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Example 2.7 We measure the period ofoscillation of a simple pendulumInsuccessive measurements, the readingsturn out to be 2 63 s 2.56 s 242 s 2.71sand 2.80 s. Calculate the absolute errors.relative error or percentage error8

Answer»

The mean period of oscillation will be

T=2.63+2.56+2.42+2.71+2.8/5

=2.62 sec

The errors in the measurement wil be

2.63-2.62=.01sec

2.56-2.62=-.06sec

2.42-2.62=-.2sec

2.71-2.62=.09sec

2.80-2.62=.18 sec

so mean of absolute error will be .11sec

Hence Period of oscillation will be (2.62±0.11)

Therefore T=2.6±.1s

relative error or percentage error is given by

=0.1/2.6*100

4%



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