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the period of oscillation of a simple pendulum is T=2??L/g,L is about 10 cm and is known to 1 mm accuracy.The period of oscillation is about .5s.The time for 100 oscillation is measured with a wrist watch of 1 s resolution.what is the accuracy in the determination of g?

Asked by 24th February 2013, 3:36 PM
Answered by Expert
The only factors concerned with T are of L and g
thus we can write
g = K L/T2
where K is a constant
taking logatithm on both sides
log g = log K + log L +2log T
∆g/g = ∆L/L + 2* ∆T/T
In terms of percentage,
(∆g/g)*100 = (∆L/L)*100 + 2* (∆T/T)*100
Here the resolution or accuracy are that values that could occur as errors .
Time period T= Time of one oscillation= t/n=Total time/ no of oscillation-
Percentage error in L =
(∆L/L)*100 = 100 *(0.1/10) =1%
Percentage error in T =
(∆T/T)*100 = 100 *(1/50) =2%
Percentage error in g =
(∆g/g)*100 = 1% +2 *2%= 5%
Answered by Expert 24th February 2013, 6:59 PM
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