Pendulum errors ?

2 Answers
Apr 11, 2017

Answer:

#T = 2.9954#[s]

Explanation:

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#.

Apr 11, 2017

Answer:

The new period is #=2.99541s#

Explanation:

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#