The water level in a vertical glass tube $1.0m$ long can be adjusted to any position in the tube. A tuning fork vibrating at $660Hz$ is held just over the open top end of the tube. At what positions of the water level measured from the open end  will there be  resonance? Speed of sound is $330m/s$.

Resonance corresponds to a pressure antinode at closed end and pressure node at open end. Further, the distance between a pressure node and a pressure antinode is $\frac{\lambda }{4}$, the condition of resonance would be,

Length of air column $l=n\frac{\lambda }{4}=n\left(\frac{v}{4f}\right)$

Here, $n=1,3,5,\dots$

$l_1=\left(1\right)\left(\frac{330}{4\times 660}\right)=0.125m$

$l_2=3l_1=0.375m$

$l_3=5l_1=0.625m$

$l_4=7l_1=0.875m$

$l_5=9l_1=1.125m$

Since, $l_5>1m$ (the length of tube), the length of air columns can have the values from $l_1$ to $l_4$ only.

Therefore, level of water at resonance will be

In all the four cases shown in figure, the resonance frequency is $660Hz$ but first one is the fundamental tone or first harmonic. Second is first overtone or second harmonic and so on.