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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum in the experiment is recorded as `2.63 s , 2.56 s , 2.42 s , 2.71 s , and 2.80 s`. Find the average absolute error.

A

0.11 s

B

0.12 s

C

0.13 s

D

0.14 s

Text Solution

Verified by Experts

The correct Answer is:
A
the mean period of oscillation of the pendulum is
`T_("mean")=(underset(i=1)overset(n)(sum)T_(i))/(n),T_("mean")=((2.63+2.56+2.42+2.71+2.80))/(5)s`
=`(13.12)/(5)s=2.624s=2.62s`
(Rounded off to two decimal places)
The absolute errors in the measurement are
`DeltaT_(1)=2.62s-2.63s=-0.01s,DeltaT_(2)=2.62s-2.56s=0.06s`
`DeltaT_(3)=2.62s-2.42s=0.20s,DeltaT_(4)=2.62s-2.71s=-0.09s`
`DeltaT_5=2.62s-2.80s=-0.18s` Mean absolute error is
`DeltaT_("mean")=(underset(i=1)overset(n)sum|DeltaT_(i)|)/(n)`
`DeltaT_("mean")=((0.01+0.06+0.20+0.09+0.18))/(5)s`
`=(0.54)/(5)s=0.11s`
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