Question

A simple pendulum of length $l$ and bob of mass $m$ is displaced from its equilibrium position $O$ to a position $P$ so that height of $P$ above $O$ is $h$. It is then released. What is the tension in the string when the bob passe through the equilibrium position $O$? Neglect friction. $v$ is the velocity of the bob at $O$.

• A. $m(g+\frac{v^2}{l})$
• B. $\frac{2mgh}{l}$
• C. $mg(1+\frac{h}{l})$
• D. $mg(1+\frac{2h}{l})$

Answer 1

The correct option is D $mg(1+\frac{2h}{l})$

Apply conservation of energy at A and B, $\frac{1}{2}m{v}^2+0=0+mgh$ ($v=0$ at B)

$⟹v^2=2gh$

Now from $e{q}^n$ of motion at A, $m{a}_c=T-mg$ where $a_c=\frac{v^2}{l}=\frac{2gh}{l}$

$T=mg+m\frac{2gh}{l}$

$T=mg(1+\frac{2h}{l})$