Two simple pendulums `A` and `B` having lengths `l` and `(l)/(4)` respectively are released from the position as shown in Fig. Calculate the time (in seconds) after which the two strings become parallel for the first time. (Take `l=(90)/(pi^2)`m and `g=10(m)/(s^2)`.

Answer

1

Explanation

Let `T = 2pi sqrt((l)/(g)), theta_(1) = theta_(2)`
`theta cos ((4pi)/(T)t) =- theta cos ((2pi)/(T)t), (4pi)/(T)t =2npi +- ((2pi)/(T)t+pi)`
for `n=0, t = T//2` for `n =1, tT//6, 3T//2`
the first time meeting, `t = (T)/(6) = (pi)/(3) sqrt((l)/(g)) =1s`