First slide

Simple hormonic motion

Question

Two pendulums of lengths 121cm and 100cm start vibrating. At some instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position?

Moderate

A

10
B

11
C

20
D

21

Solution

$T_{1} = 2 π \sqrt{\frac{121}{g}}, T_{2} = 2 π \sqrt{\frac{100}{g}}$
Let shorter pendulum makes n vibrations; then the longer will make one less than n to come in phase again.
$n T_{2} = (n - 1) T_{1}$
OR
$n .2 π \sqrt{\frac{100}{g}} = (n - 1) 2 π \sqrt{\frac{121}{g}}$
Solving, we get; $n = 11$

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