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Today, 01:04

The period of a simple pendulum in a grandfather clock on another planet is 1.80 s. What is the acceleration due to gravity (in m/s2) on this planet if the length of the pendulum is 1.00?

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  1. Today, 01:16
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    12.17 m/s²

    Explanation:

    The formula of period of a simple pendulum is given as,

    T = 2π√ (L/g) ... Equation 1

    Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie

    making g the subject of the equation,

    g = 4π²L/T² ... Equation 2

    Given: T = 1.8 s, l = 1.00 m

    Constant: π = 3.14

    Substitute into equation 2

    g = (4*3.14²*1) / 1.8²

    g = 12.17 m/s²

    Hence the acceleration due to gravity of the planet = 12.17 m/s²
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