The readings of a length come out to be 2.63 m , 2.56 m , 2.42 m , 2.71 m and 2.80 m . Calculate the absolute errors and relative errors or percentage errors. What do you think of the actual value of the length and its limits ?

Explanation

The mean period of oscillation of the pendulum
`T = ((2.63+2.56+2.42+2.71+2.80)s)/(5)`
`=(13.12)/(5)`s
=2.624s
=2.62s
As the periods are measured to a resolution of 0.01s, all times are to the second decimal, it is proper to put this mean period also to the second decimal.
The errors in the measurements are
`2.63s=2.62s = 0.01s`
256s - 2.62s = -0.06s
2.42s-2.62s = -0.20s
2.71s-2.62s = 0.09s
2.80s-2.62s = 0.18s
Note that the errors have the same units as the quantity to be measured.
The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes ) is
`Delta T_("mean") =[(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]//5`
`=0.54 s//5`
`=0.11s`
That means , the period of oscillation of the simple pendulum is `(2.62 pm 0.11) ` s i.e. . It lies between (2.62 + 0.11) s and (2.62 - 0.11) s or between 2.73 s and 2.51 s . As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second . Hence there is no point in giving the period to a hundredth. A more correct way will be to write .
`T = 2.6 pm 0.1s`
Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7 . We indicate this by saying that the measurement has two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7 .
For this example, the relative error or the percentage error is
`deltaa =(0.1)(2.6) xx 100 = 4%`