GRE Physics Companion: Acoustic Resonance in Pipes and Helmholtz Resonators
For an ideal open-open pipe,
For an ideal closed-open pipe,
For a Helmholtz resonator,
Figure 1. A compact resonance strategy: identify the boundary conditions, determine whether the
system is distributed or lumped, choose the correct resonance formula, then include end
correction or damping when required.
1 High-value GRE facts
- Open end: pressure node, particle-velocity antinode.
- Rigid closed end: pressure antinode, particle-velocity node.
- Open-open pipe: fn = nc∕(2L).
- Closed-open pipe: fn = (2n − 1)c∕(4L).
- Closed-open pipes support odd harmonics in the ideal model.
- Real open ends make the effective acoustic length longer than the geometric length.
- Increasing effective length lowers resonance frequency.
- Helmholtz resonance is lumped, not a quarter-wave standing-wave mode.
- fH ∝
.
- For an isolated weakly damped resonance, Q = f0∕Δf.
Part I: Original GRE-style problems
Problem 1: open-open fundamental
An ideal pipe of length
is open at both ends. Its fundamental frequency is
- c∕(4L)
- c∕(2L)
- c∕L
- 2c∕L
- 4c∕L
Problem 2: closed-open fundamental
A pipe of length
is rigidly closed at one end and open at the other. Its fundamental frequency is
- c∕(4L)
- c∕(2L)
- c∕L
- 2c∕L
- 4c∕L
Problem 3: open-end boundary
At an ideal open pipe end,
- pressure has an antinode and particle velocity a node
- pressure has a node and particle velocity an antinode
- both pressure and particle velocity have nodes
- both have antinodes
- neither has a standing-wave condition
Problem 4: closed-open harmonic content
The first three ideal resonances of a closed-open pipe are proportional to
- 1, 2, 3
- 1, 3, 5
- 2, 4, 6
- 1, 4, 9
- 1, 2, 4
Problem 5: end correction
Including a positive open-end correction generally
- increases effective length and lowers frequency
- increases effective length and raises frequency
- decreases effective length and lowers frequency
- does not affect resonance frequency
- changes sound speed rather than acoustic length
Problem 6: numerical open pipe
For
the ideal open-open fundamental is closest to
- 57.2 Hz
- 114 Hz
- 229 Hz
- 343 Hz
- 457 Hz
Problem 7: Helmholtz scaling with volume
If the cavity volume of a Helmholtz resonator is increased by a factor of
with all other parameters fixed, its resonance frequency becomes
- four times as large
- twice as large
- unchanged
- one half as large
- one fourth as large
Problem 8: Helmholtz scaling with neck area
If neck area is increased by a factor of
while cavity volume and effective neck length remain fixed, the Helmholtz frequency
becomes
- one fourth as large
- one half as large
- unchanged
- twice as large
- four times as large
Problem 9: effective neck mass
The effective moving mass in the elementary Helmholtz model is
- ρ0V
- ρ0SLeff
- ρ0S∕Leff
- V Leff∕ρ0
- ρ0cS
Problem 10: acoustic compliance
The acoustic compliance of the cavity is
- ρ0c2∕V
- V∕(ρ0c2)
- ρ0V∕c2
- c2∕(ρ
0V )
- S∕(ρ0c)
Problem 11: numerical Helmholtz resonance
A resonator has
and
Take
The resonance frequency is closest to
- 34 Hz
- 69 Hz
- 137 Hz
- 274 Hz
- 686 Hz
Problem 12: quality factor
A resonance has
and half-power frequencies
and
Its quality factor is
- 5
- 10
- 20
- 25
- 50
Part II: Complete worked solutions
Solution 1
For an open-open pipe,
Answer: (B).
Solution 2
For a closed-open quarter-wave resonator,
Answer: (A).
Solution 3
An ideal open end is approximately a pressure node and particle-velocity antinode. Answer:
(B).
Solution 4
The closed-open spectrum is
Thus the sequence is
times the fundamental. Answer: (B).
Solution 5
A positive end correction increases the effective acoustic length. Since
the resonance frequency decreases. Answer: (A).
Solution 6
| f1 | =  | (21)
|
| ≈ 228.7 Hz. | (22) |
Answer: (C).
Solution 7
For a Helmholtz resonator,
Therefore multiplying
by
divides the frequency by
Answer: (D).
Solution 8
Since
multiplying
by
doubles the resonance frequency. Answer: (D).
Solution 9
The neck plug has approximate volume
Thus
Answer: (B).
Solution 10
The cavity pressure-volume relation gives
Answer: (B).
Solution 11
| fH | =   | (33)
|
| ≈ 136.8 Hz. | (34) |
Answer: (C).
Solution 12
The bandwidth is
Thus
Answer: (D).
2 GRE checklist
- Translate open and closed ends into pressure and particle-velocity node conditions first.
- Use c∕(2L) only for half-wave fundamentals such as open-open pipes.
- Use c∕(4L) for an ideal closed-open quarter-wave fundamental.
- Remember the odd-harmonic sequence for a closed-open pipe.
- End correction increases acoustic length and therefore lowers resonance frequency.
- Distinguish a pipe standing-wave resonance from a Helmholtz lumped resonance.
- Read Helmholtz scaling directly from fH ∝
.
- For an isolated weakly damped resonance, use Q = f0∕Δf.
References
References
[1] PhysicsLibrary, M06-11, Acoustic Waves, Sound Intensity, and Acoustic Impedance.
[2] PhysicsLibrary, M06-03, Driven Harmonic Motion and Resonance.
[3] A. P. French, Vibrations and Waves, W. W. Norton, 1971.
[4] L. E. Kinsler, A. R. Frey, A. B. Coppens, and J. V. Sanders, Fundamentals of
Acoustics, 4th ed., Wiley, 2000.