How Is Tension Calculated in the One Dimensional Wave Equation?

[Telemachus]

Hi there. I was trying to understand this deduction of the one dimensional wave equation developed at the beggining of the book A first course in partial differential equations of H.F. Weinberger.

You can see it right here: http://books.google.com.ar/books?hl...2eHPnArCVnLC7Q0FtHljNwcV7g#v=onepage&q&f=fals

I'm having a problem with the part where he consideres the tension over the string. I don't know how this expression is deduced. The book says: We now consider an arbitrary portion s_1\leq s\leq s_2 of the string. Taking the components of the tensile force in the x-direction, we find that the total force in the x-direction acting on this portion is:

\frac{ T(s_2,t)\frac{\partial x}{\partial s}(s_2,t) }{ \sqrt { \left ( \frac{\partial x}{\partial s}\right )^2+\left( \frac{\partial y}{\partial s}\right )^2}}-\frac{ T(s_1,t)\frac{\partial x}{\partial s}(s_1,t) }{ \sqrt { \left ( \frac{\partial x}{\partial s}\right )^2+\left( \frac{\partial y}{\partial s}\right )^2}}

The whole problem for me now is just this thing. I think I understand a bit what this equation says, but perhaps some one could help me to figure this out more clearly. At first, T is some function, let's say a function force that depends on the material, and as we stretch it or compress, as a spring, it produces a reaction. Ok. Then the denominator is the arc length formula, so its saying how much we stretched or compressed the string. And the other thing is the derivative of the curve, and I think its there because the force goes tangential to the string. Is this lecture right?

 

[vela]

It's nothing terribly complicated. The slope of the string at the point s is given by tan θ = ∂y/∂x = (∂y/∂s)/(∂x/∂s). The horizontal component of the tension is therefore given by
T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}