A bob of a simple pendulum of mass ' $\mathrm{m}$ ' is displaced through $90^{\circ}$ from mean position and…
A bob of a simple pendulum of mass ' $\mathrm{m}$ ' is displaced through $90^{\circ}$ from mean position and released. When the bob is at lowest position, the tension in the string is
4mg
2mg
mg
3mg
Solution
When it is displaced through $90^{\circ}$ from mean position, it is at a height ' $r$ ' and has potential energy mgr. At the lowest position this potential energy is converted into kinetic energy.
$\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{mgr} \\
& \therefore \mathrm{mv}^2=\mathrm{mgr}
\end{aligned}$
At the lowest position the tension in the string
$\mathrm{T}=\mathrm{mg}+\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{mg}+2 \mathrm{mg}=3 \mathrm{mg}$