Question

A particle starts SHM from the mean position. Its amplitude is $A$ and total energy $E$. At an instant its kinetic energy is $\frac{3E}{4}$. Its displacement at this instant is

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

Answer 1

The correct option is B $\frac{A}{2}$

Kinetic energy can be defined as
$KE=\frac{1}{2}K(A^2-x^2)$
Now total energy of a particle executing SHM at any instant is,
$\frac{1}{2}K{A}^2=E$
Given that $KE=\frac{3E}{4}$
$\frac{3}{4}\frac{1}{2}K{A}^2=\frac{1}{2}K(A^2-x^2)$
$\frac{3}{4}{A}^2=(A^2-x^2)$
$x=\pm \frac{A}{2}$