Wave Mechanics: Average Power of a Sinusoidal WaveWaveMechanicsAveragePowerOfASinusoidalWave2026-09-12 16:49:172026-09-12 16:49:17Topicwave mechanicsaverage powersinusoidal wavestretched stringenergy transporttime averageRMS velocitypower scalingmechanical waveswave energy% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Wave Mechanics: Average Power of a Sinusoidal Wave}
WM19 derived the instantaneous power carried by a transverse wave on an ideal stretched string,
\begin{equation}
\boxed{P(x,t)=-T u_xu_t.}
\end{equation}
For the right-moving sinusoidal wave
\begin{equation}
u(x,t)=A\cos(kx-\omega t+\phi),
\end{equation}
WM19 obtained
\begin{equation}
P(x,t)=T A^2k\omega\sin^2(kx-\omega t+\phi).
\end{equation}
The instantaneous power oscillates between zero and a maximum value. In many measurements, however, the quantity of greatest interest is the power averaged over one or many complete cycles. WM20 derives that average carefully and develops several equivalent forms and physical interpretations.
The central result is
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{2}T A^2k\omega
=\frac{1}{2}\mu A^2\omega^2c
=2\pi^2\mu A^2f^2c.}
\end{equation}
For fixed string properties, average power therefore scales as the square of both amplitude and frequency \cite{French1971,Crawford1968,OpenStax164}.
\section{Instantaneous power is not constant}
Define the phase
\begin{equation}
\theta=kx-\omega t+\phi.
\end{equation}
Then the instantaneous power is
\begin{equation}
P=P_{\max}\sin^2\theta,
\end{equation}
where
\begin{equation}
\boxed{P_{\max}=T A^2k\omega.}
\end{equation}
Because
\begin{equation}
0\leq \sin^2\theta\leq 1,
\end{equation}
we have
\begin{equation}
0\leq P\leq P_{\max}
\end{equation}
for this right-moving wave.
\begin{center}
\includegraphics{WM20_fig01_instantaneous_and_average_power.png}
\vspace{0.45em}
\textbf{Figure.}
Normalized instantaneous power varies as $\sin^2\theta$. The horizontal dashed line shows the cycle average, one-half of the peak power.
\end{center}
Notice an important point: the average displacement of a sinusoidal wave over one cycle is zero, but its average power is not zero. Power depends quadratically on the wave amplitude through products of derivatives, not linearly on the displacement itself.
\section{Definition of the time average}
Let the temporal period be
\begin{equation}
T_0=\frac{2\pi}{\omega}.
\end{equation}
The average power at a fixed position $x$ over one period is
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{T_0}
\int_{t_0}^{t_0+T_0}P(x,t)\,dt.}
\end{equation}
For a periodic steady wave, the result does not depend on the starting time $t_0$ as long as the averaging interval spans a complete period.
Substitute
\begin{equation}
P=T A^2k\omega\sin^2\theta.
\end{equation}
Then
\begin{equation}
\langle P\rangle
=\frac{T A^2k\omega}{T_0}
\int_{t_0}^{t_0+T_0}\sin^2\theta\,dt.
\end{equation}
\section{Why the average of sine squared is one-half}
At fixed $x$,
\begin{equation}
\theta=kx-\omega t+\phi,
\end{equation}
so
\begin{equation}
d\theta=-\omega\,dt.
\end{equation}
During one temporal period, the phase changes by $2\pi$. Therefore averaging over time is equivalent to averaging $\sin^2\theta$ over one complete phase cycle:
\begin{equation}
\left\langle\sin^2\theta\right\rangle
=\frac{1}{2\pi}\int_0^{2\pi}\sin^2\theta\,d\theta.
\end{equation}
Use
\begin{equation}
\sin^2\theta=\frac{1-\cos(2\theta)}{2}.
\end{equation}
Then
\begin{align}
\left\langle\sin^2\theta\right\rangle
&=\frac{1}{2\pi}
\int_0^{2\pi}\frac{1-\cos(2\theta)}{2}\,d\theta\\
&=\frac{1}{2\pi}\left[\pi\right]\\
&=\boxed{\frac{1}{2}}.
\end{align}
\begin{center}
\includegraphics{WM20_fig02_average_sin_squared.png}
\vspace{0.45em}
\textbf{Figure.}
The cycle average of $\sin^2\theta$ is $1/2$. Geometrically, the area under one complete cycle equals the area of a rectangle of the same width and height $1/2$.
\end{center}
\section{Average power in the first useful form}
Since
\begin{equation}
\left\langle\sin^2\theta\right\rangle=\frac{1}{2},
\end{equation}
we obtain
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{2}T A^2k\omega.}
\end{equation}
The peak and average powers are therefore related by
\begin{equation}
\boxed{P_{\max}=2\langle P\rangle.}
\end{equation}
This factor of two is specific to the sinusoidal $\sin^2$ variation.
\section{Equivalent form using linear mass density and wave speed}
For an ideal string,
\begin{equation}
c^2=\frac{T}{\mu},
\end{equation}
so
\begin{equation}
T=\mu c^2.
\end{equation}
Also,
\begin{equation}
\omega=ck,
\end{equation}
or
\begin{equation}
k=\frac{\omega}{c}.
\end{equation}
Substitute into
\begin{equation}
\langle P\rangle=\frac{1}{2}T A^2k\omega:
\end{equation}
\begin{align}
\langle P\rangle
&=\frac{1}{2}(\mu c^2)A^2
\left(\frac{\omega}{c}\right)\omega\\
&=\boxed{\frac{1}{2}\mu A^2\omega^2c}.
\end{align}
This is one of the most useful engineering forms because it separates the wave amplitude and frequency from the medium properties \cite{OpenStax164,French1971}.
\section{Frequency form}
Using
\begin{equation}
\omega=2\pi f,
\end{equation}
we obtain
\begin{align}
\langle P\rangle
&=\frac{1}{2}\mu A^2(2\pi f)^2c\\
&=\boxed{2\pi^2\mu A^2f^2c}.
\end{align}
For a fixed string, meaning fixed $\mu$ and $c$,
\begin{equation}
\boxed{\langle P\rangle\propto A^2f^2.}
\end{equation}
\begin{center}
\includegraphics{WM20_fig03_square_law_scaling.png}
\vspace{0.45em}
\textbf{Figure.}
For fixed string properties, doubling amplitude multiplies average power by four, and doubling frequency also multiplies average power by four.
\end{center}
Thus:
\begin{itemize}
\item $A\rightarrow 2A$ gives $\langle P\rangle\rightarrow4\langle P\rangle$;
\item $f\rightarrow2f$ gives $\langle P\rangle\rightarrow4\langle P\rangle$;
\item doubling both gives a factor of $16$.
\end{itemize}
\section{RMS transverse velocity form}
The transverse material velocity is
\begin{equation}
u_t=A\omega\sin\theta.
\end{equation}
Its root-mean-square value is
\begin{equation}
v_{\text{rms}}
=\sqrt{\langle u_t^2\rangle}.
\end{equation}
Since
\begin{equation}
\langle\sin^2\theta\rangle=\frac{1}{2},
\end{equation}
we have
\begin{equation}
\boxed{v_{\text{rms}}=\frac{A\omega}{\sqrt{2}}.}
\end{equation}
Therefore
\begin{align}
\mu c v_{\text{rms}}^2
&=\mu c\frac{A^2\omega^2}{2}\\
&=\boxed{\langle P\rangle}.
\end{align}
So another useful form is
\begin{equation}
\boxed{\langle P\rangle=\mu c v_{\text{rms}}^2.}
\end{equation}
The quantity $\mu c$ will later appear naturally in the discussion of mechanical wave impedance.
\section{Average power and average energy density}
WM18 found for a sinusoidal traveling wave
\begin{equation}
\boxed{
\langle\mathcal{E}\rangle
=\frac{1}{2}\mu A^2\omega^2.}
\end{equation}
Comparing with
\begin{equation}
\langle P\rangle
=\frac{1}{2}\mu A^2\omega^2c,
\end{equation}
we obtain
\begin{equation}
\boxed{
\langle P\rangle=c\langle\mathcal{E}\rangle.}
\end{equation}
This is the average version of the more general traveling-wave relation developed in WM19.
\section{Energy in one wavelength crosses in one period}
The average energy contained in one wavelength is
\begin{equation}
E_{\lambda}=\langle\mathcal{E}\rangle\lambda.
\end{equation}
The wave travels one wavelength in one period, so
\begin{equation}
\lambda=cT_0.
\end{equation}
Therefore
\begin{align}
E_{\lambda}
&=\langle\mathcal{E}\rangle cT_0\\
&=\langle P\rangle T_0.
\end{align}
Hence
\begin{equation}
\boxed{E_{\lambda}=\langle P\rangle T_0.}
\end{equation}
This has a direct interpretation: during one oscillation period, one wavelength of the traveling pattern moves past a fixed observation point, carrying with it the energy associated with that wavelength.
\begin{center}
\includegraphics{WM20_fig04_energy_per_wavelength.png}
\vspace{0.45em}
\textbf{Figure.}
Because $cT_0=\lambda$, the energy crossing a fixed point during one period equals the average energy stored in one wavelength of a steady sinusoidal traveling wave.
\end{center}
\section{Time average and spatial average}
At a fixed time, the sinusoidal power varies through space as
\begin{equation}
P(x,t)=P_{\max}\sin^2(kx-\omega t+\phi).
\end{equation}
A spatial average over one wavelength is
\begin{equation}
\langle P\rangle_x
=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}P(x,t)\,dx.
\end{equation}
Since the phase changes by $2\pi$ over one wavelength, the same $\sin^2$ average appears. Thus
\begin{equation}
\boxed{\langle P\rangle_x=\langle P\rangle_t.}
\end{equation}
This equality holds for the steady sinusoidal traveling wave because one complete wavelength in space corresponds to one complete phase cycle, just as one period in time does.
\section{Direction and sign}
For a right-moving sinusoidal wave under the WM19 sign convention,
\begin{equation}
\boxed{\langle P\rangle>0.}
\end{equation}
For the corresponding left-moving wave,
\begin{equation}
\boxed{\langle P\rangle<0.}
\end{equation}
The magnitude is the same if amplitude, frequency, and medium properties are the same. In many contexts the phrase ``average power carried'' refers to the positive magnitude. When direction matters, the signed form should be stated explicitly.
\section{A perfect standing wave is different}
A perfect standing wave is formed from equal counter-propagating waves. WM19 showed that its instantaneous local power generally oscillates in sign, but
\begin{equation}
\boxed{\langle P\rangle_{\text{standing}}=0.}
\end{equation}
This does not mean that the standing wave contains no energy. It means that there is no net time-averaged energy transport through a fixed position.
\section{Worked Example 1: Compute average and peak power}
A sinusoidal wave has
\begin{equation}
A=4.0\,\text{mm},
\qquad
f=25\,\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, average power, and peak instantaneous power.
\subsection*{Solution}
First,
\begin{align}
c
&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{75}{0.012}}\\
&\simeq79.1\,\text{m/s}.
\end{align}
The angular frequency is
\begin{equation}
\omega=2\pi f=157.1\,\text{rad/s}.
\end{equation}
Convert the amplitude:
\begin{equation}
A=0.0040\,\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}(0.012)(0.0040)^2(157.1)^2(79.1)\\
&\simeq0.187\,\text{W}.
\end{align}
Therefore
\begin{equation}
\boxed{\langle P\rangle\simeq0.187\,\text{W}.}
\end{equation}
Since
\begin{equation}
P_{\max}=2\langle P\rangle,
\end{equation}
we obtain
\begin{equation}
\boxed{P_{\max}\simeq0.375\,\text{W}.}
\end{equation}
\section{Worked Example 2: Use the $T A^2k\omega$ form}
A right-moving sinusoidal wave has
\begin{equation}
A=3.0\,\text{mm},
\qquad
k=4.0\,\text{rad/m},
\qquad
\omega=200\,\text{rad/s},
\end{equation}
on a string under tension
\begin{equation}
T=60\,\text{N}.
\end{equation}
Find the average power.
\subsection*{Solution}
Use
\begin{equation}
\langle P\rangle
=\frac{1}{2}T A^2k\omega.
\end{equation}
With
\begin{equation}
A=0.0030\,\text{m},
\end{equation}
we obtain
\begin{align}
\langle P\rangle
&=\frac{1}{2}(60)(0.0030)^2(4.0)(200)\\
&=0.216\,\text{W}.
\end{align}
Thus
\begin{equation}
\boxed{\langle P\rangle=0.216\,\text{W}.}
\end{equation}
\section{Worked Example 3: Required amplitude for a specified average power}
A sinusoidal wave travels on a string with
\begin{equation}
\mu=0.0080\,\text{kg/m},
\qquad
c=120\,\text{m/s}.
\end{equation}
At
\begin{equation}
f=40\,\text{Hz},
\end{equation}
what amplitude is required to carry
\begin{equation}
\langle P\rangle=5.0\,\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(40)=251.3\,\text{rad/s}.
\end{equation}
Therefore
\begin{align}
A
&=\sqrt{\frac{2(5.0)}{(0.0080)(251.3)^2(120)}}\\
&\simeq1.28\times10^{-2}\,\text{m}.
\end{align}
Thus
\begin{equation}
\boxed{A\simeq12.8\,\text{mm}.}
\end{equation}
\section{Worked Example 4: Scaling without recomputing from scratch}
A wave initially carries average power $P_0$. Its amplitude is changed to
\begin{equation}
0.60A_0
\end{equation}
and its frequency is changed to
\begin{equation}
1.50f_0,
\end{equation}
while the string itself is unchanged. Find the new average power as a fraction of $P_0$.
\subsection*{Solution}
For a fixed string,
\begin{equation}
\langle P\rangle\propto A^2f^2.
\end{equation}
Therefore
\begin{align}
\frac{P_{\text{new}}}{P_0}
&=(0.60)^2(1.50)^2\\
&=0.36(2.25)\\
&=0.81.
\end{align}
Hence
\begin{equation}
\boxed{P_{\text{new}}=0.81P_0.}
\end{equation}
Even though the frequency increased, the reduction in amplitude was large enough that the average power decreased overall.
\section{Worked Example 5: RMS transverse velocity}
At one location in a sinusoidal traveling wave, the transverse material velocity has
\begin{equation}
v_{\text{rms}}=0.35\,\text{m/s}.
\end{equation}
The string has
\begin{equation}
\mu=0.020\,\text{kg/m},
\qquad
c=70\,\text{m/s}.
\end{equation}
Find the average power.
\subsection*{Solution}
Use
\begin{equation}
\langle P\rangle=\mu c v_{\text{rms}}^2.
\end{equation}
Then
\begin{align}
\langle P\rangle
&=(0.020)(70)(0.35)^2\\
&=0.1715\,\text{W}.
\end{align}
Thus
\begin{equation}
\boxed{\langle P\rangle\simeq0.172\,\text{W}.}
\end{equation}
\section{Worked Example 6: Energy per wavelength and energy per period}
A sinusoidal traveling wave has average energy density
\begin{equation}
\langle\mathcal{E}\rangle=0.15\,\text{J/m},
\end{equation}
wavelength
\begin{equation}
\lambda=2.4\,\text{m},
\end{equation}
and speed
\begin{equation}
c=48\,\text{m/s}.
\end{equation}
Find the period, the average power, the energy in one wavelength, and verify that the same amount of energy crosses a point during one period.
\subsection*{Solution}
The period follows from
\begin{equation}
c=\frac{\lambda}{T_0}.
\end{equation}
Thus
\begin{equation}
T_0=\frac{2.4}{48}=0.050\,\text{s}.
\end{equation}
Average power is
\begin{equation}
\langle P\rangle=c\langle\mathcal{E}\rangle,
\end{equation}
so
\begin{equation}
\langle P\rangle=(48)(0.15)=7.2\,\text{W}.
\end{equation}
The energy in one wavelength is
\begin{equation}
E_{\lambda}
=(0.15)(2.4)
=0.36\,\text{J}.
\end{equation}
The energy crossing a point during one period is
\begin{equation}
\langle P\rangle T_0
=(7.2)(0.050)
=0.36\,\text{J}.
\end{equation}
Therefore
\begin{equation}
\boxed{E_{\lambda}=\langle P\rangle T_0=0.36\,\text{J}.}
\end{equation}
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} averaging the displacement and concluding that zero mean displacement means zero mean power. Power is quadratic in wave derivatives.
\item \textbf{Mistake:} forgetting the factor $1/2$ from $\langle\sin^2\theta\rangle$.
\item \textbf{Mistake:} confusing peak instantaneous power with average power. For a sinusoidal traveling wave, $P_{\max}=2\langle P\rangle$.
\item \textbf{Mistake:} applying $\langle P\rangle\propto A^2f^2$ while simultaneously changing the string properties. That scaling assumes $\mu$ and $c$ remain fixed.
\item \textbf{Mistake:} forgetting that signed power is negative for a left-moving wave under the WM19 convention.
\item \textbf{Mistake:} assuming a standing wave carries zero energy because its average power is zero. A standing wave stores and exchanges energy locally even though its net time-averaged transport vanishes.
\end{itemize}
\section{What WM20 adds to the sequence}
WM19 established the instantaneous conservation law and the signed power flow
\begin{equation}
P=-T u_xu_t.
\end{equation}
WM20 turns that instantaneous quantity into a cycle-averaged transport rate for sinusoidal waves. The key result is
\begin{equation}
\boxed{
\langle P\rangle
=\frac{1}{2}T A^2k\omega
=\frac{1}{2}\mu A^2\omega^2c
=2\pi^2\mu A^2f^2c.}
\end{equation}
This is the form that will later connect naturally to intensity, impedance, reflection/transmission coefficients, and harmonic-wave treatments in other physical systems.
\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},
Prentice Hall, 1993.
\bibitem{MIT803}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
MIT OpenCourseWare, Fall 2016.
\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}