Question

The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

Answer 1

`g=4pi^(2)L//T^(2)`
Here, `T=t/n` and `Delta T=(Delta t)/n`. Therefore, `(Delta T)/T=(Delta t)/t`.
The errors in both L and t are the least count errors. Therefore,
`(Delta g//g)=(Delta L//L)+2(Delta T//T)`
`=0.1/20.0+2(1/90)=0.027`
thus, the percentage error in g is
`100(Delta g//g)=100 (Delta L//L)+2xx100 (Delta T//T)`
`=3 %`