question_answer

The angular amplitude of a simple pendulum is \[{{}_{0.}}\]The maximum-tension in its string will be

A)  \[\text{mg(1-}{{}_{0.}})\]                           

B)  \[\text{mg}\,\text{(1+}{{}_{0.}})\]

C)  \[\text{mg}\,\text{(1-}_{0}^{2})\]          

D)         \[\text{mg}\,\text{(1+}_{0}^{2})\]

Correct Answer: D

Solution :

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                 \[{{T}_{\max }}=mg+\frac{2mg}{l}l(1-\cos {{\theta }_{0}})\]                 \[=mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right]\]