Question #70d85

The equation of simple pendulum

`T=2pisqrt(L/g)`

`=>g=4pi^2*L/T^2`

Taking logarithm on both sides

`lng=ln(4pi^2)+lnL-2lnT`

Taking differential

`(Deltag)/g=(DeltaL)/L-(2DeltaT)/T....(1)`

Now for calculation of maximum error of g every error term should be taken positive.

Hence equation (1) becomes

`(Deltag)/g=(DeltaL)/L+(2DeltaT)/T`

In percent form it becomes

`(Deltag)/gxx100=(DeltaL)/Lxx100+(2DeltaT)/Txx100..(2)`

Now it is given that in measuring `L=20cm` the minimum length measurable by scale being 1mm or 0.1cm the error `DeltaL=0.1cm`

`:.(DeltaL)/L=0.1/20`

Again the resolution of watch used for the measurement of T being 1s (the smallest graduation on the face of the watch),error for the measurement of time for 100 oscillations will be 1s i.e.`DeltaT=1/100s=0.01s`

Now time measured for 100 oscillations is 90s. So time period `T=90/100=0.9s`

Hence `(DeltaT)/T=0.01/0.9`

Now the percent error in measurement of g by equation(2)

`(Deltag)/gxx100=(DeltaL)/Lxx100+(2DeltaT)/Txx100`

`=>(Deltag)/gxx100=0.1/20xx100+(2xx0.01)/0.9xx100`

`~~(0.5+2.2)%=2.7%~~3%`