Two simple pendulum `A` and `B` of lengths `1.69m` and `1.44m` start swinging at the time from a location where acceleration due to gravity is `10ms^(-1)`. Answer the following question.
How many vibratinos pendulum `A` will compelte when it will be out of phase with `B` ?

Options

A. 4
B. 5
C. 6
D. 8

Answer

C

Explanation

Let `T_(1)` and `T_(2)` be the time period of vibrations of pendulums `A` and `B` respectively.
Then, `T_(1)=2pisqrt((l_(1))/(y))` and `T_(2)=2pisqrt((l_(2))/(g))`
or `(T_(1))/(T_(2))=sqrt((l_(1))/(l_(2)))=sqrt((1.69)/(1.44))=(13)/(12)`
If the two pendulums go out of phase in time `t`, then in time `t`, if pendulum `A` completes `n` vibrations, the pendulum `B` will complete `(n+1//2)`vibrations. Therefore,
`t=nT_(1)=(n+1//2)T_(2)`
or `(T_(1))/(T_(2))=((n+1//2))/(n)=(13)/(12)`
or `12n+6=13n` or `n=6`