Question 78467
We have the equation

{{{T=k*sqrt(L)}}}  "The period of a simple pendulum is directly proportional to the square root of its length"

Substitute L=6 and T=2 and solve for k

{{{2=k*sqrt(6)}}}
{{{2/sqrt(6)=k*cross(sqrt(6)/sqrt(6))}}} Divide both sides by {{{sqrt(6)}}}

So our constant is 

{{{k=2/sqrt(6)}}}

Now the equation becomes

{{{T=2/sqrt(6)*sqrt(L)}}}

Now plug in T=1 to solve for L

{{{1=2/sqrt(6)*sqrt(L)}}}
{{{1/(2/sqrt(6))=sqrt(L)}}} Divide both sides by {{{2/sqrt(6)}}}
{{{sqrt(6)/2=sqrt(L)}}} Divide
{{{(sqrt(6)/2)^2=(sqrt(L))^2}}} Square both sides
{{{6/4=L}}}
{{{L=3/2}}} reduce

So our length must be 1.5 feet to have a period of 1 second.