Question

in a successive measurement of the period of oscillation of a simple pendulum , the readings turn out to be 2.63s ,2.56s , 2.42s , 2.71s , 2.80s . calculate absolute error , relative error percentage error ?

Answer 1

The observed or measured values of time period are: $t_1=2.63s,t_2=2.56s,t_3=2.42s,t_4=2.71s,t_5=2.80s$.
The mean period of oscillation of the pendulum

$t_m=\frac{2.63+2.56+2.42+2.71+2.80}{5}=\frac{13.12}{5}=2.624s$

As the periods are measured to a resolution of 0.01 s, all time periods are given to the second decimal place. Therefore, it is proper to put this mean time period also to the second decimal place.

So, $t_m=2.62s$

The absolute error in the measurement are :

$$\begin{gathered} t_m-t_1=2.62-2.63=-0.01s \\ {t}_m-t_2=2.62-2.56=0.06s \\ {t}_m-t_1=2.62-2.42=0.20s \\ {t}_m-t_1=2.62-2.71=-0.09s \\ {t}_m-t_1=2.62-2.80=-0.18s \end{gathered}$$

Mean absolute error, $\Delta t_{mean}=\frac{0.01+0.06+0.20+0.09+0.18}{5}=\frac{0.54}{5}=0.11s$

So, the time period of simple pendulum is

$t=\left(2.62\pm 0.11\right)s$

It means, t lies between $\left(2.62+0.11\right)s$ and $\left(2.62-0.11\right)s$ or between 2.73 s and 2.51 s.

The relative error, $\delta a=\pm \frac{0.11}{2.62}=\pm 0.041$

The percentage error,

$\delta a\times 100=\pm 0.041\times 100=\pm 4.1\%$