Explanation

(d) The simple pendulum at angular amplitude $\theta_{0}$ is shown in the figure.
Maximum tension in the string is
$T_{\max} = mg + \frac{mv^{2}}{l}$ …(i)
When bob of the pendulum comes from $A$ to $B$, it covers a vertical distance $h$
[image:14_image76.emf]
$\therefore$ $\cos{\theta_{0} = \frac{l - h}{l}}$
$\Rightarrow$ $h = l(1 - \cos\theta_{0})$ …(ii)
Also during $A$ to $B$, potential energy of bob converts into kinetic energy $ie$, $mgh = \frac{1}{2}\ mv^{2}$
$\therefore$ $v = \sqrt{2gh}$ …(iii)
Thus, using Eqs. (i),(ii) and (iii), we obtain
$T_{\max} = mg + \frac{2mg}{l}l(1 - \cos{\theta)}$
$= mg + 2mg\left\lbrack 1 - 1 + \frac{\theta_{0}^{2}}{2} \right\rbrack$
$= mg(1 + \theta_{0}^{2})$