question_answer

   A simple pendulum with a bob of mass m swings with an angular amplitude of \[60{}^\circ \] , when its angular displacement is \[30{}^\circ \], the tension of string would be

A) \[3\sqrt{3}\,mg\]                     

B) \[\frac{1}{2}\,mg(3\sqrt{3}-2)\]

C) \[\frac{1}{2}\,mg(3\sqrt{3}+2)\]      

D) \[\frac{1}{2}\,mg(3-\sqrt{2})\]   

Correct Answer: B

Solution :

As shown in the figure in going from position C to B, the bob loses a vertical height ED =h Loss in \[PE=mgl\] \[mgl[(1-\cos \cos {{\theta }_{1}})-(1-\cos \cos {{\theta }_{2}})]\] \[=mgl[coscos{{\theta }_{2}}-coscos{{\theta }_{1}}]\] Gain in \[KE=\frac{1}{2}m{{v}^{2}}\] As gain in KE=loss in PE \[\therefore \]\[\frac{1}{2}m{{v}^{2}}=mg(coscos{{\theta }_{2}}-coscos{{\theta }_{1}})\] \[\frac{m{{v}^{2}}}{l}=2mg(\cos \cos {{\theta }_{2}}-\cos \cos {{\theta }_{1}})\]       (1) At the position B, \[T=mg\cos {{\theta }_{2}}+\frac{m{{v}^{2}}}{l}\] Using (i), \[T=mg\cos {{\theta }_{2}}+2mg(\cos \cos {{\theta }_{2}}-\cos \cos {{\theta }_{1}})\] \[=mg[3cos{{\theta }_{2}}-2cos{{\theta }_{1}}]\] \[=mg[3\cos \cos {{30}^{o}}-2\cos \cos {{60}^{o}}]\] \[=mg\left[ \frac{3\sqrt{3}}{2}-2\times \frac{1}{2} \right].\] \[T=\frac{1}{2}mg\left[ 3\sqrt{3}-2 \right]\]newton Hence, the correction option is [b].