= Solution
For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of <tension> at its two ends $T y_x(x+\Delta x,t)$ and $-Ty_x(x,t)$. Its <mass> is $\rho\Delta x$. <Newton's second law> gives
$$
\rho\Delta x\,y_{tt}=T\left[y_x(x+\Delta x,t)-y_x(x,t)\right].
$$
Divide by $\Delta x$ and take the limit to obtain the <wave equation on a string>,
$$
\boxed{y_{tt}=c^2y_{xx},\qquad c=\sqrt{\frac T\rho}}.
$$
Here $\rho$ is <linear mass density>, and the derivation assumes small transverse slopes, uniform <tension>, and a flexible string.
At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-Ty_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeys
$$
\boxed{M y_{tt}(L,t)=-Ty_x(L,t)}.
$$
This is the dynamic endpoint law for a <string with an endpoint mass>.
For a separated oscillatory solution, write $y=X(x)q(t)$ with $q''=-\omega^2q$ and $X''+(\omega/c)^2X=0$. The fixed left end removes the spatial <cosine>, so the nonzero <normal modes> have
$$
y_n(x,t)=\left[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)\right]
\sin\frac{\omega_nx}{c}.
$$
The endpoint law gives $M\omega_n^2\sin(\omega_nL/c)=T(\omega_n/c)\cos(\omega_nL/c)$. Since $\sin(\omega_nL/c)=0$ would also force the <cosine> to vanish, it can be divided out for a nonzero <mode>. Using $T=\rho c^2$ gives
$$
\boxed{\cot\frac{\omega_nL}{c}=\frac{M\omega_n}{\rho c}}.
$$
For physical $M\geq0$, these give all positive modal frequencies. There is no nonzero static linear <mode>: the left value and right slope conditions would force it to vanish. Positivity also follows by <integration by parts> from
$$
\omega^2\left(\rho\int_0^L X^2\,dx+MX(L)^2\right)
=T\int_0^L(X')^2\,dx.
$$
For $M=0$, the right endpoint is free, with the <Neumann boundary condition> $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\geq0$, and $\omega_n=ck_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies gives
$$
\boxed{y(x,t)=\frac2c\sin\frac{ct}{2}\sin\frac x2
+\frac{2}{5c}\sin\frac{5ct}{2}\sin\frac{5x}{2}}.
$$
At $x=0$ each term vanishes, at $x=\pi$ each spatial <derivative> vanishes, and at $t=0$ the displacement and its time <derivative> reproduce the two <initial conditions>.
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