College Physics (4th Edition)

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