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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