Question

Two simple pendulums of length `1m` and `16m` respectively are both given small displacements in the same direction at the same instant. After how many oscillations of the shorter pendulum will, the two pendulums vibrate in the same phase ?

Answer 1

Correct Answer - 4
Time period, `T=2pisqrt((l)/(g))` or `T propsqrt(l)`
`:. (T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt((16)/(1))=4` or `T_(2)=4T_(1)`
It means, when a pendulum of smaller length will complete 4 oscillations, the pendulum of large length will complet 1 oscillation. It means, the two pendulums will be in the same phase, when shorter has completed 4 oscillations.