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%?
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Answered by
247
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%
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%
Answered by
146
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%
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%
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