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Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 0.971 m and finds that it makes 319 complete oscillations in 846 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/s^2?

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Karabo99

Answer:

the gravitational acceleration of this planet is 5.46 m/s^2.

Explanation:

The time taken per oscilation period, T = 846/319 = 2.65s

The length L, period T and acceleration of gravity g of the pendulum are given by:

T = 2×π×√(L/g)

T^2 = 4×π^2×L/g

g = 4×π^2×L/(T^2)

  = 4×π^2×(0.971)/((2.65s)^2)

  = 5.46 m/s^2

Therefore, the gravitational acceleration of this planet is 5.46 m/s^2.

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