Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum - Physics MCQ NEET 2022 2026 | ExamTest.live

hardMCQNEET 20222026Physics

1 mark

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

Solution & Step-by-step Explanation

Time period T \propto l . \frac{T _{1}}{T _{2}} = \frac{121}{100} = \frac{11}{10} Let n_{1} be vibrations of longer pendulum and n_{2} be vibrations of shorter pendulum. For them to be in phase: n_{1} T_{1} = n_{2} T_{2} ⟹ \frac{n _{2}}{n _{1}} = \frac{T _{1}}{T _{2}} = \frac{11}{10} The minimum vibrations of the shorter pendulum (n_{2}) is 11 .

Try it yourself before checking the explanation above.

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

11

9

10

8

Practice with timed mock tests and track your performance across Physics.