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2 January, 11:24

Suppose that you are on an unknown planet in a distant galaxy. Let the length of a physical pendulum be 1.02 m. In the physical pendulum experiment, the period is determined to be 0.25 s.

What is the gravitational acceleration of this planet?

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  1. 2 January, 11:37
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    Answer: acceleration due gravity = 619.5 m/s²

    Explanation: The relationship between the period of a pendulum, length and acceleration due to gravity is given as

    T = 2π * (√l/g)

    T = period = 0.25s

    l = length of pendulum = 1.02m

    g = acceleration due to gravity on the unknown planet

    0.25 = 2π * (√1.02/g)

    By squaring both sides, we have that

    0.25² = 4π² * (1.02/g)

    0.25² * g = 4π² * 1.02

    g = 4π² * 1.02 / 0.25²

    g = 40.26756 / 0.0625

    g = 619.5m/s²
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