Question:

A particle is executing simple harmonic motion with amplitude \(A\). The position at which kinetic energy and potential energy are equal is given by

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In SHM, \[ U=\frac{1}{2}kx^2, \qquad K=\frac{1}{2}k(A^2-x^2). \] When kinetic energy equals potential energy, \[ x=\pm\frac{A}{\sqrt{2}}. \] This is a frequently asked SHM result.
Updated On: Jul 9, 2026
  • \(\dfrac{A}{2}\)
  • \(2A\)
  • \(A\)
  • \(\dfrac{A}{\sqrt{2}}\) 

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The Correct Option is D

Solution and Explanation

Concept: In simple harmonic motion, \[ \text{Total Energy} = \frac{1}{2}kA^2. \] Potential energy at displacement \(x\) is \[ U = \frac{1}{2}kx^2. \] Kinetic energy is \[ K = \frac{1}{2}k(A^2-x^2). \]

Step 1:
Use the condition that kinetic energy equals potential energy. \[ K=U. \] Therefore, \[ \frac{1}{2}k(A^2-x^2) = \frac{1}{2}kx^2. \]

Step 2:
Simplify the equation. Cancelling \(\dfrac12 k\), \[ A^2-x^2=x^2. \] \[ A^2=2x^2. \] \[ x^2=\frac{A^2}{2}. \]

Step 3:
Calculate the displacement. \[ x=\pm\frac{A}{\sqrt{2}}. \] Hence the position from the mean position is \[ \frac{A}{\sqrt{2}}. \]

Step 4:
Write the final answer. \[ \boxed{x=\frac{A}{\sqrt{2}}} \] \[ \boxed{\text{Answer = (D)}} \]
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