NEETPhysicsOscillationsMCQ+4 / −1
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 :
- A11
- B9
- C10
- D8
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Correct answer: A
$$T = 2\pi \sqrt {{L \over g}} $$
Let n1 and n2 be integer.
$${n_1}{T_1} = {n_2}{T_2}$$
$$2\pi {n_1}\sqrt {{{1.21} \over g}} = 2\pi {n_2}\sqrt {{{1.00} \over g}} $$
$$ \Rightarrow {{{n_2}} \over {{n_1}}} = {{11} \over {10}}$$
$\therefore$ After completion of 11th oscillation of shorter pendulum, it will be in phase with longer pendulum.
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