Wave equation

[CPL.Luke]

So I was reading the derivation of the wave equation in Halliday's physics, and I saw that they made the simplification that the wave had a very small amplitude, in essence they assumed that the amplitude of the wave is very small and so an infintesimal segment of length on the string (dL) could be said to be equal to an infinetessimal segment of the x axis (dx).

 

Now I decided that I didn't like this assumption and seeing as I ca't sleep at the moment I decided to re-derive the formula without the assumption that the amplitude is small. I also don't have a physics professor who could really comment on anything thats not clearly written in the book with any sense of accuracy (not really his fault though because he's been teaching algebra based physics for the past 2 years) so I was wondering if anybody could comment on whether or not this is accurate.

 

so to start, you have a string of some length under some tension ? and some mass density ?, and some wave y(x,t) travelling through it. First select some element on the string ?L and set the sum the forces acting on both sides of the string element equal to mass (of the element) x acceleration

 

[MATH]

F_{{1}}-F_{{2}}={\frac {\mbox {{\tt `\Δm`}}\,{d}^{2}y}{{{\it dt}

}^{2}}}[/MATH]

 

or using delta m=? delta l

 

[MATH]F_{{1}}-F_{{2}}={\frac {\mbox {{\tt `\μ\ΔL`}}\,{d}^{2}y}{{{

\it dt}}^{2}}}[/MATH]

^where F_1 is the force acting on one side, and F_2 is the force acting on the other (it really doesn't matter which one is which)

 

 

we can then analyze the vectors by using the trig identity tan ? = S (slope) so that

[MATH]{\frac {F_{{y}}}{F_{{-1}}}}=S_{{1}}[/MATH]

 

Then we can substitute in the tension tau for F_1x as the horizontal component of force must be equal to to the tention. and get

 

[MATH]

F_{{y}}=\mbox {{\tt

`\#msub(mi("\τS",fontstyle = "italic"),mo("\!"))`}}

[/MATH]

 

doing the same thing to F_2 and plugging back into the first equation we get

 

[MATH]\tau\, \left( S_{{1}}-S_{{2}} \right) ={\frac {\mu\,\mbox {\Delta}\,{d}^{2}y}{{{\it dt}}^{2}}}[/MATH]

 

subbing in ?S for S_1-S_2 and dividing by ?L we get

 

[MATH]{\frac {\mbox {{\tt `\τ\ΔS`}}}{\mbox {{\tt `\ΔL`}}}}={

\frac {\mu\,{d}^{2}y}{{{\it dt}}^{2}}}[/MATH]

 

now this is the tricky part, but if you take the limit of both sides as delta L goes to zero, the left hand side turns into a derivative, but its in the wrong variable ie S is equal to dy/dx, so we can't directly take the derivative of that with respect to L, but (I think, I'm not 100%sure if this works) we can convert dL to a function of dx by the equation

 

 

[MATH]{\it dL}={\it dx}\,\sqrt {1+({\frac {{{\it dy}}}{{{\it dx}}})^2}}[/MATH]

 

plug that in and out pops the differential equation

 

[MATH]\tau\ \left( S_{{1}}-S_{{2}} \right) ={\frac {\mu\mbox {{\&DeltaL`}}\,{d}^{2}y}{{{\it dt}}^{2}}}[/MATH]

 

 

any comments would be appreciated, like I said at the moment I don't have any professors to go to with things like this.

 

EDIT: hmm looks like my latex is a bit broken I'll work on fixing that

[CPL.Luke]

hmm seems I can't edit anymore, so I hope that someone can get the gist of what I said, an the final solution is supposed to be

 

[MATH] /frac{1}{/squrt{1+(/frac{dy}{dx})^2} /frac{d^2y}{dx^2}=/frac{/mu}{/tau} /frac{d^2y}{dt^2}[/MATH]