#### Answer

(a) The height of the chimney is $5.05~m$
(b) The fundamental frequency is $16.3~Hz$

#### Work Step by Step

(a) We can find the frequency of a note that is three octaves below middle C:
$f = (\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})(261.6~Hz) = 32.7~Hz$
We can model the chimney as a pipe that is open at both ends. We can find the height $L$ of the chimney:
$\lambda = \frac{v}{f}$
$2L = \frac{v}{f}$
$L = \frac{v}{2f}$
$L = \frac{330~m/s}{(2)(32.7~Hz)}$
$L = 5.05~m$
The height of the chimney is $5.05~m$
(b) We can find the fundamental frequency when the chimney is closed at the bottom:
$f = \frac{v}{\lambda}$
$f = \frac{v}{4L}$
$f = \frac{330~m/s}{(4)(5.05~m)}$
$f = 16.3~Hz$
The fundamental frequency is $16.3~Hz$