Question

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

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

Answer 1

The correct option is C $\frac{A}{\sqrt{2}}$

Potential energy of particle undergoing SHM is $\left(PE\right)=\frac{1}{2}k{x}^2$
Kinetic energy of particle undergoing SHM is $\left(KE\right)=\frac{1}{2}k(A^2-x^2)$
According to the question,
$PE=KE$
$\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2-\frac{1}{2}k{x}^2$
$x^2=\frac{1}{2}{A}^2\Rightarrow x=\pm \frac{A}{\sqrt{2}}$