Q69 / 90 · JEE Main 2019 (09 Jan Shift 2) · Physics

A particle is executing simple harmonic motion $(S H M)$ of amplitude $A,$ along the $x$ -axis, about $x = 0 .$ When its potential Energy $(P E)$ equal kinetic energy $(K E),$ the position of the particle will be:

A
$A$
B
$\frac{A}{2}$
C
$\frac{A}{2 \sqrt{2}}$
Correct answer
D
$\frac{A}{\sqrt{2}}$

Solution

$\begin{matrix} P E = \frac{1}{2} k x^{2} & & K E = \frac{1}{2} K (A^{2} - x^{2}) \end{matrix}$

As given in question

$K E = P E$

$\Rightarrow \frac{1}{2} K (A^{2} - x^{2}) = \frac{1}{2} K x^{2}$

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

$\Rightarrow x = \pm \frac{A}{\sqrt{2}}$

Hence, the correct answer is option: 'D'.

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

Solution

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A particle is executing simple harmonic motion $(S H M)$ of amplitude $A,$ along the $x$ -axis, about $x = 0 .$ When its potential Energy $(P E)$ equal kinetic energy $(K E),$ the position of the particle will be: