Two simple pendulums of lengths 1.21m and 1.00 m start swinging at the same time from a location. After how much time, they will be (a) out of phase and (b) in phase again? Take `g=10m//s^(2)`.

Explanation

Let `T_(1)`and `T_(2)` be the time periods of the two pendulums of lengths `l_(1)` and `l_(2)` , where `l_(1)=1.21m` and `l_(2)=1.00m` Then,
`T_(1)=2pisqrt((l)/(g))` and `T_(2)=2pisqrt((l_(2))/(g))`
`:. (T_(1))/(T_(2))=sqrt((l_(1))/(l_(2)))=sqrt((1.21)/(1.00))=1.1 ` ...(i)
(a) Let the two pendulums go out of phase after time t.Then, during this time t, if the pendulum of loonger length `l_(1)` (smaller frequency ) makes n vibrations, the other pendulum of smaller lendgh `l_(2)` (larger frequency) should make `(n+1//2)` vibrations, i.e., `1//2` vibration extra. Therefore,
`t=nT_(1)=(n+1//2)T_(2)`
or `(T_(1))/(T_(2))=(n+1//2)/(n)` ...(ii)
Form (i) and (ii),
`1.1=(n+1//2)/(n) or n=5` vibrations
`:.t=nT_(1)=5xx2pisqrt((1.21)/(10))=10.93s`
(b) Let the two pendulums are again in phase after time `t^(')`. Then during this time `t^(')` , the pendulum of length `l_(1)` if makes `n^(')` vibrations, the pendulum of length `l^(2)` will make `(n^(')+1)` vibrations. Therefore,
`t^(')=n^(')T_(1)=(n^(')+1)T_(2)`
or `(T_(1))/(T_(2))=(n^(')+1)/(n^('))` ...(iii)
From (i) and (iii) , we have `(n^(')+1)/(n^('))=1.1`
or `n^(')=10` vibrations
`:. t^(')=n^(')T_(1)=10xx2pisqrt((1.21)/(10))=21.86s`