( 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

Copyright © 2009-2010, John Forkosh Associates, Inc.
email: john@forkosh.com



      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...

     
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 (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:

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

Due to the small– assumption, horizontal components of tension cancel, and the net horizontal force vanishes

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

where you'll notice just defines the derivative.

This net force acts on the element whose mass is where is the string's linear density. And . Therefore

where , which is the typical form of your one–dimensional wave equation.

Image Tests (ignore)

For comp.text.tex ng readers:
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Uncropped test image (note the two dots in upper left- and right-corners)...

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Copyright © 2009-2010, John Forkosh Associates, Inc.
email: john@forkosh.com