question_answer

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

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

B)         \[mg(1+{{\theta }_{0}})\]           

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

D)         \[mg(1+\theta _{0}^{2})\]

Correct Answer: D

Solution :

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 vertical distance \[h\] \[\therefore \]  \[\cos {{\theta }_{0}}=\frac{l-h}{l}\] \[\Rightarrow \]               \[h=l(1-cos{{\theta }_{0}})\]       ?(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 \[{{T}_{\max }}=mg+\frac{2mg}{l}l(1-\cos \,{{\theta }_{0}})\] \[=mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right]\]                 \[=mg(1+\theta _{0}^{2})\]