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 moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s-2)
Acceleration due to gravity on the surface of moon, g′=1.7ms−2 Acceleration due to gravity on the surface of earth, g=9.8ms−2 Time period of a simple pendulum on earth, T=3.5s T=2πgl where, l is the length of the pendulum ∴l=(2π)2T2×2 =4×(3.14)2(3.5)2×9.8m The length of pendulum remains constant On moon's surface, time period, T′=2πg′l =2π1.74×(3.14)2(3.5)2×9.8=8.4 s Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
