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 \[{{T}_{\max }}=mg+\frac{2mg}{l}l(1-\cos \,{{\theta }_{0}})\] \[=mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right]\] \[=mg(1+\theta _{0}^{2})\]You need to login to perform this action.
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