Question

The time period of a simple pendulum of length L measured in an elevator descending with acceleration g/3 is

A. 2π√3l/3g
B. π√3l/g
C. 2π√3l/g
D. 2π√2l/3g

Answer 1

The time period of the simple pendulum of length L measured in an elevator descending with acceleration g/3 is
`T = 2\pi\sqrt{(3L)/(2g)}` (option C)

Calculation on period of pendulum

First, we shall determine the effective acceleration acting on the descending elevator. This is shown below:

• Acceleration due to gravity = g
• Descending acceleration = g/3
• Effective acceleration (
  `g_(eff)`) =?

Effective acceleration = Acceleration due to gravity - Descending acceleration

`g_(eff) = g\ -\ (g)/(3) \\\\g_(eff) = (3g\ -\ g)/(3)\\\\g_(eff) = (2g)/(3)`

Finally, we shall determine the period of the simple pendulum. Details below:

Period of pendulum is given as:

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

Where

• T is the period
• L is the length
• g is the acceleration due to gravity

Now, replace g with the value of
`g_(eff)` in the equation above, we have

`T = 2\pi\sqrt{(L)/(g)}\\\\T = 2\pi\sqrt{(L)/(g_(eff))}\\\\But,\\\\g_(eff) = (2g)/(3) \\\\Thus,\ we\ have\\\\T = 2\pi\sqrt{(L\ /\ (2g)/(3) }\\\\T = 2\pi\sqrt{(L\ *\ (3)/(2g)}\\\\T = 2\pi\sqrt{(3L)/(2g)}`

Thus, the correct answer is option C

Complete question:

The time period of a simple pendulum of length L measured in an elevator descending with acceleration g/3 is

A. 2π√3l/3g

B. π√3l/g

C. 2π√3l/2g

D. 2π√2l/3g