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 aft…

Question

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 [NEET 2022]

Accepted Answer

Correct answer: D. (d) We known that, T=2 l/g ∵ T_1T_2= l_1l_2= 1.211=1.1= 1110 ⇒10T_1=11T_2 ∴10 vibrations of longer pendulum =11 vibrations of shorter pendulum ∴ Minimum number of vibrations of shorter pendulum =11

$Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is [NEET 2022]$

Solution

More Simple Harmonic Motion PYQs