Answer
(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 $1.1~m$
Work Step by Step
(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 $1.1~m$
