Question

Two simple pendulums of lengths $1$ meter and $16$ meter respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed $n$ oscillations where $n$ is -

Answer 1

Answer: 1.33

Given ${\lambda }_1=1m$ and ${\lambda }_2=16m$
$T_1=2\pi \sqrt{\frac{1}{g}},T_2=2\pi \sqrt{\frac{16}{g}}$
$\Rightarrow \frac{T_2}{T_1}=4$ or $\frac{n_1}{n_2}=4$
$\frac{n_1}{n_2}-1=3$
$\Rightarrow \frac{n_1-n_2}{n_2}=3$
Two pendulums will swing together if
$n_1-n_2=1.$
$\Rightarrow \frac{1}{n_2}=3$
or $n_2=\frac{1}{3}$
$n_1=1+n_2=1+\frac{1}{3}$
$=\frac{4}{3}$