Answer
(a) $101Hz$
(b) $1.70m$
Work Step by Step
(a) The fundamental frequency of the pipe is given as
$f_n=nf_1$
This can be rearranged as:
$f_1=\frac{f_n}{n}$
We plug in the known values to obtain:
$f_1=\frac{202}{2}=101Hz$
(b) The length of the pipe can be determined as
$f_1=\frac{v}{2L}$
This can be rearranged as:
$L=\frac{v}{2f_1}$
We plug in the known values to obtain:
$L=\frac{343}{2(101)}=1.70m$
