Question

We 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, 2.71 S,

and 2.80 S.

Calculate the absolute errors, relative error and 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 %

Explanation: