• # question_answer                 Two simple pendulums of length 0.5 m and 2.0 in respectively are given small linear displacement in one direction at die same time. They will again be in the same phase when the pendulum of shorter length has completed oscillations: A)                 5                                              B)                 1                              C)                 2              D)                 3

Key Idea: For the pendulum to be again in the same phase, there should be difference of 1 complete oscillation.                 If smaller pendulum completes n oscillations the larger pendulum will complete $(n-1)$ oscillations, so                 Time period of n oscillations of first = Time period of $(n-1)$ oscillations of second                 i.e.,        $n{{T}_{1}}=(n-1){{T}_{2}}$ or            $n\,2\,\pi \,\sqrt{\frac{{{L}_{1}}}{g}}=(n-1)\,2\,\pi \sqrt{\frac{{{L}_{2}}}{g}}$ or            $n\sqrt{{{L}_{1}}}=(n-1)\,\sqrt{{{L}_{2}}}$ or            $\frac{n}{n-1}=\sqrt{\frac{{{L}_{2}}}{{{L}_{1}}}}=\sqrt{\frac{2.0}{0.5}}$ or            $\frac{n}{n-1}=2$ or            $n=2n-2$ $\therefore$  n = 2