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JEE Mains · Physics · STD 11 - 13. oscillations

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

  1. A \(\sqrt{2\,A }\)
  2. B \(2\,A\)
  3. C \(\frac{1}{\sqrt{2}}\,A\)
  4. D \(\frac{1}{2}\,A\)
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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(KE = PE\) \(\frac{1}{2} M \omega^2\left( A ^2- x ^2\right)=\frac{1}{2} M \omega^2 x ^2\) \(A ^2- x ^2= x ^2 \Rightarrow A ^2=2 \times 2\) \(\Rightarrow x = \pm \frac{ A }{\sqrt{2}}\)
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