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Q.
The two nearest harmonics of a tube closed at one end and open at other end are $220\, Hz$ and $260\, Hz$ . What is the fundamental frequency of the system ?
Sol. Two successive frequencies of closed pipe
$\frac{n v}{4 l}=220$...(i)
$\frac{(n+2) v}{4 l}=260$...(ii)
Dividing (ii) by (i), we get
$\frac{n+2}{n}=\frac{260}{220}=\frac{13}{11}$
$11 n+22=13\, n$
$n=11$
So, $11 \frac{v}{4 I}=220$
$\frac{v}{4 l}=20$
So fundamental frequency is $20\, Hz$.