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Question:
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)

Answer:

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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Question:
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)

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Question: 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)

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Question:
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)