Pendulum errors ?

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Answer 1 · 2017-04-11T14:13:45

$T = 2.9954$ [s]

The pendulum period for small oscillations is given by

$T=2pisqrt(l/g)$

applying the logarithm $log$ to both sides

$log(T)=log(2pi)+1/2(logl-logg)$

deriving totally both sides

$(dT)/T = 1/2((dl)/l-(dg)/g)$ or for finite differences

$(DeltaT)/T = 1/2((Deltal)/l-(Deltag)/g)$

so we have:

$Delta l = 0$
$Deltag = 9.82-9.79=0.03$

and consequently

$DeltaT = -T/2((Delta g)/g) = -0.00459653$

so at the new location the period was shortened and now is

$T = 2.9954$ [s]

The claimed precision regarding $T$ has not sense because the precision in $g$ is much lesser than the demanded precision on $T$ .

Answer 2 · 2017-04-11T14:16:30

The new period is $=2.99541s$

The period of a pendulum is

$T=2pisqrt(l/g)$

In the first place, we have

$T_1=3.00000=2pisqrt(l/9.79)$

In the second place, we have

$T_2=2pisqrt(l/9.82)$

Therefore,

$T_2/3.00000=sqrt(9.79/9.82)$

$T_2=3.00000*sqrt(9.79/9.82)$

$=2.99541$