Solution:
Time period $ T \propto \sqrt{l} $. $ \frac{T_1}{T_2} = \sqrt{\frac{2}{0.5}} = 2 $, so $ T_1 = 2T_2 $. They return to the same phase when the longer pendulum completes 1 oscillation and the shorter one completes 2 oscillations.
Two pendulums suspended from same point having length 2 m and 0.5 m. If they displaced slightly and released then they will be in same phase, when small pendulum will have completed: (1998)
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