How do you solve for #g# in #T = 2pi sqrt(L/g)#?

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Answer 1 · 2018-03-08T13:54:16

#g=(4pi^2l)/T^2#

We have #T=2pisqrt(l/g)#

Divide both sides by #2pi#:

#T/(2pi)=sqrt(l/g)#

Square both sides:

#l/g=T^2/(4pi^2)#, or:

#g/l=(4pi^2)/T^2#

#g=(4pi^2l)/T^2#