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 wristwatch 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.

The only factors concerned with T are of L and g

thus we can writeg = K L/T^{2}

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%