The acceleration due to gravity on the surface of moon is 1.7 m s^-2 .
What is the time period of a simple pendulum on the surface of the
moon, if its period on the Earth is 3.5 s ?
Answers
Answered by
2
T = 2pi(l/g)^1/2
T2/T1 = (g1/g2)^1/2
T2/3.5 = (9.8/1.7)^1/2
T2 = 8.4s.
T2/T1 = (g1/g2)^1/2
T2/3.5 = (9.8/1.7)^1/2
T2 = 8.4s.
Similar questions