The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ?

Explanation

Here `(Deltal)/(l)=(0.1)/(10)`
`Deltat=1` second, and
time of 100 oscillations, `t=100xx0.5=50s`
`therefore" "(Deltat)/(t)=(1)/(50)`
From`" "t^(2)=4pi^(2)(l)/(g)`
`t^(2)=4pi^(2)(l)/(g)" "g=4pi^(2)(l)/(t^(2))`
`therefore" "(Deltag)/(g)=pm ((Deltal)/(l)+2(Deltat)/(t))`
`therefore" Error, "(Deltag)/(g)xx100=pm((0.1)/(10)+(2xx1)/(50))xx100=pm5%`