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)

Open in App

Solution

It is given that the acceleration due to gravity on the surface of Moon, g m =1⋅7 ms −2 , the acceleration due to gravity on the surface of Earth, g e =9⋅8 ms −2 and the time period of a simple pendulum on the surface of Earth, T e =3⋅5 s −1 .

The formula of time period is,

T=2π 1 g

The equation of time period on Earth is,

T e =2π 1 g e (1)

The equation of time period on Moon is,

T m =2π 1 g m (2)

Divide equation (2) by equation (1).

T m T e = g e g m

Substitute the values.

T m 3⋅5 = 9⋅8 1⋅7 T m =8⋅4 s

Hence, the time period of a simple pendulum on the surface of the Moon is 8⋅4 s.

Suggest Corrections

Similar questions

1.7 m s^{- 2}

3.5

g = 9.8 m s^{- 2})

^{- 2}

^{- 2}

5 s

4

1.7 m s^{- 2}

0.27

9.8 m s^{- 2}

Join BYJU'S Learning Program