The acceleration due to gravity on the surface of the moon is 1.7m/s^2 .What is the time period of a simple pendulum on the moon if its time period on earth is 3.5 sec? (take g on earth as 9.8m/s^2.
Answers
Answered by
0
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.
Similar questions