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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 |
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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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