Two simple pendulums A and B having lengths l and, l/4 respectively are released from the position as shown in figure. The time after which the release of the two strings become parallel for the first time is_______. [Take $l=\frac{90}{{\pi }^2} and g=10m{s}^{-2}$]

The angular position of pendulum I and 2 are(taking angles to the right of reference line xx' to be positive)

$\begin{array}{l}{\theta }_1=\theta cos\left(\frac{4\pi }{T}t\right) \left[\text{ where }T=2\pi \sqrt{\frac{l}{g}}\right]\\ {\theta }_2=-\theta cos\left(\frac{2\pi }{T}t\right)=cos\left(\frac{2\pi }{T}t+\pi \right)\end{array}$

For the strings to be parallel for the first time, ${\theta }_1={\theta }_2$

$\begin{array}{l}\cos \left(\frac{4\pi }{T}t\right)=\cos \left(\frac{2\pi }{T}t+\pi \right)\\ \therefore \frac{4\pi }{T}t=2n\pi \pm \left(\frac{2\pi }{T}t+\pi \right)\\ \text{ for }n=0,t=\frac{T}{2};\text{ for }n=1,t=\frac{T}{6},\frac{3T}{2}\end{array}$

Both the pendulum are parallel to each other for the first time

after $t=\frac{T}{6}=\frac{\pi }{3}\sqrt{\frac{l}{g}}=1\mathrm{s}$