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 speed of the bob and the tension in the thread in the lowest position of the bob will be respectively.

• A. $\sqrt{2gL}$ and $3mg$
• B. $\sqrt{gL}$ and $3mg$
• C. $\sqrt{2gL}$ and $4mg$
• D. $\sqrt{gL}$ and $2mg$

Answer 1

The correct option is A $\sqrt{2gL}$ and $3mg$

The given condition can be shown as

Let the speed of the bob at lowest position be $v$
By the conservation of energy
Potential energy at point $A=$ Kinetic energy at point $B$
$\Rightarrow mgL=\frac{1}{2}m{v}^2$
$\Rightarrow v=\sqrt{2gL}$
Also, for tension in the string we have force equation as
$T=mg+\frac{m{v}^2}{L}=mg+\frac{m}{L}(\sqrt{2gL}{)}^2=3mg$