Question

If the length and time period of an oscillating pendulum have errors of 1%and2%respectively.what is the error in estimate of g

Answer 1

Answer:

Explanation:

Attachment for reference

Answer 2

Explanation:

given:

time period of simple pendulum=2π(√L/g)

where l is the length of Pendulum

g= acceleration due to gravity

∆L/L=1℅=1/100=0.01

∆T/T=2℅=2/100=0.02

T=2π(√L/g)

g=4π^2(L/T^2)

∆g/g=∆L/L+2∆T/T

∆g/g=0.01+2×0.02=0.05

∆(g/g) ℅=0.05×100=5℅