Home
Class 12
PHYSICS
The acceleration due to gravity on the s...

The acceleration due to gravity on the surface of the moon is 1.7 `ms^(-2)`. The time period of a simple pendulum on the moon if its time period on the earth is 3.5 s is (Given, `g=9.8ms^(-2)`)2.2 s

Text Solution

Verified by Experts

Acceleration due to garavity on the surface of moon,`g=1.7ms^(2)`
Acceleration due to garavity on the surface of earth, `g=9.8ms^(-2)`
Time period of a simple pendulum on earth, T=3.5s
`T=2pisqrt((I)/(g))`
Where ,
// is the lenth of the pendulum
` thereforel=(T^(2))/((2pi)^(2))xxg=((3.5)^(2))/(4xx(3.14)^(2)xx9.8m)`
The length of the pendulum remains constant.
`T'=2pisqrt((I)/(g))`
`2pisqrt((((3.5)^(2))/(4xx(3.14)^(2))xx9.8)/(1.7))=8.4`s
Hence , the time period of the simple pendulum on the surface of moon is 8.4s.
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    VMC MODULES ENGLISH|Exercise LEVEL 0 LONG ANSWER TYPE|2 Videos

  • SIMPLE HARMONIC MOTION

    VMC MODULES ENGLISH|Exercise LEVEL (1)|75 Videos

  • SIMPLE HARMONIC MOTION

    VMC MODULES ENGLISH|Exercise LEVEL 0 SHORT ANSWER TYPE - I|8 Videos

  • ROTATIONAL MOTION

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive) (True/False Type)|3 Videos

  • SYSTEM OF A PARTICLES & ROTATIONAL MOTION

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE F|10 Videos

Similar Questions

Explore conceptually related problems

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 moon if its time period on the earth is 3.5 s ? (g on earth = 9.8 ms^(-2) )

The acceleration due to gravity on the surface of the moon is 1.7ms^(-2) . What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is 3.5s ? Take g=9.8ms^(-2) on the surface of the earth.

The acceleration due to gravity on the surface of the moon is 1.7ms^(-2) . What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is 3.5s ? Take g=9.8ms^(-2) on the surface of the earth.

The time period of a simple pendulum is 2 s. Find its frequency .

The time period of a simple pendulum is 2 s. What is its frequency?

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2) .Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)

The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

The acceleration due to gravity at height R above the surface of the earth is g/4 . The periodic time of simple pendulum in an artifical satellie at this height will be :-

The acceleration due to gravity is 9.8 m s^(-2) . Give its value in ft s^(-2)

The acceleration due to gravity at the moon's surface is 1.67 ms^(-2) . If the radius of the moon is 1.74 xx 10^(6) m , calculate the mass of the moon.