Question

we read the period of oscillation of a simple pendulum .In successive measurements ,the reading turn out to be 2.63s,2.56s,2.42s,2.71s and 2.80. calculate the absolute errors relative error or percentage error.

Answer 1

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 %

Answer 2

Answer:

The mean period of oscillation of the pendulum

T=(2.63+2.56+2.42+2.71+2.80)s5

=13.125s

=2.624s

=2.62s

As the periods are mesured to a resolution of 0.01 s, 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

2.56s−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 arithmatic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is

ΔTmean=[(0.01+0.06+0.20+0.09+0.18)s]/5

=0.54s/5

=0.11s

That means, the period of oscillation of the simple pendulum is (2.62±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 hyndredth. A more correct way will be to write

T=2.6±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. In this case, the 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

δa=0.12.6×100=4%