A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9928 m at Tokyo and 0.9943 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

Question

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9928 m at Tokyo and 0.9943 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
(g Cambridge / g Tokyo)?
any help?

bobpursley · Answer

T=2PI(L/g) so the ratio is ... 1=sqrt(.9928/gtokio*gengland/.9943) squaring both sides.. gengland/gtokyo= .9943/.9928

sisca · Answer

awesome, Thanks a lot...