The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is

Potential energy of a simple harmonic oscillator
\(
U=\frac{1}{2} m \omega^{2} y^{2}
\)
Kinetic energy of a simple harmonic oscillator
\(
K=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)
\)
Here \(y=\) displacement from mean position \(A=\) maximum \(\quad\) displacement \(\quad\) (or amplitude) from mean position Total energy is
\(
\begin{aligned}
E &=U+K \\
&=\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right) \\
&=\frac{1}{2} m \omega^{2} A^{2}
\end{aligned}
\)
When the particle is half way to its end point ie, at half of its amplitude then
\(
y=\frac{A}{2}
\)
Hence, potential energy
\(
\begin{aligned}
U &=\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2} \\
&=\frac{1}{4}\left(\frac{1}{2} m \omega^{2} A^{2}\right) \\
U &=\frac{E}{4}
\end{aligned}
\)