What is the relationship between mass and displacement in a vibrating string?

[Mechdude]

Homework Statement

hi I am reading Guenther & Lee, "partial differential equations of mathematical physics and integral equations" on the
first chapter second section i think, on "Small vibrations of an elastic string" they give this argument:
1. consider a string of x length L, the string is assumed to be a continuum and tied to posts at $x=0$
and $x=L$. a continuous density function, $\rho$, with its integral over any segment of the
string gives the mass of the segment. the string is perfectly elastic, vibrations are very small.
2. an axis perpendicular to the x-axis is constructed at $x=0$ , the equilibrium position of the string
is the horizontal segment $0 \leq x \leq L$ . the position of a given point which was at x during equilibrium
will be $u(x,t)$ at time t. if time is kept constant the function gives the shape of the string at that instant.
3. the function $\rho_0 (x)$ denotes the density at equilibrium, and $\rho (x,t)$ at time t.
as the string stretches the density will change; if we focus on an arbitrary interval between $x=x_1$ &
$x=x_2$ along the string we find that the mass m in this interval satisfies:
$$\int_{x_1}^{x_2} \rho_0 (x) dx = \int^{x_2}_{x_1} \rho (x,t) [ 1 + u^{2}_{x} (x,t) ]^{\frac{1}{2}} dx$$

this last expression has me stumped, it seems to me like he pulled it right from under his sleeve, why would the mass satisfy that
expression? i think the squared term is a partial derivative with respect to x imho.

Homework Equations

The Attempt at a Solution

 

[jbsteele]

I think the squared term is a partial derivative, which means that the mass is a function of the displacement of the string. This is because when the string stretches, the density of the string changes and thus the mass changes as well. This is why the mass is a function of the displacement of the string.