KCET 2025 Physics Paper

To understand the displacement $x_0$ where kinetic energy (KE) equals potential energy (PE) in simple harmonic motion (SHM), consider the following explanation:At the position $x_0$, the kinetic energy is equal to the potential energy. Mathematically, this is represented by:$\;\frac{1}{2} m \omega^2 (A^2-x^2) =\;\frac{1}{2} m \omega^2 x^2$Here, the left side of the equation represents the kinetic energy, and the right side represents the potential energy. Simplifying the equation:$A^2-x^2=x^2$ Rearranging terms, we get:$A^2=2 x^2$Solving for $x$ :$\;x^2=\;\frac{A^2}{2}$$\;x=\;\frac{A}{\sqrt{2}}$Thus, the value of $\left| x_0 \right|$ is $\;\frac{A}{\sqrt{2}}$.