A simple pendulum has a string of length $l$ and bob of mass $m$. When the bob is at its lower position, it is given the maximum horizontal speed necessary for it to move in a circular path about the point of suspension. The tension in the string at the lowest position of the bob is

Let A be the topmost point of the circle and B be the lowest point of the circle.

Let $v_2$ and $v_1$ be the velocities at B and A respectively.

Applying the principle of conservation of energy between A and B.

$\displaystyle \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2 = mg2l$

or $\displaystyle \frac{mv_2^2}{l} = \frac{mv_1^2}{l} + 4mg$ .....(1)

At B, $\displaystyle\frac{mv_2^2}{l} = T -mg$ .....(2)

At A. $\displaystyle \frac{mv_1^2}{l} = mg$ ....(3)

Solving (1), (2) and (3)

or $mg+4mg=T-mg$ or $T=6mg$