Question

In an experiment for determining the value of acceleration due to gravity `(g)` using a simpe pendulum , the following observations were recorded:
Length of the string `(l) = 98.0 cm`
Diameter of the bob `(d) = 2.56 cm`
Time for `10 oscillations (T) = 20.0 s`
Calculate the value of `g` with maximum permissible absolute error and the percentage relative error.

Answer 1

Correct Answer - ` ( 9.80 +- 0.11) ms^(-2)`
Time period for a simple pendulum is `T = 2 pi sqrt((l_(eff))/(g))` …..(i)
where `l_(eff)` is the efective length of the pendulum equal to `(l + (d)/(2))` and time period equals `T = ( 20.0)/( 10) = 2.00 s`
From (i) , we get `g = (4 pi^(2) (_(eff)))/(T^(2))`
To calculate actual value of `g`,
Since `g = ( 4 pi^(2) (98 + 1.28))/((2.00)^(2) = 980 cm s^(-2) = 9.80 ms^(-2)`
Error int he value of `g`:
`(Delta g)/( g) = (Delta l_(eff))/(l_(eff)) + 2 ((Delta T)/( T)) = (Delta l + Delta r )/( l + r) + 2 ((Delta T)/(T))`
Further , since errors can never exceed the least count of he measuring instrument. So , `Delta l = 0.1 cm and Delta r = 0.01 cm`.
`(Delta g)/( g) = (( 0.1 + 0.01)/(98.0 + 1.28)) + 2 ((0.1)/( 20.0))`
`rArr = 0.0011 + 0.01 = 0.0111`
Thus , percentage error `(Delta g)/( g) xx 100 % = 1.1%`
and absolute error `= Delta g = g (0.011) = 0.11 ms^(-2)`
So , `g = ( 9.80 ms^(-2) +- 1.1 %) = ( 9.80 +- 0.11) m s^(-2)`