Question

The total energy of the body executing $SHM$ is $E$. Then the kinetic energy when the displacement is half of the amplitude, is:

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

Answer 1

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

$\begin{array}{c}TotalenergyisSHME=\frac{1}{2}m{w}^2{a}^2\left(wherea=amplitude\right)\\ potentialenergyU=\frac{1}{2}m{w}^2\left(a^2-y^2\right)\\ =E-\frac{1}{2}m{w}^2{y}^2\\ Wheny=\frac{a}{2}\\ U=E-\frac{1}{2}m{w}^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}=\frac{3E}{4}\end{array}$

Hence,

option $\left(A\right)$ is correct answer.