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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2021
  • Medium
  • MCQ

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = \(a\)) is

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Answer & Solution

Correct Answer

(C) \(\frac{a}{{\sqrt 2 }}\)

Step-by-step Solution

Detailed explanation

(c) Suppose at displacement \(y\) from mean position potential energy = kinetic energy 

==> \(\frac{1}{2}m({a^2} - {y^2}){\omega ^2} = \frac{1}{2}m{\omega ^{\rm{2}}}{y^2}\) 

==> \({a^2} = 2{y^2}\) 

==> \(y = \frac{a}{{\sqrt 2 }}\)

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