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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

A particle is executing simple harmonic motion \((SHM)\) of amplitude \(A,\) along the \(x-\) axis, about \(x = 0.\) When its potential energy \((PE)\) equals kinetic energy \((KE),\) the position of the particle will be

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

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\(\mathrm{PE}=\mathrm{KE}\)

\(\Rightarrow \quad \frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}\)

\(\Rightarrow \quad x=\frac{A}{\sqrt{2}}\)

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