Two pendulums of length 121 ,cm and 100 ,cm start vibrating in ph… - Vidyadip

MCQ

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
$8$
✓
$11$

✓

Answer

Correct option: D.

$11$

d
$( n ) T _{\ell}=( n +1) T _{ s }$

$( n ) 2 \pi \sqrt{\frac{1.21}{ g }}=( n +1) 2 \pi \sqrt{\frac{1}{ g }}$

$( n )(1.1)=( n +1)$

$0.1( n )=1$

$n =10$

No. of oscillation of smaller one

$= n +1$

$=10+1$

$=11$

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