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. $m g\left(1+\theta_{0}\right)$
• 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. ${{T}_{\max }}=mg+\frac{m{{v}^{2}}}{l} \,\,\,...$ (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 ${{T}_{\max }}=mg+\frac{2mg}{l}l(1-\cos {{\theta }_{0}})$ $=mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right]$ $=m g\left(1+\theta_{0}^{2}\right)$