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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 is suspended from the ceiling of a...

A simple pendulum of length ll is suspended from the ceiling of a car travelling with a speed vv on a circular horizontal road of radius rr. Determine the time period of small oscillations.

Options

  1. A2\pi \sqrt{\dfrac{l}{\left( g^2 + \dfrac{v^4}{r^2} \right)^{1/2}}}
  2. B2\pi \sqrt{\dfrac{l}{\left( g^2 + \dfrac{v^2}{r} \right)^{1/2}}}
  3. C2πlg2\pi \sqrt{\dfrac{l}{g}}
  4. D2\pi \sqrt{\dfrac{l}{\left( g^2 - \dfrac{v^4}{r^2} \right)^{1/2}}}

Correct answer

A. 2\pi \sqrt{\dfrac{l}{\left( g^2 + \dfrac{v^4}{r^2} \right)^{1/2}}}

Step-by-step solution

In the frame of the car, the pendulum bob experiences a pseudo force (centrifugal force) in the horizontal direction due to the circular motion of the car.

The centrifugal acceleration is ac=v2ra_c = \dfrac{v^2}{r} directed radially outward.

The acceleration due to gravity gg acts vertically downward.

Since the horizontal and vertical accelerations are perpendicular to each other, the effective acceleration due to gravity geffg_{eff} is given by:

geff=g2+ac2=g2+(v2r)2=(g2+v4r2)1/2g_{eff} = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + \left(\dfrac{v^2}{r}\right)^2} = \left( g^2 + \dfrac{v^4}{r^2} \right)^{1/2}

The time period of small oscillations 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 = 2\pi \sqrt{\dfrac{l}{\left( g^2 + \dfrac{v^4}{r^2} \right)^{1/2}}}
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