Question

To find the value of `g` using simple pendulum , `T = 2.00 s and l = 1.00 m` were measured . Estimate maximum permissible error in `g` . Also find the value of `g`.

Answer 1

` T = 2 pi sqrt((l)/(g)) rArr = ( 4 pi^(2) l)/( T^(2))`
`(( dg)/( g))_(max) = ( Delta l)/( l) + 2 ( Delta T)/(T) = ((0.01)/(1.00) + 2 (0.01)/(2.00)) xx 100% = 2%`
`g = ( 4 pi^(2) l) /( T^(2)) = ( 4 xx 10 xx 1.00)/((2.00)^(2)) = 10.0 ms^(-2)`
`((dg)/( g))_(max) = ( 2) /(100) or (dg_(max))/( 10.0) = ( 2) /(100)`
(dg)max = `0.2` = max error in `g`
So `g = ( 10.0 +- 0.2 ) ms^(-2)`