Question

Two pendulums of lengths $121\text{cm}$ and $100\text{cm}$ 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 at the mean position?

Answer 1

Time period of pendulum of length $121\text{cm}$ is
$T_1=2\pi \sqrt{\frac{121}{g}},$
Time period of pendulum of length $100\text{cm}$ is
$T_2=2\pi \sqrt{\frac{100}{g}}$
Let shorter pendulum makes n vibrations; then the longer will make one less than $n$ to come in phase again.
$n{T}_2=(n-1)T_1$
$n.2\pi \sqrt{\frac{100}{g}}=(n-1)2\pi \sqrt{\frac{121}{g}}$
$n=11$