Answer
The fundamental frequency of Pipe B is $~~143~Hz$
Work Step by Step
We can use the information about pipe A to find the frequency:
$f = \frac{nv}{2L},~~$ where $~~n = 3$
$f = \frac{(3)(343~m/s)}{(2)(1.20~m)}$
$f = 428.75~Hz$
We can find the length of Pipe B:
$f = \frac{3v}{4L} = 428.75~Hz$
$L = \frac{3v}{(4)(428.75~Hz)}$
$L = \frac{(3)(343~m/s)}{(4)(428.75~Hz)}$
$L = 0.60~m$
We can find the fundamental frequency of Pipe B:
$f = \frac{(1)(v)}{4L}$
$f = \frac{(1)(343~m/s)}{(4)(0.60~m)}$
$f = 143~Hz$
The fundamental frequency of Pipe B is $~~143~Hz$
