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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

A simple pendulum of length l hangs from the ceiling of an...

A simple pendulum of length ll hangs from the ceiling of an elevator. Calculate the time period of small oscillations assuming the elevator is going down with an acceleration a0a_0.

Options

  1. A2πlg2\pi \sqrt{\dfrac{l}{g}}
  2. B2πlg+a02\pi \sqrt{\dfrac{l}{g + a_0}}
  3. Cπlg−a0\pi \sqrt{\dfrac{l}{g - a_0}}
  4. D2πlg−a02\pi \sqrt{\dfrac{l}{g - a_0}}

Correct answer

D. 2πlg−a02\pi \sqrt{\dfrac{l}{g - a_0}}

Step-by-step solution

In the frame of reference of the elevator, a pseudo force ma0m a_0 acts on the pendulum bob in the upward direction, opposite to the downward acceleration of the elevator.

The true gravitational force mgmg acts in the downward direction.

The net effective force on the bob is Feff=mg−ma0=m(g−a0)F_{eff} = mg - m a_0 = m(g - a_0).

The effective acceleration due to gravity is geff=g−a0g_{eff} = g - a_0.

The time period of a simple pendulum is given by T=2πlgeffT = 2\pi \sqrt{\dfrac{l}{g_{eff}}}.

Substituting the value of geffg_{eff}, we get T=2πlg−a0T = 2\pi \sqrt{\dfrac{l}{g - a_0}}.
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