= Solution
{c}
The lowest positive root belongs to the symmetric family and lies in $(0,\pi/2)$; all antisymmetric roots are at least $\pi$. Let $\alpha=m/(2\rho\ell)\gg1$. The equation $\cot z=\alpha z$ forces its first root to tend to zero: for any fixed $\varepsilon>0$, the right side eventually exceeds $\cot\varepsilon$ at $z=\varepsilon$, so the decreasing-minus-increasing difference has its zero below $\varepsilon$.
The <Taylor series> $\cot z=z^{-1}-z/3+O(z^3)$ then gives
$$
1=\left(\alpha+\frac13\right)z^2+O(z^4),\qquad
z\sim\alpha^{-1/2}=\sqrt{\frac{2\rho\ell}{m}}.
$$
Since $\omega=cz/\ell$ and $c=\sqrt{T/\rho}$, the lowest <angular frequency> is
$$
\boxed{\omega\sim\frac1\ell\sqrt{\frac T\rho}\sqrt{\frac{2\rho\ell}{m}}
=\sqrt{\frac{2T}{m\ell}}=\omega_0.}
$$
Physically, at such a low <frequency> each half-string is nearly straight. If the mass moves up by $Y$, the two slopes are approximately $Y/\ell$ and $-Y/\ell$, so the restoring <force> is $-2TY/\ell$. The resulting <harmonic oscillator equation>, $m\ddot Y+(2T/\ell)Y=0$, has exactly the leading <angular frequency> above. The next correction from the same expansion is $\omega/\omega_0=1-\rho\ell/(3m)+O((\rho\ell/m)^2)$.
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