Energy in S.H.M.: Kinetic and Potential Energy

A particle executing S.H.M. continuously exchanges energy between two different forms as it oscillates back and forth: KINETIC energy, associated with its motion, and POTENTIAL energy, stored in the restoring force itself (for a spring, this stored energy is elastic potential energy).\n\nUsing the displacement-speed relation $v = \omega\sqrt{A^2-x^2}$ derived in Section 13.6, the kinetic energy of the particle when its displacement is $x$ works out to be\n\n$K(x) = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2\left(A^2 - x^2\right)$\n\nand, taking the potential energy to be zero at the mean position ($x=0$, a natural and conventional choice) and using the relation $k = m\omega^2$ from Section 13.4, the potential energy stored when the displacement is $x$ works out to be\n\n$U(x) = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2$\n\nAdding these two together gives the TOTAL mechanical energy of the oscillator at displacement $x$:\n\n$E = K(x) + U(x) = \frac{1}{2}m\omega^2\left(A^2-x^2\right) + \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}kA^2$\n\nStrikingly, the two $x^2$ terms cancel COMPLETELY when $K(x)$ and $U(x)$ are added together, leaving a total energy that does NOT depend on $x$ at all -- it is a fixed constant, set once and for all by the amplitude $A$ and angular frequency $\omega$ (equivalently, by $A$ and the force constant $k$) of the particular oscillation, and it stays exactly the same throughout the entire motion. This is exactly what is expected physically: the spring force (or, more generally, any S.H.M.-producing restoring force $F=-kx$) is a CONSERVATIVE force, and no energy is lost to friction anywhere in an idealised S.H.M., so the total mechanical energy must be conserved at every instant.\n\nAt the two extreme positions of the motion ($x=\pm A$), the particle is momentarily at rest ($v=0$), so ALL of the energy at that instant is potential: $U=E$ and $K=0$. At the mean position ($x=0$), the particle is moving at its own maximum speed $v_{\max}=A\omega$, so ALL of the energy at that instant is kinetic: $K=E$ and $U=0$. At every intermediate point of t …