s,$ $2.56s,$ $2.42s,$ $2.71s,$ and $2.80s$. Calculate the absolute errors, relative error or percentage error.

Answer 1

Hint
To solve this question, we need to find out the mean of the values given in the problem. From the mean, we can find out the absolute errors. Then from the mean of the absolute errors, we will get the relative error or the percentage error.
The formula used in solving this question is
$\Rightarrow \Delta {T_{rel}} = \dfrac{{\Delta {T_{avg}}}}{\mu } \times 100$, where $\Delta {T_{rel}}$ is the relative error or the percentage error, $\Delta {T_{avg}}$ is the mean of the absolute errors, and $\mu $ is the mean of the measurements.

Complete step by step answer
To find the absolute errors of each of the measurements, we need to find out the mean of all the measurements given in the problem.
The mean $\mu $of the measurements is given by
$\Rightarrow \mu = \dfrac{{{T_1} + {T_2} + {T_3} + {T_4} + {T_5}}}{5}$
$\Rightarrow \mu = \dfrac{{2.63 + 2.56 + 2.42 + 2.71 + 2.80}}{5}$
On solving we get
$\Rightarrow \mu = 2.624$
We need to round off this value to the same number of decimal places as the values are given in the problem
$\therefore \mu = 2.62$
Now, to find the absolute errors, we need to subtract the mean from each of the values given.
For the first measurement
$\Rightarrow \Delta {T_1} = |2.63 - 2.62| = 0.01s$
For the second measurement
$\Rightarrow \Delta {T_2} = |2.56 - 2.62| = 0.06s$
For the third measurement
$\Rightarrow \Delta {T_3} = |2.42 - 2.62| = 0.20s$
For the fourth measurement
$\Rightarrow \Delta {T_4} = |2.71 - 2.62| = 0.09s$
For the fifth measurement
$\Rightarrow \Delta {T_5} = |2.80 - 2.62| = 0.18s$
For calculating the relative error, we have to find the mean of the absolute errors.
$\therefore \Delta {T_{avg}} = \dfrac{{\Delta {T_1} + \Delta {T_2} + \Delta {T_3} + \Delta {T_4} + \Delta {T_5}}}{5}$
Putting the above values, we get
$\Rightarrow \Delta {T_{avg}} = \dfrac{{0.01 + 0.06 + 0.20 + 0.09 + 0.18}}{5}$
$\Rightarrow \Delta {T_{avg}} = 0.108s$
Rounding off to two decimal places, we get
$\Rightarrow \Delta {T_{avg}} = 0.11s$
Now, the relative error, or the percentage error is
$\Rightarrow \Delta {T_{rel}} = \dfrac{{\Delta {T_{avg}}}}{\mu } \times 100$
$\Rightarrow \Delta {T_{rel}} = \dfrac{{0.11}}{{2.62}} \times 100$
On solving, we get
$\Rightarrow \Delta {T_{rel}} = 0.04 \times 100$
$\therefore \Delta {T_{rel}} = 4\% $

Note
While calculating the absolute errors, do not forget to take the modulus of the differences. As the name suggests, the absolute error is the magnitude of the deflection of a measurement from its mean value. If the modulus is not taken, then we may get the mean of the absolute errors as incorrect, which will make the relative error value incorrect.