NCERT · Class 11 · Physics · OscillationsAt what displacement from the mean position is the kinetic energy equal to potential energy in SHM?
Step-by-Step Solution
$ \text{KE} = \frac{1}{2} m \omega^2 (A^2 - x^2), \quad \text{PE} = \frac{1}{2} m \omega^2 x^2 $\nSetting $\text{KE} = \text{PE} \implies A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = \pm \frac{A}{\sqrt{2}}$.
Detailed Options Breakdown
Option : $x = \pm A$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
Option 1: $x = \pm \frac{A}{2}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
Option 2: $x = \pm \frac{A}{\sqrt{2}}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 3: $x = 0$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Oscillations.
💡 Study Guide: This question tests core syllabus concepts from Oscillations. For formulas, key summaries, and mock exam reference guides, read the full Oscillations Revision Notes.