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Energy Transport in Waves

Level 1 - Physics topic page in Mechanical Waves.

Principle

Waves transport energy, and for a sinusoidal string wave that transported energy scales with the square of the amplitude.

The key point is that each small piece of the string oscillates, but the energy associated with that motion flows along the string with the wave.

Notation

\(\mu\)
mass per unit length
\(\mathrm{kg\,m^{-1}}\)
\(T\)
string tension
\(\mathrm{N}\)
\(A\)
wave amplitude
\(\mathrm{m}\)
\(\omega\)
angular frequency
\(\mathrm{rad\,s^{-1}}\)
\(k\)
wavenumber
\(\mathrm{rad\,m^{-1}}\)
\(u\)
energy per unit length
\(\mathrm{J\,m^{-1}}\)
\(P\)
power carried by the wave
\(\mathrm{W}\)
\(v_w\)
wave speed
\(\mathrm{m\,s^{-1}}\)

Method

Derivation 1: Build the energy density terms

For a short string element \(dx\), the kinetic energy comes from the element's transverse speed. The elastic part comes from the extra stretch created by the slope of the string.

Element mass
\[dm=\mu\,dx\]
Kinetic density
\[u_K=\frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^2\]
Extra stretch
\[ds-dx\approx \frac{1}{2}\left(\frac{\partial y}{\partial x}\right)^2dx\]
Elastic density
\[u_U=\frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^2\]

Derivation 2: Substitute a sinusoidal wave

Take \(y=A\\cos(kx-\\omega t)\). Differentiate it with respect to time and position, then substitute into the energy densities.

Time derivative
\[\frac{\partial y}{\partial t}=A\omega\sin(kx-\omega t)\]
Position derivative
\[\frac{\partial y}{\partial x}=-Ak\sin(kx-\omega t)\]
Kinetic density
\[u_K=\frac{1}{2}\mu A^2\omega^2\sin^2(kx-\omega t)\]
Elastic density
\[u_U=\frac{1}{2}TA^2k^2\sin^2(kx-\omega t)\]

Derivation 3: Use the wave-speed relation to simplify

For a string wave, \(v_w^2=T/\\mu\) and \(v_w=\\omega/k\). Together they give \(Tk^2=\\mu\\omega^2\), so the two energy contributions are equal.

Match coefficients
\[Tk^2=\mu\omega^2\]
Equal parts
\[u_K=u_U=\frac{1}{2}\mu A^2\omega^2\sin^2(kx-\omega t)\]
Total density
\[u=\mu A^2\omega^2\sin^2(kx-\omega t)\]
Average density
\[\langle u\rangle=\frac{1}{2}\mu A^2\omega^2\]
The average of \(\sin^2\) over a full cycle is \(1/2\).
Average power
\[\langle P\rangle=\langle u\rangle v_w=\frac{1}{2}\mu A^2\omega^2 v_w\]

Rules

These are the compact results from the method above.

Kinetic density
\[u_K=\frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^2\]
Elastic density
\[u_U=\frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^2\]
Average energy density
\[\langle u\rangle=\frac{1}{2}\mu A^2\omega^2\]
Average power
\[\langle P\rangle=\frac{1}{2}\mu A^2\omega^2 v_w\]

Examples

Question
A sinusoidal string wave has
\[\mu=0.020\,\mathrm{kg\,m^{-1}}\]
amplitude
\[A=0.030\,\mathrm{m}\]
frequency
\[12\,\mathrm{Hz}\]
and speed
\[40\,\mathrm{m\,s^{-1}}\]
Find the average power.
Answer
First find
\[\omega=2\pi f=75.4\,\mathrm{rad\,s^{-1}}\]
Then
\[\langle P\rangle=\frac{1}{2}\mu A^2\omega^2 v_w=\frac{1}{2}(0.020)(0.030)^2(75.4)^2(40)=2.04\,\mathrm{W}\]

Checks

  • Doubling amplitude makes the average power four times larger.
  • The average transported energy is nonzero even though each element of the string oscillates about equilibrium.
  • Kinetic and elastic contributions are equal on average for a sinusoidal string wave.
  • This result is specific to linear string waves, not a universal formula for every medium.

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