The acceleration due to gravity on the s | Class 11 Physics Chapter Oscillations, Oscillations NCERT Solutions

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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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 15: The acceleration due to gravity on the surface of moon is 1.7 ms-2. What is the time period of a sim....