A simple pendulum suspended in a rocket ship has a period T0. Assume that the rocket ship is near the earth in a uniform gravitational field.

true or false:
a) If the mass of the pendulum is halved, the period decreases.
b) If the ship accelerates downward at 9.81 m/s2, the pendulum will no longer oscillate.
c) If the ship accelerates upward, the period increases.
d) If the ship moves upward with a constant velocity, the period decreases.
e) If the length of the pendulum is doubled, the new period will be: square root of 2 times T0.

a) True: The period of a simple pendulum is given by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Since only the mass is halved, and it does not appear in the formula for the period, the period will decrease.

b) False: The motion of the simple pendulum is independent of the acceleration of the ship. As long as gravity is present, the pendulum will continue to oscillate with its original period T0.

c) False: The period of a simple pendulum depends only on the length of the pendulum and the acceleration due to gravity. The ship's acceleration does not affect the period directly.

d) True: When the ship moves upward with a constant velocity, there is still a net force of gravity acting on the pendulum. However, the effective gravitational field is reduced, resulting in a decrease in the period of the pendulum.

e) True: The period of a simple pendulum is given by the formula T = 2π√(L/g). If the length of the pendulum is doubled while the gravitational field remains the same, the new period will be square root of 2 (approximately 1.414) times the original period T0.