A simple pendulum is mounted in an elevator. What happens to the period of the pendulum (does it increase, decrease, or remain the same) if the elevator (a) accelerates upward at 5.0 m/s2; (b) moves upward at a steady 5.0 m/s; (c) accelerates downward at 5.0 m/s2; (d) accelerates downward at 9.8 m/s2? Justify your answers.

Solution 11DQ Step 1: a) In the first case, the elevator accelerates upward with an acceleration, a = 5 m/s So, the force acting on the bob initially, F = mg (Gravitational force) acting gravitational downward Where, m- mass of the bob g - acceleration due to gravity If the elevator accelerates upward with an acceleration, a = 5 m/s Upward force, F upward- ma (minus sign represents the opposite direction) Resultant force acting on the pendulum, F = F gravitationaupward F = mg - ma = m (g - a) So, the acceleration of the pendulum, a’ = (g-a) l So, the time period, T = 2 (ga) From the equation we can see that, the time period is increasing, since the acceleration is decreasing. Step 2: b) In the second case, the elevator moves upward with a steady speed, v = 5 m/s 2 So, the upward acceleration, aupward0 m/s Therefore, T = 2 g So, there is no change in time period if the elevator is moving in a steady speed.