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:                                                                                                          [AIPMT 2000]

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

\[v=\sqrt{2gl(1-\cos \theta )}\]