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?

Solution :
Frequency of n^th harmonic in a closed end tube
$$ \Rightarrow f = \frac{{\left( {2n - 1} \right)v}}{{4l}}\,\,n = 1,2,3,\,....$$
Also, only odd harmonics exists in a closed end tube.
Now, given two nearest harmonics are of frequency $$220\,Hz$$  and $$260\,Hz.$$
$$\therefore \frac{{\left( {2n - 1} \right)v}}{{4l}} = 220\,Hz\,......\left( {\text{i}} \right)$$
Next harmonic occurs at
$$\frac{{\left( {2n + 1} \right)v}}{{4l}} = 260\,Hz\,......\left( {{\text{ii}}} \right)$$
On subtracting Eq. (i) from Eq. (ii), we get
$$\eqalign{ & \frac{{\left\{ {\left( {2n + 1} \right) - \left( {2n - 1} \right)} \right\}v}}{{4l}} = 260 - 220 \cr & 2\left( {\frac{v}{{4l}}} \right) = 40 \Rightarrow \frac{v}{{4l}} = 20\,Hz \cr} $$
$$\therefore $$ Fundamental frequency of the system
$$ = \frac{v}{{4l}} = 20\,Hz$$