Question

What is the error in the estimation of g if the length and time period of an oscillating pendulum have errors of 1% and 2%?

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%

Answer 2

T = 2 π √(L / g)

Rearrange the above equation to get “g”

g = 4 π² L / T²

From above,
(Δg/g × 100) = (ΔL/L × 100) × (2∆T/T × 100)
= 1% + (2 × 2%)
= 5%

∴ Percentage error in estimation of g is 5%