Question

the time period of simple pendulum is t=2π√l/g .measured value of l is 20cmknown to 1mm accuracy and time for 100 oscilations of pendulum is found to be 90s using a wrist watch of 1 s resolution the accuracy in determination ogf g is
a)2%
b)1%
c)3%
d)5%

Answer 1

T= Time period of one oscillation

t= Time period of 100 oscillation

For 100 oscillation, t=100T

Take ln and differentiate.

tΔt=TΔT=902.....(1)

Time period of one oscillation

T=2πgL

g=4π2T2L

Take ln on both side and differentiate

gΔg=±(LΔL+2TΔT)

gΔg=±(200.1+902)==±2.7%

Hence, error in gravity is ±2.7%= 3 %