Question

The angular amplitude of simple pendulum is ${\theta }_0$. The maximum tension in its string will be (assume ${\theta }_0$ is small)

• A. $mg(1+{\theta }_0)$
• B. $mg(1-{\theta }_0)$
• C. $mg(1+{\theta }_0^2)$
• D. $mg(1-{\theta }_0^2)$

Answer 1

The correct option is C $mg(1+{\theta }_0^2)$

$\text{Given- * Small angular displacement of Pendulum =}{\theta }_0$

$\begin{array}{cc}\text{we get}h& =l(1-\cos {\theta }_0)=l[1-(1-\frac{{\theta }_0^2}{2})]\\ & h\approx \frac{\ell {\theta }_0^2}{2}\end{array}$

$\begin{array}{c}\text{at the lowest point the pendulum will have some}\\ \text{velocity (v) let.}\end{array}$

$\begin{array}{c}\Rightarrow \text{By energy conservation}\\ \begin{array}{c}\frac{1}{2}m{v}^2=mgh=mg\frac{\ell {\theta }_0^2}{2}\\ \Rightarrow v^2=gl\cdot {\theta }_0^2\end{array}\end{array}$

$\begin{array}{c}\text{Tension in the string}=mg+\frac{m{v}^2}{l}=mg+mg{\theta }_0^2\\ \Rightarrow T_{max}=mg[1+{\theta }_0^2]\end{array}$