= String with an endpoint mass
For a uniform string with <tension> $T>0$, <linear mass density> $\rho>0$, left end fixed, and a freely transverse-moving point mass $M\geq0$ at $x=L$, the <wave equation> and endpoint law are
$$
\rho y_{tt}=Ty_{xx},\qquad y(0,t)=0,\qquad
M y_{tt}(L,t)=-Ty_x(L,t).
$$
The last equation is <Newton's second law>: the string pulls the mass back toward the interior, giving the minus sign. With <wave speed> $c=\sqrt{T/\rho}$, a nonzero <normal mode> has spatial part $X(x)=\sin(\omega x/c)$ and satisfies
$$
\cot\frac{\omega L}{c}=\frac{M\omega}{\rho c}.
$$
For $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi,n\pi+\pi/2)$, $n\geq0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous <Neumann boundary condition>. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.
Multiplying the spatial equation by $X$ and using <integration by parts> gives
$$
\omega^2=\frac{T\int_0^L(X')^2\,dx}
{\rho\int_0^L X^2\,dx+M X(L)^2}>0.
$$
For two different modes, the same calculation gives the <weighted inner product> orthogonality $\rho\int_0^L X_mX_n\,dx+M X_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.
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