## Doubling Period of Pendulum By Increasing Length

If the length of a simple pendulum increases from
$L$
to
$L+0.1$
the period doubles. Find
$L$
.
The period of a pendulum is given by
$T= 2 \pi \sqrt{ \frac{l}{g}}$
.
Originally
$T= 2 \pi \sqrt{ \frac{L}{g}}$
, and then
$2T= 2 \pi \sqrt{ \frac{L+0.1}{g}}$
.
Dividing the second equation by the first gives
$2= \sqrt{ \frac{L+0.1}{L}}$
.
Then
$4= \frac{L+0.1}{L} \rightarrow 4L=L+0.1 \rightarrow 3L=0.1 \rightarrow L=\frac{0.1}{3}= \frac{1}{30}$
m.