The kinetic energy and potential energy of a particle execu…

Physics Simple Harmonic Motion (Oscillations) Energy of Simple Harmonic Motion Single Correct MCQ

Published on: September 11, 2026

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is

$(a) \frac{a}{2} \quad (b) a\sqrt{2} \quad (c) \frac{a}{\sqrt{2}} \quad (d) \frac{a\sqrt{2}}{2}$

Text Solution

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The correct answer is:

Suppose at displacement y from mean position potential energy = kinetic energy

$\Rightarrow \frac{1}{2} m (a^{2} - y^{2}) \omega^{2} = \frac{1}{2} m \omega^{2} y^{2}$

⇒ ⇒ $a^{2} = 2y^{2}$ ⇒ ⇒ $y = \frac{a}{\sqrt{2}}$

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