Question

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

• A. $mg\left(1-{\theta }_0\right)$
• 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. Maximum tension in the string is
$T_{\max }=mg+\frac{m{v}^2}{l}...(i)$
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, $mgh=\frac{1}{2}m{v}^2$
$\therefore v=\sqrt{2gh}...(iii)$
Thus, using Eqs. (i), (ii) and (iii), we obtain
$T_{\max }=mg+\frac{2mg}{l}l\left(1-\cos {\theta }_0\right)$
$=mg+2mg\left[1-1+\frac{{\theta }_0^2}{2}\right]$
$=mg\left(1+{\theta }_0^2\right)$