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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Medium
  • MCQ

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is on

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Answer & Solution

Correct Answer

(C) \(\frac{1}{\sqrt{2}}\,A\)

Step-by-step Solution

Detailed explanation

\(KE = PE\)

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

\(A ^2- x ^2= x ^2 \Rightarrow A ^2=2 \times 2\)

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