Question

The angular amplitude of a simple pendulum is $\theta_{0}$. The maximum tension in its string will be

• A. $m g\left(1-\theta_{0}\right)$
• B. $m g\left(1+\theta_{0}\right)$
• C. $m g\left(1-\theta_{0}^{2}\right)$
• D. $m g\left(1+\theta_{0}^{2}\right)$

Answer 1

Answer: (D)

$m g\left(1+\theta_{0}^{2}\right)$

Answer 2

The simple pendulum at angular amplitude $\theta_{0}$ is shown in the figure. Maximum tension in the string is
$T_{\max }=m g+\frac{m v^{2}}{l}\,\,\,...(i)$
When bob of the pendulum comes from $A$ to $B$, it covers a vertidal distance $h$

$\therefore \cos \theta_{0}=\frac{l-h}{l}$
$\Rightarrow h=l\left(1-\cos \theta_{0}\right)\,\,\,...(ii)$
Also during $A$ to $B$, potential energy of bob converts into kinetic energy ie, $m g h=\frac{1}{2} m v^{2}$
$\therefore v=\sqrt{2 g h}\,\,\,...(iii)$
Thus, using Eqs. (i), (ii) and (iii), we obtain
$T_{\max } =m g+\frac{2 m g}{l} l\left(1-\cos \theta_{0}\right)$
$=m g+2 m g\left[1-1+\frac{\theta_{0}^{2}}{2}\right]$
$=m g\left(1+\theta_{0}^{2}\right)$