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)), theta(1) = theta(2)`.

Answer

`(pi)/(3)sqrt((l)/(g))`

Explanation

The angular position of pendulum `1` and `2` are (taking angles to the right of reference line `xx'` to be positive)
`theta_(1) = theta cos((4pi)/(T)t)`, [where `T = 2pi sqrt((l)/(g))`]
`theta_(2) = -theta cos ((2pi)/(T))t = cos ((2pi)/(T) t + pi)`
`F` for the strings to be parallel for the first time
or `cos((4pi)/(T)t) = cos ((2pi)/(T)t + pi)` ltbr. `:. (4pi)/(T)t = 2n pi +- ((2pi)/(T)t + pi)`
for `n = 0, t = (T)/(2)`
for `n = 1, t = (T)/(6), (3T)/(2)`
`:.` Both the pendulum are parallel to each other for the first time after `t = (T)/(6) = sqrt((l)/(g))`