Question

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

Answer 1

Step-by-step explanation:

Time period of Simple pendulum is given by

`T=2\pi\sqrt{(L)/(g_(effective))}`

where L=length of Pendulum

g=acceleration due to gravity

(a)When elevator is accelerating upward with
`5 m/s^2`

effective g will be g+a=9.8+5

therefore time period will decrease

(b)When elevator is moving upward with 5 m/s

effective g will be g as there is no added acceleration

therefore time period will remain same

(c)When elevator is accelerating downward with
`5 m/s^2`

effective g will be g-a=9.8-5

therefore time period will increase

(d)When elevator is accelerating downward with
`9.8 m/s^2`

effective g will be g-9.8=0

therefore time period will become infinite

Answer 2

Step-by-step explanation:

The time period of the simple pendulum is given by

`T=2\pi \sqrt{(L)/(g_(effective))}`

(a) when it accelerates upwards, the effective value of g is g + a where, a is the acceleration of the elevator in upwards direction, so effective value of g increases, thus, the time period of the simple pendulum decreases.

(b) As it moves with constant speed so the effective g remains same, so that the time period of the simple pendulum remains same.

(c) when it accelerates downwards, the effective value of g is g - a where, a is the acceleration of the elevator in downwards direction, so effective value of g decreases, thus, the time period of the simple pendulum increases.

(d) when it accelerates downwards, the effective value of g is g - a where, a is the acceleration of the elevator in downwards direction, here, a = g so the effective g is zero, thus the time period of the simple pendulum becomes infinity.