Question

The total energy of a particle performing $\text{SHM}$ is $E$. What will be the kinetic energy of the particle, when displacement is half of the amplitude?

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

Answer 1

The correct option is C $\frac{3E}{4}$

Kinetic energy of particle executing SHM is given by
$\text{KE}=\frac{1}{2}m{v}^2$
$\because v=\omega \sqrt{A^2-y^2}$
$\text{KE}=\frac{1}{2}m{\omega }^2(A^2-y^2)$
When $y=\frac{A}{2}$
$\text{KE}=\frac{1}{2}m{\omega }^2(A^2-{\left(\frac{A}{2}\right)}^2)$
$=\frac{1}{2}m{\omega }^2\left(\frac{3A^2}{4}\right)$
$=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$
Total energy of particle executing SHM is $\left(E\right)=\frac{1}{2}m{\omega }^2{A}^2$
$\therefore \text{KE}=\frac{3E}{4}$
Hence, option (c) is correct.