Acoustic Resonance in Pipes and Helmholtz Resonators

Acoustic Resonance in Pipes and Helmholtz Resonators

M06-11 developed acoustic pressure, particle velocity, sound intensity, and acoustic impedance.

Those ideas now let us understand why pipes, bottles, ducts, and cavities resonate at particular frequencies.

Two physically different resonance mechanisms will be developed.

The first is a distributed standing-wave resonance in a pipe. Its frequency is determined by how an acoustic wave fits between boundaries.

The second is a lumped Helmholtz resonance. A plug of air in a narrow neck acts approximately as an inertial mass, while compression of the air in the cavity acts approximately as a spring.

For an ideal uniform pipe, the open-open resonance frequencies are

(1)

For an ideal pipe closed at one end and open at the other,

(2)

A Helmholtz resonator with cavity volume

(3)

neck cross-sectional area

(4)

and effective neck length

(5)

has the approximate resonance frequency

(6)

These formulas look simple, but each contains important assumptions and boundary physics.

1 Acoustic standing waves in one dimension

For a uniform lossless duct of constant cross section, the acoustic pressure satisfies

(7)

Seek a separated harmonic solution

(8)

Substitution gives

(9)

where

(10)

The spatial pressure amplitude therefore has the form

(11)

The allowed values of

(12)

are selected by the acoustic boundary conditions.

2 Pressure and particle velocity

The linearized momentum equation is

(13)

For harmonic motion, pressure and particle velocity are spatially shifted relative to one another.

Where pressure has a standing-wave antinode, particle velocity has a node.

Where pressure has a standing-wave node, particle velocity has an antinode.

This quarter-wavelength spatial offset is the central geometry of pipe resonances.

3 Ideal closed end

At a rigid end, the fluid cannot move Normal to the wall.

Therefore

(14)

From the momentum relation, this corresponds to a pressure extremum.

Thus a rigid closed end is:

• a particle-velocity node,
• an acoustic-pressure antinode.

For the spatial pressure function, the ideal boundary condition can be written

(15)

4 Ideal open end

An open pipe end communicates with a much larger external region.

In the simplest approximation, the acoustic pressure perturbation at the opening is small:

(16)

Thus an ideal open end is:

• an acoustic-pressure node,
• a particle-velocity antinode.

This ideal condition is not exact at the physical pipe edge.

The effective pressure node lies slightly outside the pipe, producing the end correction developed later.

5 Open-open pipe

Let a pipe of length

(17)

be open at both ends.

Choose

(18)

and

(19)

at the ideal pressure nodes.

The first boundary condition is

(20)

Since

(21)

this gives

(22)

Therefore

(23)

The second boundary condition is

(24)

Hence

(25)

Therefore

(26)

Since

(27)

(28)

The open-open pipe supports all positive integer harmonics of the fundamental.

6 Open-open fundamental

For

(29)

(30)

Thus

(31)

The fundamental frequency is

(32)

The pressure has:

• nodes at both ends,
• an antinode at the midpoint.

Particle velocity has the complementary pattern:

• antinodes at both ends,
• a node at the midpoint.

7 Closed-closed pipe

A pipe rigidly closed at both ends has

(33)

Using

(34)

the condition at

(35)

gives

(36)

Therefore

(37)

The condition at

(38)

requires

(39)

Thus the nonzero acoustic modes have

(40)

and

(41)

The nonzero resonance frequencies therefore match those of the ideal open-open pipe, although the pressure and particle-velocity node patterns are interchanged.

8 Closed-open pipe

Now close the left end and open the right end.

At

(42)

the rigid boundary requires

(43)

Therefore

(44)

At the open end,

(45)

Thus

(46)

Therefore

(47)

so

(48)

The resonance frequencies are

(49)

Only odd multiples of the fundamental appear in this ideal one-dimensional model.

9 Quarter-wave fundamental

For the closed-open fundamental,

(50)

Therefore

(51)

The corresponding fundamental is

(52)

This is why a closed-open pipe is called a quarter-wave resonator.

An open-open pipe is a half-wave resonator because

(53)

10 Pressure nodes and velocity nodes

Pressure and particle velocity do not have nodes at the same places.

For an ideal standing wave in a uniform duct:

• a pressure node coincides with a particle-velocity antinode,
• a pressure antinode coincides with a particle-velocity node.

The spatial separation between a neighboring pressure node and pressure antinode is

(54)

11 Example 1: open-open pipe

Take

(55)

and

(56)

The ideal fundamental is

f1	=	(57)
	≈ 228.7 Hz.	(58)

The next two resonances are

(59)

and

(60)

12 Example 2: closed-open pipe

For the same physical length,

f1	=	(61)
	≈ 114.3 Hz.	(62)

The next ideal resonances are

(63)

and

(64)

13 Why the open-end condition is not exact

At a real open end, oscillating air just outside the physical pipe also participates in the motion.

The acoustic field therefore does not terminate abruptly at the geometric edge.

Instead, the effective pressure node lies a short distance outside the pipe.

This makes the acoustic resonator behave as though it were slightly longer than its measured geometric length.

Define

(65)

Here

(66)

is the open-end correction.

14 Approximate end corrections

The exact end correction depends on geometry.

For a circular pipe of radius

(67)

a commonly used low-frequency approximation for one unflanged open end is

(68)

For an idealized flanged circular opening, a larger value near

(69)

is often used.

These are approximations, not universal constants.

The correction depends on flange geometry, wall thickness, frequency, and nearby boundaries.

15 Open-open pipe with end correction

If both ends are approximately identical and unflanged,

(70)

The corrected fundamental is approximately

(71)

Since

(72)

the corrected resonance frequency is lower than the ideal geometric-length prediction.

16 Closed-open pipe with end correction

If only the open end requires a correction,

(73)

Then

(74)

17 Example 3: effect of end correction

Consider an open-open circular pipe with

(75)

and radius

(76)

Use the unflanged estimate

(77)

Then

(78)

With two open ends,

(79)

The corrected fundamental is

f1	≈	(80)
	≈ 223.3 Hz.	(81)

The end correction lowers the prediction by about

(82)

for this example.

18 Driven acoustic resonance

A pipe resonance is observable when the acoustic system is driven near one of its natural frequencies.

As in M06-03, damping prevents unbounded growth and produces a finite resonance peak.

A simple single-mode approximation has the same mathematical form as a driven damped oscillator:

(83)

where

(84)

is now a modal coordinate.

Near an isolated acoustic resonance, the response is therefore approximately oscillator-like.

19 Quality factor

For a weakly damped isolated resonance, define

(85)

where

(86)

is the full width between the half-power frequencies.

Equivalently,

(87)

Large

(88)

means:

• a narrow resonance,
• relatively slow energy loss,
• longer ring-down time.

20 From distributed pipe resonance to lumped resonance

A long pipe resonance requires a spatial standing-wave pattern.

A Helmholtz resonator is different.

If the cavity dimensions are small compared with the acoustic wavelength, the pressure inside the cavity is approximately uniform at any instant.

The resonator can then be approximated by:

• an oscillating plug of air in the neck,
• a nearly uniform compressible air volume in the cavity.

This is a lumped-element approximation.

21 Helmholtz resonator geometry

Let:

• V be the cavity volume,
• S be the neck cross-sectional area,
• L_eff be the effective neck length,
• ρ₀ be the equilibrium air density.

Let

(89)

be the displacement of the neck air plug.

The volume displaced into the cavity is approximately

(90)

22 Effective mass of the neck air

The mass of the moving neck plug is approximated by

(91)

The use of

(92)

rather than the geometric neck length accounts approximately for fluid outside the neck that participates in the oscillation.

23 Cavity pressure as a spring

The cavity air is compressed by the volume change.

For a small adiabatic compression,

(93)

Since

(94)

(95)

The pressure force on the neck plug is

(96)

Therefore

(97)

The effective acoustic spring constant is

(98)

24 Helmholtz resonance frequency

Newton’s second law for the neck plug is

(99)

Thus

(100)

Substitute

(101)

and

(102)

Then

ωH2	=	(103)
	= .	(104)

Therefore

(105)

The ordinary frequency is

(106)

25 Acoustic compliance

The cavity can also be described through its acoustic compliance.

Define volume displacement

(107)

From

(108)

the magnitude of the compliance is

(109)

The neck provides acoustic inertance

(110)

Then

(111)

This is exactly the mathematical structure of a mass-spring oscillator.

26 Example 4: bottle-like Helmholtz resonator

Take

(112)

which is

(113)

Let the neck radius be

(114)

so

(115)

Take

(116)

and

(117)

Then

fH	=	(118)
	≈ 136.8 Hz.	(119)

27 How geometry changes Helmholtz frequency

From

(120)

we can read off the trends.

Increasing neck area:

(121)

Increasing cavity volume:

(122)

Increasing effective neck length:

(123)

These trends have direct physical interpretations:

• larger neck area strengthens the pressure restoring force relative to neck inertia,
• larger cavity volume makes the cavity more compliant,
• a longer neck increases the moving air mass.

28 Neck end correction

The effective neck length is longer than the geometric neck length because some air just outside the neck participates in the motion.

Write

(124)

The exact correction depends on:

• whether each end is flanged or unflanged,
• neck radius,
• wall thickness,
• cavity geometry,
• frequency.

For introductory calculations, the correction is often represented by a geometry-dependent multiple of the neck radius.

It should not be treated as a universal single number.

29 Helmholtz resonance versus pipe resonance

The distinction is important.

A pipe resonance is a distributed standing-wave phenomenon.

Its frequency scales primarily as

(125)

A Helmholtz resonance is a lumped inertance-compliance oscillation.

Its frequency scales as

(126)

A cavity can possess both a low-frequency Helmholtz resonance and higher-frequency distributed acoustic modes.

30 Validity of the Helmholtz approximation

The lumped Helmholtz model requires the cavity to be acoustically compact.

A useful condition is that the cavity’s characteristic dimensions be small compared with the acoustic wavelength:

(127)

Then pressure can be approximated as nearly uniform throughout the cavity.

If the frequency becomes high enough that significant spatial pressure variation develops inside the cavity, distributed cavity modes must be included.

31 Losses and resonance width

Real acoustic resonators lose energy through:

• viscous boundary layers,
• thermal conduction,
• radiation from openings,
• turbulence,
• material losses,
• leakage.

These effects broaden the resonance and reduce its peak amplitude.

A practical Helmholtz resonator is therefore better modeled as a damped oscillator:

(128)

The same resonance concepts developed in M06-03 apply.

32 Example 5: bandwidth and quality factor

Suppose an acoustic resonance has

(129)

and measured half-power frequencies

(130)

Then

(131)

Thus

(132)

33 Connection to acoustic impedance

At resonance, the reactive parts of an acoustic system can cancel.

In a lumped Helmholtz model:

• neck inertance contributes an inertia-like reactance,
• cavity compliance contributes a spring-like reactance.

At

(133)

their magnitudes balance.

This impedance viewpoint is the acoustic analogue of resonance in mechanical and electrical oscillators.

34 Common mistakes

1. Calling an open end a particle-velocity node rather than a pressure node.
2. Calling a rigid closed end a pressure node rather than a pressure antinode.
3. Forgetting that pressure and particle-velocity nodes are interchanged.
4. Using the open-open spectrum for a closed-open pipe.
5. Forgetting that a closed-open pipe supports odd harmonics in the ideal model.
6. Writing the quarter-wave fundamental as c∕(2L) instead of c∕(4L).
7. Treating the physical pipe length as exactly equal to the acoustic length.
8. Applying a fixed numerical end correction without specifying geometry.
9. Adding only one open-end correction to an open-open pipe with two similar ends.
10. Confusing a Helmholtz resonance with a quarter-wave pipe resonance.
11. Omitting the effective neck length from the Helmholtz formula.
12. Treating the cavity pressure as uniform when the cavity dimensions are not small compared with wavelength.
13. Forgetting that resonance peaks have finite width in real systems.
14. Confusing the half-power bandwidth with half the bandwidth.
15. Using a high-frequency distributed cavity mode formula when the intended model is a lumped Helmholtz resonator.

35 Practice exercises

1. Starting from P′′ + k²P = 0, derive the open-open pipe spectrum.
2. Derive the nonzero closed-closed pipe spectrum.
3. Derive the closed-open pipe spectrum.
4. Draw the pressure and particle-velocity node patterns for the first three open-open modes.
5. Draw the pressure and particle-velocity patterns for the first three closed-open modes.
6. Reproduce Examples 1 and 2 numerically.
7. Apply a 0.6a end correction to a specified circular open pipe and compare the corrected and uncorrected fundamentals.
8. Explain physically why end correction lowers resonance frequency.
9. Starting from a neck air plug of mass ρ₀SL_eff, derive the Helmholtz frequency.
10. Derive the acoustic compliance C_a = V∕(ρ₀c²).
11. Derive the acoustic inertance M_a = ρ₀L_eff∕S.
12. Show that ω_H = 1∕.
13. Reproduce the bottle-like Helmholtz numerical example.
14. Determine how f_H changes if cavity volume is quadrupled while all other parameters remain fixed.
15. Given f₀ and half-power frequencies, compute Q.

36 Summary

For a uniform ideal open-open pipe,

(134)

For a uniform ideal closed-open pipe,

(135)

An open end is approximately:

• a pressure node,
• a particle-velocity antinode.

A rigid closed end is:

• a pressure antinode,
• a particle-velocity node.

Real open ends require an effective length

(136)

For a Helmholtz resonator,

(137)

and

(138)

Therefore

(139)

and

(140)

The acoustic-compliance form is

(141)

Thus

(142)

Pipe resonance and Helmholtz resonance are therefore two different realizations of the same broader idea: energy repeatedly exchanges between inertial and restoring mechanisms while boundary conditions select the allowed motion.

References

References