The resonance tube

Making a tube boom

In the school lab there is a long glass tube that stands in water. You can raise or lower the water level. Strike a tuning fork and hold it over the open top. Now lower the water slowly. At first you hear only the fork. Then, at one level, the tube suddenly booms. The sound is much louder. Lower the water further, and the boom fades. Further still, it comes back.

This is the resonance tube. With it we can measure the speed of sound in air.

Why does it boom?

The air in the tube is a column closed at the bottom by the water and open at the top. Sound from the fork travels down, reflects from the water and comes back up. At certain lengths the reflected wave and the new wave build a standing wave. Then the air column vibrates strongly at the fork's frequency. That is resonance.

At the water the air cannot move, so there is a displacement node. At the open end the air moves most, so there is an antinode. The antinode is not exactly at the top of the tube. It is a little above it, a distance $e$ called the end correction. For a round tube of inside diameter $d$, $e \approx 0.3d$.

From a node to the next antinode is a quarter of a wavelength. So resonance comes when

and so on. A column closed at one end gives only odd multiples of $\lambda/4$.

The speed of sound

We do not know $e$ exactly. So we take two resonances and subtract:

The end correction cancels. So $\lambda = 2(l_2 - l_1)$, and the speed of sound is

To find the end correction, put $\lambda/4$ back in. From the two equations, $3(l_1 + e) = l_2 + e$, so

A worked example. A fork of $450\ \text{Hz}$ gives resonances at $l_1 = 18.0\ \text{cm}$ and $l_2 = 55.6\ \text{cm}$.

• Wavelength: $\lambda = 2(55.6\ \text{cm} - 18.0\ \text{cm}) = 75.2\ \text{cm} = 0.752\ \text{m}$.
• Speed of sound: $v = f\lambda = 450\ \text{Hz} \times 0.752\ \text{m} = 338\ \text{m/s}$.
• End correction: $e = \dfrac{55.6\ \text{cm} - 3 \times 18.0\ \text{cm}}{2} = 0.8\ \text{cm}$.

Find the two loud places for a fork of $420\ \text{Hz}$. They are about $40.5\ \text{cm}$ apart, which is half a wavelength. Now make the tube wider. Both loud places move down by the same amount, so their gap does not change.

Why use two resonances?

• One resonance gives $\lambda = 4(l_1 + e)$. We would need $e$, which we do not know well.
• Two resonances give $\lambda$ without $e$. That is why the method is good.
• A higher fork gives a shorter wavelength. Its resonances come closer to the top.

The speed of sound grows with the temperature of the air. On a hot day the same fork gives resonances a little lower down the tube.

Practise with new numbers

Try it in Python

The program takes the readings of three forks. It finds the speed of sound and the end correction from each, then the average speed.

Common mistakes

• Taking $l_2 - l_1$ as the wavelength. It is half a wavelength.
• Using $\lambda = 4l_1$. That leaves out the end correction.
• Thinking the second resonance is at $2l_1$. It is near $3l_1$, because only odd quarter waves fit.
• Putting a node at the open end. The open end has an antinode. The node is at the water.

Remember these

• First resonance: $l_1 + e = \lambda/4$. Second: $l_2 + e = 3\lambda/4$.
• $\lambda = 2(l_2 - l_1)$ and $v = 2f(l_2 - l_1)$.
• End correction: $e = \dfrac{l_2 - 3l_1}{2}$, about $0.3d$.
• Node at the water, antinode just above the open end.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A fork gives resonances at $26.0\ \text{cm}$ and $80.0\ \text{cm}$. What is the wavelength?

2. In the same tube, what is the end correction?

3. Why is there a node at the water surface?

4. A tube is used with a fork of a higher frequency. Do the resonances come nearer to the top or further down?