A tuning fork of frequency 340 Hz is vibrated just above a tube of 120 cm height. Water is slowly poured in the tube. What is the minimum height of water necessary for resonance?

Because the tuning fork is in resonance with air column with a pipe closed at one end, the frequency is \(\mathrm{n}=\frac{(2 \mathrm{~N}-1) \mathrm{v}}{4 l}\) where \(\mathrm{N}=1,2,3 \ldots\). corresponds to different modes of vibration Substituting \(\mathrm{n}=340 \mathrm{~Hz}, \mathrm{v}=340 \mathrm{~m} / \mathrm{s}\) (speed of sound in air), the length of air column in the pipe can be
\(l=\frac{(2 \mathrm{~N}-1) 340}{4 \times 340}=\frac{(2 \mathrm{~N}-1)}{4} \mathrm{~m}=\frac{(2 \mathrm{~N}-1) \times 100}{4} \mathrm{~cm}\)
For \(\mathrm{N}=1,2,3, \ldots\) we get \(\cdot l=25 \mathrm{~cm}, 75 \mathrm{~cm}\), \(125 \mathrm{~cm} . .\). etc.
As the tube is only 120 cm long, length of air column after water is poured in it may be 25 cm or 75 cm only. Hence, the corresponding length of water column in the tube will be \((120-25) \mathrm{cm}=95 \mathrm{~cm}\) or \((120-75) \mathrm{cm}=45 \mathrm{~cm}\).
Thus minimum length of water column is 45 cm.