Answer
$0.328\ m,1.31\ m$
Work Step by Step
The lengths of the two pipes can be determined as
$f_1=\frac{v}{4L}$
This can be rearranged as:
$L=\frac{v}{4f_1}$
We plug in the known values to obtain:
$L=\frac{343}{4(261.6)}=0.328m$
Similarly $f_2=2(\frac{v}{2L})$
This can be rearranged as:
$L=\frac{v}{f_2}$
We plug in the known values to obtain:
$L=\frac{343}{261.6}$
$L=1.31m$
