Physics/Lessons/Oscillations/Energy in SHM
Physics · Oscillations · A2 Level

Energy in SHM

Watch a pendulum swing and you can see energy visibly changing form: fastest at the bottom, momentarily frozen at each extreme. Nothing is lost in an ideal, undamped oscillation — kinetic and potential energy just continuously trade places, and the total never changes.

8 min read

Total energy is conserved (and constant) in SHM

Ignoring damping (resistive forces such as air resistance or friction, covered in a later lesson), an oscillating system's total mechanical energy — kinetic plus potential — stays exactly constant throughout the motion. This follows directly from conservation of energy: with no resistive forces doing negative work, no mechanical energy is lost to heat.

What changes continuously is how that fixed total is split between kinetic and potential forms. For a mass-spring system, potential energy is elastic strain energy in the spring; for a pendulum, it's gravitational potential energy as the bob rises above its lowest point. In both cases, the same pattern holds:

  • At the extremes of the oscillation ($x=\pm x_0$), the object is momentarily stationary ($v=0$), so all the energy is potential and kinetic energy is zero.
  • At the equilibrium position ($x=0$), the object moves at its maximum speed, so all the energy is kinetic and potential energy is zero (taking the equilibrium position as the reference level for potential energy).
  • At every point in between, the energy is split between the two forms, with kinetic and potential energy continuously exchanging as the object moves.

The total energy formula

Since total energy equals kinetic energy at the instant the object passes through equilibrium (where speed is maximum, $v_{max}=\omega x_0$):

$E_{total} = \tfrac12mv_{max}^2 = \tfrac12m(\omega x_0)^2 = \tfrac12m\omega^2x_0^2$

This is the same value at every point, since total energy is constant — it depends only on mass, angular frequency, and amplitude. Total energy is proportional to the square of amplitude: doubling amplitude quadruples total energy, not just doubles it.

Kinetic and potential energy as functions of displacement

Using $v = \pm\omega\sqrt{x_0^2-x^2}$, kinetic energy at displacement $x$ is:

$E_k = \tfrac12mv^2 = \tfrac12m\omega^2(x_0^2-x^2)$

Since total energy is fixed and equals kinetic plus potential at every point, potential energy follows by subtraction:

$E_p = E_{total} - E_k = \tfrac12m\omega^2x_0^2 - \tfrac12m\omega^2(x_0^2-x^2) = \tfrac12m\omega^2x^2$

$E_k$ is greatest at $x=0$ and falls to zero at $x=\pm x_0$, while $E_p$ is zero at $x=0$ and rises to its maximum, $\tfrac12m\omega^2x_0^2$ (equal to total energy), at $x=\pm x_0$.

Kinetic energy, potential energy and total energy plotted against displacement for SHM. The two parabolas are mirror images of each other and always sum to the constant total energy line, at every value of x.
Reading the energy-displacement graph

At any displacement $x$, the vertical gap between the total-energy line and the potential-energy curve equals kinetic energy at that point, and vice versa — the two parabolas always sum to the same constant total. Where they cross in the middle ($x=0$), kinetic energy is at its sole maximum and potential is zero; at the far edges ($x=\pm x_0$), the roles fully reverse.

Worked exampleFinding kinetic energy at a given displacement
1
A mass of $0.40\,\text{kg}$ oscillates with SHM of amplitude $0.050\,\text{m}$ and angular frequency $6.0\,\text{rad}\,\text{s}^{-1}$. Find its kinetic energy at a displacement of $0.030\,\text{m}$ from equilibrium.
2
$E_k = \tfrac12m\omega^2(x_0^2-x^2) = \tfrac12\times0.40\times6.0^2\times(0.050^2-0.030^2)$.
3
$= 0.20 \times 36 \times (0.0025 - 0.0009) = 7.2 \times 0.0016 = 0.01152\,\text{J} \approx 1.2\times10^{-2}\,\text{J}$.
4
Check against the total energy: $E_{total} = \tfrac12m\omega^2x_0^2 = 0.20\times36\times0.0025 = 0.018\,\text{J}$. Since $E_k \approx 0.0115\,\text{J} < E_{total} = 0.018\,\text{J}$, the remainder, $E_p \approx 0.0065\,\text{J}$, is the potential energy at this displacement — a useful sanity check that $E_k$ alone can't exceed $E_{total}$.
This cross-check is worth doing whenever time allows — an $E_k$ value larger than $E_{total}$ signals an arithmetic slip immediately.
▪ solved

Key takeaways

  • Ignoring damping, total mechanical energy in SHM is constant, continuously exchanging between kinetic and potential forms as the object oscillates.
  • $E_{total} = \tfrac12m\omega^2x_0^2$ — proportional to the square of the amplitude, so doubling amplitude quadruples total energy.
  • $E_k = \tfrac12m\omega^2(x_0^2-x^2)$ is maximum at $x=0$ and zero at $x=\pm x_0$; $E_p=\tfrac12m\omega^2x^2$ is zero at $x=0$ and maximum at $x=\pm x_0$ — the two always sum to $E_{total}$ at any displacement.
  • On an energy-against-displacement graph, $E_k$ and $E_p$ appear as two mirror-image parabolas, always summing to the flat, constant total-energy line.
← Previous
SHM Equations & Graphs