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#