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)

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

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