Easy MCQ +4 / -1 PYQ · JEE Mains 2022

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :

  1. A Circular
  2. B Elliptical Correct answer
  3. C Sinusoidal
  4. D Straight line

Solution

<p>Let $x = A\sin \omega t$</p> <p>$\Rightarrow v = A\omega \cos \omega t$</p> <p>$\Rightarrow v = \, \pm \,\omega \sqrt {{A^2} - {x^2}}$</p> <p>$\Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}$</p> <p>$\Rightarrow$ Ellipse</p>

About this question

Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion

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