( u n d e r   c o n s t r u c t i o n . . . )

 nrows=75&ncols=151&ncoefs=12& chksz=3&fgblue=135&f=0,0,1,.25,1,0,-1,-.75,-1,0,0 O n e – D i m e n s i o n a l W a v e   E q u a t i o n Click for:   download nrows=75&ncols=151&ncoefs=64&dt=.02& chksz=3&chkoff=2&fgblue=135&f=0,1,.25,1,0,0,0

email: john@forkosh.com

$\parstyle\usepackage{color} \large\color{blue}\begin{center}\today\\\time\end{center}$

 C o n t e n t s

 O n e – D i m e n s i o n a l   W a v e   E q u a t i o n   S o l v e r   B o x Enter any waveform parameters you like, as explained below, in the top box. Then press Submit to render its solution. I've started you out with a little example already in the box. Or you can Click any waveform on this page to place its corresponding parameterization in the box.
 First enter one-dimensional wave equation parameters... ncoefs=12&fgblue=135&f=0,0,1,.25,1,0,-1,-.75,-1,0,0
Now click "Submit" to see your waveform rendered below...

You should see   nrows=33&ncols=65& ncoefs=12&fgblue=135&f=0,0,1,.25,1,0,-1,-.75,-1,0,0   if you submit the sample expression already in the box.
Or try clicking this image   nrows=33&ncols=65&ncoefs=128&dt=.03& fgblue=135&f=0,,,,,,,,0,1,,,,,,,1,0,,,,,,,,,,,,,,,,,,,,,,0   and see what happens.

## (1) Preliminaries

Suppose you pluck a string at a point, say, one-quarter way from its left end, so that it looks something like nrows=33&ncols=65&ncoefs=99& fgblue=135&f=1,.75,.5,.25>imestep=1&bigf=1 just before you release it. Exactly how does the string behave after you release it? The one-dimensional wave equation is essentially the plucked string problem. Our plucked string behaves like nrows=33&ncols=65&ncoefs=99& fgblue=135&f=1,.75,.5,.25>imestep=0&bigf=1&dt=.05 , and this page discusses how we arrive at that solution.

### Derivation

All problems in classical mechanics boil down to $\normalsize\ \mathbf{F}=m\mathbf{a}\$ (plus conservation). Point masses subject to constant forces are typically trivial. Extended masses subject to forces that vary over space and time can be more complicated to treat. Reasonable simplifying assumptions often make such complicated problems more tractable. We make two assumptions:

• The string is displaced only a small amount, i.e., its initial displacement looks more like nrows=15&ncols=65&ncoefs=99& fgblue=135&f=1,.75,.5,.25>imestep=1&bigf=1 than like nrows=33&ncols=65&ncoefs=99& fgblue=135&f=1,.75,.5,.25>imestep=1&bigf=1 . In particular, the angle $\definecolor{blueblack}{RGB}{0,0,135} \color{blueblack} \setlength{\unitlength}{1.0in} \ \begin{picture}(1.0,.25) \thicklines \put(0,0){\line(1,0){.5}} \put(0,0){\line(3,1){.3}} \put(.35,.06){\makebox(0,0)[r]{\footnotesize\theta}} \end{picture}\$ remains small, so that the approximations $\normalsize\ \sin\theta\approx\theta\approx \tan\theta\$ and $\normalsize\ \cos\theta\approx1\$ remain valid at all times and at all points along the string. Our little diagrams exaggerate the displacement for illustrative purposes, but don't forget that it's actually small.

• The magnitude of the tension force $\normalsize\ \mathbf{T}\$ remains constant (at all times and at all points) regardless of how much the string is stretched. Before the initial displacement, all points (more correctly, all differential elements $\normalsize\ ds$ ) along the string feel equal and opposite $\definecolor{blueblack}{RGB}{0,0,135} \color{blueblack} \setlength{\unitlength}{1.0in} \ \begin{picture}(1.0,.25) \thicklines \put(.20,0){\vector(-1,0){.20}} \put(.30,0){\vector(1,0){.20}} \put(.24,0){\makebox(0,0){\bullet}} \put(.26,0){\makebox(0,0){\bullet}} \put(0.25,0.075){\makebox(0,0){\footnotesizeds}} \put(0.05,0.075){\makebox(0,0){\footnotesizeT}} \put(0.50,0.075){\makebox(0,0){\footnotesizeT}} \end{picture}\$ and remain at rest. After a typical displacement like nrows=33&ncols=65&ncoefs=2& fgblue=135&f=.15,1,.75,.5,.25>imestep=1&bigf=1 , an average point along the stretched string looks more like $\usepackage{eepic} \definecolor{blueblack}{RGB}{0,0,135} \color{blueblack} \setlength{\unitlength}{1.0in} \ \begin{picture}(1.0,.25) \thicklines \put(.19,0.16){\vector(-2,-1){.20}} \put(.31,0.21){\vector(4,1){.22}} \put(.27,0.19){\makebox(0,0){\bullet}} \put(.25,0.18){\makebox(0,0){\bullet}} \put(.23,0.17){\makebox(0,0){\bullet}} \put(0.25,0.11){\makebox(0,0){\footnotesizeds}} \put(0.05,0.15){\makebox(0,0){\footnotesizeT}} \put(0.45,0.15){\makebox(0,0){\footnotesizeT}} \end{picture}\$ . The horizontal components $\normalsize\ T\cos\theta\approx T\$ still cancel out despite slightly different (but both still small) $\normalsize\ \theta$ 's on either side, so there's no horizontal motion. The vertical components $\normalsize\ T\sin\theta\$ don't cancel, and we'll deal with the resulting vertical motion below.

Keeping these assumptions in mind, the forces acting on an element $\normalsize\ ds\$ of the stretched string are illustrated in more detail by

$\png \usepackage{epic} \definecolor{blueblack}{RGB}{0,0,135} \color{blueblack} \begin{picture}(4,1.75) \thicklines \put(2,0.01){\arc{3}{3.53588}{5.8888}} \put(.375,.575){\line(1,0){3.25}} \put(1.22,1.375){\makebox(0,0){\footnotesizeds}} \put(.6,.5){\makebox(0,0){\footnotesizex=0}} \put(3.36,.5){\makebox(0,0){\footnotesizex=\ell}} \dottedline{.05}(1.0,.575)(1.0,1.10) \put(1.0,.5){\makebox(0,0){\footnotesizex}} \dottedline{.05}(1.5,.575)(1.5,1.40) \put(1.5,.5){\makebox(0,0){\footnotesizex+dx}} \put(1.22,.65){\makebox(0,0){\footnotesizedx}} \dottedline{.04}(0.6,1.12)(1.25,1.12) \put(1.0,1.14){\vector(-1,-1){.45}} \put(.58,0.83){\makebox(0,0){\footnotesizeT}} \put(.77,1.05){\makebox(0,0){\scriptsize\theta(x)}} \put(1.18,1.16){\makebox(0,0){\scriptsize\theta(x)}} \dottedline{.04}(1.5,1.41)(2.1,1.41) \put(1.5,1.44){\vector(4,1){.67}} \put(2.22,1.59){\makebox(0,0){\footnotesizeT}} \put(1.95,1.45){\makebox(0,0){\scriptsize\theta(x+dx)}} \end{picture}$

Due to the small–$\normalsize\theta\$ assumption, horizontal components of tension $\normalsize\ \mathbf{T}\$ cancel, and the net horizontal force $\normalsize\ F_x\$ vanishes

\parstyle\normalsize\begin{align*} F_x &= T\cos\theta_{x+dx} - T\cos\theta_x \\ &\approx 0 \end{align}

so there's no $\normalsize\ x$ –direction motion. But there's a non–zero net vertical force

\parstyle\normalsize\begin{align*} F_y &= T\sin\theta_{x+dx} - T\sin\theta_x \\ &\approx T\tan\theta_{x+dx} - T\tan\theta_x \\ &\equiv T\left(\left.\frac{\partial y}{\partial x}\right|_{x+dx} -\left.\frac{\partial y}{\partial x}\right|_{x}\right) \end{align}

where you'll notice $\normalsize\ \tan\theta=\Delta y/\Delta x\$ just defines the derivative.

This net $\normalsize\ F_y\$ force acts on the $\normalsize\ ds\$ element whose mass is $\normalsize\ \lambda ds\$ where $\normalsize\ \lambda\$ is the string's linear density. And $\normalsize\ dx=ds\cos\theta\approx ds$ . Therefore

\parstyle\normalsize\begin{align*} F_y &= ma_y \hspace{.15in} \mbox{means} \\ T\left(\left.\frac{\partial y}{\partial x}\right|_{x+dx} -\left.\frac{\partial y}{\partial x}\right|_{x}\right) &= \lambda ds\,a_y \\ &= \lambda dx \frac{\partial^2 y}{\partial t^2} \\ T\,\frac{\left.\frac{\partial y}{\partial x}\right|_{x+dx} -\left.\frac{\partial y}{\partial x}\right|_{x}}{dx} &= \lambda \frac{\partial^2 y}{\partial t^2} \\ T\frac{\partial^2 y}{\partial x^2} &= \lambda \frac{\partial^2 y}{\partial t^2}\ \textbf{,} \hspace{.15in} \mbox{or} \\ \frac{\partial^2 y}{\partial x^2} - \frac1{c^2}\frac{\partial^2 y}{\partial t^2} &= 0 \end{align}

where $\normalsize\ c=\sqrt{T/\lambda}$ , which is the typical form of your one–dimensional wave equation.

## Image Tests (ignore)

Cropped test image (dvips bounding box determined by A and B)...

start(latex)...$\png \begin{picture}(4,1.75) \put(1.5,.25){\spline(0,0)(.25,.5)(.5,.5)(.75,.5)(1,0)} \put(1.0,.5){\spline(0,0)(.5,1.5)(.75,.5)(2,0)} \put(2,.5){\arc{2}{3.14}{0}} \put(2,.5){\arc{2}{3.14}{0}} \put(2,.5){\arc{2}{2.7473}{.3943}} \put(.75,.5){\line(1,0){2.5}} \put(1,1.5){\circle{.5}} \put(1,1.5){\makebox(0,0){A}} \put(2,0.5){\circle{.6}} \put(2,0.5){\circle{.30}} \put(3,1.5){\circle{.5}} \put(3,1.5){\makebox(0,0){B}} \put(1.2,1.3){\line(1,-1){0.6}} \put(1.4,1.0){\makebox(0,0)[r]{E}} \put(2.1,.6){\line(1,1){0.72}} \put(2.6,1.0){\makebox(0,0)[l]{F}} \put(2.0,0.1){\makebox(0,0){A implies B.}} \end{picture}$ ...end

start(pdflatex)...$\latex\png \begin{picture}(4,1.75) \put(1,1.5){\circle{.5}} \put(1,1.5){\makebox(0,0){A}} \put(2,0.5){\circle{.6}} \put(2,0.5){\circle{.30}} \put(3,1.5){\circle{.5}} \put(3,1.5){\makebox(0,0){B}} \put(1.2,1.3){\line(1,-1){0.6}} \put(1.4,1.0){\makebox(0,0)[r]{E}} \put(2.1,.6){\line(1,1){0.72}} \put(2.6,1.0){\makebox(0,0)[l]{F}} \put(2.0,0.1){\makebox(0,0){A implies B.}} \end{picture}$ ...end

start(latex)...$\gif \begin{picture}(4,1.75) \put(1,1.5){\circle{.5}} \put(1,1.5){\makebox(0,0){A}} \put(2,0.5){\circle{.6}} \put(2,0.5){\circle{.30}} \put(3,1.5){\circle{.5}} \put(3,1.5){\makebox(0,0){B}} \put(1.2,1.3){\line(1,-1){0.6}} \put(1.4,1.0){\makebox(0,0)[r]{E}} \put(2.1,.6){\line(1,1){0.72}} \put(2.6,1.0){\makebox(0,0)[l]{F}} \put(2.0,0.1){\makebox(0,0){A implies B.}} \end{picture}$ ...end

Uncropped test image (note the two dots in upper left- and right-corners)...

start...$\begin{picture}(4,1.75) \put(0.75,1.75){\makebox(0,0){.}} \put(3.25,1.75){\makebox(0,0){.}} \put(1,1.5){\circle{.5}} \put(1,1.5){\makebox(0,0){A}} \put(2,0.5){\circle{.6}} \put(2,0.5){\circle{.30}} \put(3,1.5){\circle{.5}} \put(3,1.5){\makebox(0,0){B}} \put(1.2,1.3){\line(1,-1){0.6}} \put(1.4,1.0){\makebox(0,0)[r]{E}} \put(2.1,.6){\line(1,1){0.72}} \put(2.6,1.0){\makebox(0,0)[l]{F}} \put(2.0,0.1){\makebox(0,0){A implies B.}} \end{picture}$ ...end

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