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A simple pendulum is set up at an unknow...

A simple pendulum is set up at an unknown planet.The acceleration due to gravity on the surface is `6.2ms^(-2)`.Find the time period of the pendulum on the surface of the planet,if its time period on the surface of the earth is `3.5s`,Take `g`=9.8ms^(-2)`

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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 period of as simple pendulum on the moon, if its time period on the earth is 3.5s ? [g = 9.8ms^(-2)]

Knowledge Check

  • 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

    A
    4s
    B
    8s
    C
    10s
    D
    12s

  • The acceleration due to gravity at a place is pi^(2)m//s^(2) . Then, the time period of a simple pendulum of length 1 m is

    A
    `(2)/(pi)s`
    B
    `2pis`
    C
    `2s`
    D
    `pis`

  • As the acceleration due to gravity increases, the time period of the simple pendulum_______

    A
    increases
    B
    decreases
    C
    remains same
    D
    first increases and then decreases.

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