Wave Mechanics: Power Carried by a 1D WaveWaveMechanicsPowerCarriedByA1DWave2026-09-12 16:38:162026-09-12 16:38:16Topicwave mechanicswave powerenergy fluxenergy conservationstretched stringtraveling wavestanding wavesinusoidal waveaverage powertensionlinear mass density% this is the default PhysicsLibrary preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics).
% If you use \includegraphics{figure} with no file extension, PhysicsLibrary
% will prefer figure.png before falling back to figure.eps.
\usepackage{graphicx}
\DeclareGraphicsExtensions{.png,.jpg,.jpeg,.pdf,.eps}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here\section*{Wave Mechanics: Power Carried by a 1D Wave}
WM18 developed the mechanical energy density of an ideal stretched string,
\begin{equation}
\mathcal{E}(x,t)
=\frac{1}{2}\mu u_t^2+\frac{1}{2}T u_x^2.
\end{equation}
Energy density answers the question, ``How much mechanical energy is stored near this position?'' The next question is different:
\begin{center}
\textbf{How rapidly does wave energy cross a fixed position?}
\end{center}
That rate of energy transfer is the wave power. With positive power defined as energy flow toward increasing $x$, the ideal-string result is
\begin{equation}
\boxed{P(x,t)=-T u_x u_t.}
\end{equation}
The sign matters. A positive value means net mechanical energy crosses the chosen position toward $+x$; a negative value means the net flow is toward $-x$. This local expression is the one-dimensional energy-flux law associated with the string wave equation and is consistent with standard wave-energy treatments \cite{French1971,Crawford1968,OpenStax164,Georgi1992}.
WM19 derives the result mechanically and from local energy conservation, applies it to arbitrary traveling profiles and sinusoidal waves, derives the time-averaged sinusoidal power, and explains what happens when right- and left-moving waves are both present.
\section{Power is energy crossing a position per unit time}
Choose a fixed position $x$. During a short time interval $dt$, mechanical energy can cross that position even though the material of the string merely oscillates transversely.
We define $P(x,t)$ so that
\begin{equation}
P>0
\end{equation}
means energy transport toward increasing $x$, and
\begin{equation}
P<0
\end{equation}
means energy transport toward decreasing $x$.
The SI unit of power is
\begin{equation}
\text{W}=\frac{\text{J}}{\text{s}}.
\end{equation}
This is not yet intensity. Intensity is power per unit area and will be introduced later when waves spread through higher-dimensional space \cite{OpenStax164}.
\section{Mechanical derivation at a cut in the string}
Imagine cutting the mathematical description of the string at one fixed position. The material just to the left exerts a tension force on the material just to the right.
For a small-slope string, the transverse component of the force exerted by the left side on the right side is approximately
\begin{equation}
F_u=-T u_x.
\end{equation}
The transverse velocity of the material at the cut is
\begin{equation}
v_u=u_t.
\end{equation}
Mechanical power delivered from the left portion into the right portion is force times velocity:
\begin{align}
P
&=F_u v_u\\
&=(-T u_x)u_t.
\end{align}
Therefore
\begin{equation}
\boxed{P=-T u_xu_t.}
\end{equation}
The geometry and sign convention are summarized below.
\begin{center}
\includegraphics{WM19_fig01_power_at_cut.png}
\vspace{0.45em}
\textbf{Figure.}
At a fixed cut, the local slope determines the transverse component of tension and $u_t$ gives the local material velocity. Their product determines the mechanical work rate across the cut. Positive $P$ is defined as energy transport toward $+x$.
\end{center}
\section{Dimensional check}
The tension has units of newtons,
\begin{equation}
[T]=\text{N},
\end{equation}
while $u_x$ is dimensionless and
\begin{equation}
[u_t]=\text{m/s}.
\end{equation}
Hence
\begin{align}
[T u_xu_t]
&=\text{N}\frac{\text{m}}{\text{s}}\\
&=\frac{\text{J}}{\text{s}}\\
&=\text{W}.
\end{align}
So the expression has the correct dimensions for power.
\section{The same result from local energy conservation}
The energy-density formula from WM18 is
\begin{equation}
\mathcal{E}=\frac{1}{2}\mu u_t^2+\frac{1}{2}T u_x^2.
\end{equation}
Differentiate with respect to time:
\begin{equation}
\mathcal{E}_t
=\mu u_tu_{tt}+T u_xu_{xt}.
\end{equation}
For the ideal string,
\begin{equation}
\mu u_{tt}=T u_{xx}.
\end{equation}
Substitute this into the first term:
\begin{align}
\mathcal{E}_t
&=T u_tu_{xx}+T u_xu_{xt}\\
&=T\frac{\partial}{\partial x}(u_tu_x).
\end{align}
Therefore
\begin{equation}
\mathcal{E}_t
+\frac{\partial}{\partial x}(-T u_tu_x)=0.
\end{equation}
Identifying
\begin{equation}
\boxed{P=-T u_xu_t}
\end{equation}
gives the local conservation law
\begin{equation}
\boxed{\mathcal{E}_t+P_x=0.}
\end{equation}
This compact equation says that a local change in stored energy is caused by an imbalance of power flow into and out of the region.
\section{Energy balance on a finite interval}
Integrate the local conservation law from $x=a$ to $x=b$:
\begin{equation}
\int_a^b\mathcal{E}_t\,dx
+\int_a^b P_x\,dx=0.
\end{equation}
Thus
\begin{equation}
\frac{d}{dt}\int_a^b\mathcal{E}\,dx
+P(b,t)-P(a,t)=0.
\end{equation}
Equivalently,
\begin{equation}
\boxed{
\frac{d}{dt}\int_a^b\mathcal{E}\,dx
=P(a,t)-P(b,t).}
\end{equation}
Power entering at the left boundary increases the energy stored in the interval; power leaving at the right boundary decreases it.
\begin{center}
\includegraphics{WM19_fig02_energy_control_volume.png}
\vspace{0.45em}
\textbf{Figure.}
Energy conservation on a fixed interval: rate of change of stored energy equals incoming power minus outgoing power.
\end{center}
\section{Power in a pure right-moving wave}
Consider
\begin{equation}
u(x,t)=F(x-ct).
\end{equation}
With
\begin{equation}
\xi=x-ct,
\end{equation}
we have
\begin{equation}
u_x=F'(\xi)
\end{equation}
and
\begin{equation}
u_t=-cF'(\xi).
\end{equation}
Therefore
\begin{align}
P
&=-T u_xu_t\\
&=-T F'(\xi)[-cF'(\xi)]\\
&=Tc[F'(\xi)]^2.
\end{align}
Hence
\begin{equation}
\boxed{P_{\rightarrow}=Tc[F'(x-ct)]^2\geq0.}
\end{equation}
The power is nonnegative because this wave carries energy toward $+x$.
WM18 showed that the energy density of the same wave is
\begin{equation}
\mathcal{E}=T[F'(x-ct)]^2.
\end{equation}
Therefore
\begin{equation}
\boxed{P_{\rightarrow}=c\mathcal{E}.}
\end{equation}
This is physically intuitive: a packet of energy density $\mathcal{E}$ translating at speed $c$ carries energy past a fixed point at rate $c\mathcal{E}$.
\section{Power in a pure left-moving wave}
For
\begin{equation}
u(x,t)=G(x+ct),
\end{equation}
we have
\begin{equation}
u_x=G'
\end{equation}
and
\begin{equation}
u_t=cG'.
\end{equation}
Thus
\begin{equation}
\boxed{P_{\leftarrow}=-Tc[G'(x+ct)]^2\leq0.}
\end{equation}
Since
\begin{equation}
\mathcal{E}=T[G'(x+ct)]^2,
\end{equation}
we can write
\begin{equation}
\boxed{P_{\leftarrow}=-c\mathcal{E}.}
\end{equation}
The sign is a direction marker: the energy moves toward decreasing $x$.
\section{Sinusoidal traveling wave}
Consider the right-moving sinusoidal wave
\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_x=-Ak\sin\theta
\end{equation}
and
\begin{equation}
u_t=A\omega\sin\theta.
\end{equation}
The instantaneous power is therefore
\begin{align}
P
&=-T(-Ak\sin\theta)(A\omega\sin\theta)\\
&=T A^2k\omega\sin^2\theta.
\end{align}
Hence
\begin{equation}
\boxed{
P(x,t)=T A^2k\omega\sin^2(kx-\omega t+\phi).}
\end{equation}
It is always nonnegative for this right-moving wave.
\begin{center}
\includegraphics{WM19_fig03_sinusoid_power.png}
\vspace{0.45em}
\textbf{Figure.}
For a right-moving sinusoidal wave, normalized instantaneous power varies as $\sin^2\theta$. Power is largest where the displacement passes through zero and vanishes at displacement extrema.
\end{center}
\section{Average power of a sinusoidal wave}
Over one phase cycle,
\begin{equation}
\left\langle\sin^2\theta\right\rangle=\frac{1}{2}.
\end{equation}
Therefore
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{2}T A^2k\omega.}
\end{equation}
Use
\begin{equation}
\omega=ck
\end{equation}
and
\begin{equation}
T=\mu c^2.
\end{equation}
Then
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{2}\mu A^2\omega^2c.}
\end{equation}
This is the standard time-averaged power of a sinusoidal transverse wave on an ideal string \cite{OpenStax164,French1971}.
Because WM18 found
\begin{equation}
\langle\mathcal{E}\rangle
=\frac{1}{2}\mu A^2\omega^2,
\end{equation}
we again have
\begin{equation}
\boxed{\langle P\rangle=c\langle\mathcal{E}\rangle.}
\end{equation}
\section{Square-law scaling}
For a fixed string and a sinusoidal traveling wave,
\begin{equation}
\langle P\rangle
=\frac{1}{2}\mu A^2\omega^2c.
\end{equation}
Thus
\begin{equation}
\boxed{\langle P\rangle\propto A^2\omega^2}
\end{equation}
when the string properties are fixed.
Consequently:
\begin{itemize}
\item doubling $A$ multiplies average power by $4$;
\item doubling $f$ or $\omega$ multiplies average power by $4$;
\item doubling both amplitude and frequency multiplies average power by $16$.
\end{itemize}
OpenStax emphasizes this same amplitude-squared and frequency-squared scaling for sinusoidal mechanical waves \cite{OpenStax164}.
\section{Both propagation directions at once}
For the general two-direction solution
\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}
Substitute into the power law:
\begin{align}
P
&=-T[F'+G'][-cF'+cG']\\
&=Tc\left([F']^2-[G']^2\right).
\end{align}
Thus
\begin{equation}
\boxed{
P=Tc\left([F'(x-ct)]^2-[G'(x+ct)]^2\right).}
\end{equation}
The net power is the right-moving contribution minus the left-moving contribution.
\begin{center}
\includegraphics{WM19_fig04_directional_power_balance.png}
\vspace{0.45em}
\textbf{Figure.}
Directional power adds with a sign. Right-moving energy contributes positive power and left-moving energy contributes negative power under the chosen convention.
\end{center}
\section{Standing waves and zero average transport}
A standing wave can be written
\begin{equation}
u(x,t)=B\sin(kx)\cos(\omega t).
\end{equation}
Then
\begin{equation}
u_x=Bk\cos(kx)\cos(\omega t)
\end{equation}
and
\begin{equation}
u_t=-B\omega\sin(kx)\sin(\omega t).
\end{equation}
Therefore
\begin{align}
P
&=-T u_xu_t\\
&=T B^2k\omega\cos(kx)\sin(kx)\cos(\omega t)\sin(\omega t).
\end{align}
Using double-angle identities,
\begin{equation}
\boxed{
P(x,t)=\frac{1}{4}TB^2k\omega
\sin(2kx)\sin(2\omega t).}
\end{equation}
The instantaneous local power is generally not zero. Its sign reverses as energy moves back and forth within the standing-wave pattern.
However, over a complete time cycle,
\begin{equation}
\left\langle\sin(2\omega t)\right\rangle=0,
\end{equation}
so
\begin{equation}
\boxed{\langle P\rangle_{\text{standing}}=0.}
\end{equation}
A perfect standing wave therefore has no net time-averaged transport through a fixed position, even though energy can flow locally and instantaneously.
\section{Worked Example 1: Power from local slope and velocity}
A string has tension
\begin{equation}
T=100\,\text{N}.
\end{equation}
At one event,
\begin{equation}
u_x=0.020,
\qquad
u_t=-0.30\,\text{m/s}.
\end{equation}
Find the instantaneous power and interpret its sign.
\subsection*{Solution}
Use
\begin{equation}
P=-T u_xu_t.
\end{equation}
Then
\begin{align}
P
& =-(100)(0.020)(-0.30)\\
& =0.60\,\text{W}.
\end{align}
Thus
\begin{equation}
\boxed{P=+0.60\,\text{W}.}
\end{equation}
The positive sign means net energy is crossing the selected position toward increasing $x$ at that instant.
\section{Worked Example 2: Average power of a sinusoidal wave}
A right-moving sinusoidal wave has
\begin{equation}
A=5.0\,\text{mm},
\qquad
f=10\,\text{Hz},
\end{equation}
on a string with
\begin{equation}
T=80\,\text{N},
\qquad
\mu=0.020\,\text{kg/m}.
\end{equation}
Find $c$, $\omega$, $k$, and the time-averaged power.
\subsection*{Solution}
The wave speed is
\begin{align}
c
&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{80}{0.020}}\\
&\simeq63.25\,\text{m/s}.
\end{align}
The angular frequency is
\begin{equation}
\omega=2\pi f=62.83\,\text{rad/s}.
\end{equation}
Then
\begin{align}
k
&=\frac{\omega}{c}\\
&\simeq0.993\,\text{rad/m}.
\end{align}
Convert the amplitude:
\begin{equation}
A=0.0050\,\text{m}.
\end{equation}
Now use
\begin{equation}
\langle P\rangle
=\frac{1}{2}\mu A^2\omega^2c.
\end{equation}
Thus
\begin{align}
\langle P\rangle
&=\frac{1}{2}(0.020)(0.0050)^2(62.83)^2(63.25)\\
&\simeq6.24\times10^{-2}\,\text{W}.
\end{align}
Therefore
\begin{equation}
\boxed{\langle P\rangle\simeq0.0624\,\text{W}.}
\end{equation}
\section{Worked Example 3: Required amplitude for a desired average power}
A sinusoidal traveling wave moves on a string with
\begin{equation}
\mu=0.010\,\text{kg/m},
\qquad
c=100\,\text{m/s}.
\end{equation}
At frequency
\begin{equation}
f=50\,\text{Hz},
\end{equation}
what amplitude is required to carry an average power of
\begin{equation}
\langle P\rangle=10\,\text{W}?
\end{equation}
\subsection*{Solution}
Start with
\begin{equation}
\langle P\rangle
=\frac{1}{2}\mu A^2\omega^2c.
\end{equation}
Solve for $A$:
\begin{equation}
A=\sqrt{\frac{2\langle P\rangle}{\mu\omega^2c}}.
\end{equation}
The angular frequency is
\begin{equation}
\omega=2\pi(50)=314.16\,\text{rad/s}.
\end{equation}
Therefore
\begin{align}
A
&=\sqrt{\frac{2(10)}{(0.010)(314.16)^2(100)}}\\
&\simeq1.42\times10^{-2}\,\text{m}.
\end{align}
Thus
\begin{equation}
\boxed{A\simeq14.2\,\text{mm}.}
\end{equation}
\section{Worked Example 4: Power carried by a translating pulse from its energy density}
At one location in a pure right-moving pulse, the instantaneous energy density is
\begin{equation}
\mathcal{E}=0.12\,\text{J/m}.
\end{equation}
The pulse speed is
\begin{equation}
c=50\,\text{m/s}.
\end{equation}
Find the instantaneous power at that event.
\subsection*{Solution}
For a pure right-moving wave,
\begin{equation}
P=c\mathcal{E}.
\end{equation}
Therefore
\begin{align}
P
&=(50)(0.12)\\
&=6.0\,\text{W}.
\end{align}
Hence
\begin{equation}
\boxed{P=+6.0\,\text{W}.}
\end{equation}
If the same energy-density profile were traveling toward $-x$, the power would instead be $-6.0\,\text{W}$.
\section{Worked Example 5: Net power with two propagation directions}
At one event in a two-direction wave field, suppose
\begin{equation}
T=50\,\text{N},
\qquad
c=40\,\text{m/s},
\end{equation}
and the directional profile derivatives have values
\begin{equation}
F'=0.060,
\qquad
G'=0.040.
\end{equation}
Find the net power.
\subsection*{Solution}
Use
\begin{equation}
P=Tc\left([F']^2-[G']^2\right).
\end{equation}
Then
\begin{align}
P
&=(50)(40)\left[(0.060)^2-(0.040)^2\right]\\
&=2000(0.0036-0.0016)\\
&=4.0\,\text{W}.
\end{align}
Thus
\begin{equation}
\boxed{P=+4.0\,\text{W}.}
\end{equation}
The right-moving contribution is larger than the left-moving contribution, so the net energy transport is toward $+x$.
\section{Worked Example 6: Instantaneous power in a standing wave}
Consider
\begin{equation}
u(x,t)=B\sin(kx)\cos(\omega t)
\end{equation}
with
\begin{equation}
T=60\,\text{N},
\qquad
B=0.010\,\text{m},
\qquad
k=\pi\,\text{rad/m},
\qquad
c=30\,\text{m/s}.
\end{equation}
Find the power at
\begin{equation}
x=0.25\,\text{m}
\end{equation}
when
\begin{equation}
\omega t=\frac{\pi}{8}.
\end{equation}
Also state the time-averaged power at that position.
\subsection*{Solution}
Because
\begin{equation}
\omega=ck,
\end{equation}
we have
\begin{equation}
\omega=30\pi\,\text{rad/s}.
\end{equation}
The standing-wave power is
\begin{equation}
P=\frac{1}{4}TB^2k\omega\sin(2kx)\sin(2\omega t).
\end{equation}
At $x=0.25\,\text{m}$,
\begin{equation}
2kx=2\pi(0.25)=\frac{\pi}{2},
\end{equation}
so
\begin{equation}
\sin(2kx)=1.
\end{equation}
Also
\begin{equation}
2\omega t=\frac{\pi}{4},
\end{equation}
so
\begin{equation}
\sin(2\omega t)=\frac{\sqrt{2}}{2}.
\end{equation}
Therefore
\begin{align}
P
&=\frac{1}{4}(60)(0.010)^2(\pi)(30\pi)
\frac{\sqrt{2}}{2}\\
&\simeq0.314\,\text{W}.
\end{align}
Thus
\begin{equation}
\boxed{P\simeq+0.314\,\text{W}}
\end{equation}
at this instant.
Over a complete oscillation cycle,
\begin{equation}
\boxed{\langle P\rangle=0.}
\end{equation}
The local energy flow reverses later in the cycle, so there is no net time-averaged transport through the point.
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} dropping the minus sign in $P=-T u_xu_t$. The sign carries propagation-direction information.
\item \textbf{Mistake:} assuming positive displacement means positive power. Power depends on slope and velocity, not displacement alone.
\item \textbf{Mistake:} treating a standing wave as having zero instantaneous power everywhere. Its time-averaged net transport is zero, but instantaneous local power can be nonzero.
\item \textbf{Mistake:} confusing power with energy density. Their units differ: watts versus joules per meter.
\item \textbf{Mistake:} confusing one-dimensional power with intensity. Intensity requires division by an area.
\item \textbf{Mistake:} applying $P=c\mathcal{E}$ to an arbitrary superposition. That relation holds directly for a single pure right-moving wave; a left-moving wave gives $P=-c\mathcal{E}$, and a two-direction field requires the signed difference of the directional contributions.
\end{itemize}
\section{Summary}
For an ideal stretched string, the instantaneous mechanical power crossing a fixed position is
\begin{equation}
\boxed{P=-T u_xu_t.}
\end{equation}
Together with
\begin{equation}
\mathcal{E}=\frac{1}{2}\mu u_t^2+\frac{1}{2}T u_x^2,
\end{equation}
it satisfies the local conservation law
\begin{equation}
\boxed{\mathcal{E}_t+P_x=0.}
\end{equation}
For pure traveling waves,
\begin{equation}
\boxed{P_{\rightarrow}=+c\mathcal{E},
\qquad
P_{\leftarrow}=-c\mathcal{E}.}
\end{equation}
For a right-moving sinusoidal wave,
\begin{equation}
\boxed{P=T A^2k\omega\sin^2(kx-\omega t+\phi)}
\end{equation}
and
\begin{equation}
\boxed{\langle P\rangle
=\frac{1}{2}T A^2k\omega
=\frac{1}{2}\mu A^2\omega^2c.}
\end{equation}
The next step is to turn these results into a systematic treatment of average power, intensity, and wave impedance.
\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{Georgi1992}
Howard Georgi,
\emph{The Physics of Waves},
Benjamin/Cummings, 1992,
continuum and traveling-wave chapters; also distributed through MIT OpenCourseWare 8.03SC.
\bibitem{MIT803}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
Fall 2016,
mechanical-wave lectures, notes, and Problem Set 5,
MIT OpenCourseWare.
\bibitem{Feynman47}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Chapter 47, ``Sound. The wave equation.''
\end{thebibliography}