The total energy of the body executing simple harmonic motion is $E$. When the displacement is half of the…
The total energy of the body executing simple harmonic motion is $E$. When the displacement is half of the amplitude then the kinetic energy is
$\frac{E}{4}$
$\frac{3 E}{4}$
$\frac{\sqrt{3} E}{4}$
$\frac{E}{2}$
Solution
Total energy is SHM $E=\frac{1}{2} m \omega^2 a^2$ (where $a=$ amplitude)
The potential energy at displacement $y$ is $U=\frac{1}{2} m \omega^2 y^2$
The kinetic energy is $K=E-U=\frac{1}{2} m \omega^2\left(a^2-y^2\right)=E-\frac{1}{2} m \omega^2 y^2$
Now at $y=\frac{a}{2}$
$K=E-\frac{1}{2} m \omega^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}=\frac{3 E}{4}$