Question

Please I want someone to answer it clearly and step by step

It is desired to construct a simple pendulum with a mass of 3 kg that has a period of 2 seconds. (a) How long must the pendulum be? (b) If the amplitude of the displacement from equilibrium is doubled, what will the period of the pendulum be? (c) If the pendulum is now placed on a elevator that accelerates upward at 4.0 m/s^2, what is the period of the pendulum? (d) If the pendulum is placed on an earth satellite that is circling the earth, what is the period of the pendulum?

Answer 1

T = 2 pi squareroot L/g T/2 pi = squareroot L/g T^2/4 pi^2 = L/g (T^2/4 pi^2)g = L ((2)^2/4 times (3.14)^2)9.8 = L L = (4/39.438)9.8 = 0.9939 m. If the amplitude of the displacement from equilibrium is doubled if period of the pendulum does not change because pendulum period does not depend on Amplitude T = 2 pi squareroot L/g T = 2 pi squareroot L/g + a = 2 pi squareroot 0.9939/9.8 + 4 = 2 pi squareroot 0.9939/13.8 = 1.679 sec If the pendulum is placed on an earth satellite that is circulating the earth, the periods of the pendulum is T = 2 sec (does not change).