Imagine an infinite string with constants $\rho$ and $T$. Then $\rho u_{t t}=T u_{x x}$ for $-\infty<x<+\infty$. From physics we know that the kinetic energy is $\frac{1}{2} m v^{2}$ which in our case takes the form $\mathrm{KE}=\frac{1}{2} \rho \int u_{t}^{2} d x .$
This makes no sense as I think we should instead have $$KE(t)=\dfrac{1}{2}\rho u_t^2=\dfrac{mv^2} {2}$$
where $$\rho \approx m$$
The formula $\mathrm{KE} = \frac{1}{2} \rho \int u_t^2\,\mathrm{d}x$ is derived using the following physical argument.
Consider a bit of the string at position $x$ of a very small length $\mathrm{d}x$. The displacement of this little bit of string is $u(x,t)$, and so its velocity is $u_t(x,t)$. The mass of this little bit is $\rho \,\mathrm{d}x$. Thus the kinetic energy of this little bit is $\frac{1}{2}\rho u_t^2 \,\mathrm{d}x$. "Summing up" the kinetic energy contributed by each little bit of string yields the desired integral.