Suspended from an elevator's ceiling, a simple pendulum of length 1 feet requires /3 seconds to complete one oscillation. Determine the acceleration of the elevator.

Step-by-step solution

The time period of a simple pendulum in an accelerating elevator is given by T = 2 l g_ eff . Given T = 3 s and l = 1 ft, substituting these values: 3 = 2 1 g_ eff Squaring both sides: 1 9 = 4 g_ eff g_ eff = 36 ft/s ^2 The standard acceleration due to gravity is g = 32 ft/s ^2 . Since g_ eff > g , the elevator must be accelerating upwards. For an elevator accelerating upwards with acceleration a , g_ eff = g + a . 36 = 32 + a a = 4 ft/s ^2 The acceleration of the elevator is 4 ft/s ^2 upwards.