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220

8 Acoustic Waveguides

Cutoff Frequency

Because

Therefore

The cutoff

 

wavenumber is the

 

LOWEST wavenumber

 

in both the x and y

units:

directions

Note that any pure tone with a frequency smaller than the cutoff frequency will have only a direct wave (m ¼ 0) since the corresponding wavenumber is too small to form a circle that intercepts any transverse eigenmode kym with m > 0.

As a result, any pure tone with a frequency smaller than the cutoff frequency have only direct wave and no echoes. When there are no echoes for any frequencies, the sound will have no distortion after passing through a waveguide and will be suitable for transmitting signals or sounds.

8.5Traveling Waves in Acoustic Waveguides

The traveling time of a wave from one end of the waveguide to the other end depends on the eigenmode which carries the pure tone. Larger mode numbers have larger propagation angles and will result in longer traveling times. On the other hand, smaller mode numbers have smaller propagation angles and will result in shorter traveling times. When the pure tone is not carried by any mode (m>0), it will travel parallel to the axial direction and result in the shortest possible traveling time.

The traveling time for the wave (direct or reected) to arrive at the other end of the waveguide can be formulated as:

Traveling Time ¼ DistanceVelocity

8.5 Traveling Waves in Acoustic Waveguides

221

where:

Velocity ¼ c kkz

Example 8.4: (Traveling Time in Waveguides)

An 850 [Hz] pure tone is traveling through a waveguide. Assume the waveguide is 340 [m] long and has a rectangular cross-section of Lx ¼ 13 [m] and Ly ¼ 12 [m].

(a)Calculate the traveling time of the pure tone moving in the axial direction as a direct wave (l ¼ m ¼ 0 mode) to arrive at the other end of the waveguide along the z-axis.

(b)Calculate the traveling times of the pure tone carried by all the eigenmodes to arrive at the other end of the waveguide. Use eigenmodes calculated in Example 8.3.

Example 8.4: Solution

Part (a)

The traveling time for the direct wave to arrive at the other end of the waveguide is:

 

Traveling Time

Distance

340 ½m&

1 s

 

 

 

 

!

 

¼ Velocity ¼

340

m

¼

½ &

 

s

Part (b)

The eigenmodes calculated in Example 8.3 are:

!

 

 

 

ðdirect waveÞ

 

k 00 ¼ 5πbez

 

!

 

 

 

 

p

 

 

 

 

k 01 ¼ 2πey þ

 

 

 

z

 

!

 

 

b

 

 

 

b

 

 

 

 

 

 

 

 

21πe

 

 

k

 

 

4πe

 

3πe

 

 

 

!02 ¼

 

by þ

 

bz

 

 

k

3πe

 

 

2bπe

 

 

b

12πe

 

10 ¼

3πex þ

4πez

 

 

!

 

x

þ

 

y

 

p

 

z

k 11 ¼

 

 

þ

 

b

 

b

 

 

 

b

Traveling time of (l, m, n) ¼ (0, 0, ) for the direct wave is:

time

 

distance

 

340 ½m&

 

1 s

 

 

 

 

 

 

 

 

kz

 

 

m

 

5π

 

 

 

¼

c

k

¼

340

s

 

5π

¼

½ &

Traveling time of (l, m, n) ¼ (0, 1, ) for the rst eigenmode in the y-direction is:

222 8 Acoustic Waveguides

time

 

distance

 

 

340 ½m&

 

 

5

s

 

¼

 

c kkz

¼

 

 

 

 

 

5

 

 

 

½ &

 

 

 

340

m

 

p21π

¼ p21

 

 

 

 

 

 

 

 

 

 

 

s

 

π

 

 

 

Traveling time of (l, m, n)

¼ (0, 2, )

for

the second eigenmode in the

y-direction is:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

time

 

 

distance

 

340 ½m&

5

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kz

 

 

 

 

m

3π

 

 

 

 

 

¼

 

 

c

k

¼

340

s

5π

¼ 3

½ &

 

Traveling time of (l, m, n) ¼ (1, 0, ) for the rst eigenmode in the x-direction is:

time

 

 

distance

 

 

 

340 ½m&

5

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kz

 

 

 

 

 

 

m

4π

 

 

 

 

¼

 

 

c

k

 

¼

340

s

5π

¼ 4

½ &

 

Traveling time of (l, m, n) ¼ (1, 1, ) is:

 

 

 

 

 

 

 

 

time

distance

 

 

 

340 ½m&

 

 

5

s

 

 

 

 

kz

 

 

 

 

 

 

 

 

5

 

 

 

 

¼

 

c

k

 

 

¼

340

m

 

12π

¼ p12

½ &

 

 

 

 

 

 

 

 

 

 

 

s

 

π

 

 

 

This pure tone will travel in this waveguide with ve modes. The difference in travel times between different eigenmodes and the direct wave will create echoes of the sound. A sound is usually containing many different frequencies.

When a sound is made of multiple frequencies, since the traveling times are different for different frequencies, even by the same eigenmode, the sound transmitted in the waveguide will be distorted.

Example 8.5: (Cutoff Frequency of Waveguides)

This example shows how to calculate the frequency range of an undisturbed pure tones in a narrow waveguide. Assume that a narrow waveguide has a square crosssection with the dimensions of Lx ¼ Ly¼ 0 .0085 [m].

Example 8.5: Solution

Since the dimensions in the x- and y-direction are the same, the cutoff frequencies (the lowest rst transverse rst mode) are the same for both directions.

The transverse frequency of the rst mode can be calculated as:

 

 

 

2π

 

2π

 

 

2

 

 

 

 

 

 

 

 

 

 

f y1

¼

ωy1

¼

 

c

ky1

¼

340

ms

 

 

π

1

¼

20000

½

Hz

&

 

 

 

 

 

 

π

 

 

0:0085 ½m&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where the wavenumber ky1 is calculated using:

8.6 Homework Exercises

223

 

 

π

 

kym ¼

 

m; m ¼ 1

 

Ly

The transverse frequency of the rst mode is called the cutoff frequency of the waveguide because it is the maximum frequency that could travel in the waveguide without an echo. Any frequency above the cutoff frequency will disturb the direct sound due to the echo carried by the transverse eigenmode, as shown in Sect. 8.3.

Therefore, the non-echo effect frequency range of this waveguide is 020 [kHz]. And the frequency range of sounds that will not disturb the direct sound due to the echo of the sound carried by the transverse eigenmode of the waveguide is 020 [kHz].

Remarks: Stethoscopes Are Waveguides with a Very High Cutoff Frequency

Typical stethoscopes have diameters around 0 .0085 [m]. Based on the example above, the cutoff frequency of stethoscopes is around 20 [kHz].

Since the human audible range is 20 [Hz] to 20 [kHz] and there are no echoes with frequencies below 20 [kHz] in this waveguide, any sound waves such as heartbeats can be clearly heard on the other side of a stethoscope. Frequencies above the cutoff frequency of 20 [kHz] will still result in echoes, but these echoes cannot be heard by a human. Stethoscopes are designed with a diameter less than 8.5 [mm] in order to eliminate/disable echo sound with frequencies below the cutoff frequency of 20 [kHz].

8.6Homework Exercises

Exercise 8.1: (2D Traveling Wave Velocity: Forward)

!

Derive the 2D forward ow velocity vector u þðx, y, tÞ of a 2D plane forward traveling wave from the 2D acoustic pressure p+(x, y, t) using Eulers force equation:

 

þð

pþðx, y, tÞ ¼ Aþ cos

ωt

 

kxx þ kyy þ θ

 

 

 

Þ ¼

b

ρ0c b

 

þ

 

 

 

 

x

 

þ

y

þ

!u

 

x, y, t

 

nxex þ nyey

A

 

cos

 

ωt

k

x

 

k

y

θ

 

 

 

 

 

where:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

nx ¼

 

kx

;

ny

 

 

 

 

ky

 

 

 

 

 

 

 

 

q

¼

q

 

 

 

 

 

 

 

kx2 þ ky2

 

kx2 þ ky2

 

 

 

Eulers force equation:

pðx, y, tÞ ¼ ρo

!

ðx, y, tÞ

 

 

 

 

t

u

 

 

 

 

Exercise 8.2: (Calculations of Wavenumber Vectors and Natural Frequencies)

The dimensions of a resonant cavity are Lx ¼ 1 [m], Ly

¼ 31

[m] and Lz ¼ 21 [m].

224

8 Acoustic Waveguides

!

(a)Calculate the wavenumber vectors k lmn of the eigenmodes (l, m, n) ¼ (2, 1, 3).

(b)Calculate the natural frequencies flmn of the eigenmodes (l, m, n) ¼ (2, 1, 3).

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

(Answers): (a) !k 213

¼ 2π m1

ex þ 3π m1

ey þ 6π m1

ez; (b) f 213 ¼

27 m1 c

 

 

 

b

 

b

 

b

 

 

 

Exercise 8.3: (3D Mode Shapes in a Rectangular Cavity)

The dimensions of a rectangular resonant cavity are given as (Lx, Ly, Lz) ¼ (3, 9, 24) [m]:

= 9 [ ]

(a)Plot nodal lines and indicate peaks and valleys of eigenmode (l, m, n) ¼ (0, 3, 6).

(b)Plot peak lines and valley lines, and draw four wavenumber vectors (without

values) of eigenmode (l, m, n) ¼ (0, 3, 6).

!

(c)Calculate the wavenumber vector k 036 and natural frequency f036 of eigenmode (l, m, n) ¼ (0, 3, 6).

!

(d)Sketch a wavenumber grid, and draw the wavenumber vector k 036 of eigenmode (l, m, n) ¼ (0, 3, 6) in the wavenumber grid.

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

Exercise 8.4: (Wavenumber Vectors in a Waveguide)

Awaveguide has a rectangular cross-section of Lx ¼ 0.75 [m] and Ly ¼ 1.5 [m].

(a)Calculate all the possible transverse eigenmodes that can carry a 283.33 [Hz]

!

pure tone in the waveguide. State the wavenumber vectors as k lm ¼

kxlbex þ kymbey þ kz bez.

(b)Calculate the cutoff frequency of the waveguide.

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

8.6 Homework Exercises

225

 

 

 

 

 

!

 

¼ 3

 

þ q

!

¼

4

 

þ

 

!

¼

3

 

þ

 

(Answers): (a)

k 01

 

 

πey

9

πez; k 02

 

3

πey

 

πez;

k 10

 

 

πex

 

πez;

!

 

¼

3

 

þ 3

 

 

þ q

 

 

 

 

b

 

b

 

 

 

b

 

b

 

 

4

b

2

 

b

 

5

b

113.33 [Hz]

 

 

 

 

 

k

11

 

 

πex

 

πey

 

 

 

 

 

 

 

 

 

 

 

 

 

b9πez; (b) b

 

 

 

 

 

 

 

Exercise 8.5: (Cutoff Frequency in Ventilation Fans)

A ventilation fan is installed on a long pipeline with a rectangular cross-section, generating a single-frequency noise that can be adjusted by modifying the speed of the fan motor. If the cross-section dimensions of the pipeline are Lx ¼0.75 [m], Ly ¼1.5[m]:

(a)Which transverse mode of the pipeline will carry the acoustic energy if the fan produces a noise signal at 115 [Hz]? State the mode number.

(b)What should the maximum frequency of the fan be so that the noise is not carried to long distances with any one of the transverse modes of the pipeline?

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

(Answers): (a) (l, m) ¼ (0, 1) mode; (b) 113.3 [Hz]

Exercise 8.6: (Echo Effect in Waveguides)

A 170 Hz pure tone travels in a cross-section waveguide with the dimensions of (Lx, Ly) ¼ (3, 4) [m].

Assuming that the length of the waveguide is 170 [m], calculate the:

(a)Traveling time of a direct wave (l, m) ¼ (0, 0) of the pure tone to arrive at the other end of the waveguide along the x-axis.

(b)Traveling time of the pure tone carried by the rst transverse eigenmode in the x-direction (l, m) ¼ (1, 0) to arrive at the other end of the waveguide.

(c)Traveling time of the pure tone carried by the rst transverse eigenmode in the y-direction (l, m) ¼ (0, 1) to arrive at the other end of the waveguide.

(d)What is the cutoff frequency of this waveguide?

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

(Answers): (a) 0.5 [s]; (b) 0.530 [s]; (c) 0.516 [s]; (d) 42.5 [Hz]

Exercise 8.7: (Echo Effect in Waveguide)

An 85 Hz pure tone travels in a cross-section waveguide with the dimensions of (Lx, Ly) ¼ (4, 5) [m].

Assume the length of the waveguide is 240 [m], calculate the:

(a)Traveling time of a direct wave ( m ¼ n ¼ 0 mode) of the pure tone to arrive at the other end of the waveguide along the x-axis.

(b)Traveling time of the pure tone carried by the rst transverse eigenmode in the x-direction (l, m) ¼ (1, 0) to arrive at the other end of the waveguide.

226

8 Acoustic Waveguides

(c)Traveling time of the pure tone carried by the rst transverse eigenmode in the y-direction (l, m) ¼ (0, 1) to arrive at the other end of the waveguide.

(d)What is the cutoff frequency of this waveguide?

Use 340 [m/s] for the speed of sound (c) in air. Show units in the Meter-Kilogram- Second (MKS) system.

(Answers): (a) 0.706 [s]; (b) 0.815 [s]; (c) 0.770 [s]; (d) 34 [Hz]

Chapter 9

Sound Pressure Levels and Octave Bands

In the eld of noise control, noise is quantied by sound pressure level (SPL). SPL can be either unweighted or weighted. In this chapter, you will learn how to calculate both unweighted SPL and weighted SPL. This chapter is divided into four sections and is summarized as follows:

Section 9.1 introduces and denes the decibel scale. The decibel scale is used for sound pressure level (SPL) and other power-like physical quantities in vibrations. In this section, you will learn how to calculate the combined level from separated levels of n incoherent radiating source using the formula below:

Ltot ¼ 10 log 10

10

L1 ½dB&

þ 10

L2 ½dB&

þ þ 10

Ln10½dB&

½dB&

10

10

Section 9.2 introduces and denes the sound pressure level (SPL). Parsevals theorem will be used to derive the formula for calculating SPL in the frequency domain. Parsevals theorem is important because it allows us to formulate RMS pressure in the frequency domain as shown in the owchart below:

TIME DOMAIN

 

 

 

 

FREQUENCY DOMAIN

 

 

 

FFT

 

 

 

 

 

 

 

 

 

 

 

 

Pressure in Time Domain

 

 

Pressure in Frequency Domain

 

 

 

 

 

+ ∑

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Unweighted RMS Pressure

 

 

Unweighted RMS Pressure

 

=

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Parsevals Theorem

Sound Pressure Level (SPL)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021

227

H. Lin et al., Lecture Notes on Acoustics and Noise Control, https://doi.org/10.1007/978-3-030-88213-6_9