Question
Sat February 09, 2013 By: Pramod R

# The period of oscillation of a simple pendulum is T = 2pi(L/g)^1/2 , L is about 10 cm and is known to 1 mm accuracy. The period of oscillation is about 0.5 s. The time for 100 oscillation is measured with a wrist watch of 1 s resolution . What is the accuracy in the determination of g ? In this question asked by me last time , please give the answer also, as the answer given by you is not full understandable. eg; what is delta L , L , delta T,T etc..and how accuracy can be connected to relative error. pls explain.

Wed February 13, 2013
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%
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