Question

Example 2.7 Wc measure the period of
oscillation of a simple pendulum. In
successive measurements, the readings
turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s
and 2.80 s. Calculate the absolute errors,
relative error or percentage error​

Answer 1

Answer:

Arithmetic mean,a mean = 2.63+2.56+2.42+2.71+2.80 / 5

                                        = 13.12 / 5

                           a mean = 2.624 s

Absolute error,

 delta a1 = 2.63 - 2.62

               = 0.01 s

 delta a2 = 2.56 - 2.62

               = 0.06 s

 delta a3 = 2.42 - 2.62

               =0.20 s

 delta a4 = 2.71 - 2.62

               = 0.09 s

 delta a5 = 2.80 - 2.62

               = 0.18 s

delta a mean =|delta a1|+|delta a2|+|delta a3|+|delta a4|+|delta a5|  / 5

                      = 0.108

                      = 0.11 s

 a = a mean (+/-) delta a mean

    = 2.62 (+/-) 0.11

    =2.51 < 2.62 < 2.73

Relative error = 0.11 / 2.6

                       = 0.04 s

Percentage error = delta a mean / a mean x 100

                            = 0.11 / 2.6 x 100

                            =  4 %