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)`.

Explanation

The angular positions of pendulums 1 and 2 are (takeing angles to the right of reference line `xx'` to be positive)
`theta_1=thetacos((4pi)/(T)t)` (where `T=2pisqrt((l)/(g))`
`theta_2=-thetacos((2pi)/(T)t)=cos((2pi)/(T)t+pi)`
For the strings to be parallel for the first time
`theta_1=theta_2`
`cos((4pi)/(T)t)=cos((2pi)/(T)t+pi)`
`(4pi)/(T)t=2npi+-((2pi)/(T)t+pi)`
for `n=0`,`t=(T)/(2)`
for `n=1` ,`t=(T)/(6)`,`(3T)/(2)`
Both the pendulums are parallel to each other for the first time after
`t=(T)/(6)=(pi)/(3)sqrt((l)/(g))=1s`