Question

Two pendulums of lengths 100 cm and 110.25 cm start oscillating in phase. After how many oscillations will they again be in the same phase?

Answer 1

$T=2\pi \sqrt{\frac{l}{g}},l_1=100cm,l_2=110.25cm.$
For smaller pendulum, $T_1=2\pi \sqrt{\frac{100}{g}}$
For larger pendulum, $T_2=2\pi \sqrt{\frac{110.25}{g}}$
Let these pendulums oscillate in phase. If larger pendulum completes one oscillations, it means smaller pendulum must complete (n + 1) oscillations.
$\therefore n{T}_2=(n+1)T_1$
Or $\frac{n+1}{n}=\frac{T_2}{T_1}=\sqrt{\frac{110.25}{100}}=1.05$
or $1+\frac{1}{n}=1.05or\frac{1}{n}=0.05=\frac{5}{100}=\frac{1}{20}$
$\therefore n=20$
Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.