Energy In Shm

In SHM the total energy stays constant at $E=\frac{1}{2}k{A}^2$, continuously trading between potential and kinetic forms. The kinetic energy can be written as $K=\frac{1}{2}k(A^2-x^2)$, which shows the energy of motion shrinks steadily as the displacement grows toward the amplitude. The potential energy at displacement $x$ is $U=\frac{1}{2}k{x}^2$, so at $x=A/2$ we get $U=\frac{1}{2}k(A/2{)}^2=\frac{1}{4}\left(\frac{1}{2}k{A}^2\right)=\frac{1}{4}E$. By conservation of energy the kinetic energy is the remainder, $$\begin{gathered} K=E- \\ cup=E-\frac{1}{4}E=\frac{3}{4}E \end{gathered}$$. The option $\frac{1}{4}E$ actually gives the potential energy and confuses the two stores. The option $\frac{1}{2}E$ would only hold at a special displacement of $A/\sqrt{2}$, not at $A/2$. The option $\frac{1}{3}E$ has no consistent basis in the quadratic energy relations. This uses the NCERT energy treatment of SHM where potential energy scales as the square of displacement. As a sanity check, since the body is still well inside its turning point at $x=A/2$, it is moving fast and should carry most of the energy as kinetic, consistent with the dominant $\frac{3}{4}E$ fraction.