Why does a tube suddenly ring out at one water level and go quiet at the next?

How to use this simulation

1. Start with the 512 Hz fork at 20 °C and a tube of radius 2 cm. The true speed of sound reads 343.0 m/s and the wavelength 0.670 m.
2. Drag the air column length L slowly from the top. Watch the blue loudness curve on the right and the Loudness reading. Most of the time it sits near zero.
3. Stop where the loudness jumps to 100%. The tube turns green and the air particles swing in big arcs. That is the first resonance, L₁ = 0.155 m.
4. Keep going. The next loud spot is L₂ = 0.490 m, and a third one follows at 0.825 m. Notice that the gap between L₁ and L₂ is 0.335 m, which is exactly λ/2 = 0.335 m.
5. Change the fork. A higher frequency means a shorter wavelength, so every loud spot moves up the tube and they crowd together.
6. Raise the temperature and the loud spots move down the tube, because sound travels faster and the wavelength grows. Widen the tube and all of them shift up slightly, because the end correction grows.

The experiment in one picture

The apparatus is plain: a tall glass tube standing in a jar of water, a clamp, and a tuning fork of known frequency. By lifting or lowering the tube, or by draining the water, you change how long the column of air above the water is. You strike the fork on a rubber pad and hold it just over the mouth of the tube.

Most of the time the sound you hear is the faint hum of the fork itself. At certain heights the air in the tube joins in, and the whole thing becomes much louder. That joining in is resonance. The fork keeps pushing on the air at its own frequency, and when the column of air has just the right length, each push arrives in step with the wave already bouncing inside the tube, so the wave grows.

Why do we care? Because the loud spots tell us the wavelength, and wavelength times frequency is the speed of sound. A school lab can measure the speed of sound to within a few percent with a fork, a tube and a ruler.

The air column is a pipe closed at one end

The water surface acts like a closed end: air cannot move through it, so the particles there cannot swing. That is a displacement node. The open mouth of the tube is the opposite: the air is free to move, so it is a displacement antinode. This is exactly the closed pipe from the standing-waves page.

A node and its nearest antinode are a quarter of a wavelength apart. So the shortest column that fits a standing wave is λ/4 long. The next fits another half wavelength, giving 3λ/4, then 5λ/4, and so on. Each extra loud spot adds half a wavelength, because the pattern has to end in a node at the water and an antinode at the mouth every time.

In the animation the blue dots are air particles. At resonance, the ones near the water barely move and the ones near the mouth swing the most. The dashed green line just above the mouth marks where the antinode really sits, and the next section explains why it is not exactly at the glass.

The end correction, in plain words

If you look closely, the first loud spot comes slightly lower than λ/4. That surprises many students on the first try. The reason: the air does not stop swinging exactly at the mouth of the tube. A little blob of air just outside the mouth is dragged along, so the antinode sits a short distance above the mouth.

That distance is the end correction e. It depends on how wide the tube is, and a good rule of thumb is e ≈ 0.6 r, where r is the inner radius. For the default 2 cm tube, e = 1.2 cm = 0.012 m. Try the radius slider: a wider tube pushes the antinode further out and the loud spots move to shorter columns.

The measured air column is from the mouth to the water, so each resonance is shorter than the ideal value by e. That gives L₁ = λ/4 − e and L₂ = 3λ/4 − e. With the default fork, λ/4 = 0.167 m, so L₁ = 0.155 m and L₂ = 0.490 m. The animation draws the loudness curve with exactly these peaks.

From two readings to the speed of sound

Subtract L₁ from L₂: L₂ − L₁ = (3λ/4 − e) − (λ/4 − e) = λ/2. The e vanishes. So λ = 2 (L₂ − L₁), and since v = f λ, the speed of sound is v = 2 f (L₂ − L₁).

With the default settings the two loud spots are 0.155 m and 0.490 m apart by 0.335 m, so v = 2 × 512 × 0.335 = 343.0 m/s. The true value at 20 °C is 343.0 m/s, so the method works. You can check this yourself in the readings panel: it computes v from the same L₁ and L₂.

If you also know L₁ and the tube radius, you can use v = 4 f (L₁ + e). It needs only one resonance, but only as good as your estimate of e. Many lab sheets ask for both methods so that you can compare them.

What the fork frequency does

Swap the fork and the whole pattern squeezes or stretches. A high-pitched fork has a short wavelength, so its loud spots are close to the mouth and close together. A low-pitched fork has a long wavelength, and its loud spots sit far down the tube and far apart.

The table uses the five forks of a standard set at 20 °C and a 2 cm tube. Notice that the 256 Hz fork has its second resonance near the bottom of the tube, so a short tube might miss it. That is why a lab always picks a fork whose L₂ fits the tube available.

What temperature does

The speed of sound in air rises by about 0.6 m/s for every degree Celsius, so v ≈ 331 + 0.6 T. At 0 °C it is 331.0 m/s, and at 40 °C it is 355.0 m/s. The fork frequency does not change, so a faster wave means a longer wavelength and the loud spots move down the tube.

The table uses the 512 Hz fork. The first resonance moves by about a centimetre across the whole range, the second by about three times as much, and the speed you compute follows the temperature faithfully. This is why the lab sheet asks you to note the room temperature before you begin.

Solved problems

Four problems in the exam style, using the same functions as the animation.

What if you forget the end correction?

Suppose a student treats L₁ as exactly λ/4 and calculates v = 4 f L₁ with the 512 Hz fork. The ideal first length is 0.167 m, but the tube really rings at 0.155 m, because the antinode sits 0.012 m above the mouth. So 4 × 512 × 0.155 = 318.4 m/s, which is 7.2% away from the true 343.0 m/s.

That is a small error, but it is systematic: it is always in the same direction. Using the difference L₂ − L₁ removes it, which is why a good report always shows both resonances.

Mistakes that cost marks

Check this list before the practical exam. Each one has taken marks from someone.

• Writing L₁ = λ/2. A closed pipe's first resonance is a quarter wavelength, not a half.
• Dividing L₂ − L₁ by λ instead of equating it to λ/2. The gap between consecutive resonances is half a wavelength.
• Forgetting the factor of 2 in v = 2f(L₂ − L₁).
• Mixing units: reading the lengths in centimetres and the speed in metres per second. Convert first.
• Using the speed at the wrong temperature. Always note the room temperature and use v = 331 + 0.6 T as a check.
• Using a tube that is too short to show L₂, then guessing the second reading.
• Thinking the tube makes sound out of nothing. Resonance only lets the fork's small push build up a big wave in the air column, so the fork dies away sooner.

Where else you see air-column resonance

Blow across the top of a bottle and it hums: the air in the bottle is a closed pipe, and the note drops as you add more air by pouring out water. A clarinet and an organ pipe closed at the top work the same way. A bamboo flute is closer to an open pipe, which has all harmonics rather than odd ones.

Resonance matters for the ear too: the ear canal is a short closed pipe about 2.5 cm long, so it boosts sounds around 3,000 Hz, near where speech consonants live. For the general picture of nodes, antinodes and harmonics, see the standing-waves animation next.

Frequently asked questions