Wave Mechanics: Energy in a 1D WaveWaveMechanicsEnergyInA1DWave2026-09-12 16:01:062026-09-12 16:01:06Topicwave mechanicswave energyenergy densitykinetic energy densityelastic potential energy densitystretched stringtraveling wavestanding wavelinear mass densitytensionamplitude squaredfrequency squared% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Wave Mechanics: Energy in a 1D Wave}
The previous articles developed the one-dimensional string wave equation,
\begin{equation}
\boxed{\mu u_{tt}=T u_{xx}},
\end{equation}
and its traveling-wave solutions. The string does more than move, however. A disturbance also carries mechanical energy.
For the ideal stretched string, the local mechanical energy per unit equilibrium length is
\begin{equation}
\boxed{
\mathcal{E}(x,t)
=\frac{1}{2}\mu u_t^2
+\frac{1}{2}T u_x^2.}
\end{equation}
The first term is kinetic energy density and the second is elastic potential energy density. This expression is standard for transverse waves on an ideal stretched string and follows directly from the same small-slope model used to derive the wave equation \cite{French1971,Crawford1968,OpenStax164,MIT803PS5}.
WM18 develops that result from the mechanics of a short string element, checks its dimensions, applies it to traveling waves and standing waves, and shows why wave energy scales as the square of amplitude. The next article will ask how this energy crosses a fixed position and will derive the corresponding power flow.
\section{Energy is distributed along the string}
A string is a continuous system. Rather than assigning one energy to one point, we describe the energy contained in a short equilibrium-length interval $dx$.
Write
\begin{equation}
dE=\mathcal{E}(x,t)\,dx,
\end{equation}
where $\mathcal{E}$ has units of energy per unit length.
For an ideal string there are two mechanical contributions:
\begin{equation}
dE=dK+dU.
\end{equation}
The local bookkeeping is summarized below.
\begin{center}
\includegraphics{WM18_fig01_local_energy_bookkeeping.png}
\vspace{0.45em}
\textbf{Figure.}
A moving string element has kinetic energy through $u_t$. A sloped element is slightly longer than its equilibrium projection $dx$, producing elastic potential energy through $u_x$.
\end{center}
\section{Kinetic energy density}
Consider a short string element of equilibrium length $dx$. Its mass is
\begin{equation}
dm=\mu\,dx,
\end{equation}
where $\mu$ is the linear mass density.
The transverse velocity of the material point at position $x$ is
\begin{equation}
v_u=u_t=\frac{\partial u}{\partial t}.
\end{equation}
Therefore the kinetic energy of the element is
\begin{align}
dK
&=\frac{1}{2}dm\,v_u^2\\
&=\frac{1}{2}(\mu\,dx)u_t^2.
\end{align}
Divide by $dx$:
\begin{equation}
\boxed{
\mathcal{K}(x,t)=\frac{dK}{dx}=\frac{1}{2}\mu u_t^2.}
\end{equation}
This expression is exact within the transverse-motion model used here: kinetic energy is controlled by the local material velocity, not by the propagation speed alone.
\section{Elastic potential energy density}
The potential-energy term requires the string geometry.
At a fixed time, a short displaced string element has arclength
\begin{equation}
ds=\sqrt{dx^2+du^2}.
\end{equation}
Since
\begin{equation}
du=u_x\,dx,
\end{equation}
we obtain
\begin{equation}
ds=\sqrt{1+u_x^2}\,dx.
\end{equation}
For the small-slope model,
\begin{equation}
|u_x|\ll 1.
\end{equation}
Use the expansion
\begin{equation}
\sqrt{1+q}\simeq 1+\frac{q}{2}
\end{equation}
for small $q$. With $q=u_x^2$,
\begin{equation}
ds\simeq\left(1+\frac{1}{2}u_x^2\right)dx.
\end{equation}
The extra length is therefore
\begin{equation}
d\ell=ds-dx\simeq\frac{1}{2}u_x^2\,dx.
\end{equation}
If the equilibrium tension magnitude is approximately constant at $T$, the work required to create this additional length is
\begin{equation}
dU=T\,d\ell.
\end{equation}
Thus
\begin{equation}
dU\simeq\frac{1}{2}T u_x^2\,dx,
\end{equation}
and the elastic potential energy density is
\begin{equation}
\boxed{
\mathcal{U}(x,t)=\frac{dU}{dx}=\frac{1}{2}T u_x^2.}
\end{equation}
The same small-slope approximation that linearized the force law in WM14 also gives this quadratic elastic-energy expression.
\section{Total local energy density}
Adding the kinetic and potential terms gives
\begin{equation}
\boxed{
\mathcal{E}(x,t)
=\mathcal{K}+\mathcal{U}
=\frac{1}{2}\mu u_t^2
+\frac{1}{2}T u_x^2.}
\end{equation}
This formula contains an important physical separation:
\begin{itemize}
\item $u_t$ measures how rapidly the material is moving through equilibrium space;
\item $u_x$ measures how much the string is locally tilted and therefore stretched relative to its equilibrium projection.
\end{itemize}
A point can have zero displacement while still carrying energy. Energy depends on velocity and deformation, not simply on the value of $u$.
\section{Dimensional check}
The kinetic term has dimensions
\begin{equation}
[\mu u_t^2]
=\frac{\text{kg}}{\text{m}}
\frac{\text{m}^2}{\text{s}^2}
=\frac{\text{J}}{\text{m}}.
\end{equation}
For the potential term, $u_x$ is dimensionless because both $u$ and $x$ have dimensions of length. Hence
\begin{equation}
[T u_x^2]
=\text{N}
=\frac{\text{J}}{\text{m}}.
\end{equation}
Both terms therefore have the required units of energy per unit length.
\section{Pure traveling waves have equal kinetic and potential densities}
Consider a right-moving traveling profile
\begin{equation}
u(x,t)=F(x-ct).
\end{equation}
Let
\begin{equation}
\xi=x-ct.
\end{equation}
Then
\begin{equation}
u_x=F'(\xi)
\end{equation}
and
\begin{equation}
u_t=-cF'(\xi).
\end{equation}
The kinetic energy density becomes
\begin{equation}
\mathcal{K}
=\frac{1}{2}\mu c^2[F'(\xi)]^2.
\end{equation}
For the ideal string,
\begin{equation}
c^2=\frac{T}{\mu},
\end{equation}
so
\begin{equation}
\mu c^2=T.
\end{equation}
Therefore
\begin{equation}
\mathcal{K}
=\frac{1}{2}T[F'(\xi)]^2.
\end{equation}
But
\begin{equation}
\mathcal{U}
=\frac{1}{2}T u_x^2
=\frac{1}{2}T[F'(\xi)]^2.
\end{equation}
Hence
\begin{equation}
\boxed{\mathcal{K}=\mathcal{U}}
\end{equation}
for every point of a pure right-moving wave. The same proof holds for a pure left-moving wave $G(x+ct)$.
The total energy density of a pure traveling wave can therefore be written as
\begin{equation}
\boxed{
\mathcal{E}
=T[F'(x-ct)]^2
=\mu c^2[F'(x-ct)]^2.}
\end{equation}
A pulse example is shown below.
\begin{center}
\includegraphics{WM18_fig02_traveling_pulse_energy.png}
\vspace{0.45em}
\textbf{Figure.}
For a Gaussian pure traveling pulse, with normalized coordinate $\zeta=(x-ct)/\sigma$, the kinetic and elastic potential energy densities are equal point by point. The energy density depends on the square of the pulse slope, so a smooth pulse can have zero energy density exactly at its displacement maximum.
\end{center}
\section{Why the center of a smooth pulse can have zero energy density}
This result can initially seem surprising. Consider a smooth pulse at its maximum. At the exact peak,
\begin{equation}
F'=0.
\end{equation}
For a shape-preserving traveling pulse,
\begin{equation}
u_x=F'=0
\end{equation}
and
\begin{equation}
u_t=-cF'=0.
\end{equation}
Therefore both local energy-density terms vanish at the exact pulse peak.
The pulse still carries finite total energy because energy is distributed over the regions where the profile changes. Large displacement alone is not the same as large local wave energy.
\section{Sinusoidal traveling wave}
Consider
\begin{equation}
u(x,t)=A\cos(kx-\omega t+\phi).
\end{equation}
Define
\begin{equation}
\theta=kx-\omega t+\phi.
\end{equation}
Then
\begin{equation}
u_t=A\omega\sin\theta
\end{equation}
and
\begin{equation}
u_x=-Ak\sin\theta.
\end{equation}
The kinetic energy density is
\begin{equation}
\mathcal{K}
=\frac{1}{2}\mu A^2\omega^2\sin^2\theta.
\end{equation}
The potential energy density is
\begin{equation}
\mathcal{U}
=\frac{1}{2}T A^2k^2\sin^2\theta.
\end{equation}
The ideal-string dispersion relation is
\begin{equation}
\omega=ck
\end{equation}
with
\begin{equation}
c^2=\frac{T}{\mu}.
\end{equation}
Thus
\begin{equation}
T k^2=\mu\omega^2,
\end{equation}
so again
\begin{equation}
\boxed{\mathcal{K}=\mathcal{U}}.
\end{equation}
The total energy density is
\begin{equation}
\boxed{
\mathcal{E}(x,t)
=\mu A^2\omega^2\sin^2(kx-\omega t+\phi).}
\end{equation}
\begin{center}
\includegraphics{WM18_fig03_sinusoid_energy_density.png}
\vspace{0.45em}
\textbf{Figure.}
The displacement varies as $\cos\theta$, while the energy density varies as $\sin^2\theta$. Energy density is nonnegative and repeats twice during one displacement phase cycle.
\end{center}
\section{Average energy density of a sinusoidal traveling wave}
Over one full phase cycle,
\begin{equation}
\left\langle\sin^2\theta\right\rangle=\frac{1}{2}.
\end{equation}
Therefore
\begin{equation}
\boxed{
\langle\mathcal{E}\rangle
=\frac{1}{2}\mu A^2\omega^2.}
\end{equation}
The average kinetic and average potential contributions are each half of this:
\begin{equation}
\boxed{
\langle\mathcal{K}\rangle
=\langle\mathcal{U}\rangle
=\frac{1}{4}\mu A^2\omega^2.}
\end{equation}
The energy contained in one wavelength, averaged over the sinusoidal phase pattern, is
\begin{equation}
E_{\lambda}
=\langle\mathcal{E}\rangle\lambda.
\end{equation}
Hence
\begin{equation}
\boxed{
E_{\lambda}
=\frac{1}{2}\mu A^2\omega^2\lambda.}
\end{equation}
This result agrees with standard treatments of sinusoidal mechanical waves \cite{OpenStax164}.
\section{Energy scales as amplitude squared and frequency squared}
For fixed $\mu$, the average sinusoidal energy density obeys
\begin{equation}
\langle\mathcal{E}\rangle
\propto A^2\omega^2.
\end{equation}
Therefore:
\begin{itemize}
\item doubling the amplitude multiplies average energy density by $4$;
\item tripling the angular frequency at fixed amplitude multiplies it by $9$;
\item changing the sign of the displacement amplitude does not change the energy.
\end{itemize}
This square-law behavior is one of the most important recurring features of wave physics. Similar quadratic energy measures will appear later for acoustic, electromagnetic, and quantum-wave contexts, although the detailed physical meaning of the field variables changes.
\section{A general two-direction field}
WM16 showed that a sufficiently smooth whole-line solution can be written as
\begin{equation}
u(x,t)=F(x-ct)+G(x+ct).
\end{equation}
Let
\begin{equation}
\xi=x-ct,
\qquad
\eta=x+ct.
\end{equation}
Then
\begin{equation}
u_x=F'(\xi)+G'(\eta)
\end{equation}
and
\begin{equation}
u_t=-cF'(\xi)+cG'(\eta).
\end{equation}
Use $T=\mu c^2$. The total energy density is
\begin{align}
\mathcal{E}
&=\frac{1}{2}\mu c^2[-F'+G']^2
+\frac{1}{2}T[F'+G']^2\\
&=\frac{T}{2}[-F'+G']^2
+\frac{T}{2}[F'+G']^2.
\end{align}
Expanding the squares, the cross terms cancel:
\begin{equation}
\boxed{
\mathcal{E}=T[F'(\xi)]^2+T[G'(\eta)]^2.}
\end{equation}
This is a useful result. The displacement fields interfere linearly, but in the ideal string model the total mechanical energy density separates into the sum of the directional component energy densities.
The kinetic and potential parts individually need not be equal when both directions are present.
\section{Standing waves exchange kinetic and potential energy}
A standing wave can be written
\begin{equation}
u(x,t)=B\sin(kx)\cos(\omega t),
\end{equation}
with fixed ends chosen so that
\begin{equation}
k=\frac{n\pi}{L}.
\end{equation}
The derivatives are
\begin{equation}
u_t=-B\omega\sin(kx)\sin(\omega t)
\end{equation}
and
\begin{equation}
u_x=Bk\cos(kx)\cos(\omega t).
\end{equation}
Thus
\begin{equation}
\mathcal{K}
=\frac{1}{2}\mu B^2\omega^2
\sin^2(kx)\sin^2(\omega t)
\end{equation}
and
\begin{equation}
\mathcal{U}
=\frac{1}{2}T B^2k^2
\cos^2(kx)\cos^2(\omega t).
\end{equation}
Unlike a single traveling wave, these two densities are not generally equal point by point.
Integrating over a normal-mode length $0\leq x\leq L$ gives
\begin{equation}
K(t)=\frac{1}{4}\mu B^2\omega^2L\sin^2(\omega t)
\end{equation}
and
\begin{equation}
U(t)=\frac{1}{4}\mu B^2\omega^2L\cos^2(\omega t).
\end{equation}
Therefore
\begin{equation}
\boxed{
E=K+U=\frac{1}{4}\mu B^2\omega^2L,}
\end{equation}
which is constant in the ideal lossless model.
\begin{center}
\includegraphics{WM18_fig04_standing_wave_energy_exchange.png}
\vspace{0.45em}
\textbf{Figure.}
For one ideal standing-wave normal mode, total energy is constant while the integrated kinetic and elastic potential energies exchange periodically.
\end{center}
\section{Worked Example 1: Local energy density from measured slope and velocity}
A string has
\begin{equation}
\mu=0.020\,\text{kg/m},
\qquad
T=80\,\text{N}.
\end{equation}
At one event $(x,t)$, suppose
\begin{equation}
u_t=1.5\,\text{m/s}
\end{equation}
and
\begin{equation}
u_x=0.080.
\end{equation}
Find the kinetic, potential, and total energy densities.
\subsection*{Solution}
The kinetic density is
\begin{align}
\mathcal{K}
&=\frac{1}{2}\mu u_t^2\\
&=\frac{1}{2}(0.020)(1.5)^2\\
&=0.0225\,\text{J/m}.
\end{align}
The potential density is
\begin{align}
\mathcal{U}
&=\frac{1}{2}T u_x^2\\
&=\frac{1}{2}(80)(0.080)^2\\
&=0.256\,\text{J/m}.
\end{align}
Therefore
\begin{equation}
\boxed{
\mathcal{E}=0.2785\,\text{J/m}\simeq0.279\,\text{J/m}.}
\end{equation}
The kinetic and potential densities are not equal here. Equality is guaranteed point by point only for a pure traveling wave moving at the string wave speed.
\section{Worked Example 2: Sinusoidal average energy density and energy per wavelength}
A sinusoidal traveling wave has
\begin{equation}
A=4.0\,\text{mm},
\qquad
f=50\,\text{Hz},
\end{equation}
on a string with
\begin{equation}
\mu=0.012\,\text{kg/m},
\qquad
T=75\,\text{N}.
\end{equation}
Find the wave speed, wavelength, average energy density, and average energy contained in one wavelength.
\subsection*{Solution}
First convert the amplitude:
\begin{equation}
A=0.0040\,\text{m}.
\end{equation}
The wave speed is
\begin{align}
c
&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{75}{0.012}}\\
&=79.1\,\text{m/s}.
\end{align}
The wavelength is
\begin{align}
\lambda
&=\frac{c}{f}\\
&=\frac{79.1}{50}\\
&=1.58\,\text{m}.
\end{align}
The angular frequency is
\begin{equation}
\omega=2\pi f=314.2\,\text{rad/s}.
\end{equation}
The average energy density is
\begin{align}
\langle\mathcal{E}\rangle
&=\frac{1}{2}\mu A^2\omega^2\\
&=\frac{1}{2}(0.012)(0.0040)^2(314.2)^2\\
&\simeq9.47\times10^{-3}\,\text{J/m}.
\end{align}
Thus
\begin{equation}
\boxed{\langle\mathcal{E}\rangle\simeq9.47\,\text{mJ/m}.}
\end{equation}
The energy per wavelength is
\begin{align}
E_{\lambda}
&=\langle\mathcal{E}\rangle\lambda\\
&=(9.47\times10^{-3})(1.58)\\
&\simeq1.50\times10^{-2}\,\text{J}.
\end{align}
Therefore
\begin{equation}
\boxed{E_{\lambda}\simeq0.0150\,\text{J}.}
\end{equation}
\section{Worked Example 3: Total energy of a Gaussian traveling pulse}
Consider the right-moving pulse
\begin{equation}
u(x,t)=A\exp\left[-\frac{(x-ct)^2}{2\sigma^2}\right].
\end{equation}
Let
\begin{equation}
A=1.0\,\text{cm},
\qquad
\sigma=0.080\,\text{m},
\qquad
T=60\,\text{N}.
\end{equation}
Find the total mechanical energy of the pulse.
\subsection*{Solution}
Let
\begin{equation}
z=x-ct.
\end{equation}
Then
\begin{equation}
F(z)=A\exp\left(-\frac{z^2}{2\sigma^2}\right)
\end{equation}
and
\begin{equation}
F'(z)
=-\frac{Az}{\sigma^2}
\exp\left(-\frac{z^2}{2\sigma^2}\right).
\end{equation}
For a pure traveling wave,
\begin{equation}
\mathcal{E}=T[F'(z)]^2.
\end{equation}
Hence
\begin{equation}
\mathcal{E}(z)
=T\frac{A^2z^2}{\sigma^4}
\exp\left(-\frac{z^2}{\sigma^2}\right).
\end{equation}
The total energy is
\begin{equation}
E=\int_{-\infty}^{\infty}\mathcal{E}(z)\,dz.
\end{equation}
Using
\begin{equation}
\int_{-\infty}^{\infty}
z^2e^{-z^2/\sigma^2}\,dz
=\frac{\sqrt{\pi}}{2}\sigma^3,
\end{equation}
we obtain
\begin{align}
E
&=\frac{TA^2}{\sigma^4}
\frac{\sqrt{\pi}}{2}\sigma^3\\
&=\boxed{\frac{TA^2\sqrt{\pi}}{2\sigma}}.
\end{align}
Now use
\begin{equation}
A=0.010\,\text{m}.
\end{equation}
Then
\begin{align}
E
&=\frac{(60)(0.010)^2\sqrt{\pi}}{2(0.080)}\\
&\simeq6.65\times10^{-2}\,\text{J}.
\end{align}
Therefore
\begin{equation}
\boxed{E\simeq0.0665\,\text{J}.}
\end{equation}
The result scales as $A^2$ and inversely with the pulse width parameter $\sigma$: for the same peak displacement, a sharper pulse has larger slopes and therefore more energy.
\section{Worked Example 4: Amplitude and frequency scaling}
Wave 1 and wave 2 travel on the same string. Suppose
\begin{equation}
A_2=2A_1
\end{equation}
and
\begin{equation}
\omega_2=3\omega_1.
\end{equation}
Find the ratio of their average sinusoidal energy densities.
\subsection*{Solution}
Since
\begin{equation}
\langle\mathcal{E}\rangle
=\frac{1}{2}\mu A^2\omega^2,
\end{equation}
and both waves are on the same string,
\begin{align}
\frac{\langle\mathcal{E}_2\rangle}
{\langle\mathcal{E}_1\rangle}
&=\frac{A_2^2\omega_2^2}{A_1^2\omega_1^2}\\
&=(2)^2(3)^2\\
&=36.
\end{align}
Thus
\begin{equation}
\boxed{
\langle\mathcal{E}_2\rangle
=36\langle\mathcal{E}_1\rangle.}
\end{equation}
The square-law scaling compounds rapidly when both amplitude and frequency increase.
\section{Worked Example 5: Energy exchange in a standing mode}
A fixed-fixed string normal mode is
\begin{equation}
u(x,t)=B\sin(kx)\cos(\omega t)
\end{equation}
with
\begin{equation}
B=6.0\,\text{mm},
\qquad
\mu=0.015\,\text{kg/m},
\qquad
f=30\,\text{Hz},
\qquad
L=1.0\,\text{m}.
\end{equation}
Find the total mode energy, and find the integrated kinetic and potential energies when
\begin{equation}
\omega t=\frac{\pi}{6}.
\end{equation}
\subsection*{Solution}
Convert the amplitude:
\begin{equation}
B=0.0060\,\text{m}.
\end{equation}
The angular frequency is
\begin{equation}
\omega=2\pi f=188.5\,\text{rad/s}.
\end{equation}
The constant total mode energy is
\begin{align}
E
&=\frac{1}{4}\mu B^2\omega^2L\\
&=\frac{1}{4}(0.015)(0.0060)^2(188.5)^2(1.0)\\
&\simeq4.80\times10^{-3}\,\text{J}.
\end{align}
Therefore
\begin{equation}
\boxed{E\simeq4.80\,\text{mJ}.}
\end{equation}
At $\omega t=\pi/6$,
\begin{equation}
\sin^2\left(\frac{\pi}{6}\right)=\frac{1}{4}
\end{equation}
and
\begin{equation}
\cos^2\left(\frac{\pi}{6}\right)=\frac{3}{4}.
\end{equation}
Hence
\begin{equation}
K=\frac{1}{4}E\simeq1.20\,\text{mJ}
\end{equation}
and
\begin{equation}
U=\frac{3}{4}E\simeq3.60\,\text{mJ}.
\end{equation}
So
\begin{equation}
\boxed{K\simeq1.20\,\text{mJ},\qquad U\simeq3.60\,\text{mJ}.}
\end{equation}
Their sum remains the same total mechanical energy.
\section{Worked Example 6: Energy density of simultaneous right- and left-moving components}
At one event, suppose a two-direction solution has
\begin{equation}
F'=0.10,
\qquad
G'=-0.040,
\qquad
T=50\,\text{N}.
\end{equation}
Find the total energy density. Also find the kinetic and potential parts separately.
\subsection*{Solution}
For the two-family solution,
\begin{equation}
\mathcal{E}=T[(F')^2+(G')^2].
\end{equation}
Therefore
\begin{align}
\mathcal{E}
&=50[(0.10)^2+(-0.040)^2]\\
&=50(0.0116)\\
&=\boxed{0.580\,\text{J/m}}.
\end{align}
For the kinetic part,
\begin{equation}
u_t=c(-F'+G').
\end{equation}
Using $\mu c^2=T$,
\begin{align}
\mathcal{K}
&=\frac{T}{2}(-F'+G')^2\\
&=\frac{50}{2}(-0.10-0.040)^2\\
&=25(0.0196)\\
&=\boxed{0.490\,\text{J/m}}.
\end{align}
For the potential part,
\begin{align}
\mathcal{U}
&=\frac{T}{2}(F'+G')^2\\
&=25(0.10-0.040)^2\\
&=25(0.0036)\\
&=\boxed{0.0900\,\text{J/m}}.
\end{align}
As a check,
\begin{equation}
0.490+0.0900=0.580\,\text{J/m}.
\end{equation}
The kinetic and potential densities are not equal because this field contains both propagation directions at once.
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} treating displacement amplitude itself as energy. Energy depends on squared velocity and squared slope.
\item \textbf{Mistake:} using the propagation speed $c$ in place of the material velocity $u_t$ in the kinetic energy.
\item \textbf{Mistake:} assuming $\mathcal{K}=\mathcal{U}$ for every wave field. Pointwise equality holds for a pure traveling wave, not for an arbitrary superposition or standing wave.
\item \textbf{Mistake:} forgetting that $u_x$ is dimensionless for transverse displacement $u$ measured in meters against coordinate $x$ measured in meters.
\item \textbf{Mistake:} assuming the displacement maximum of a pulse must also be the local energy-density maximum. For a smooth pure traveling pulse, the exact displacement peak can have zero slope and zero transverse velocity.
\item \textbf{Mistake:} applying the ideal-string energy density outside the small-slope, constant-tension model without checking whether the approximations remain valid.
\end{itemize}
\section{What WM18 establishes}
The ideal one-dimensional string contains mechanical energy with local density
\begin{equation}
\boxed{
\mathcal{E}
=\frac{1}{2}\mu u_t^2
+\frac{1}{2}T u_x^2.}
\end{equation}
For a pure traveling wave,
\begin{equation}
\boxed{\mathcal{K}=\mathcal{U}}
\end{equation}
and, for a sinusoidal traveling wave,
\begin{equation}
\boxed{
\langle\mathcal{E}\rangle
=\frac{1}{2}\mu A^2\omega^2.}
\end{equation}
The next question is no longer how much energy is locally present, but how rapidly that energy passes a fixed position. That leads naturally to wave power and energy flux.
\section*{References}
\begin{thebibliography}{9}
\bibitem{French1971}
A.~P. French,
\emph{Vibrations and Waves},
M.I.T. Introductory Physics Series,
W. W. Norton \& Company, 1971.
\bibitem{Crawford1968}
Frank S. Crawford, Jr.,
\emph{Waves},
Berkeley Physics Course, Volume 3,
McGraw-Hill, 1968.
\bibitem{OpenStax164}
William Moebs, Samuel J. Ling, and Jeff Sanny,
\emph{University Physics, Volume 1},
OpenStax, 2016,
Section 16.4, ``Energy and Power of a Wave.''
\bibitem{Georgi1993}
Howard Georgi,
\emph{The Physics of Waves},
Benjamin/Cummings, 1992,
continuum and traveling-wave chapters; also distributed through MIT OpenCourseWare 8.03SC.
\bibitem{MIT803PS5}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
Fall 2016, Problem Set 5 and Lecture 10 materials on traveling waves and string energy,
MIT OpenCourseWare.
\end{thebibliography}