Two pendulums suspended from same point having lengths 2m and 0.5m. If they displaced slightly and released, then they will be in same phase, when small pendulum will have completed :

A
2 oscillations
B
4 oscillations
C
3 oscillations
D
5 oscillations

Solution:

$(N + 1) T_{s} = {NT}_{l} becuase T \propto \sqrt{l} \frac{N + 1}{N} = \sqrt{\frac{l_{l}}{l_{s}}} = \sqrt{\frac{2}{0.5}} = 2 \Rightarrow N = 1 and N + 1 = 2 .$

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Solution:

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