An astronaut lands on an unknown planet and sets up a simple pendulum of length 0.5 m. She observes that the pendulum takes 40 s to complete 20 oscillations. What is the acceleration due to gravity on the surface of this planet? (Take ^2 = 10 )

Step-by-step solution

Time period of the simple pendulum is given by T = t n , where t is the total time and n is the number of oscillations. T = 40 20 = 2 s The formula for the time period of a simple pendulum is T = 2 l g . Squaring both sides, we get T^2 = 4 ^2 l g . Rearranging for g , we have g = 4 ^2 l T^2 . Substituting l = 0.5 m, T = 2 s, and ^2 = 10 : g = 4 10 0.5 (2)^2 g = 20 4 = 5 m/s ^2 Answer: 5 m/s ^2