Class 11 Physics - Chapter Oscillations NCERT Solutions | The acceleration due to gravity on the s
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)
Given, Acceleration due to gravity on the surface of moon, g’ = 1.7 m s-2
Acceleration due to gravity on the surface of earth, g = 9.8 m s-2
Time period of a simple pendulum on earth, T = 3.5 s
We know, T=2π √l/g
Where,
l is the length of the pendulum
∴ l = T2 x g / (2π)2
= (3.5)2 x 9.8m / 4 x (3.14)2
The length of the pendulum remains constant.
On moon's surface, time period,
= T '= 2π √l/g'
= 2π √ [ (3.5)2 x 9.8m / 4 x (3.14)2] / 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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