Question

The mass of the bob of a simple pendulum of length $L$ is $m$. If the bob is left from its horizontal position then the speedof the bob and the tension in the thread in the lowest position of the bob will be respectively:

• A. $\sqrt{2gL}and3mg$
• B. $3mgand\sqrt{2gL}$
• C. $2mgand\sqrt{2gL}$
• D. $2gLand3mg$

Answer 1

The correct option is A $\sqrt{2gL}and3mg$

Velocity of the bob at point A is zero, thus its kinetic energy at A is zero.

Let the velocity of the bob at point B be $v$ and tension in the string be $T$.

Using work-energy theorem from A to B : $W=\Delta K.E$

$\therefore$ $mgL=\frac{1}{2}m{v}^2-0$ $⟹v=\sqrt{2gL}$

Circular motion equation at B : $\frac{m{v}^2}{L}=T-mg$

$\therefore$ $T=\frac{m{v}^2}{L}+mg=\frac{m\left(2gL\right)}{L}+mg=3mg$