 Solution

The lowest positive root belongs to the symmetric family and lies in $(0,\pi /2)$; all antisymmetric roots are at least $\pi$. Let $\alpha =m/(2\rho \ell )\gg 1$. The equation $\cot z=\alpha z$ forces its first root to tend to zero: for any fixed $\epsilon >0$, the right side eventually exceeds $\cot \epsilon$ at $z=\epsilon$, so the decreasing-minus-increasing difference has its zero below $\epsilon$.

The Taylor series $\cot z=z^{-1}-z/3+O(z^3)$ then givesSince $\omega =cz/\ell$ and $c=\sqrt{T/\rho }$, the lowest angular frequency isPhysically, 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 $\omega /{\omega }_0=1-\rho \ell /(3m)+O((\rho \ell /m{)}^2)$.