Question:medium
A particle is executing simple harmonic motion with amplitude \(A\). The position at which kinetic energy and potential energy are equal is given by
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
- \(\dfrac{A}{2}\)
- \(2A\)
- \(A\)
\(\dfrac{A}{\sqrt{2}}\)
Show Solution
The Correct Option is D
Solution and Explanation
Concept: SHM: KE = PE \(\Rightarrow \frac12 k(A^2-x^2) = \frac12 kx^2 \Rightarrow A^2 = 2x^2 \Rightarrow x = A/\sqrt2\).
Step 1: Write the final answer. \(\boxed{x=\frac{A}{\sqrt{2}}}\)
Step 1: Write the final answer. \(\boxed{x=\frac{A}{\sqrt{2}}}\)
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