Two simple pendulums of length l m and 16 m start at t =0 from mean position. The minimum time after which they will be again in phase will be

 The phase of simple pendulum is given by $\mathrm{\theta }=\mathrm{\omega t}$
$$\begin{gathered} \text{ where }\mathrm{\omega }\text{ is angular speed. } \\ \text{ Phase of first pendulum }={\mathrm{\omega }}_1\mathrm{t} \\ \text{ Phase of second pendulum }={\mathrm{\omega }}_2\mathrm{t} \\ \text{ Phase difference }=2\mathrm{n\pi } \\ {\mathrm{\omega }}_1\mathrm{t}-{\mathrm{\omega }}_2\mathrm{t}=2\mathrm{n\pi } \\ \mathrm{or} \frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}-\frac{2\mathrm{\pi }}{4\mathrm{T}}\mathrm{t}=2\mathrm{n\pi } \\ \left(\because {\mathrm{\omega }}_1=2\mathrm{\pi }/\mathrm{T}\text{ and }{\mathrm{\omega }}_2=2\mathrm{\pi }/4\mathrm{T}\right) \\ \mathrm{Puttirg} \mathrm{n} = 1, \mathrm{we} \mathrm{get} \\ \mathrm{t}\left[\frac{1}{\mathrm{T}}-\frac{1}{4\mathrm{T}}\right]=1\text{ or }\mathrm{t}\left[\frac{3}{4\mathrm{T}}\right]=1\text{ or }\mathrm{t}=\frac{4\mathrm{T}}{3} \end{gathered}$$