0.21 m 9planet = = T2 (0.92 s)² 2 9.8 m/s² Constants In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, 9Mars = 0.389 earth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on? He is on planet earth! REFLECT In principle, an astronaut can use this method to measure the acceleration due to gravity of any planet or satellite while traveling around the solar system. Part A Practice Problem: The acceleration due to gravity near the surface of a distant planet is about 3.6 m/s2. If the astronaut arrives there and swings the same pendulum bob, what will be the period of the pendulum? Express your answer in seconds to two significant figures. Η ΑΣΦ Submit Request Answer ? S

The block is at rest as shown. What is the peiod of the oscillation if the block is pulled down by 10 cm, in seconds? Use g = 10 m/s². Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. 100 g 10 20 30 40 50 60 70 80 90

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5 kilogram object is suspended from a string. The string is 4 meters long. What is the period of the object ting as a pendulum? Express your answer a number of seconds, omitting the unit.