Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 . The expl? After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 . The explorer finds that the pendulum completes 107 full swing cycles in a time of 132 What is the magnitude of the gravitational acceleration on this planet?

Answer 1

We can find the period P of one cycle, and then we can use the period to find the gravitational acceleration g on this planet. P = (132 s) / (107 cycles) = 1.2336 s/cycle The period P is 1.2336 seconds. This means that it takes 1.2336 seconds for the pendulum to swing back and forth one. Now we can use the period P to find the gravitational acceleration g. The equation for the period of a pendulum is as follows: P = 2 pi \sqrt{L/g} P^2 = (4 pi^2) L / g g = (4 pi^2) L / P^2 g = (4)(pi^2)(0.540 m) / (1.2336 s)^2 g = 14.0 m/s^2 The acceleration of gravity on the planet is 14.0 m/s^2.