The acceleration due to gravity on the surface of moon is 1.7 ms-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 ms-2
The acceleration due to gravity on the surface of moon is 1.7 ms-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 ms-2
Solution and Explanation
Acceleration due to gravity on the surface of moon, g'=1.7 ms-2
Acceleration due to gravity on the surface of earth, g=9.8 ms-2
Time period of a simple pendulum on earth, T=3.5 s
\(T=2π\sqrt\frac{l}{g}\)
Where,
l is the length of the pendulum
\(∴l=\frac{T^2}{(2π)^2}×g\)
\(=\frac{(3.5)^2}{4×(3.14)^2}×9.8 m\)
The length of the pendulum remains constant.
On moon’s surface, time period, \(T'=2π\sqrt\frac{l}{g'}\)
\(=2π\sqrt\frac{\frac{(3.5)^2}{4×(3.14)^2}×9.8}{1.7}=8.4 s\)
Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
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