question_answer

                A pendulum is displaced to an angle \[\theta \] from its equilibrium position; then it will pass through its mean position with a velocity v equal to:                                                                                                              

A)                 \[\sqrt{2gl}\]     

B)                 \[\sqrt{2gl\sin \theta }\]              

C)                 \[\sqrt{2gl\cos \theta }\]             

D)                 \[\sqrt{2gl\,\,(1-\cos \theta )}\]

Correct Answer: D

Solution :

                Key Idea: The potential energy at an angular displacement \[\theta \] will be converted to kinetic energy at mean position.                 If \[l\] be is the length of pendulum and \[\theta \] die angular amplitude, then height                 \[h=AB-AC\]                 \[=l-l\cos \theta \]                 \[=l(1-\cos \theta )\]                 At point P (maximum displacement position i.e., extreme position), potential energy is maximum and kinetic energy is zero. At point B (mean or equilibrium position) potential energy is minimum and kinetic energy is maximum, so from principle of conservation of energy.                 (PE + KE) at P = (KE + PE) at B or            \[mgh+0=\frac{1}{2}m{{v}^{2}}+0\] or            \[v=\sqrt{2gh}\]                                               .....(ii)                 Substituting the value of h from Eq. (i) into Eq. (ii), we get                 \[v=\sqrt{2gl(1-\cos \theta )}\]