= String with a central point mass
{title2=$\sin(k\ell)=0\quad\text{or}\quad\cot(k\ell)=mk/(2\rho)$}
A uniform string with fixed ends $x=\pm\ell$, <tension> $T$, <linear mass density> $\rho$, and attached central <point mass> $m>0$ obeys the <wave equation> away from zero, continuous displacement at zero, and $m y_{tt}(0,t)=T[y_x(0^+,t)-y_x(0^-,t)]$. The <normal modes> split into antisymmetric modes that leave the mass at rest and symmetric modes that move it.
To prove the spectral equations, use $X=A\sin(k(x+\ell))$ on the left and $X=B\sin(k(\ell-x))$ on the right, with $\omega^2=(T/\rho)k^2$. Continuity gives $(A-B)\sin(k\ell)=0$ and the force law gives $Tk(A+B)\cos(k\ell)=m\omega^2A\sin(k\ell)$. If $\sin(k\ell)=0$, then $k\ell=n\pi$, $n\geq1$, and $B=-A$. Otherwise $A=B$ and cancellation gives $\cot(k\ell)=mk/(2\rho)$. In dimensionless variables $z=k\ell$, $\alpha=m/(2\rho\ell)$, strict <monotonicity> gives one symmetric root in each $(n\pi,n\pi+\pi/2)$, $n\geq0$. The apparent zero root of the sine equation has only the zero profile and is excluded.
For $m/(\rho\ell)\gg1$, the lowest root tends to zero. The <Taylor series> $\cot z=1/z-z/3+O(z^3)$ gives $z\sim\alpha^{-1/2}$ and hence $\omega\sim\sqrt{2T/(m\ell)}$. The two nearly straight half-strings supply a restoring <force> $-2TY/\ell$, explaining this <harmonic oscillator> limit.
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