If T=2π√L/8

Question

Then given that L=100.0±0.1cm(limit standard errors) and T=2.01±0.01s(limit standard error)
Calculate the value of g and its standard error.

Pls I need answer to that

Answer 1

I guess you mean a pendulum
(Please type entire problem in the future)
let theta = A sin 2pi t/T
h = L (1 - cos theta)
for small theta that is
h = L theta^2/2
h max = (L/2)A^2
Max potential energy = m g h max
= m g (L/2)A^2

theta= A sin 2 pi t/T
v = A L (2 pi/T) cos 2 pi t/T
max V = 2 pi AL/T
so
max Ke = (1/2) m (2pi)^2 A^2 L^2/T^2
so
m g L/2 = (1/2) m (2 pi)^2 L^2/T^2
T^2 = (2 pi)^2 L/g
g = (2 pi)^2 L T^-2
so calculate your g =
then do
dg/dL = 2 pi^2 T^-2
dg/dT = 2 pi (-2)T^-3
now do dg = |dL dg/dL | + |dT dg/dT|

Answer 2

It is a problem on error analysis

Answer 3

That is how the question is stated