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 \[v=\sqrt{2gl(1-\cos \theta )}\]You need to login to perform this action.
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