Doubling Period of Pendulum By Increasing Length
\[L\]
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. 
