Wave Mechanics: Mechanical Wave ImpedanceWaveMechanicsMechanicalWaveImpedance2026-09-12 18:29:362026-09-12 18:29:36Topicwave mechanicsmechanical impedancecharacteristic impedancestring impedanceforce velocity ratiopower flowreflectiontransmissionimpedance matchingtraveling waves% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Wave Mechanics: Mechanical Wave Impedance}
WM18--WM20 developed energy and power transport on an ideal string, and WM21 generalized the idea of energy flux to higher-dimensional waves. We now return to the one-dimensional string and ask a different but closely related question:
\begin{quote}
How much transverse force is associated with a given transverse velocity in a traveling wave?
\end{quote}
The answer is the \emph{characteristic mechanical wave impedance} of the string.
For an ideal string with tension $T$, linear mass density $\mu$, and wave speed
\begin{equation}
c=\sqrt{\frac{T}{\mu}},
\end{equation}
the characteristic impedance is
\begin{equation}
\boxed{
Z_0=\frac{T}{c}=\mu c=\sqrt{T\mu}.}
\end{equation}
Its SI unit is
\begin{equation}
\boxed{[Z_0]=\text{N}\,\text{s}/\text{m}=\text{kg}/\text{s}.}
\end{equation}
The impedance $Z_0$ is a property of the medium and its tension. It tells us how transverse force and transverse velocity are related for a \emph{one-way traveling wave} \cite{French1971,Crawford1968,Georgi1993,MIT803}.
\section{Mechanical impedance and characteristic impedance}
In vibration theory, a general mechanical impedance is often defined in harmonic steady state as a ratio of force amplitude to velocity amplitude,
\begin{equation}
Z_{\text{mech}}(\omega)=\frac{\widetilde F(\omega)}{\widetilde v(\omega)}.
\end{equation}
For a mass-spring-damper system or another reactive load, this quantity can depend on frequency and can be complex.
A lossless uniform string has a particularly simple traveling-wave result. Its \emph{characteristic impedance}
\begin{equation}
Z_0=\sqrt{T\mu}
\end{equation}
is real and, within the ideal nondispersive string model, independent of frequency.
This article uses $Z_0$ for that traveling-wave property. The ratio of force to velocity in an arbitrary standing-wave field need not equal $Z_0$ point by point.
\section{Force and velocity at a cut in the string}
Let $u(x,t)$ be the transverse displacement. The local transverse velocity is
\begin{equation}
v_\perp=u_t.
\end{equation}
For small slopes, define the transverse force transmitted across a cut in the positive $x$ direction as
\begin{equation}
\boxed{F_\perp^{(+x)}=-T u_x.}
\end{equation}
With this sign convention, the instantaneous power flowing in the positive $x$ direction is exactly the WM19 result
\begin{equation}
P=F_\perp^{(+x)}v_\perp=-T u_xu_t.
\end{equation}
\begin{center}
\includegraphics{WM22_fig01_force_velocity_at_cut.png}
\vspace{0.45em}
\textbf{Figure.}
At a cut in the string, the traveling wave carries a transverse force and transverse velocity. Their ratio defines the characteristic mechanical impedance for a one-way wave.
\end{center}
The sign convention is useful because positive $P$ means energy transport toward increasing $x$.
\section{Right-moving wave: derive the impedance}
For a right-moving profile
\begin{equation}
u(x,t)=F(x-ct),
\end{equation}
WM15 showed that
\begin{equation}
u_t=-c u_x.
\end{equation}
Therefore
\begin{equation}
u_x=-\frac{u_t}{c}.
\end{equation}
The transmitted transverse force becomes
\begin{align}
F_\perp^{(+x)}
&=-T u_x\\
&=\frac{T}{c}u_t.
\end{align}
Define
\begin{equation}
\boxed{Z_0=\frac{T}{c}.}
\end{equation}
Then a right-moving wave satisfies
\begin{equation}
\boxed{F_\perp^{(+x)}=Z_0u_t.}
\end{equation}
Because
\begin{equation}
c=\sqrt{\frac{T}{\mu}},
\end{equation}
we may rewrite the impedance in two equivalent forms:
\begin{align}
Z_0
&=\frac{T}{\sqrt{T/\mu}}\\
&=\sqrt{T\mu},
\end{align}
and
\begin{equation}
\mu c
=\mu\sqrt{\frac{T}{\mu}}
=\sqrt{T\mu}.
\end{equation}
Thus
\begin{equation}
\boxed{Z_0=\frac{T}{c}=\mu c=\sqrt{T\mu}.}
\end{equation}
\section{Left-moving wave and the sign of power}
For a left-moving profile
\begin{equation}
u(x,t)=G(x+ct),
\end{equation}
we have
\begin{equation}
u_t=+c u_x.
\end{equation}
Hence
\begin{equation}
u_x=\frac{u_t}{c}
\end{equation}
and
\begin{align}
F_\perp^{(+x)}
&=-T u_x\\
&=-\frac{T}{c}u_t\\
&=-Z_0u_t.
\end{align}
Therefore
\begin{equation}
\boxed{
\begin{array}{ll}
\text{right-moving:} & F_\perp^{(+x)}=+Z_0u_t,\\[0.35em]
\text{left-moving:} & F_\perp^{(+x)}=-Z_0u_t.
\end{array}}
\end{equation}
The medium has the same positive characteristic impedance $Z_0$ in either direction. The sign change records the direction of power flow.
\section{Power written in impedance form}
For a right-moving wave,
\begin{align}
P
&=F_\perp^{(+x)}u_t\\
&=Z_0u_t^2.
\end{align}
Thus
\begin{equation}
\boxed{P_{\rightarrow}=+Z_0u_t^2.}
\end{equation}
For a left-moving wave,
\begin{equation}
\boxed{P_{\leftarrow}=-Z_0u_t^2.}
\end{equation}
This reproduces the directional power result from WM19 because
\begin{equation}
Z_0=\mu c.
\end{equation}
The impedance language therefore compresses the force, velocity, and power relations into a compact set of equations.
\section{Sinusoidal average power}
Consider a right-moving sinusoidal wave
\begin{equation}
u(x,t)=A\cos(kx-\omega t+\phi).
\end{equation}
Its transverse velocity is
\begin{equation}
u_t=A\omega\sin(kx-\omega t+\phi).
\end{equation}
The RMS transverse velocity is
\begin{equation}
v_{\text{rms}}=\frac{A\omega}{\sqrt{2}}.
\end{equation}
Averaging
\begin{equation}
P=Z_0u_t^2
\end{equation}
over one cycle gives
\begin{equation}
\boxed{\langle P\rangle=Z_0v_{\text{rms}}^2.}
\end{equation}
Equivalently,
\begin{equation}
\boxed{\langle P\rangle=\frac12 Z_0A^2\omega^2.}
\end{equation}
Since $Z_0=\mu c$, this is exactly the WM20 formula
\begin{equation}
\langle P\rangle=\frac12\mu A^2\omega^2c.
\end{equation}
This is the mechanical-wave analogue of the familiar statement that power depends on the square of a wave amplitude multiplied by a characteristic impedance or admittance factor.
\section{How $T$ and $\mu$ affect speed and impedance}
The speed and impedance depend differently on tension and linear density:
\begin{equation}
\boxed{c=\sqrt{\frac{T}{\mu}},}
\qquad
\boxed{Z_0=\sqrt{T\mu}.}
\end{equation}
Thus increasing tension with $\mu$ fixed increases both $c$ and $Z_0$ as $\sqrt T$.
Increasing linear density with $T$ fixed has opposite effects on the two quantities:
\begin{equation}
c\propto\frac{1}{\sqrt\mu},
\qquad
Z_0\propto\sqrt\mu.
\end{equation}
\begin{center}
\includegraphics{WM22_fig02_speed_impedance_parameter_map.png}
\vspace{0.45em}
\textbf{Figure.}
Wave speed depends on the ratio $T/\mu$, while characteristic impedance depends on the product $T\mu$. Two strings can therefore have the same wave speed but different impedances, or the same impedance but different wave speeds.
\end{center}
This distinction becomes central at an interface.
\section{Why impedance matters at an interface}
Suppose a harmonic wave traveling in medium 1 reaches an ideal junction with medium 2. Let the characteristic impedances be
\begin{equation}
Z_1
\qquad\text{and}\qquad
Z_2.
\end{equation}
At an ideal massless junction, two conditions are imposed:
\begin{enumerate}
\item the transverse displacement is continuous;
\item the transverse force is continuous.
\end{enumerate}
Write the incident, reflected, and transmitted displacement amplitudes as
\begin{equation}
A_i,\qquad A_r,\qquad A_t.
\end{equation}
Displacement continuity gives
\begin{equation}
\boxed{A_i+A_r=A_t.}
\end{equation}
For harmonic waves, the transverse-force condition gives
\begin{equation}
\boxed{Z_1(A_i-A_r)=Z_2A_t.}
\end{equation}
The minus sign occurs because the reflected wave travels in the negative $x$ direction.
\begin{center}
\includegraphics{WM22_fig03_impedance_interface.png}
\vspace{0.45em}
\textbf{Figure.}
At an ideal interface, an incident wave generally produces both reflected and transmitted waves. The relative impedances determine their amplitudes.
\end{center}
Solving the two equations gives the displacement-amplitude reflection coefficient
\begin{equation}
\boxed{
r=\frac{A_r}{A_i}
=\frac{Z_1-Z_2}{Z_1+Z_2},}
\end{equation}
and the displacement-amplitude transmission coefficient
\begin{equation}
\boxed{
t=\frac{A_t}{A_i}
=\frac{2Z_1}{Z_1+Z_2}.}
\end{equation}
These formulas use the displacement-amplitude convention. Other wave variables can have different amplitude-coefficient formulas even though the physical power balance is the same \cite{French1971,Crawford1968,Georgi1993}.
\section{Power reflection and transmission}
For a harmonic traveling wave, the magnitude of the average power is
\begin{equation}
\langle P\rangle=\frac12 Z_0\omega^2A^2.
\end{equation}
The fraction of incident power reflected is therefore
\begin{equation}
\boxed{
R=\frac{|\langle P_r\rangle|}{\langle P_i\rangle}=r^2.}
\end{equation}
Using the transmitted amplitude coefficient $t$,
\begin{equation}
\boxed{
\mathcal T
=\frac{\langle P_t\rangle}{\langle P_i\rangle}
=\frac{Z_2}{Z_1}t^2.}
\end{equation}
Substituting the expression for $t$ gives
\begin{equation}
\boxed{
\mathcal T
=\frac{4Z_1Z_2}{(Z_1+Z_2)^2}.}
\end{equation}
For a lossless ideal junction,
\begin{equation}
\boxed{R+\mathcal T=1.}
\end{equation}
Note that $t$ itself can exceed $1$ without violating energy conservation. Power depends on both amplitude and impedance.
\section{Matched, fixed-like, and free-like limits}
The reflection coefficient can be written in terms of the impedance ratio
\begin{equation}
q=\frac{Z_2}{Z_1}
\end{equation}
as
\begin{equation}
r=\frac{1-q}{1+q}.
\end{equation}
Three limits are especially important.
\subsection*{Matched impedance}
If
\begin{equation}
Z_2=Z_1,
\end{equation}
then
\begin{equation}
\boxed{r=0,\qquad R=0,\qquad \mathcal T=1.}
\end{equation}
No reflected wave is required.
\subsection*{Very large terminating impedance}
If
\begin{equation}
Z_2\gg Z_1,
\end{equation}
then
\begin{equation}
r\longrightarrow-1.
\end{equation}
The displacement reflection is inverted, reproducing the fixed-end behavior introduced in WM12.
\subsection*{Very small terminating impedance}
If
\begin{equation}
Z_2\ll Z_1,
\end{equation}
then
\begin{equation}
r\longrightarrow+1.
\end{equation}
The displacement reflection is not inverted, reproducing the free-end behavior.
\begin{center}
\includegraphics{WM22_fig04_reflection_vs_impedance_ratio.png}
\vspace{0.45em}
\textbf{Figure.}
Displacement reflection coefficient $r$, reflected-power fraction $R$, and transmitted-power fraction $\mathcal T$ versus the impedance ratio $q=Z_2/Z_1$. Perfect matching occurs at $q=1$.
\end{center}
Thus the fixed and free boundaries from WM12 can be understood as limiting cases of an impedance mismatch.
\section{Impedance matching does not require equal wave speed}
A subtle but important point is that
\begin{equation}
c=\sqrt{\frac{T}{\mu}}
\end{equation}
depends on a ratio, while
\begin{equation}
Z_0=\sqrt{T\mu}
\end{equation}
depends on a product.
Therefore two media can satisfy
\begin{equation}
Z_1=Z_2
\end{equation}
while still having
\begin{equation}
c_1\neq c_2.
\end{equation}
A harmonic wave can cross such an ideal matched interface without reflection even though its wavelength changes because
\begin{equation}
\lambda=\frac{c}{f}.
\end{equation}
The frequency remains fixed by the source while the wavelength adjusts to the new wave speed.
\section{Impedance across wave physics}
The basic idea of impedance is broader than the string:
\begin{quote}
impedance relates a wave's generalized effort variable to its generalized flow variable.
\end{quote}
For the string, these variables are transverse force and transverse velocity. In acoustics, a characteristic impedance relates acoustic pressure to particle velocity. In electromagnetism, wave impedance relates electric and magnetic fields.
The formulas and units differ between physical systems, but the recurring ideas are the same:
\begin{itemize}
\item a traveling medium has a characteristic relation between paired wave variables;
\item that relation determines how much power a given amplitude transports;
\item changes in characteristic impedance cause reflection;
\item matching impedances suppresses reflection.
\end{itemize}
This is one reason the string is such a useful first model for later acoustic, optical, RF, and transmission-line wave physics.
\section{Worked Example 1: Compute characteristic impedance}
An ideal string has
\begin{equation}
T=100\,\text{N},
\qquad
\mu=0.010\,\text{kg/m}.
\end{equation}
Find the wave speed and characteristic impedance.
\subsection*{Solution}
The wave speed is
\begin{align}
c
&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{100}{0.010}}\,\text{m/s}\\
&=100\,\text{m/s}.
\end{align}
The impedance is
\begin{align}
Z_0
&=\mu c\\
&=(0.010)(100)\,\text{kg/s}\\
&=1.00\,\text{kg/s}.
\end{align}
Thus
\begin{equation}
\boxed{c=100\,\text{m/s},\qquad Z_0=1.00\,\text{kg/s}.}
\end{equation}
As a check,
\begin{equation}
\sqrt{T\mu}=\sqrt{(100)(0.010)}=1.00\,\text{kg/s}.
\end{equation}
\section{Worked Example 2: Force and instantaneous power}
A right-moving wave travels on a string with
\begin{equation}
Z_0=1.50\,\text{kg/s}.
\end{equation}
At one instant and position, the transverse velocity is
\begin{equation}
u_t=0.30\,\text{m/s}.
\end{equation}
Find the transverse force transmitted in the positive $x$ direction and the instantaneous power. Then repeat for a left-moving wave with the same local transverse velocity.
\subsection*{Solution}
For the right-moving wave,
\begin{equation}
F_\perp^{(+x)}=Z_0u_t.
\end{equation}
Therefore
\begin{equation}
F_\perp^{(+x)}=(1.50)(0.30)=0.45\,\text{N}.
\end{equation}
The power is
\begin{equation}
P=Z_0u_t^2=(1.50)(0.30)^2=0.135\,\text{W}.
\end{equation}
Thus
\begin{equation}
\boxed{F_\perp^{(+x)}=+0.45\,\text{N},\qquad P=+0.135\,\text{W}.}
\end{equation}
For a left-moving wave,
\begin{equation}
F_\perp^{(+x)}=-Z_0u_t=-0.45\,\text{N}
\end{equation}
and
\begin{equation}
P=-Z_0u_t^2=-0.135\,\text{W}.
\end{equation}
The magnitude of the characteristic impedance is unchanged; the sign of the power identifies the transport direction.
\section{Worked Example 3: Average power from impedance}
A string has
\begin{equation}
\mu=0.012\,\text{kg/m},
\qquad
c=100\,\text{m/s}.
\end{equation}
A right-moving sinusoid has amplitude
\begin{equation}
A=2.0\,\text{mm}
\end{equation}
and frequency
\begin{equation}
f=40\,\text{Hz}.
\end{equation}
Find $Z_0$, the RMS transverse velocity, and the average power.
\subsection*{Solution}
First,
\begin{equation}
Z_0=\mu c=(0.012)(100)=1.20\,\text{kg/s}.
\end{equation}
The angular frequency is
\begin{equation}
\omega=2\pi f=80\pi\,\text{rad/s}.
\end{equation}
With
\begin{equation}
A=0.0020\,\text{m},
\end{equation}
the RMS transverse velocity is
\begin{align}
v_{\text{rms}}
&=\frac{A\omega}{\sqrt2}\\
&=\frac{(0.0020)(80\pi)}{\sqrt2}\,\text{m/s}\\
&\simeq0.355\,\text{m/s}.
\end{align}
Then
\begin{align}
\langle P\rangle
&=Z_0v_{\text{rms}}^2\\
&=(1.20)(0.355)^2\,\text{W}\\
&\simeq0.152\,\text{W}.
\end{align}
Therefore
\begin{equation}
\boxed{Z_0=1.20\,\text{kg/s},}
\end{equation}
\begin{equation}
\boxed{v_{\text{rms}}\simeq0.355\,\text{m/s},}
\end{equation}
\begin{equation}
\boxed{\langle P\rangle\simeq0.152\,\text{W}.}
\end{equation}
This is the same numerical result obtained earlier from $\frac12\mu A^2\omega^2c$.
\section{Worked Example 4: Required amplitude for a desired average power}
A string has characteristic impedance
\begin{equation}
Z_0=2.0\,\text{kg/s}.
\end{equation}
What displacement amplitude is required for a right-moving sinusoidal wave of frequency
\begin{equation}
f=50\,\text{Hz}
\end{equation}
to carry average power
\begin{equation}
\langle P\rangle=0.50\,\text{W}?
\end{equation}
\subsection*{Solution}
Use
\begin{equation}
\langle P\rangle=\frac12Z_0A^2\omega^2.
\end{equation}
Solve for $A$:
\begin{equation}
A=\sqrt{\frac{2\langle P\rangle}{Z_0\omega^2}}.
\end{equation}
The angular frequency is
\begin{equation}
\omega=2\pi(50)=100\pi\,\text{rad/s}.
\end{equation}
Hence
\begin{align}
A
&=\sqrt{\frac{2(0.50)}{(2.0)(100\pi)^2}}\,\text{m}\\
&\simeq2.25\times10^{-3}\,\text{m}.
\end{align}
Thus
\begin{equation}
\boxed{A\simeq2.25\,\text{mm}.}
\end{equation}
\section{Worked Example 5: Reflection from an impedance change}
A sinusoidal wave travels from a string with
\begin{equation}
Z_1=1.0\,\text{kg/s}
\end{equation}
into a second string with
\begin{equation}
Z_2=4.0\,\text{kg/s}.
\end{equation}
Find the displacement reflection coefficient, displacement transmission coefficient, reflected-power fraction, and transmitted-power fraction.
\subsection*{Solution}
The displacement reflection coefficient is
\begin{align}
r
&=\frac{Z_1-Z_2}{Z_1+Z_2}\\
&=\frac{1-4}{1+4}\\
&=-0.60.
\end{align}
The negative sign means the reflected displacement is inverted.
The displacement transmission coefficient is
\begin{align}
t
&=\frac{2Z_1}{Z_1+Z_2}\\
&=\frac{2}{5}\\
&=0.40.
\end{align}
The reflected-power fraction is
\begin{equation}
R=r^2=(-0.60)^2=0.36.
\end{equation}
The transmitted-power fraction is
\begin{align}
\mathcal T
&=\frac{4Z_1Z_2}{(Z_1+Z_2)^2}\\
&=\frac{4(1)(4)}{25}\\
&=0.64.
\end{align}
Thus
\begin{equation}
\boxed{r=-0.60,\qquad t=0.40,}
\end{equation}
\begin{equation}
\boxed{R=0.36,\qquad \mathcal T=0.64.}
\end{equation}
The power check is
\begin{equation}
R+\mathcal T=0.36+0.64=1.
\end{equation}
\section{Worked Example 6: Match impedances while changing wave speed}
Medium 1 has
\begin{equation}
T_1=81\,\text{N},
\qquad
\mu_1=0.010\,\text{kg/m}.
\end{equation}
Medium 2 has tension
\begin{equation}
T_2=144\,\text{N}.
\end{equation}
Choose $\mu_2$ so that the two characteristic impedances match. Then find both wave speeds.
\subsection*{Solution}
For medium 1,
\begin{equation}
Z_1=\sqrt{T_1\mu_1}
=\sqrt{(81)(0.010)}
=0.90\,\text{kg/s}.
\end{equation}
For matching,
\begin{equation}
Z_2=Z_1=0.90\,\text{kg/s}.
\end{equation}
Since
\begin{equation}
Z_2^2=T_2\mu_2,
\end{equation}
we need
\begin{align}
\mu_2
&=\frac{Z_2^2}{T_2}\\
&=\frac{(0.90)^2}{144}\,\text{kg/m}\\
&=5.625\times10^{-3}\,\text{kg/m}.
\end{align}
Thus
\begin{equation}
\boxed{\mu_2=0.005625\,\text{kg/m}.}
\end{equation}
Now compute the speeds:
\begin{align}
c_1
&=\sqrt{\frac{81}{0.010}}\\
&=90\,\text{m/s},
\end{align}
while
\begin{align}
c_2
&=\sqrt{\frac{144}{0.005625}}\\
&=160\,\text{m/s}.
\end{align}
Therefore
\begin{equation}
\boxed{c_1=90\,\text{m/s},\qquad c_2=160\,\text{m/s}.}
\end{equation}
The impedances match even though the wave speeds do not. An ideal harmonic wave can therefore have zero reflection while its wavelength changes across the interface.
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} confusing impedance with wave speed. For a string, $c$ depends on $T/\mu$, while $Z_0$ depends on $T\mu$.
\item \textbf{Mistake:} dropping the propagation-direction sign. $Z_0$ is positive, but the force-velocity relation changes sign between right- and left-moving waves under the chosen positive-$x$ convention.
\item \textbf{Mistake:} using $Z_0=F/u$ instead of force divided by \emph{velocity}. Mechanical impedance pairs force with velocity.
\item \textbf{Mistake:} applying the one-way relation $F=Z_0u_t$ to an arbitrary standing wave. That relation assumes a pure right-moving component.
\item \textbf{Mistake:} treating the displacement transmission coefficient $t$ as a power fraction. Power transmission also depends on the impedance ratio.
\item \textbf{Mistake:} assuming impedance matching requires equal wave speeds. Equal $Z_0$ does not imply equal $c$.
\end{itemize}
\section{What WM22 adds to the wave-mechanics picture}
The string-wave sequence now has a compact force-velocity transport relation:
\begin{equation}
\boxed{Z_0=\frac{T}{c}=\mu c=\sqrt{T\mu}.}
\end{equation}
For one-way waves,
\begin{equation}
\boxed{
F_\perp^{(+x)}=\pm Z_0u_t,}
\end{equation}
and
\begin{equation}
\boxed{P=\pm Z_0u_t^2.}
\end{equation}
For a sinusoid,
\begin{equation}
\boxed{\langle P\rangle=Z_0v_{\text{rms}}^2.}
\end{equation}
At an ideal interface,
\begin{equation}
\boxed{r=\frac{Z_1-Z_2}{Z_1+Z_2}.}
\end{equation}
Impedance therefore connects three ideas that were previously introduced separately:
\begin{equation}
\boxed{
\text{force-velocity relation}
\longleftrightarrow
\text{power transport}
\longleftrightarrow
\text{reflection at interfaces}.}
\end{equation}
This framework will later transfer naturally to acoustic and electromagnetic wave impedance. It also prepares the way for characteristic impedance in transmission lines.
\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{Georgi1993}
Howard Georgi,
\emph{The Physics of Waves},
Prentice Hall, 1993.
\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{OpenStax165}
William Moebs, Samuel J. Ling, and Jeff Sanny,
\emph{University Physics, Volume 1},
OpenStax, 2016,
Section 16.5, ``Interference of Waves,'' including reflection at boundaries.
\bibitem{MIT803}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
MIT OpenCourseWare, Fall 2016.
\end{thebibliography}