E. Waves and Sound · Waves and SoundResonance in Air Columns: Measuring the Speed of Sound
E2.3 E2.6 E3.4 A1.8 A1.10About 75 minutes
Learning goals- I can describe standing sound waves in air columns closed at one end and open at both ends.
- I can use L = (2n − 1)λ/4 (closed) and L = nλ/2 (open) to find resonant lengths and frequencies.
- I can use a resonance tube to determine the speed of sound and compare it with the theoretical value.
- I can explain sources of error, including the end correction, and how the L₂ − L₁ method avoids it.
Words to know
air column
The air inside a tube or pipe, which can vibrate as a standing wave.
closed air column
A tube closed at one end and open at the other. Resonant lengths are odd multiples of λ/4.
open air column
A tube open at both ends. Resonant lengths are whole multiples of λ/2.
resonance (in an air column)
A large, sustained vibration of the air in a tube when the tube's length matches a standing-wave pattern for the frequency of the source.
end correction
A small extra length (roughly 0.3 times the tube's diameter at each open end) because the antinode forms slightly outside the open end.
percent difference / percent error
|measured − accepted| ÷ accepted × 100%: a way to compare an experimental result with a theoretical value.
The pop-bottle trick
Blow gently across the top of an empty bottle and it hoots. Pour in some water and blow again: the note goes up. You have just made an air column resonate. Every flute, clarinet, trumpet and church organ works this way, and the same idea gives you a neat way to measure the speed of sound in a lab.
Standing waves in air
Sound reflects at both ends of a tube, so standing waves can form in the air inside, just like on a string. For air columns we describe the displacement of the air molecules:
- At a closed end, the air cannot move: a displacement node (N).
- At an open end, the air moves freely: a displacement antinode (A).
Tubes closed at one end
The simplest pattern has a node at the closed end and an antinode at the open end: that is one quarter of a wavelength. The next pattern fits three quarters, then five quarters:
L = λ/4, 3λ/4, 5λ/4, … or L = (2n − 1) λ/4
The resonant frequencies are f = (2n − 1) v/(4L): f₁, 3f₁, 5f₁, … Only odd harmonics are possible.
Tubes open at both ends
Now both ends are antinodes. The simplest pattern has one node in the middle, so the tube holds half a wavelength:
L = λ/2, λ, 3λ/2, … or L = n λ/2
The resonant frequencies are fₙ = n v/(2L): f₁, 2f₁, 3f₁, … All harmonics are possible, as for a string.
Compare: a closed tube's fundamental is v/(4L), while an open tube of the same length has v/(2L), twice as high. A closed pipe sounds an octave lower than an open pipe of the same length.
| Closed at one end | Open at both ends |
|---|
| Ends | node at closed end, antinode at open end | antinodes at both ends |
| Resonant lengths | L = (2n − 1)λ/4 | L = nλ/2 |
| Fundamental | f₁ = v/(4L) | f₁ = v/(2L) |
| Harmonics present | odd only: f₁, 3f₁, 5f₁ … | all: f₁, 2f₁, 3f₁ … |
| Spacing between resonant lengths | λ/2 | λ/2 |
Measuring the speed of sound with a resonance tube
In the classic lab, a tube open at both ends is lowered into a tall cylinder of water. The water surface acts as the closed end, so moving the tube changes the length L of the air column. A tuning fork of known frequency is held over the top.
- Predict: using v ≈ 331 + 0.6T, calculate λ and the expected resonant lengths λ/4 and 3λ/4. (Predicting before testing is what scientists do: expectation E2.6.)
- Test: start with a short air column and slowly lengthen it while the fork sounds. The sound suddenly gets loud at the first resonance, L₁. Keep going to find the second, L₂.
- Calculate: L₂ − L₁ = λ/2, so λ = 2(L₂ − L₁) and v = fλ.
- Compare with the theoretical value and calculate the percent difference.
Why use L₂ − L₁? The antinode at the open top actually forms a little above the tube (the end correction, about 0.3 times the tube's diameter). So L₁ is slightly less than λ/4. But the same correction affects L₂ too, so it cancels when you subtract.
Worked exampleExample 1: Speed of sound from two resonances
A 512 Hz fork gives resonances at L₁ = 0.160 m and L₂ = 0.495 m. The room is 22 °C. Find the speed of sound, the percent difference from theory, and the end correction.
- L₂ − L₁ = 0.495 − 0.160 = 0.335 m = λ/2, so λ = 0.670 m.
- v = fλ = 512 × 0.670 = 343 m/s.
- Theory: v = 331 + 0.6(22) = 344.2 m/s. Percent difference = |343 − 344.2| ÷ 344.2 × 100% ≈ 0.35%.
- Ideally L₁ + e = λ/4 = 0.1675 m, so the end correction e ≈ 0.1675 − 0.160 = 0.0075 m (about 0.8 cm).
v ≈ 343 m/s, within about 0.4% of theory; end correction ≈ 0.8 cm.
Worked exampleExample 2: Predict the resonant lengths
Predict the first two resonant lengths of a closed air column for a 384 Hz fork at 20 °C (ignore the end correction).
- v = 331 + 0.6(20) = 343 m/s.
- λ = v/f = 343 ÷ 384 = 0.893 m.
- L₁ = λ/4 = 0.223 m; L₂ = 3λ/4 = 0.670 m.
About 22 cm and 67 cm. (Measured values will be slightly shorter because of the end correction.)
Worked exampleExample 3: Closed versus open pipe
An organ pipe is 0.60 m long. Air is at 20 °C. Find the fundamental and the next harmonic if the pipe is (a) open at both ends, (b) closed at one end.
- (a) Open: f₁ = v/(2L) = 343 ÷ 1.20 = 286 Hz; next harmonic f₂ = 2f₁ = 572 Hz.
- (b) Closed: f₁ = v/(4L) = 343 ÷ 2.40 = 143 Hz; next possible is the 3rd harmonic, 3f₁ = 429 Hz (no even harmonics).
Open: 286 Hz and 572 Hz. Closed: 143 Hz and 429 Hz.
Just for funWhat did the trombone say to the tuba?
"Stop blowing things out of proportion!" (The tuba's long tubing gives it a very low fundamental frequency.)
Hands-on activityA kitchen resonance tube
You need: a tall glass, vase or plastic cylinder, water, a cardboard tube or a rigid plastic tube that fits inside it (a paper-towel tube wrapped in plastic wrap so it stays dry, or a length of PVC pipe), a ruler, a thermometer or a weather app, and a free tone-generator app on a phone.
- Fill the container almost to the top with water. Stand the tube in the water.
- Play a steady tone of 512 Hz from the phone, held just above the tube.
- Predict L₁ = λ/4 using v = 331 + 0.6T.
- Slowly lift the tube (making the air column longer) and listen for the sudden loud spot. Measure L₁ from the top of the tube to the water surface.
- If your tube is long enough, keep lifting to find L₂ (about 3 times L₁).
- Calculate v. If you only have L₁, use v = 4fL₁ and compare with theory; explain the difference using the end correction (about 0.3 × the tube diameter).
- Bonus: blow across bottles holding different amounts of water and order them from lowest to highest pitch.
What to look for: at 20 °C and 512 Hz, L₁ ≈ 16 cm to 17 cm and L₂ ≈ 50 cm. The bottle with the most water (shortest air column) gives the highest pitch.
Your turn
1Is a closed end of an air column a displacement node or antinode? An open end?
Closed end: node. Open end: antinode.
2A tube closed at one end is 0.25 m long. Find the wavelength of its fundamental.
L = λ/4, so λ = 4L = 1.0 m.
3A tube open at both ends is 0.40 m long. Find the wavelength of its fundamental.
L = λ/2, so λ = 2L = 0.80 m.
4Which harmonics can a closed air column produce?
Only odd harmonics: f₁, 3f₁, 5f₁, …
5In a resonance tube lab with a 440 Hz fork, L₁ = 0.18 m and L₂ = 0.57 m. Find λ and v.
λ = 2(0.57 − 0.18) = 0.78 m; v = 440 × 0.78 ≈ 343 m/s.
6Find the fundamental of a 0.50 m open pipe at 20 °C.
f₁ = 343 ÷ (2 × 0.50) = 343 Hz.
7Why does pouring water into a bottle raise the note you get by blowing across it?
The air column gets shorter. For a closed column f₁ = v/(4L), so a smaller L gives a higher frequency.
More practice: warm-up
1What is the length of the first resonance of a closed air column, in terms of λ?
2What is the spacing between successive resonant lengths for any air column?
3Write the resonant-frequency formula for an open air column.
4What acts as the closed end in the resonance-tube lab?
5What is the end correction?
A small extra length because the antinode forms slightly outside the open end (about 0.3 × diameter).
More practice: on your own
1Predict L₁ and L₂ for a closed air column with a 256 Hz fork at 15 °C (ignore end correction).
v = 331 + 0.6(15) = 340 m/s; λ = 340 ÷ 256 = 1.33 m. L₁ = λ/4 ≈ 0.332 m; L₂ = 3λ/4 ≈ 0.996 m.
2A student finds L₁ = 0.150 m with a 550 Hz fork and uses v = 4fL₁. What value does she get, and why is it lower than the theoretical 343 m/s?
v = 4 × 550 × 0.150 = 330 m/s. It is low because the true quarter-wavelength is L₁ plus the end correction; ignoring it makes λ (and v) too small.
3A 0.85 m pipe is closed at one end. At 20 °C, find its first three resonant frequencies.
f₁ = 343 ÷ (4 × 0.85) ≈ 101 Hz; then 3f₁ ≈ 303 Hz and 5f₁ ≈ 504 Hz.
4A pipe resonates at 300 Hz, 500 Hz and 700 Hz, with no resonances in between. Is it open or closed? Find f₁.
The gaps are 200 Hz, so 2f₁ = 200 → f₁ = 100 Hz. The resonances 300, 500, 700 are 3f₁, 5f₁, 7f₁: odd harmonics only, so the pipe is closed at one end.
5Explain why the L₂ − L₁ method gives a better speed of sound than using L₁ alone.
Both L₁ and L₂ are short by the same end correction, so subtracting cancels it: L₂ − L₁ is exactly λ/2. Using L₁ alone includes the error from the end correction.
Challenge
1A tube 0.80 m long is closed at one end. What is the lowest frequency of sound at 20 °C that it resonates with? If the tube is opened at both ends, what is the new lowest frequency? Explain the musical relationship.
Closed: f₁ = 343 ÷ 3.2 ≈ 107 Hz. Open: f₁ = 343 ÷ 1.6 ≈ 214 Hz. The open pipe's fundamental is twice as high: one octave higher.
2In a resonance-tube lab the measured lengths are L₁ = 0.320 m and L₂ = 0.985 m with a 256 Hz fork; the thermometer reads 12 °C. Find v, the percent difference, and the end correction. Then suggest one improvement.
λ = 2(0.985 − 0.320) = 1.33 m; v = 256 × 1.33 = 340.5 m/s ≈ 341 m/s. Theory: 331 + 0.6(12) = 338.2 m/s. Percent difference ≈ |340.5 − 338.2| ÷ 338.2 × 100% ≈ 0.7%. End correction: λ/4 − L₁ = 0.3325 − 0.320 ≈ 0.013 m. Improvement (sample): move the tube in both directions to find each loud point and average the readings, or use several forks and graph.
3A flute (an open pipe) and a clarinet (behaves roughly like a closed pipe) are the same length. Use air-column physics to explain why the clarinet can play notes lower than the flute and why the two sound different.
For the same L, the closed pipe's fundamental v/(4L) is half the open pipe's v/(2L), so the clarinet can play about an octave lower. The closed pipe produces mainly odd harmonics, while the open pipe produces all harmonics, so the mix of harmonics (timbre) differs.
Quiz
1. At the closed end of an air column there is always a displacement…
2. The first resonant length of a tube closed at one end is…
3. A tube open at both ends has a fundamental of 200 Hz. Its next resonance is…
4. A tube closed at one end has a fundamental of 200 Hz. Its next resonance is…
5. In a resonance-tube lab, L₂ − L₁ equals…
6. Why is L₁ usually a bit shorter than λ/4?
Big idea: expectation E2.3 asks students to determine the speed of sound in a medium (classically with an air column), compare the measurement with the theoretical value, and explain the difference. E2.6 asks them to predict the conditions for resonance and test the predictions; E3.4 covers standing-wave conditions.
How to help: The kitchen version works surprisingly well with a phone tone generator. Ask your teen to write their prediction before they test. Blowing across bottles with different water levels is a quick, fun check.
Common difficulties:
- Mixing up closed (λ/4, odd harmonics) and open (λ/2, all harmonics) formulas. Drawing the node/antinode pattern first prevents this.
- Forgetting that L₂ − L₁ is λ/2, not λ.
- Explaining error vaguely ("human error"). Encourage specific sources: end correction, judging the loudest point, temperature, fork frequency.