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CGP EDU Academic Team
Published on: September 12, 2026
In a simple harmonic oscillator, at the mean position
Text Solution
Verified by ExpertsThe correct answer is:
C
Kinetic energy of particle of mass m in SHM at any point is,
$= \frac{1}{2} m \omega^{2} (a^{2} - x^{2})$ and potential energy $= \frac{1}{2} m \omega^{2} x^{2}$ where, a is amplitude of particle and x is the distance from mean position. So, at mean position, $x = 0$ KE $= \frac{1}{2} m \omega^{2} a^{2}$ (maximum) PE $= 0$ (minimum)
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