Energy in Classical Waves

Waves carry energy

A mechanical wave — say, a transverse wave on a stretched string — moves energy from one place to another without transporting matter. Where does that energy live? In the restoring forces and the motion of the medium.

Consider a sinusoidal traveling wave whose displacement is

$y(x, t) = A\cos(kx - \omega t),$

where $A$ is the amplitude, $k$ the wave number, and $\omega$ the angular frequency. A small segment of string of length $dx$ and mass $dm = \mu\,dx$ ( $\mu$ is the linear mass density) oscillates with transverse velocity

$v_y = \frac{\partial y}{\partial t} = A\omega\sin(kx - \omega t).$

Its kinetic energy is $dK = \tfrac{1}{2}\,dm\,v_y^2$ . Averaged over a full cycle, $\langle\sin^2\rangle = \tfrac{1}{2}$ , so

$\langle dK \rangle = \tfrac{1}{4}\mu\omega^2 A^2\,dx.$

The elastic (potential) energy stored in the segment equals the kinetic energy on average — this is a general feature of linear oscillators. Adding both contributions gives the time-averaged energy per unit length:

$\frac{\langle E \rangle}{L} = \tfrac{1}{2}\mu\omega^2 A^2.$

The key result is clear: the energy is proportional to the square of the amplitude, $E \propto A^2$ . Doubling the amplitude quadruples the energy; halving it cuts the energy to one quarter.

Power and intensity

The wave carries this energy along the string at the wave speed $v = \omega/k$ . The time-averaged power (energy per unit time passing a fixed point) is therefore

$\langle P \rangle = \tfrac{1}{2}\mu\omega^2 A^2 v.$

For a three-dimensional wave (e.g., a sound wave expanding spherically from a point source) the corresponding quantity is the intensity $I = \langle P \rangle / (\text{area})$ , which also scales as $A^2$ .

Why this matters for quantum mechanics

Classical wave physics predicts that the energy radiated by a vibrating system is continuous and can take any value — you just need the right amplitude. This prediction works for sound and water waves. For light, however, experiments on blackbody radiation and the photoelectric effect showed that energy comes in discrete chunks $E = h\nu$ , not as a continuously adjustable amplitude. The $A^2$ rule is correct within each mode; what changes is that the amplitude is not free to vary continuously but must correspond to an integer number of photons.

Understanding the classical result first makes it easier to see exactly what breaks down and what Einstein and Planck had to replace it with.

Try it

This is a numerical exercise — your code should return a number. Wave 1 has amplitude $A_1 = 3\,\text{cm}$ and Wave 2 has amplitude $A_2 = 6\,\text{cm}$ . Both waves travel through the same medium at the same frequency. By how much does the energy of Wave 2 exceed that of Wave 1? Return the ratio $E_2/E_1$ .

Run your code to see the quantum state.