Physics - W.E.P. Question with Solution | TestHub

At the highest point, let the speed be $v_H=2\sqrt{gL}$ and tension be $T_H$. The equation of motion is $T_H+mg=m{v}_H^2/L$.$$T_H=\frac{m(2\sqrt{gL}{)}^2}{L}-mg=\frac{4mgL}{L}-mg=4mg-mg=3mg$$ By conservation of mechanical energy between the highest and lowest points:$$\frac{1}{2}m{v}_H^2+mg(2L)=\frac{1}{2}m{v}_L^2$$ $$\frac{1}{2}m(4gL)+2mgL=\frac{1}{2}m{v}_L^2⟹2mgL+2mgL=\frac{1}{2}m{v}_L^2⟹v_L^2=8gL$$

At the lowest point, let the speed be $v_L$ and tension be $T_L$. The equation of motion is $T_L-mg=m{v}_L^2/L$.$$T_L=mg+\frac{m(8gL)}{L}=mg+8mg=9mg$$ The ratio of tension at the lowest point to the highest point is $T_L/T_H=(9mg)/(3mg)=3$.