question_answer

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle \[\theta \] and then released. The bob will then pass through its equilibrium position with a speed v, where v equals                                                  [Haryana CEE 1996]

A)            \[U=K{{X}^{2}}\]                

B)            \[\sqrt{2gl(1+\cos \theta )}\]

C)            \[U=KX\]                               

D)            \[\frac{a}{2}\]

Correct Answer: C

Solution :

                   If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position                     \[\Rightarrow mgh=\frac{1}{2}mv_{\max }^{2}\]\[\Rightarrow {{v}_{\max }}=\sqrt{2gh}\]                                Also, from figure \[\cos \theta =\frac{l-h}{l}\]                    \[\Rightarrow h=l(1-\cos \theta )\] So, \[{{v}_{\max }}=\sqrt{2gl(1-\cos \theta )}\]