Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Question 1.
To assess the environmental noise in a residential area, within the duration of different
periods the equivalent noise levels measured are as below:
Between 7 am and 19 pm:
First period :
Second period :
Third period :
Fourth period :
2 hours, 65 dBA
3 hours, 54 dBA
4 hours, 50 dBA
3 hours, 60 dBA
Between 19 pm and 23 pm:
First period :
Second period :
2 hours, 50 dBA
2 hours, 48 dBA
Between 23 pm and 07 am:
First period :
Second period :
4 hours, 45 dBA
4 hours, 43 dBA
Find the values;
a) day equivalent noise level, 𝐿𝑑𝑎𝑦 ,
b) evening equivalent noise level, 𝐿𝑒𝑣𝑒𝑛𝑖𝑛𝑔 ,
c) night equivalent noise level, 𝐿𝑛𝑖𝑔ℎ𝑡 ,
d) 24 hours equivalent noise level, 𝐿(24ℎ).
Answer 1.
1
𝐿𝑑𝑎𝑦 = 10 log10 [12 (2 × 1065⁄10 + 3 × 1054⁄10 + 4 × 1050⁄10 + 3 × 1060⁄10 )]
𝐿𝑑𝑎𝑦 = 59.4 dBA
1
𝐿𝑒𝑣𝑒𝑛𝑖𝑛𝑔 = 10 log10 [2 (1050⁄10 + 1048⁄10 )] = 49.1 dBA
1
𝐿𝑛𝑖𝑔ℎ𝑡 = 10 log10 [2 (1045⁄10 + 1043⁄10 )] = 44.1 dBA
1
𝐿(24ℎ) = 10 log10 [24 (12 × 1059.4⁄10 + 4 × 1049.1⁄10 + 8 × 1044.1⁄10 )] = 56.6 dBA
29.11.2022 / Rev. 3
Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Question 2.
In a free field, the sound pressure level, 𝐿𝑝 , that a point sound source with sound power level,
𝐿𝑊 , will create at a distance 𝑟 from the source can be found from the equation:
𝐿𝑝 = 𝐿𝑊 + 10 log10 [
𝑄
],
4𝜋𝑟 2
where directivity factor Q = 1 for free-space, factor Q = 2 for source centered in a large flat
surface, factor Q = 4 for source centered at the edge formed by the junction of two large flat
surfaces, and Q = 8 for source centered at the corner formed by the junction of three large
flat surfaces.
A machine with a total sound power level of 88 dB is placed outdoors on a concrete floor. Find
the sound pressure level at a receiver point 3 m from the machine. What can we say about
the sound level at the same receiver point? (Q
Answer 2.
𝐿𝑊 = 88 dB, 𝑟 = 3 m, 𝑄 = 2.
𝐿𝑝 = 88 + 10 log10 [
2
] = 70.5 dB
4𝜋32
Question 3.
A 3000kg machine is placed on 12 support members. Each support element is on an
elastomeric pad which is produced by superimposing two elastic wedges with spring
coefficients of 4 MN/m and 2 MN/m. Obtain the simplest model of the system in which we
can examine vertical vibrations, ignoring damping. What is the disturbing frequency that this
system will perform well under?
Hint: Disturbing frequency must be about 3x larger than the natural frequency of the
system.
29.11.2022 / Rev. 3
Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Answer 3.
The simplest model for such a system can be considered as a SDOF spring-mass system. Since
damping is ignored, 𝑐 = 0. Finding the equivalent spring constant is possible by considering
that a single supporting member is a series of two springs. For such an isolator, the equivalent
spring constant can be calculated from
1
1
1
1 1 3
4
= +
= + = → 𝑘𝑒 = MN/m.
𝑘𝑒 𝑘1 𝑘2 4 2 4
3
Since there are 12 supports in parallel:
𝑘 = 12𝑘𝑒 = 16 MN/m.
Since 𝑚 = 3000 kg,
𝑓𝑛 =
1 16 × 106 N/m
√
= 11.6 Hz
2𝜋
3000 kg
Usually, it is expected that the disturbing frequency of a machine to be three times higher
than the natural frequency of the system. Therefore, a machine with a disturbing frequency
of
𝑓𝑑 ≥ 30Hz,
would be recommended to be used with this system.
Question 4 / Answer 4
Two equal sources produce a …3… dB increase in sound power level.
Question 5 / Answer 5
Frequency of vibrations in which a system can store energy is called …the resonant
frequency... .
Question 6 / Answer 6
What does the free-field sound propagation equation imply in terms of distance to noise
level relation? Hint: 𝐿𝑝 = 𝐿𝑊 − 10 log10 (4𝜋𝑟 2 )
Sound pressure level decreases 6 dB per doubling of distance.
29.11.2022 / Rev. 3
Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Question 7
A room has a 10 m length, 8 m width and 3 m height. The ceiling has a suspended ceiling
system that has an absorption coefficient of 0.5 and the floor is covered with a carpet having
an absorption coefficient of 0.25. The absorption at the walls may be considered negligible.
(1) What is the average Sabine absorption coefficient of the space?
(2) What is the reverberation time in this space?
(3) What is the sound pressure level at 5m to the source if a source with sound power of
10−6 W is operated at the middle of the room on the floor?
Answer 7
(1)
Total surface area:
𝑆 = 𝑆ceiling + 𝑆floor + 𝑆walls = 2(8 × 10) + 2(10 × 3) + 2(8 × 3) = 268 m2
Average Sabine absorption coefficient:
1
1
(0.5 × 80 m2 + 0.25 × 80 m2 ) = 60⁄268
𝛼̅ = ∑ 𝛼𝑖 𝑆𝑖 =
𝑆
268 m2
𝑖
(2)
𝑇60 = 0.161
(8 × 10 × 3)
𝑉
= 0.161
= 0.644 s.
(0.5 × 80 m2 + 0.25 × 80 m2 )
𝐴
(3) Sound power level of the source is:
𝑊
10−6
) = 10 log10 ( −12 ) = 60 dB
𝐿𝑊 = 10 log10 (
𝑊ref
10
The classic relationship between source sound power level and room sound pressure
level requires:
𝑄
4(1 − 𝛼̅)
2
4(1 − 0.22)
𝐿𝑝 = 𝐿𝑊 + 10 log10 (
+
) = 60 + 10 log10 (
+
)
2
2
4𝜋𝑟
𝑆𝛼̅
4𝜋1
60
𝐿𝑝 ≅ 53.2 dB
29.11.2022 / Rev. 3
Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Question 8 / Answer 8
The unit used for expressing levels is decibels (dB).
Question 9 / Answer 9
In the figure below, you are given the time-history of two acoustic pressure signals.
From this plot find:
a. (5pt) the period and the frequency of the acoustic waves:
T = 5 / 1000 = 0.005s
f = 1/0.005s = 200Hz
b. (5pt) phase of the second acoustic pressure signal compared to the first one:
π
c. (10pt) amplitude of the acoustic pressure and the sound pressure levels:
A = 2.5 Pa
L = 20log10(2.5 Pa / 20µPa) = 101.9 dB
d. (10pt) the medium from the table below if the wavelengths are λ = 1.67 m.
c = f λ = 200 Hz x 1.67 m = 334 m/s
Substance
Carbon Dioxide
Hydrogen
Helium
Nitrogen
Oxygen
Air (21% Oxygen, 78% Nitrogen)
Air (20°C)
Water
Note: Be careful with the units!
29.11.2022 / Rev. 3
c (m/s)
259
1284
965
334
316
331
344
1493
Yeditepe University | ME 438 Engineering Acoustics | Fall Semester 2022-2023
Example Questions
Question 10 / Answer 10
Noise Criterion (NC) curves are used to determine the maximum noise levels allowed within a space.
As a mechanical engineer, you are trying to comply with the NC curve given by decreasing the noise
level within the mechanical space by removing one of the noise sources.
Octave-band frequency, f
NC
Source-1, L1
Source-2, L2
Source-3, L3
Total, LA,tot (dBA)
A-weighting
Total, Ltot (dB)
a.
b.
c.
d.
63 Hz
71 dB
39.5 dBA
40.2 dBA
38.1 dBA
44.1
-26.2
70.3
125Hz
64 dB
42.6 dBA
41.1 dBA
43.6 dBA
47.3
-16.1
63.4
250Hz
58 dB
46.7 dBA
42.1 dBA
44.6 dBA
49.6
-8.6
58.2
(5pt + 5pt) Find the total sound power levels for each octave-band given.
(5pt) Find the total sound power level for each source.
(5pt + 5pt) Find the total sound power level for all sources combined.
(5pt) Show if it is enough to remove source-1 to comply with the given NC.
Yes.
63Hz:
68.5 dB < 71 dB,
125Hz:
61.6 dB < 64 dB,
250Hz:
55.1 dB < 58 dB
29.11.2022 / Rev. 3
Total
--48.7
46.0
47.6
52.3
--71.3