The period of oscillation of a simple pendulum is T=2π√L/g where L is effective length of the pendulum and g is acceleration due to gravity.if percentage error in the measurement of T is 2% and in the measurement of L is 1%,then percentage error in measurement of g is
(1) 5%
(2) 2%
(3) 3%
(4) 1%
{kindly give explanation,meathod,formula applied and most importantly full solution}
(THANKS IN ADVANCE)
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you know, time period of simple pendulum is T = 2π√{L/g}
For error ,
±∆T/T = ±1/2 ∆L/L ± 1/2 ∆g/g
Here we have to find error in g
So, formula will be ,
1/2 × ∆g/g = ∆T/T + 1/2 × ∆L/L
For finding % error , multiply 100 both sides,
100 × 1/2 × ∆g/g = 100 × ∆T/T + 100 × 1/2 × ∆L/L
100 × ∆g/g = 2 {100 × ∆T/T} + 100 × ∆L/L
% error in g =2 × % error in T + % error in L
Given,
% error in T = 2%
% error in L = 1 %
∴ % error in g = 2 × 2 % + 1 % = 5%
Hence, option (a) is correct.
For error ,
±∆T/T = ±1/2 ∆L/L ± 1/2 ∆g/g
Here we have to find error in g
So, formula will be ,
1/2 × ∆g/g = ∆T/T + 1/2 × ∆L/L
For finding % error , multiply 100 both sides,
100 × 1/2 × ∆g/g = 100 × ∆T/T + 100 × 1/2 × ∆L/L
100 × ∆g/g = 2 {100 × ∆T/T} + 100 × ∆L/L
% error in g =2 × % error in T + % error in L
Given,
% error in T = 2%
% error in L = 1 %
∴ % error in g = 2 × 2 % + 1 % = 5%
Hence, option (a) is correct.
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