Period error ?

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Answer 1 · 2016-08-28T19:55:55

#(dT)/T = -1/2(dg)/g#

Error propagation is easily handled using the #log# transformation.
Given #T=pi/2sqrt(L/g)# after tansformation reads

#log_eT = log_e(pi/2) +1/2(log_eL-log_e g)#

After deriving we have

#(dT)/T = 1/2((dL)/L-(dg)/g)#

supposing that #L# is known with precision # dL approx 0#

so

#(dT)/T = -1/2(dg)/g#