Wave Mechanics: Deriving the 1D String Wave Equation from Newton's Second LawWaveMechanicsDerivingThe1DStringWaveEquationFromNewtonsSecondLaw2026-09-12 06:42:402026-09-12 06:42:40Topicwave mechanicswave equationstretched stringNewton's second lawtensionlinear mass densitytransverse wavesmall-slope approximationcontinuum mechanicswave speedpartial differential equation% 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: Deriving the 1D String Wave Equation from Newton's Second Law}
WM13 introduced the derivative structure
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}}
\end{equation}
and showed that translating profiles such as $F(x-ct)$ satisfy it. That was a
\emph{kinematic} result: it described a mathematical relation obeyed by a
shape-preserving traveling disturbance.
WM14 asks the dynamical question:
\begin{equation}
\boxed{\text{Why should a real stretched string obey that equation?}}
\end{equation}
The answer comes directly from Newton's second law. A curved string has
slightly different tension directions at neighboring points. The difference
between those tension directions produces a transverse force. That force is
proportional to the local curvature of the string, and Newton's second law then
turns curvature into transverse acceleration. Under the ideal assumptions
developed below, this leads to
\begin{equation}
\boxed{\mu u_{tt}=T u_{xx}}
\end{equation}
or
\begin{equation}
\boxed{u_{tt}=\frac{T}{\mu}u_{xx}.}
\end{equation}
Comparison with the standard one-dimensional wave equation gives
\begin{equation}
\boxed{c=\sqrt{\frac{T}{\mu}}.}
\end{equation}
This derivation is standard in treatments of waves on stretched strings and is
a canonical example of how a continuum partial differential equation emerges
from Newtonian mechanics \cite{French1971,Crawford1968,OpenStax163,MIT803Waves}.
\section{The physical model}
Consider a thin flexible string stretched primarily along the $x$ direction.
Let
\begin{equation}
u(x,t)
\end{equation}
denote its transverse displacement from equilibrium.
The derivation uses several assumptions. They are not merely mathematical
conveniences; they define the physical model.
\begin{itemize}
\item The string is continuous and perfectly flexible, with negligible bending stiffness.
\item The string has uniform linear mass density $\mu$.
\item The equilibrium tension magnitude $T$ is approximately constant along the string.
\item Motion is transverse; longitudinal motion is neglected to first order.
\item The slope is small:
\begin{equation}
\left|\frac{\partial u}{\partial x}\right|\ll 1.
\end{equation}
\item Damping, gravity, and distributed external transverse forces are neglected.
\end{itemize}
The most important approximation is the small-slope condition. The displacement
itself need not be zero, but neighboring pieces of the string must make only
small angles with the equilibrium $x$ direction. For a sinusoidal wave, the
small-slope requirement is roughly controlled by the dimensionless quantity
$Ak$.
These assumptions produce the \emph{linear} string wave equation. If the
slopes become large, the tension varies strongly, the string stretches
significantly, or bending stiffness matters, additional nonlinear or higher
order terms appear.
\section{Choose a short string element}
Select a small piece of string extending from $x$ to $x+\Delta x$.
If $\mu$ is the mass per unit equilibrium length, then the mass of this element
is approximately
\begin{equation}
\boxed{\Delta m=\mu\,\Delta x.}
\end{equation}
The element is pulled by the rest of the string at both ends. Because an ideal
string can sustain tension but not bending moment, the tension force at each end
acts tangent to the local string direction.
\begin{center}
\includegraphics{WM14_fig01_string_element_forces.png}
\vspace{0.45em}
\textbf{Figure.}
A short string element between $x$ and $x+\Delta x$. The neighboring string
pulls tangentially on the two ends with approximately equal tension magnitude
$T$, but the directions differ because the string is curved.
\end{center}
Let the local tangent angles be $\theta_L$ at the left end and $\theta_R$ at
the right end.
\section{Geometry connects tangent angle to spatial slope}
At any point on the string, the slope of the tangent line is
\begin{equation}
\tan\theta=\frac{\partial u}{\partial x}.
\end{equation}
This relation is geometric and exact for the graph $u(x,t)$ at a fixed time.
The small-slope assumption gives
\begin{equation}
|\theta|\ll 1,
\end{equation}
so the standard small-angle approximations apply:
\begin{equation}
\sin\theta\simeq\tan\theta\simeq\theta.
\end{equation}
Therefore
\begin{equation}
\boxed{\sin\theta\simeq\frac{\partial u}{\partial x}.}
\end{equation}
This is the step that linearizes the force law.
A more exact relation would be
\begin{equation}
\sin\theta
=\frac{u_x}{\sqrt{1+u_x^2}},
\end{equation}
which reduces to $\sin\theta\simeq u_x$ when $|u_x|\ll 1$.
\section{Resolve the tension forces}
The right-hand tension contributes a transverse component
\begin{equation}
+T\sin\theta_R,
\end{equation}
while the left-hand tension contributes
\begin{equation}
-T\sin\theta_L.
\end{equation}
The net transverse force on the element is therefore
\begin{equation}
F_u=T\sin\theta_R-T\sin\theta_L.
\end{equation}
Using the small-slope approximation,
\begin{equation}
F_u\simeq T\left[(u_x)_R-(u_x)_L\right].
\end{equation}
In coordinate form,
\begin{equation}
F_u\simeq
T\left[
u_x(x+\Delta x,t)-u_x(x,t)
\right].
\end{equation}
The horizontal components are
\begin{equation}
F_x=T\cos\theta_R-T\cos\theta_L.
\end{equation}
For small slopes,
\begin{equation}
\cos\theta\simeq 1,
\end{equation}
so the horizontal components cancel to first order. This is consistent with
the model assumption that the leading motion is transverse and that the tension
magnitude can be treated as constant.
\section{Curvature produces a transverse force}
Rewrite the transverse force as
\begin{equation}
F_u\simeq
T\left[
\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}
\right]\Delta x.
\end{equation}
As the element becomes arbitrarily short,
\begin{equation}
\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}
\longrightarrow u_{xx}(x,t).
\end{equation}
Thus
\begin{equation}
\boxed{F_u\simeq T\,u_{xx}\,\Delta x.}
\end{equation}
This equation contains the central physical idea:
\begin{equation}
\boxed{\text{curvature creates a transverse imbalance of tension}.}
\end{equation}
If the string is locally straight, $u_{xx}=0$, the tension directions balance
in the transverse direction and there is no transverse net force from tension.
If the string is curved, the two tension vectors do not cancel transversely.
\begin{center}
\includegraphics{WM14_fig02_curvature_force_acceleration.png}
\vspace{0.45em}
\textbf{Figure.}
For a concave-down string element, $u_{xx}<0$. The tension imbalance points
downward, so the transverse acceleration is also negative. The signs of
curvature and acceleration therefore agree in the linear string model.
\end{center}
\section{Apply Newton's second law}
The transverse acceleration of the material element is
\begin{equation}
a_u=u_{tt}.
\end{equation}
The element mass is
\begin{equation}
\Delta m=\mu\Delta x.
\end{equation}
Newton's second law in the transverse direction is
\begin{equation}
F_u=(\Delta m)a_u.
\end{equation}
Substitute the force and mass expressions:
\begin{equation}
T\,u_{xx}\,\Delta x
=\mu\Delta x\,u_{tt}.
\end{equation}
Cancel the nonzero element length $\Delta x$:
\begin{equation}
\boxed{\mu u_{tt}=T u_{xx}.}
\end{equation}
Finally divide by $\mu$:
\begin{equation}
\boxed{u_{tt}=\frac{T}{\mu}u_{xx}.}
\end{equation}
This is the one-dimensional linear wave equation for an ideal stretched string.
\begin{center}
\includegraphics{WM14_fig03_derivation_chain.png}
\vspace{0.45em}
\textbf{Figure.}
The derivation chain from string geometry to Newton's second law. A difference
in local slope creates a transverse tension imbalance; in the continuum limit
that slope difference becomes $u_{xx}$.
\end{center}
\section{Identify the wave speed}
The standard one-dimensional wave equation has the form
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
The string equation is
\begin{equation}
u_{tt}=\frac{T}{\mu}u_{xx}.
\end{equation}
Therefore
\begin{equation}
c^2=\frac{T}{\mu}
\end{equation}
and
\begin{equation}
\boxed{c=\sqrt{\frac{T}{\mu}}.}
\end{equation}
The positive root is used for the speed magnitude. Direction is carried by the
traveling-wave form $F(x-ct)$ or $G(x+ct)$ rather than by assigning a negative
value to the speed magnitude.
This result says:
\begin{itemize}
\item increasing the tension makes waves travel faster;
\item increasing the mass per unit length makes waves travel slower.
\end{itemize}
More precisely,
\begin{equation}
\boxed{c\propto\sqrt{T}}
\end{equation}
when $\mu$ is fixed, and
\begin{equation}
\boxed{c\propto\frac{1}{\sqrt{\mu}}}
\end{equation}
when $T$ is fixed.
\begin{center}
\includegraphics{WM14_fig04_speed_scaling.png}
\vspace{0.45em}
\textbf{Figure.}
Wave speed on an ideal string scales with the square root of tension and with
the inverse square root of linear mass density.
\end{center}
\section{Dimensional check}
A correct physical equation must have consistent units.
Tension has units of force:
\begin{equation}
[T]=\text{N}=\frac{\text{kg}\,\text{m}}{\text{s}^2}.
\end{equation}
Linear mass density has units
\begin{equation}
[\mu]=\frac{\text{kg}}{\text{m}}.
\end{equation}
Therefore
\begin{align}
\left[\frac{T}{\mu}\right]
&=\frac{\text{kg}\,\text{m}/\text{s}^2}{\text{kg}/\text{m}}\\
&=\frac{\text{m}^2}{\text{s}^2}.
\end{align}
Taking the square root gives
\begin{equation}
\left[\sqrt{\frac{T}{\mu}}\right]
=\frac{\text{m}}{\text{s}},
\end{equation}
which is the correct dimension for speed.
The differential equation is also dimensionally consistent. If $u$ is a
displacement,
\begin{equation}
[u_{tt}]=\frac{\text{m}}{\text{s}^2}
\end{equation}
and
\begin{equation}
\left[\frac{T}{\mu}u_{xx}\right]
=\frac{\text{m}^2}{\text{s}^2}\frac{1}{\text{m}}
=\frac{\text{m}}{\text{s}^2}.
\end{equation}
\section{The equation expresses a local feedback law}
The string equation can be read physically as
\begin{equation}
\boxed{\text{local acceleration}\propto\text{local curvature}.}
\end{equation}
A region that is locally concave upward has
\begin{equation}
u_{xx}>0,
\end{equation}
so the transverse acceleration is upward. A region that is locally concave
downward has
\begin{equation}
u_{xx}<0,
\end{equation}
so the transverse acceleration is downward.
At an inflection point,
\begin{equation}
u_{xx}=0,
\end{equation}
so the string has no transverse acceleration from the local tension imbalance
at that instant, even though the displacement or velocity at that point may be
nonzero.
This is more informative than viewing the wave equation as a purely symbolic
relation between second derivatives. The equation is a local Newtonian law:
curvature produces force, force produces acceleration, and the resulting motion
changes the curvature at neighboring points.
\section{Why the equation is linear}
The equation
\begin{equation}
\mu u_{tt}=T u_{xx}
\end{equation}
is linear in the field $u$. If $u_1$ and $u_2$ are solutions for the same
constant $T$ and $\mu$, then any linear combination
\begin{equation}
a u_1+b u_2
\end{equation}
is also a solution.
This mathematical linearity is the reason the superposition principle from
WM09 works for the ideal string model.
The linearity comes from the modeling assumptions. In particular, the
approximation
\begin{equation}
\sin\theta\simeq\tan\theta\simeq u_x
\end{equation}
replaces the exact geometric force relation by one that is linear in the slope.
If the slope is not small, that simplification fails and nonlinear effects can
appear.
\section{Consistency with a translating disturbance}
WM13 showed that a sufficiently smooth right-moving profile
\begin{equation}
u(x,t)=F(x-ct)
\end{equation}
obeys
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
The mechanical derivation now says that the stretched string obeys
\begin{equation}
u_{tt}=\frac{T}{\mu}u_{xx}.
\end{equation}
The two are consistent when
\begin{equation}
\boxed{c^2=\frac{T}{\mu}.}
\end{equation}
Thus the traveling-wave speed is no longer just a parameter in a chosen
function. The mechanical properties of the string determine it.
A sinusoidal wave
\begin{equation}
u(x,t)=A\cos(kx-\omega t+\phi)
\end{equation}
therefore satisfies
\begin{equation}
\omega^2=\frac{T}{\mu}k^2,
\end{equation}
or
\begin{equation}
\boxed{\omega=ck.}
\end{equation}
For the ideal string this is a nondispersive relation: all sinusoidal components
have the same phase speed $\omega/k=c$ within the model.
\section{The role of initial and boundary conditions}
The wave equation does not by itself determine one unique motion. WM12 made
that point explicit.
For a finite string we still need initial data such as
\begin{equation}
u(x,0)=f(x)
\end{equation}
and
\begin{equation}
u_t(x,0)=g(x),
\end{equation}
plus boundary conditions such as
\begin{equation}
u(0,t)=0,
\qquad
u(L,t)=0
\end{equation}
for a fixed-fixed string.
The local differential law
\begin{equation}
\mu u_{tt}=T u_{xx}
\end{equation}
tells us how the interior evolves. The initial and boundary conditions select
the particular physical solution.
\section{What changes outside the ideal-string assumptions?}
The derivation also shows where more complicated models come from.
If the tension varies with position, a more general linearized string equation
has the structure
\begin{equation}
\mu(x)u_{tt}
=\frac{\partial}{\partial x}
\left(T(x)u_x\right),
\end{equation}
rather than simply $T u_{xx}$.
If bending stiffness is important, as in a beam or stiff wire, higher spatial
derivatives appear. If damping is important, velocity-dependent terms appear.
If slopes become large, geometric nonlinearities appear. If external forcing
acts along the string, a forcing term appears on the right-hand side.
Thus
\begin{equation}
\boxed{u_{tt}=\frac{T}{\mu}u_{xx}}
\end{equation}
should be understood as the governing equation of a specific idealized physical
system, not as a universal equation for every one-dimensional object that can
vibrate.
\section{Worked example 1: compute wave speed}
A string is under tension
\begin{equation}
T=180\,\text{N}
\end{equation}
and has linear mass density
\begin{equation}
\mu=0.012\,\text{kg/m}.
\end{equation}
The wave speed is
\begin{align}
c
&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{180}{0.012}}\,\text{m/s}\\
&=\sqrt{15000}\,\text{m/s}\\
&\approx\boxed{122\,\text{m/s}}.
\end{align}
\section{Worked example 2: scaling with tension and density}
Suppose an ideal string initially has speed $c_0$.
If the tension is increased by a factor of four while $\mu$ stays fixed,
\begin{equation}
c=\sqrt{\frac{4T_0}{\mu}}=2c_0.
\end{equation}
So quadrupling tension doubles wave speed.
If instead the linear density is increased by a factor of four while $T$ stays
fixed,
\begin{equation}
c=\sqrt{\frac{T}{4\mu_0}}=\frac{c_0}{2}.
\end{equation}
So quadrupling linear density halves wave speed.
\section{Worked example 3: find the required tension}
A string has
\begin{equation}
\mu=0.020\,\text{kg/m}
\end{equation}
and must support waves at speed
\begin{equation}
c=75\,\text{m/s}.
\end{equation}
From
\begin{equation}
c^2=\frac{T}{\mu},
\end{equation}
we obtain
\begin{align}
T
&=\mu c^2\\
&=(0.020)(75)^2\,\text{N}\\
&=\boxed{112.5\,\text{N}}.
\end{align}
\section{Worked example 4: connect the wave equation to $k$ and $\omega$}
Consider
\begin{equation}
u(x,t)=0.005\,\text{m}\cos(8x-240t).
\end{equation}
The wavenumber and angular frequency are
\begin{equation}
k=8\,\text{rad/m},
\qquad
\omega=240\,\text{rad/s}.
\end{equation}
Hence
\begin{equation}
c=\frac{\omega}{k}=\frac{240}{8}=30\,\text{m/s}.
\end{equation}
If the string has
\begin{equation}
\mu=0.010\,\text{kg/m},
\end{equation}
then the required tension is
\begin{align}
T
&=\mu c^2\\
&=(0.010)(30)^2\,\text{N}\\
&=\boxed{9.0\,\text{N}}.
\end{align}
The same conclusion follows from matching the second derivatives:
\begin{equation}
u_{tt}=-\omega^2u,
\qquad
u_{xx}=-k^2u.
\end{equation}
Substitution into
\begin{equation}
\mu u_{tt}=T u_{xx}
\end{equation}
gives
\begin{equation}
\mu\omega^2=Tk^2,
\end{equation}
which is equivalent to
\begin{equation}
\frac{T}{\mu}=\left(\frac{\omega}{k}\right)^2.
\end{equation}
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} using displacement itself as the restoring-force variable. For an ideal stretched string, the local tension imbalance is controlled by curvature $u_{xx}$, not directly by $u$.
\item \textbf{Mistake:} assuming $u_x$ is the wave speed. It is spatial slope.
\item \textbf{Mistake:} assuming $u_t$ is the propagation speed. For string displacement it is the transverse material velocity at fixed $x$.
\item \textbf{Mistake:} forgetting that the mass of the small element is $\mu\Delta x$.
\item \textbf{Mistake:} adding the two vertical tension components instead of taking their signed difference.
\item \textbf{Mistake:} using $c=T/\mu$. The correct speed is the square root $c=\sqrt{T/\mu}$.
\item \textbf{Mistake:} treating constant tension as exact for arbitrary large slopes. It is part of the linear ideal-string approximation.
\item \textbf{Mistake:} thinking the wave equation alone fixes the motion. Initial and boundary conditions are still required.
\end{itemize}
\section{What WM14 adds to the wave-mechanics language}
The earlier lessons built the mathematical structure of a wave. WM14 supplies
the mechanical cause for that structure in an ideal string.
The chain is
\begin{equation}
\boxed{
\text{curvature}
\longrightarrow
\text{tension imbalance}
\longrightarrow
\text{transverse force}
\longrightarrow
\text{acceleration}.
}
\end{equation}
Quantitatively,
\begin{equation}
\boxed{\mu u_{tt}=T u_{xx}}
\end{equation}
and therefore
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}},
\qquad
\boxed{c=\sqrt{\frac{T}{\mu}}.}
\end{equation}
This is the first point in the series where the one-dimensional wave equation
has been obtained from a physical law rather than merely recognized as a
relation satisfied by a chosen traveling waveform.
\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{OpenStax163}
William Moebs, Samuel J. Ling, and Jeff Sanny,
\emph{University Physics, Volume 1},
OpenStax, 2016,
Section 16.3, ``Wave Speed on a Stretched String.''
\bibitem{OpenStax162}
William Moebs, Samuel J. Ling, and Jeff Sanny,
\emph{University Physics, Volume 1},
OpenStax, 2016,
Section 16.2, ``Mathematics of Waves.''
\bibitem{MIT803Waves}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves - The Physics of Waves},
Fall 2016, MIT OpenCourseWare,
material on the one-dimensional wave equation and transverse waves on a string.
\end{thebibliography}