Question 147969
{{{t = 2*PI*sqrt(L/g)}}}, solve for L
Same thing, get L by itself on one side of the equal sign.
Square both sides to get rid of the radical
{{{t^2 = 4*PI^2*(L/g)}}}
Divide thru by {{{4*PI^2}}}
{{{(t^2)/(4*PI^2) = L/g}}}
Multiply thru by g
{{{(t^2)*g/(4*PI^2) = L}}}
L = {{{(t^2)*g/(4*PI^2)}}}
or, L = {{{g*(t/2PI)^2}}}, same thing different format
This is the formula for the length of a pendulum (L), with a period of t seconds.