Question

If the length and the time period of an oscillating pendulum have errors 1% and 2% respectively. What is the error in the estimate of g?

Answer 1

Given :
Time period of Simple pendulum =2π(√L/g)
where L is the length of pendulum
g =acceleration due to gravity.

Δ L/L=1%=1/100=0.01
ΔT/T=2%=2/100 =0.02
T=2π(√L/g)

g=4π² (L/T²)

Δg/g =ΔL/L +2Δ T/T

Δg/g=0.01 +2x0.02

=0.05
Δ(g/g)%=0.05x100=5%
∴The error in the estimation of g is 5%