Question

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

• A. $\frac{A}{2}$
• B. $\frac{A}{2\sqrt{2}}$
• C. $\frac{A}{\sqrt{2}}$
• D. A

Answer 1

Answer: (C)

$\frac{A}{\sqrt{2}}$

Answer 2

Potential energy $(U) = \frac{1}{2} kx^2$
Kinetic energy $(K) = \frac{1}{2} kA^2 - \frac{1}{2} kx^2$
According to the question, U = k
$\therefore \frac{1}{2} kx^{2} = \frac{1}{2} kA^{2} - \frac{1}{2} kx^{2}$
$x = \pm \frac{A}{\sqrt{2}}$