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