Energy in SHM

In a mass–spring system, energy transfers between kinetic energy of the mass and elastic potential energy of the spring. In a pendulum, it transfers between kinetic and gravitational potential energy.

At maximum displacement, the object is momentarily at rest: kinetic energy is zero and potential energy is maximum. At equilibrium, potential energy is at its minimum (taken as zero) and kinetic energy is maximum.

With no damping, no energy is lost, so the sum stays constant throughout the motion.

At equilibrium all energy is kinetic, and the speed is $v_{\max }=A\omega$. So the total energy is $\frac{1}{2}m(A\omega {)}^2$.

Total energy is proportional to $A^2$. Halving the amplitude of an oscillation removes three-quarters of its energy.

For a mass on a spring, ${\omega }^2=k/m$, so this is equivalent to $\frac{1}{2}k{A}^2$ — exactly the elastic energy stored at maximum extension.

Potential energy at displacement $x$ is $E_p=\frac{1}{2}m{\omega }^2{x}^2$. Kinetic energy is the remainder: $E_k=\frac{1}{2}m{\omega }^2(A^2-x^2)$.

Plotted against displacement, $E_p$ is an upward parabola from zero at the centre, $E_k$ is an inverted parabola, and the total is a horizontal line.

Kinetic and potential energy are equal when $x^2=A^2/2$, that is, at $x=A/\sqrt{2}\approx 0.71A$ — not at half the amplitude.

Starting from maximum displacement, $E_p=E_{\text{total}}{\cos }^2\omega t$ and $E_k=E_{\text{total}}{\sin }^2\omega t$.

Both are always positive and each reaches its maximum twice per oscillation, so the energy graphs have $\text{twice the frequency}$ of the displacement.

At every instant the two curves add to the constant total — they are mirror images about half the total energy.