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

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Solution

Q.
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 [2022]

1 11

2 9

3 10

4 8

Ans.
(1)

Let $L_{1} = 121$ $cm = \frac{121}{100} = 1.21$ $m$

$L_{2} = 100$ $cm = \frac{100}{100} = 1$ $m$

Let $T_{1} =$ longer pendulum ; $T_{2} =$ shorter pendulum

We know, $T = 2 π \sqrt{\frac{L}{g}} \Rightarrow T \propto \sqrt{L}$

$\frac{T_{1}}{T_{2}} \propto \sqrt{\frac{L_{1}}{L_{2}}}$

$\frac{T_{1}}{T_{2}} \propto \sqrt{\frac{1.21}{1}} = \frac{1.1}{1}$

$10 T_{1} = 11 T_{2}$

10 vibrations of longer pendulum = 11 vibrations of shorter pendulum.

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