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Oscillations

JEE Main level, single correct, medium

95 of 194

Question 95 of 194

A particle is executing simple harmonic motion $(SHM)$ of amplitude $A,$ along the $x$ -axis, about $x=0.$ When its potential Energy $\left(PE\right)$ equal kinetic energy $\left(KE\right),$ the position of the particle will be:

Show answer and solution

Answer: (D) $\dfrac{A}{\sqrt{2}}$

\begin{matrix}PE=\dfrac{1}{2}k{x}^{2} & &KE=\dfrac{1}{2}K\left({A}^{2}-{x}^{2}\right)\end{matrix}

As given in question

KE=PE

\Rightarrow \dfrac{1}{2}K\left({A}^{2}-{x}^{2}\right)=\dfrac{1}{2}K{x}^{2}

\Rightarrow 2{x}^{2}={A}^{2}

\Rightarrow x=\pm \dfrac{A}{\sqrt{2}}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Oscillations (Energy of Simple Harmonic Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A particle is executing simple harmonic motion (SHM)(SHM)(SHM) of amplitude A,A,A, along the xxx -axis, about x=0.x=0.x=0. When its potential Energy (PE)\left(PE\right)(PE) equal kinetic energy (KE),\left(KE\right),(KE), the position of the particle will be:

JEE Main question paper PDF

More Physics practice

A particle is executing simple harmonic motion $(SHM)$ of amplitude $A,$ along the $x$ -axis, about $x=0.$ When its potential Energy $\left(PE\right)$ equal kinetic energy $\left(KE\right),$ the position of the particle will be: