14.15 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)

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Vishal Baghel | Contributor-Level 10

8 months ago

Acceleration due to gravity on Moon surface, g' = 1.7 m/ $s^{2}$

Acceleration due to gravity on Earth surface, g = 9.8 m/ $s^{2}$

Time period on Earth, T = 3.5 s

We know T = 2 $π \sqrt{\frac{l}{g}}$ where l = length of the pendulum

l = $\frac{T^{2}}{{(2 π}^{2})} * g$ = $\frac{{3.5}^{2}}{{(2 π}^{2})}$ $* 9.8$ = 3.041 m

On Moon surface, the length of the pendulum remained same = 3.041 m

So time period on moon

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14.15 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)

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14.15 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)