Question

If the length and time period of oscillation pendulum have a errors 3% and 2% respectively what is the estimated g

Answer 1

The time period of an oscillating pendulum is given by;

T = 2 π √(L/g)

or

T² = 4 π² (L/g)

or

g = 4 π² (L/T²)

Taking logarithms of both sides we get

log g = 2 log π + log L - 2 log T

Differentiating

(∆g/g) = (∆ L/L) + 2 (∆T/T)

(∆g/g) = (3/100) + 2(2/100) = 3% + 2 × 2% = 7%

The error in g = 7%