question_answer

A simple pendulum of length \[l\] is moved aside till the string makes an angle \[{{\theta }_{1}}\], with the vertical. If the acceleration due to gravity is g, the kinetic energy of the bob when the string is inclined at \[{{\theta }_{2}}\] to the vertical is

A)  \[mgl\,\cos \,\,({{\theta }_{1}}-{{\theta }_{2}})\]    

B)  \[mgl\,\,(\cos \,{{\theta }_{2}}-\cos {{\theta }_{1}})\]

C)  \[mgl\,\,(\cos \,{{\theta }_{1}}-\cos \,{{\theta }_{2}})\]

D)  \[mgl\,\,\sin \,({{\theta }_{1}}-{{\theta }_{2}})\]

Correct Answer: B

Solution :

 PE at the point B = TE \[=mg\times AP=mg\,(l-l\,\cos \,{{\theta }_{1}})\] \[(\because \,KE=0\,\,at\,B)\] PE at the point \[C=mg\times AQ=mg\,(l-l\,\cos \,{{\theta }_{2}})\] If KE of the bob at the point C is E, then according to law of conservation of energy, \[E+mg\,(l-l\,\cos \,{{\theta }_{2}})=mg\,(l-l\,\cos \,{{\theta }_{1}})\] \[E=mgl\,(\cos \,{{\theta }_{2}}-\cos \,{{\theta }_{1}})\]     (Here E = KE)