Interactive physics simulator
Resonance in Sound
Explore acoustic resonance using a virtual water column apparatus. Adjust the reservoir height to find standing wave nodes and antinodes, or couple tuning forks sympathetically through the air.
Acoustic Resonance Lab
Strike the tuning fork and adjust the water level. Watch the red pressure standing wave align and swell inside the air column at resonant lengths.
Live Telemetry
What is Acoustic Resonance?
Acoustic resonance is a physical phenomenon where an air-filled cavity, column, or material vibrates at a significantly higher amplitude when it is driven by sound waves whose frequency matches one of its own natural frequencies.
This happens because the reflected waves returning from the boundaries of the cavity interfere **constructively** with the incoming waves. At these specific frequencies, the waves align peak-to-peak, building up a large standing wave with high sound pressure. This energy build-up radiates outward, resulting in a dramatic swell in volume.
In a vertical column closed by water at the bottom, sound reflects off the water surface. Since water is a rigid barrier, air molecules cannot move there, creating a **displacement node**. The open mouth at the top allows free air motion, forming a **displacement antinode**. Because one end is a node and the other is an antinode, the length of the column must contain an odd number of quarter wavelengths:
Resonance Column Method
How physicists measure the speed of sound in a laboratory:
- Finding First Resonance: A vibrating tuning fork of frequency \(f\) is held over the column. The water level is slowly lowered until a loud hum is heard. The air column length \(L_1\) is recorded. This corresponds to the fundamental mode: \(L_1 + e = \frac{\lambda}{4}\).
- Finding Second Resonance: The water level is lowered further until a second volume swell is heard at length \(L_2\). This corresponds to the third harmonic: \(L_2 + e = \frac{3\lambda}{4}\).
- Speed Calculation: Subtracting these two equations cancels the end correction term (\(e\)): \(L_2 - L_1 = \frac{\lambda}{2}\). Thus, the wavelength is \(\lambda = 2(L_2 - L_1)\). Since \(v = f\lambda\), we get:
v = 2f(L_2 - L_1)
Resonance Formulas
Acoustic standing waves inside closed and open pipes:
- Closed Tube Resonance (Odd Harmonics): f_n = n \frac{v}{4(L + e)} \quad (n = 1, 3, 5, \dots)
- End Correction (\(e\)): Due to the air inertia outside the tube opening, the antinode forms slightly above the mouth:
e \approx 0.6 \cdot r \quad (\text{where } r = \text{tube radius})
- Open Tube Resonance (All Harmonics): Tubes open at both ends form antinodes at both mouths, allowing all integer harmonics:
f_n = n \frac{v}{2(L + 2e)} \quad (n = 1, 2, 3, \dots)
Solved Examples
In a resonance column experiment, the first resonance is observed at an air column length of (17.5 ext{ cm}) and the second resonance is observed at (53.5 ext{ cm}) when using a tuning fork of frequency (512 ext{ Hz}). Calculate (a) the speed of sound in air, and (b) the end correction of the tube.
- Identify the given values: First resonance length (L_1 = 17.5 ext{ cm} = 0.175 ext{ m}), second resonance length (L_2 = 53.5 ext{ cm} = 0.535 ext{ m}), and frequency (f = 512 ext{ Hz}).
- The distance between two successive resonance points in a closed pipe is equal to half a wavelength: (L_2 - L_1 = rac{lambda}{2}).
- Calculate the wavelength (lambda): (lambda = 2 cdot (L_2 - L_1) = 2 cdot (0.535 - 0.175) = 2 cdot 0.360 = 0.720 ext{ m}).
- Calculate the speed of sound using wave velocity equation: (v = f cdot lambda = 512 ext{ Hz} cdot 0.720 ext{ m} = 368.64 ext{ m/s}).
- Calculate the end correction (e) using the fundamental resonance condition: (L_1 + e = rac{lambda}{4}).
- Substitute values: (0.175 + e = rac{0.720}{4} = 0.180 ext{ m}).
- Therefore, (e = 0.180 - 0.175 = 0.005 ext{ m} = 0.5 ext{ cm}).
Answer: Speed of sound = (368.64 ext{ m/s}) | End correction = (0.5 ext{ cm})
A tuning fork of frequency (340 ext{ Hz}) is held over a glass resonance tube. If the speed of sound is (340 ext{ m/s}) and the inner radius of the tube is (2.5 ext{ cm}), calculate the physical lengths of the tube at which the first and second resonances will occur.
- Identify the given values: Frequency (f = 340 ext{ Hz}), sound speed (v = 340 ext{ m/s}), and tube radius (r = 2.5 ext{ cm} = 0.025 ext{ m}).
- Calculate wavelength (lambda): (lambda = rac{v}{f} = rac{340}{340} = 1.00 ext{ m} = 100 ext{ cm}).
- Calculate end correction (e = 0.6 cdot r = 0.6 cdot 2.5 ext{ cm} = 1.5 ext{ cm}).
- For the first (fundamental) resonance, the column length satisfies: (L_1 + e = rac{lambda}{4}).
- Substitute values: (L_1 + 1.5 = rac{100}{4} = 25 ext{ cm}), so (L_1 = 25 - 1.5 = 23.5 ext{ cm}).
- For the second resonance (third harmonic), the column length satisfies: (L_2 + e = 3 cdot rac{lambda}{4}).
- Substitute values: (L_2 + 1.5 = rac{300}{4} = 75 ext{ cm}), so (L_2 = 75 - 1.5 = 73.5 ext{ cm}).
Answer: First resonance length = (23.5 ext{ cm}) | Second resonance length = (73.5 ext{ cm})
A closed organ pipe of length (0.5 ext{ m}) contains air. If the speed of sound is (340 ext{ m/s}), explain why a tuning fork of frequency (250 ext{ Hz}) will not cause acoustic resonance inside the pipe.
- Identify the given values: Closed pipe length (L = 0.5 ext{ m}), speed of sound (v = 340 ext{ m/s}).
- The natural resonant frequencies of a closed pipe are given by: (f_n = n cdot rac{v}{4L}) where (n = 1, 3, 5, dots) (odd integers).
- Calculate the fundamental frequency ((n=1)): (f_1 = rac{340}{4 cdot 0.5} = 170 ext{ Hz}).
- Calculate the third harmonic ((n=3)): (f_3 = 3 cdot 170 = 510 ext{ Hz}).
- Calculate the fifth harmonic ((n=5)): (f_5 = 5 cdot 170 = 850 ext{ Hz}).
- The sound source frequency is (250 ext{ Hz}). Since (250 ext{ Hz}) does not match any of the allowed discrete natural resonant frequencies of the pipe ((170 ext{ Hz}), (510 ext{ Hz}), (850 ext{ Hz}), etc.), resonance will not occur.
Answer: Resonance cannot occur because (250 ext{ Hz}) is not an allowed natural harmonic of the closed pipe.
Common Mistakes
- Forgetting End Correction: Students often calculate frequencies using \(L\) directly instead of the corrected length \(L + e\). Remember, the standing wave extends slightly outside the physical tube boundary!
- Using Even Harmonics for Closed Pipes: Believing a closed pipe can have a second harmonic (\(n=2\)). Closed pipes can only sustain odd harmonics (\(n=1,3,5\dots\)) because one boundary must be a node and the other an antinode.
- Confusing closed vs open formulas: Closed pipes use a factor of \(4\) in the denominator (\(4L\)), while open pipes (both ends open) use a factor of \(2\) (\(2L\)).
- Assuming Nodes are positions of max motion: A displacement node is where air molecules do NOT move at all. An antinode is where they move with maximum amplitude.
Practice Questions
Question 1
A student performs a resonance column experiment using a tuning fork of frequency (480 ext{ Hz}). The first resonance occurs at a column length of (17.2 ext{ cm}). If the speed of sound in air is (340 ext{ m/s}), what is the end correction of the tube?
View Solution & Answer
Wavelength (lambda = rac{v}{f} = rac{340}{480} approx 0.7083 ext{ m} = 70.83 ext{ cm}). The quarter wavelength is ( rac{lambda}{4} = rac{70.83}{4} approx 17.71 ext{ cm}). For a closed resonance tube, (L_1 + e = rac{lambda}{4}). Substituting the values: (17.2 + e = 17.71 ext{ cm}), which gives (e = 17.71 - 17.2 = 0.51 ext{ cm}).
Question 2
Explain how sympathetic acoustic resonance is demonstrated in a classroom using two tuning forks.
View Solution & Answer
Place two identical tuning forks mounted on wooden soundboxes near each other, with their open ends facing. Strike Fork A with a rubber mallet, causing it to vibrate and emit sound. After a few seconds, grasp Fork A with your hand to damp its vibration completely. You will hear a clear tone still ringing. This tone is emitted by Fork B, which was forced into vibration solely by the acoustic pressure waves traveling through the air from Fork A, due to their matching natural resonant frequencies.
Question 3
For a closed resonance tube, sketch or describe the displacement and pressure standing wave profiles for the fundamental mode and third harmonic.
View Solution & Answer
For a closed tube of length (L), the water surface behaves as a hard boundary (node for air particle displacement, antinode for air pressure) and the open top behaves as a soft boundary (displacement antinode, pressure node). In the fundamental mode ((1^{st}) harmonic), there is 1 displacement node at the water and 1 displacement antinode at the top. In the third harmonic, there are 2 displacement nodes (one at the water, one at ( rac{2}{3} L) up) and 2 displacement antinodes (one at the mouth, one at ( rac{1}{3} L) up).
Question 4
Why does the pitch of water filling a glass beaker get higher as the water level rises?
View Solution & Answer
The pouring water splashes and creates a broad noise spectrum. The air column in the glass acts as a closed organ pipe, resonating at its fundamental frequency (f = rac{v}{4L}). As the beaker fills with water, the physical length (L) of the air column above the water shrinks. Since frequency is inversely proportional to column length (L), a smaller (L) results in a higher resonant frequency, causing the pitch of the wailing sound to rise.
Frequently Asked Questions
What is acoustic resonance?
Acoustic resonance is the phenomenon where a cavity or column of air amplifies sound waves whose frequency matches one of its own natural frequencies of vibration.
What is a resonance column apparatus?
It is a laboratory instrument consisting of a vertical tube filled with water to adjustable heights, used to determine the speed of sound in air by finding the standing wave resonance points of a tuning fork.
What are the boundary conditions for a closed pipe?
A closed pipe (like a resonance tube closed by water at one end) must have a displacement node (zero air movement) at the closed boundary and a displacement antinode (maximum air movement) at the open mouth.
Why do only odd harmonics exist in a closed pipe?
Because the boundaries require a node at one end and an antinode at the other, only odd quarter-wavelength fractions ((L = rac{lambda}{4}, rac{3lambda}{4}, rac{5lambda}{4}, dots)) can form standing waves. This results in frequencies of (f_1, 3f_1, 5f_1, dots)
What is end correction?
End correction is the physical phenomenon where the standing wave antinode does not form exactly at the open mouth of the tube, but slightly outside it (by a distance (e approx 0.6r) where (r) is the tube radius) due to the inertia of the air column.
How does the speed of sound relate to temperature?
The speed of sound in gas is proportional to the square root of absolute temperature ((v propto sqrt{T})). As temperature increases, sound speed increases, which in turn increases the resonant frequencies of an air cavity.
What is sympathetic resonance?
Sympathetic resonance is a vibration excited in a passive acoustic resonator (like a tuning fork or string) by sound waves of matching frequency emitted by an active vibrator.
Why does a sound swell in volume at resonance?
Inside the resonator, reflecting sound waves align in phase with the source waves. This constructive interference creates high-amplitude standing waves, radiating a much louder sound into the room.
Can a singer really break glass with their voice?
Yes. If a singer hits the exact resonant frequency of a wine glass, and sings loudly enough, the glass begins to flex. The resonance builds up mechanical energy until the glass exceeds its elastic limit and shatters.
What is the Helmholtz resonator?
A Helmholtz resonator is a hollow sphere with a neck (like an empty bottle). Blowing across the mouth creates a single, deep resonant hum governed by the volume of the cavity and area of the neck, rather than simple column length.