The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s, 2.56s, 2.42s, 2.71s and 2.80s respectively. The average absolute error is
A
0.11s
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B
0.01s
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C
1.0s
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D
0.1s
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Solution
The correct option is A0.11s Find true or mean value of time period of oscillation. αtrue=αmean=α0 =α1+α2+α3+...+αnn=1n∑in=1ai =2.63+2.56+2.42+2.71+2.805 =2.62sec
Find mean absolute error in time period of oscillation. Δαmean=|Δalpha1|+|Δα2|+|Δα3|+...⋅+|Δαn|n Δαmean is also represented as ¯¯¯¯¯Δα⇒a =α0±¯¯¯¯¯¯¯¯Δα
Average value |ΔT1|=2.63−2.62=0.01 |ΔT2|=2.62−2.56=0.06 |ΔT3|=2.62−2.42=0.20 |ΔT4|=2.71−2.62=0.09 |ΔT5|=2.80−2.62=0.18
Mean obsolute error, ΔT=|ΔT1|+|ΔT2|+|ΔT3|+|ΔT4|+|ΔT5|5 =0.545 =0.108 =0.11sec
Final answer : (b)