NEET 2022, Physics Q36: SHM in Pendulums

NEET2022Physics

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 :

A
9
B
10
C
8
D
11

Solution

For the two pendulums to be in phase together at the mean position, the shorter one must complete one extra oscillation: $n T_{ℓ} = (n + 1) T_{s}$ .

Period ratio: $\frac{T _{ℓ}}{T _{s}} = \frac{121}{100} = \frac{11}{10}$ .

Hence $\frac{n + 1}{n} = \frac{11}{10} \Rightarrow n = 10$ (oscillations of the longer pendulum).

Number of oscillations of the shorter pendulum: $n + 1 = 11$ .

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NEET 2022, Physics Q36: SHM in Pendulums

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