A "seconds" pendulum is one that goes through its equilibriumposition once each second. (The period of the pendulum is 2.000 s.)The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 mat Cambridge, England. What is the ratio of the free-fallaccelerations at these two locations? (Cambridge / Tokyo) = 1 + A "seconds" pendulum is one that goes through its equilibriumposition once each second. (The period of the pendulum is 2.000 s.)The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 mat Cambridge, England. What is the ratio of the free-fallaccelerations at these two locations? (Cambridge / Tokyo) = 1 +
Explanation / Answer
Modified: ---------- we know T = 2 [ L / g ] for seconds pendulum time period T = 2 s i.e., period T = constant L g ( gc / g t ) =[ L c / L t ] = [ 0.9942 m / 0.9927 ] = 1.001511 =1 + 0.001511