Answer
The fundamental frequency of this string is $616~Hz$
Work Step by Step
We can find the speed of the wave in the string:
$v = \sqrt{\frac{F}{\mu}}$
$v = \sqrt{\frac{mg}{\mu}}$
$v = \sqrt{\frac{(2.20~kg)(9.80~m/s^2)}{3.55\times 10^{-6}~kg/m}}$
$v = 2464~m/s$
We can find the fundamental frequency:
$f = \frac{v}{\lambda}$
$f = \frac{v}{2L}$
$f = \frac{2464~m/s}{(2)(2.00~m)}$
$f = 616~Hz$
The fundamental frequency of this string is $616~Hz$.
