Two pendulums of lengths 16l and l are in phase at the mean position at a certain time. If ‘T’ is the time period of shorter pendulum, the minimum time after which they will be again in phase is

Time period of shorter pendulum ${\text{T}}_{\text{1}}=\text{T}$

Long pendulum length $l_2=16l$

Time period of longer pendulum ${\text{T}}_2=?$

$\text{T}\propto \sqrt{l};{\text{T}}_2=4\text{T}$

In phase means $\Delta \varphi =\text{2}\pi$  radian

$\therefore \Delta \varphi =\text{2}\pi \left(\frac{\text{1}}{{\text{T}}_{\text{1}}}+\frac{1}{{\text{T}}_{\text{2}}}\right)\text{t}$

$\text{2}\pi =\text{2}\pi \left(\frac{\text{1}}{\text{T}}-\frac{1}{\text{4T}}\right)\text{t}$

$\Rightarrow \text{t}=\frac{4}{3}\text{T}$