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)

The simple pendulum at angular amplitude $\theta_{n}$ is shown in figure.
Maximum tension in the string is
$T_{\max }=m g+\frac{m v^{2}}{l} \ldots(i)$
When bob of the pendulum comes form $A$ to $B$, it covers a vertical distance $h$
$\therefore \cos \theta=\frac{l-h}{h}$
$\Rightarrow h=l\left(1-\cos \theta_{0}\right) \ldots (ii)$
Also, during $B$ to $A$, potential energy of bob converts into kinetic energy, i.e.
$m g h=\frac{1}{2} m v^{2}$
$\therefore v=\sqrt{2 g h}\dots (iii)$
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) $
$=m g+2 m g\left[1-1+\frac{\theta_{0}^{2}}{2}\right]$
$=m g\left(1+\theta_{0}^{2}\right)$