Answer
The new fundamental frequency in each wire is $933~Hz$
Work Step by Step
We can find an expression for the original fundamental frequency:
$f = \frac{v}{\lambda}$
$f = \frac{\sqrt{\frac{mg}{\mu}}}{2L}$
$f = \frac{1}{2}\sqrt{\frac{mg}{\mu~L^2}}$
When the wire is cut in half, the length of each wire is half of the original length but the mass per unit length is still the same. Each wire has half the tension since each wire supports half the weight of the sign. We can find the new fundamental frequency:
$f' = \frac{v}{\lambda}$
$f' = \frac{\sqrt{\frac{mg/2}{\mu}}}{2(L/2)}$
$f' = \frac{2}{\sqrt{2}}\times \frac{\sqrt{\frac{mg}{\mu}}}{2L}$
$f' = \sqrt{2}\times \frac{1}{2}\sqrt{\frac{mg}{\mu~L^2}}$
$f' = \sqrt{2}\times f$
$f' = \sqrt{2}\times 660~Hz$
$f' = 933~Hz$
The new fundamental frequency in each wire is $933~Hz$.
