## College Physics (4th Edition)

(a) The pipe is closed at one end. (b) The fundamental frequency of this pipe is $78~Hz$ (c) The length of the pipe is $1.1~m$
(a) We can find the difference between the successive resonant frequencies: $390~Hz-234~Hz = 156~Hz$ $546~Hz-390~Hz = 156~Hz$ Since these three resonant frequencies are not multiples of $156~Hz$, the pipe can not be open at both ends. Then the pipe must be closed at one end. (b) The resonant frequencies of a pipe that is closed at one end have the form: $f_1, 3f_1, 5f_1, 7f_1, etc...$ The difference between successive frequencies is $2f_1$. Therefore, the fundamental frequency of this pipe is $\frac{156~Hz}{2} = 78~Hz$ (c) We can find the length of the pipe: $\lambda = \frac{v}{f}$ $4L = \frac{v}{f}$ $L = \frac{v}{4f}$ $L = \frac{343~m/s}{(4)(78~Hz)}$ $L = 1.1~m$ The length of the pipe is $1.1~m$