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Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.29 m and finds that it makes 407 complete oscillations in 698 s. The amplitude of the oscillations is very small compared to the pendulum’s length.
What is the gravitational acceleration on the surface of this planet?
Answer in units of m/s2

  • Physics -

    Use the small-amplitude formula:

    Period = 2*pi*sqrt(L/g)
    and solve for g.

    L = 1.29 m
    Period P = 698/407 = 1.715 seconds

    g/L = 4*pi^2/P^2 = 13.42 s^-2

    g = 17.31 m/s^2

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