?

\[\]
eqalign{

& T = 2\pi \left( {\sqrt {\dfrac{l}{g}} } \right) \cr

& T = 2\pi \left( {\sqrt {\dfrac{{160}}{{10}}} } \right) \cr

& T = 2\pi \left( {\sqrt {16} } \right) \cr

& T = 2\pi \left( 4 \right) \cr

& T = 2\pi \left( {0,04} \right) \cr

& T = 0,08\pi {\text{\ s}}\ \cr }\ \ \ Portanto,\ o\ período\ do\ pêndulo\ será\ de\ \boxed{T\ =\ 0,08\pi\ {\text{\ s}}}\(\).