A resonance tube closed at one end is of height 1.5 m . A tuning fork of frequency 340 Hz is vibrating above…
A resonance tube closed at one end is of height 1.5 m . A tuning fork of frequency 340 Hz is vibrating above the tube. Water is poured in the tube gradually. The minimum height of water column for which resonance is obtained is . (Neglect end correction, speed of sound in air $=340 \mathrm{~m} / \mathrm{s}$ )
75 cm
50 cm
30 cm
25 cm
Solution
For a resonance tube, In the first mode, we have $\mathrm{n}=\frac{\mathrm{V}}{4 \mathrm{~L}_{\mathrm{j}}}$ or $\mathrm{L}_1=\frac{\mathrm{V}}{4 \mathrm{n}}=\frac{340}{4 \times 340}=0.25 \mathrm{~m}=25 \mathrm{~cm}$
In this case $\mathrm{L}_1=\frac{\lambda}{4}=25 \mathrm{~cm} \Rightarrow \lambda=100 \mathrm{~cm}$
Resonance can also be obtained when the length of the air column is $\frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \frac{7 \lambda}{4}$ etc.
At $\frac{5 \lambda}{4}=\frac{5 \times 100}{4}=125 \mathrm{~cm}$ it stops as length of tube is $1.5 \mathrm{~m}=150 \mathrm{~cm}$.
Hence the minimum height of water $=150-125=25 \mathrm{~cm}$