Let $u$ be the velocity of the bead along the rod. The component of the bead’s velocity along the string must equal the speed at which the string is being pulled ($v_0$). $$ u \cos \theta = v_0 $$ $$ u = v_0 (\cos \theta)^{-1} $$

Differentiate the velocity $u$ with respect to time to find the acceleration $a$. $$ a = \frac{du}{dt} = \frac{d}{dt} [v_0 (\cos \theta)^{-1}] $$ $$ a = -v_0 (\cos \theta)^{-2} (-\sin \theta) \frac{d\theta}{dt} $$ $$ a = v_0 \frac{\sin \theta}{\cos^2 \theta} \dot{\theta} \quad \dots(1) $$

The component of the bead’s velocity perpendicular to the string is $u \sin \theta$. This component causes the string (of length $l$) to rotate. $$ l \dot{\theta} = u \sin \theta $$ Substituting $u = \frac{v_0}{\cos \theta}$: $$ \dot{\theta} = \frac{v_0 \sin \theta}{l \cos \theta} $$

Substitute $\dot{\theta}$ back into equation (1) to find acceleration: $$ a = v_0 \frac{\sin \theta}{\cos^2 \theta} \left( \frac{v_0 \sin \theta}{l \cos \theta} \right) = \frac{v_0^2 \sin^2 \theta}{l \cos^3 \theta} $$ The horizontal component of the Tension $T$ provides this acceleration: $$ T \cos \theta = ma $$ $$ T = \frac{ma}{\cos \theta} = \frac{m}{\cos \theta} \left( \frac{v_0^2 \sin^2 \theta}{l \cos^3 \theta} \right) $$