Question

If the length and time period of the oscillating pendulum have errors of 1% and 3% respectively then the error in measurement of acceleration due to gravity is

Answer 1

Given conditions ⇒
% error in Length(ΔL/L) = 1 %
% error in time period(ΔT/T)= 3 %

Using the Relation, 
T = 1/2π √(l/g)
On Squaring both the sides, 
T² = 4π² l/g
∴ g = 4π² l/T²

∴ (Δg/g × 100) = (ΔL/L × 100) + (2∆T/T × 100)
    = (1 × 100) + (2 × 3 × 100)
   = 100 + 600
   = 700

Hence, % error in the acceleration due to the gravity is 700/100 = 7 %.


Hope it helps.

Answer 2

Answer:

We know,

Time period , T = 2π√(l/g)

Taking square both sides,

T² = 4π²l/g ⇒g = 4π²l/T²

So, for finding error in accelerating due to gravity we have to take formula,

% error in g = % error in l + 2 × % error in T

Given,

% error in l = 1 %

% error in T = 3 %

∴ % error in g = 1 % + 2 × 3 % = 7 %

Hence, option (d) is correct