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

Given $\lambda_{1}=1 m$ and $\lambda_{2}=16\, m$
$T _{1} =2 \pi \sqrt{\frac{1}{ g }}, T _{2}=2 \pi \sqrt{\frac{16}{ g }}$
$\Rightarrow \frac{ T _{2}}{ T _{1}}=4$ 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 $n _{2}=\frac{1}{3}$
$n _{1}=1+ n _{2}=1+\frac{1}{3}$
$=\frac{4}{3}$