Question

A particle executing SHM with an amplitude A. The displacement of the particle when its potential energy is half of its total energy is

• A. $\frac{A}{\sqrt{2}}$
• B. $\frac{A}{2}$
• C. $\frac{A}{4}$
• D. $\frac{A}{3}$

Answer 1

Answer: (A)

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

Given : $PE=\frac{1}{2}\times \left(\text{total energy}\right)$

$⟹KE=\frac{1}{2}\times \left(\text{total energy}\right)$

Total energy of $SHM=\frac{1}{2}m{\omega }^2{A}^2$

$⟹KE=\frac{1}{4}m{\omega }^2{A}^2$

$⟹\frac{1}{2}m{v}^2=\frac{1}{4}m{\omega }^2{A}^2$

$⟹v^2=\frac{{\omega }^2{A}^2}{2}$ .........(1)

Velocity of particle in SHM.

$⟹V=(\omega \sqrt{A^2-x^2}{)}^2=\frac{{\omega }^2{A}^2}{2}$ $[v=\omega \sqrt{A^2-x^2},\text{when}x=\text{displacement from mean position.}]$

$⟹{\omega }^2(A^2-x^2)=\frac{{\omega }^2{A}^2}{2}$

$⟹A^2{x}^2=\frac{A^2}{2}$

$⟹x^2=\frac{A^2}{2}$

$⟹x=\pm \frac{A}{\sqrt{2}}$

Option A.