Question

The angular amplitude of a simple pendulum is ${\theta }_0$ The maximum-tension in its string will be

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

Answer 1

Answer: (D)

The simple pendulum at angular amplitude ${\theta }_0$ is shown in the figure.
$T_{\max }=mg+\frac{m{v}^2}{l}...$ (i)
When bob of pendulum comes from $A$ to $B$ , it covers a vertical distance h.

$\therefore \cos {\theta }_0=\frac{l-h}{l}$
$h=l(1-\cos {\theta }_0)$ ... (ii)
Also during $A$ to $B$ , potential energy of bob converts into kinetic energy ie, $mgh=\frac{1}{2}m{v}^2$
$\therefore$ $v=\sqrt{2gh}$ ... (iii)
Thus, using Eqs. (i), (ii) and (iii), we obtain
$T_{\max }=mg+\frac{2mg}{l}l(1-\cos {\theta }_0)$
$=mg+2mg\left[1-1+\frac{{\theta }_0^2}{2}\right]$
$=mg\left(1+{\theta }_0^2\right)$