Two simple pendulums of length 1 m and 16 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed n oscillations. The value of n is

$T_1 = 2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}, {\mathrm{T}}_2 = 2\mathrm{\pi }\sqrt{\frac{16}{\mathrm{g}}} = 4{\mathrm{T}}_1$

at any time t, phase of pendulums are:

$\varnothing { }_1 = {\mathrm{\omega }}_1\mathrm{t} = \frac{2\mathrm{\pi } }{{\mathrm{T}}_1}\mathrm{t}, {\varnothing }_2= {\mathrm{\omega }}_2\mathrm{t} = \frac{2\mathrm{\pi }}{{\mathrm{T}}_2}\mathrm{t}$

First pendulum is faster. Both will be in same phase again when faster pendulum has completed one oscillation more than slower pendulum

${\varnothing }_1-{\varnothing }_2 = 2\mathrm{\pi }$

$\Rightarrow \frac{2\mathrm{\pi }}{T_1}t-\frac{2\mathrm{\pi }}{4T_1}t = 2\mathrm{\pi } \Rightarrow \mathrm{t} = \frac{4{\mathrm{T}}_1}{3}$

No. of oscillations completed by shorter pendulum in time t:

$\mathrm{n} = \frac{\mathrm{t}}{{\mathrm{T}}_1 } = \frac{4}{3}$