Two simple pendulums A and B having lengths $\ell \text{ and }\ell /4,$ respectively are released from the position as shown in the following figure. Calculate the time (in seconds) after which the two strings become parallel for the first time. (assume they have same angular amplitude)

$\left(\mathrm{Take} \ell =\frac{90}{{\pi }^2}\mathrm{m}\text{ and }\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)$

The angular positions of pendulums 1 and 2 are (taking angles to the right of reference line 
x x' to be positive)

${\mathrm{\theta }}_1=\mathrm{\theta cos}\left(\frac{4\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)\left(\text{ where }\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{\ell }}{\mathrm{g}}}\right)$

${\mathrm{\theta }}_2=-\mathrm{\theta cos}\left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)=\cos \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}+\mathrm{\pi }\right)$For the strings to be parallel for the first time

${\mathrm{\theta }}_1={\mathrm{\theta }}_2\cos \left(\frac{4\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)=\cos \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}+\mathrm{\pi }\right)$

$\therefore \frac{4\mathrm{\pi }}{\mathrm{T}}\mathrm{t}=2\mathrm{n\pi }\pm \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}+\mathrm{\pi }\right)$

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

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