Final answer:
a. The period of the pendulum's motion is 0.5 seconds and the frequency is 2 Hz. b. The graphs of velocity and acceleration would be in phase with the displacement graph. c. The acceleration due to gravity on the surface of the planet is approximately 39.478 m/s², which corresponds to 4.03 g-forces.
Step-by-step explanation:
a. The period of the pendulum's motion can be determined by identifying the time it takes for one complete oscillation. From the graph, we can observe that it takes approximately 0.5 seconds to complete one cycle. Therefore, the period of the pendulum's motion is 0.5 seconds. The frequency can be calculated by taking the reciprocal of the period, which gives us a frequency of 2 Hz.
b. The displacements, velocity, and acceleration of a pendulum in simple harmonic motion can be represented by sine and cosine functions. The phase difference between the displacements, velocity, and acceleration graphs depends on where they start in their respective cycles. Since the pendulum is set at a specific phase in simple harmonic motion, the graphs of velocity and acceleration would be in phase with the displacement graph, meaning there would be no time difference.
c. The acceleration due to gravity on the surface of the planet can be determined using the formula for the period of a simple pendulum: T = 2π√(L/g), where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity. Rearranging the formula, we can solve for g: g = 4π²(L/T)². Substituting the given values of L = 0.2 m and T = 0.5 s, we calculate g to be approximately 39.478 m/s². To determine the number of g-forces, we divide this value by the acceleration due to gravity on Earth (9.8 m/s²): 39.478 m/s² ÷ 9.8 m/s² = 4.03 g-forces.