= Solution
{c}
Use the ideal transverse-motion model of the question: constant <tension>, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $Ty_x(x+\Delta x,t)$ and $-Ty_x(x,t)$. Its mass is $\rho\Delta x$, since $\rho$ is <linear mass density>. <Newton's second law> gives
$$
\rho\Delta x\,y_{tt}=T\bigl[y_x(x+\Delta x,t)-y_x(x,t)\bigr].
$$
Dividing by $\Delta x$ and taking its limit proves the <wave equation on a string> on each side of the mass:
$$
\boxed{y_{tt}=c^2y_{xx},\qquad c^2=\frac T\rho,\qquad -\ell<x<0\text{ or }0<x<\ell.}
$$
The string on the right pulls the <point mass> with vertical component $Ty_x(0^+,t)$, while the string on the left contributes $-Ty_x(0^-,t)$. Hence
$$
\boxed{m\ddot Y(t)=T\bigl[y_x(0^+,t)-y_x(0^-,t)\bigr].}
$$
The fixed supports and attachment impose the <boundary conditions>
$$
\boxed{y(-\ell,t)=y(\ell,t)=0,\qquad y(0^-,t)=y(0^+,t)=Y(t).}
$$
Together with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical <acceleration>. Requiring the slope to be continuous would incorrectly remove its dynamics.
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