Rankers Physics

miscellaneous: Practice Problem & Solution

Two pendulums of length $121 \text{ cm}$ and $100 \text{ cm}$ start vibrating in phase. At some instant, the two are at their means position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the means position is : (2022)
$8$
$11$
$9$
$10$

Solution Explained:

To solve this problem, we apply the core principles of miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:

$T \propto \sqrt{l}$. So $T_1/T_2 = \sqrt{121/100} = 11/10$. This gives $10 T_1 = 11 T_2$. The shorter pendulum ($T_2$) completes $11$ vibrations.

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