(Strauss 4.3.13) Consider a string that is fixed at the end $x = 0$ and is free at the end $x = \ell$ except that a load (weight) of given mass is attached to the right end.
Show that it satisfies the problem
$u_{tt} = c^2 u_{xx}$, for $0 < x < \ell$
$u(0,t) = 0$
$u_{tt}(\ell, t) = −ku_x(\ell, t)$
for some constant $k$.
I don't want a solution as much as some help visualizing exactly what he means and maybe a push in the right direction. I tried thinking of it vertically, like a limp pendulum, but I couldn't get anything out of it.
Here's the answer I came up with, for anyone else who's tackling this problem:
Consider a segment of string from $x^*$ to $\ell$, for some $x^* \in (0,\ell)$. The transverse mass-times-acceleration along the string is $$\int_{x^*}^\ell \rho u_{tt}(x,t) \ \text dx$$ and at the endpoint $x=\ell$ is $mu_{tt}(\ell,t)$, where $\rho$ is the linear mass density of the string and $m$ is the mass at the free end.
The transverse external forces are just the tension force with magnitude $T$ at $x=x^*$, since the other end is free, so $$F = -T\frac{u_x(x^*,t)}{\sqrt{1+u_x(x^*,t)^2}} \approx -Tu_x(x^*,t)$$ for small vibrations. Thus, by Newton's Second Law, we have that $$mu_{tt}(\ell,t)+\int_{x^*}^\ell \rho u_{tt}(x,t) \ \text dx = -Tu_x(x^*,t)$$ and taking the limit as $x^*$ goes to $\ell$, we get $$mu_{tt}(\ell,t) = -Tu_x(\ell,t) \implies u_{tt}(\ell,t) = -ku_x(\ell,t)$$ where $k = T/m > 0$.