Noise - Answer
Noise - Answer
Noise - Answer
1. Example of engineering noise control is use of sound absorbing materials NOT Use
of earmuffs NOR Reducing the number of hours of exposure NOR Audiometric
testing. Engineering controls are used to permanently eliminate or lessen the severity
of a hazardous condition. In this case, use of sound absorbing materials for control of
over-exposure to noise is a good example of engineering control
2. The long-term cumulative effect of repeated and prolonged noise exposure, which
results in permanent pathologic changes in the cochlea, is referred to as: Noise-
induced hearing loss NOT Conductive hearing loss NOR Temporary threshold shift.
NOR Presbycusis. // Sensory hearing losses are associated with irreversible damage
to the inner ear. Noise produces primarily sensory hearing loss
Ans. C
4. Which of the following type of ear protectors has maximum noise reduction
capabilities ? (a) ear plugs (b) √Ear muffs (c) No protection (d) None
Ans (b)
5. For a sound level of 115 decibel, which of the following ear protectors is
suitable. (a) √Ear plug (b) Ear muff. (c) No plug required (d) None of the above
Ans (a)
Noise measurement instruments are classified into sound level meters (SLM), integrating
sound level meters (ISLM), and spectrum analyzers (SA). The SLM, which is the most
common instrument for the measurement of sound levels, can be Type 1 for precision
measurements or Type 2 for general-purpose use. The ISLM instruments are designed for
situations where there is a need to measure the average sound levels over a period of time.
Spectrum analyzers identify the distribution of sound levels in frequency.
Decibel addition
Table 1 can be used for decibel addition. Decibel addition relies on a few simple rules.
When decibels are measured on A-scale, they are referred to as dBA.
1. When two decibel levels are equal or within 1 dB of each other, their sum is 3 dB
higher than the higher individual level. For example, 80 dBA + 80 dBA = 83
dBA; 100 dB + 101 dB = 104 dB.
2. When two decibel levels are 2 or 3 dB apart, their sum is 2 dB higher than the
higher individual level. For example, 87 dBA + 85 dBA = 89 dBA; 75 dBA + 77
dBA = 79 dBA.
3. When two decibel levels are 4 to 9 dB apart, their sum is 1 dB higher than the
higher individual level. For example, 80 dBA + 86 dBA = 87 dBA; 32 dB + 36
dB = 37 dB.
4. When two decibel levels are 10 or more dB apart, their sum is the same as the
higher individual level. For example, 80 dB + 95 dB = 95 dB.
0 3.0
1 2.6
2 2.1
3 1.8
4 1.4
5 1.2
6 1.0
7 0.8
8 0.6
9 0.5
10 0.4
11 0.3
12 0.2
Table 1 Obtaining decibel sum of two decibel levels. This table is, in
essence, a summary of adding logarithms.
In situations where several sound levels have to be added to obtain the combined effect,
the following procedure can be used:
1. Add the two lowest decibel levels.
2. Add the result of step 1 to the next higher level.
3. Continue until all decibel levels have been added.
Example
Assume that five machines are operating in a machine shop, and each is generating a
sound level as summarized below:
Machine 1: 79 dBA
Machine 2: 85 dBA
Machine 3: 87 dBA
Machine 4: 95 dBA
Machine 5: 93 dBA
Solution
In order to calculate the combined sound level, we proceed as follows:
79 dBA + 85 dBA = 86 dBA
86 dBA + 87 dBA = 90 dBA
90 dBA + 93 dBA = 95 dBA
95 dBA + 95 dBA = 98 dBA
Therefore, the combined effect from all five machines operating simultaneously is
equivalent to a sound level of 98 dBA.
8
t
29590 / 5
8 8
t 5/5
2 2
t 4 hours
Noise Dose
In many work situations, a worker may be exposed to varying levels of noise during a
given shift. If the value of the following exceeds 1, the OSHA allowable times have been
exceeded, and the worker has been exposed to excessive noise.
C1 C1 C1
1
t1 t1 t1
where:
C1, C2, C3, … = actual exposure time
t1, t2, t3, … = allowable time
Example
During a given work shift, a worker has been exposed to noise levels of 100 dB for one
hour, 95 dB for three hours, and 92 dB for four hours. Determine whether this worker has
been exposed to excessive noise.
Solution
C1 1 hour
C2 3 hours
C3 4 hours
8
t100 2 hours
210090 / 5
8
t95 9590 / 5 4 hours
2
8
t92 9290 / 5 6 hours
2
Now we must calculate the value of
C1 C2 C3
t1 t2 t 3
1 3 4
19
. 1
2 4 6
Since the above value exceeds 1, we have exceeded the allowable time, and this worker
has been exposed to excessive noise.
Physiology of Hearing
Sound vibrations are transferred through the ear canal to the eardrum and into the middle
ear. The middle ear contains three bones called the ossicles that transmit sound waves to
the inner ear. The hearing part of the inner ear is called cochlea.
a) Presbycusis
Presbycusis is loss of hearing as a result of aging.
a) cure
b) prevention
c) research
d) data collection
A. B
Q8. Epidemiological studies that attempt to note the number of cases of a specific disease
in a specific time period are generally known as:
a) inductive studies
b) biological studies
c) descriptive studies
d) risk factor studies
A. C
a) descriptive studies
b) risk factor studies
c) analytic studies
d) hypothetical studies
A. C
Q10. In an epidemiological study 2 out of 1,000 people were found to have a disease. The
prevalence for this study is:
a) 0.002
b) 0.004
c) 0.001
d) 2
A. A
Prevalence = 2 = 0.002
1,000
1. What is the noise dose for a worker who is exposed to a 95 dBA source for a period of
5 hours? (a) 165 % (b) 125% (c) 93% (d) 315%
Ans (b) The noise dose can be calculated from
Where Ci is the exposure time and the allowable time. The allowable time can be
calculated from
Where L is the exposure in decibels
For the problem at hand
2. A worker is exposed to noise level of 110 dBA for one hour, 95 dBA for 3 hours. The
noise dose for this worker is : (a) 50% (b) 275 % (c) 425 % (d) 312 %
Ans. (d_ First, we calculate allowable times for noise levels of 110, 95 and 90dB
3. Two machines each generate a noise level of 80dBA. When both machines work
simultaneously, the noise level is approximately : (a) 85dBA (b) 83 dBA (c) 86 dBA (d)
160 dBA
Ans. (b) When in a free filed and the difference between two sound levels is zero, we
add 3dB to obtain the combined effect.
The noise does expressed in percentage of the OSHA permissible exposure is : (a) 423%
(b) 87% (c) 219% (d) 215%
Ans (c) First, calculate allowable times.
5. A square ventilation duct 2 feet on each side connects two rooms through a 20feet
length of the duct. A noise reduction of 0.2 dB/ft. duct is required. What should the
sound absorption coefficient of the lining material be ? (a) 0.9716 (b) 0.612 (c) 0.526
(d) 0.186
Ans (d) We use the following equation.
Where NR is the noise reduction in decibels perimeter of the duct in inches, and A is the
cross section area of the duct in square inches.
For the problem at hand.
6. What is the noise reduction in decibels per foot of a 2ft. diameter circular duct lined
with material having a sound absorption coefficient of 0.75? (a) 1.4 (b) 2.3 (c) 3.8 (d)
2.9
Ans. (a)
8. A sound wave has a sound pressure level of 95 dB with respect to 20 micropascals. The
root mean square sound pressure of this wave expressed in units of pascals is : (a)
9500 (b 5.2 (c) 1.12 (d) 0.8
Ans. (c)
10. The sound pressure level at a distance of 4 ft. from a sound source is 96 decibels.
The sound pressure level at 12ft. away from this sound sources is : (a) 90 decibel (b) 82
decibel (c) 86 decibel (d) 94 decibel
Ans (c) The relationship between sound level and distance is expressed by the following
relationship :
Where : dB1 is sound level at distance d1
dB0 is sound level at distance do
For the problem hand, we have
dB1 = 96 + 20 log
11. If the distance from a sound source is doubled, the reduction in sound pressure
level is : (a) 4 decibel (b) 8 decibel (c) 12 decibel (d) 6 decibel
Ans. (d) We start with
iB1 = dB0 + 20 log
This means that if we double our distance from a sound source, the sound pressure
level decreases by decibels.
12. In industrial noise control, every attempt must be made to reduce noise level by
means of engineering controls. One way to accomplish this is treatment of ceiling
with sound absorbing materials such as fiberglass. In a machine shop with a ceiling area
of 6000 ft2 and sound absorption coefficient of 0.04, the ceiling was treatment with a
sound absorbing material which has a coefficient of absorption of 0.8. The amount
of noise reduction in this machine shop is : (a) 13 decibel (b) 48 decibel (c) 24 decibel
(d) 8 decibel
Ans. (a) The relationship for calculation of noise reduction is :
Where :
dB = amount of noise reduction in decibels
A1 = amount of absorption before treatment expressed in units of “sabins”
A2 = amount of absorption after treatment in “sabins.
Values of A1 and A2 can be obtained by multiplying the ceiling area by the
coefficients of absorption. For the problem at hand.
13. The time weighted noise exposure for a worker is 95 decibels in a shift. The
noise dose is : (a) 120% (b) 199 % (c) 50% (d) 190%
Ans (b) The relationship between noise dose. ATWA is
TWA = time weighted noise exposure in dB
D = noise dose in percent. For the problem at hand.
14. The noise level at an operator's position I m away from a noise source is 98 dBA. A safety
professional has proposed a design change that will keep the operator 2 m away from the
noise source. What is the expected noise level at this new location, assuming a point source
and free field radiation?
A. 49 dBA
B. 89 dBA
C. 92 dBA
D. 95 dBA
15. Find the sound pressure level in dB when the operator moves to a new location 2 m away
from a noise source after having been I m away from the noise source.
Equation:
where
TLV-C refers to that concentration of a chemical, which can not be exceeded even
momentarily. It is an abbreviation for Threshold Limit Value Ceiling. It must be
pointed out that worker exposure may exceed the Threshold Limit Value Time
Weighted Average (TLV-TWA) as long as the eight hour average does not exceed
TLV-TWA and the value has never exceeded TLV-C during an eight hour work
period. The following graph shows the relationship between instantaneous exposures,
TLV-TWA and TLV-C.
Q14. A sound wave completes 360,000 cycles every hour. The frequency of this sound
wave is:
a) 3600 hertz.
b) 3600 megahertz.
c) 100 hertz.
d) 60,000 hertz.
A. C
hr 3600 seconds
second
Q15. The frequency of a sound wave is 4000 hertz. The amount of time required for one
complete cycle is:
a) 25 seconds
b) 25 x 10-5 seconds
c) 4 seconds
d) 2500 seconds
A. B
T=1
T= 1 = 25 x 10-5
4000
Q16. The root mean square sound pressure of a sound wave is 200,000 micropascals. The
Sound Pressure Level (SPL) in describes is:
a) 90
b) 75
c) 85
d) 80
A. D
20
a) 0.632 pascals
b) 0.90 pascals
c) 1.8 pascals
d) 1.2 pascals
A. A
20 micropacal
90 = 20log rms
20 micropacal
Take antilog of 90
20
Q18. The allowable time for a continuous noise is 2 hours. The noise level expressed in
describes (dBA) is:
a) 80
b) 100
c) 75
d) 85
A. B
T= 8
2(L-90)/5
2= 8
2(L-90)/5
2(L-90)/5 = 4
log2(L-90)/5 = log 4
or:
(L -90)log2 = log4
or:
L -90 = 2
L = 100
Q19. The allowable exposure time for a sound level of 90 decibel is:
a) 2 hours
b) 6 hours
c) 1 hour
d) 8 hours
A. D
T= 8
2(L-90)/5
T= 8
2(90-90)/5
T = 8 = 8 hours
20
Q20. In measuring sound levels in the work place for evaluation of employee noise
exposure, OSHA requires that all measurements be conducted on:
a) B, or C scales
b) A scale
c) A or B scales
d) A or C scales
A. B
Ossicles are bones in the human ear, and are also the smallest bones in the human
body. The major function is to transmit the sound wave to the inner ear for
transmission to the brain by the inner ear nerve cells.
a) conductive
b) convective
c) sensorineural
d) only (a) and (c) above
A. D
The adverse effects of noise on hearing can be categorized into conductive impairment,
sensorineural, mixed and central impairment. A conductive impairment is caused by any
condition that interferes with the transmission of sound to the cochlea. Excessive noise can
contribute to this type of hearing impairment. A sensorineural impairment is caused as a
result of damage to the organ of corti and nerve cells and is generally irreversible. Excessive
noise can contribute to this type of hearing impairment as well.
In a machine shop, two machines are working simultaneously. The noise generated by the
first machine is 87 dB and the noise generated by the second machine is 85 dB. The
combined noise effect expressed in dB is:
a) 172
b) 92
c) 115
d) 89
A. D
85 dB+87 dB = 89 dB
Q40. The combined effect of two sounds, one at 80 dB and the other at 86 dB is:
a) 87 dB
b) 166 dB
c) 85 dB
d) 93 dB
A. A
Since the two decibel levels differ by more than 4, we add 1 to the higher value.
80 dB+86 dB = 87 dB
Sound absorption coefficient indicates the amount of sound energy which is absorbed,
the remainder is reflected by the material.
Q42. “Noise reduction coefficient” refers to:
”Noise reduction coefficient” refers to the arithmetic average of the sound absorption
at frequencies of 250, 500, 1000, and 2000Hz.
Which of the following devices is used to measure noise energy across a range of
frequencies?
Octave band analyzers are used to measure the actual sound energy over a range of
frequencies. This data in turn is used to calculate the noise exposure of employees
In a machine shop, the 8-hour TWA noise exposure exceeds the regulatory levels. The
management decides to solve the problem by allowing workers to work four hours in
the machine shop and four hours in a quiet area. This is an example of
a) Engineering control.
b) Management control
c) Administrative control
d) Time management control
A. C
Engineering controls offer a permanent solution to a safety problem. In this case, the problem
is solved by job rotation, which is an example of administrative control
Q22. The forms of Noise Induced Hearing Loss (HIHL) are:
a) conductive
b) convective
c) sensorineural
d) only (a) and (c) above
A. D
A sound absorbing material has a sound absorption coefficient of 0.3, 0.6, 0.8, and 0.9 at
frequencies of 250, 500, 1000, and 2000 Hertz, respectively. The noise reduction
coefficient of this material is:
a) 0.45
b) 0.55
c) 0.65
d) 0.75
A. C
Q87. During a given work shift, a worker has been exposed to noise levels of 100 dB of one
hour, 95 dB for three hours, and 92 dB for four hours. The value of noise dose is:
a) 0.9
b) 0.8
c) 1.7
d) 1.9
A. D
t1 t2 t3
1 + 3 + 4 = 1.9 > 1
2 4 6
Since the above value exceeds 1, we have exceeded the allowable, and this worker has
been exposed to excessive noise
a) Continuous
b) Impact
c) Impulse
d) Amplify
A. D
Continuous noise is defined as the type of noise in the workplace that has a frequency
of occurrence of more that once per second
Q105. The time weighted average noise exposure for a worker is 95 decibles in a shaft. The
noise dose is:
a) 120%
b) 199%
c) 50%
d) 190%
A. B
Where:
TWA = 16.61log D + 90
100
95 = 16.61log D + 90
100
log D = 5
100 16.61
or
D = 1.99
100
D = 199%
Intermittent: a given broadband sound pressure level that occurs several times
during a normal workday.
Impact Type: a sharp burst of sound, generally less than one-half second in duration
and does not repeat more often than once per second.
A sound wave has a frequency of 5000 hertz. The amount of time required for one
complete cycle (in units of seconds) is:
a) 3.
b) 2 10-4.
c) 3 10-5.
d) 5.
Select (b)
Solution
1
T
F
1
T 2 10 4 seconds
5000
Decibels
The decibel (abbreviated dB) is the unit for expressing the sound pressure level relative to
2 10-10 atmosphere. In the metric system, this reference pressure is 2 10-5 newton/m2.
The unit termed “pascal” is defined as 1 N/m2; so the sound pressure level reference is
expressed as 2 10-5 pascal or 20 micropascals. Thus, to be technically correct, one
should say, “The sound pressure level is 75 decibels relative to 20 micropascals.” As this
is a universally recognized pressure base, it often is not stated, however, and one usually
says, “The sound pressure level is 75 dB.”
The sound pressure level (SPL) for any measured sound is defined by:
2
(rms sound pressure measured )
SPL in decibels = 10 log 2
(20 micropascals )
or
If we show the sound pressure level (in decibels) with Lp, the rms sound pressure
measured with P, and our reference pressure (20 micropascals) with P0, we can write:
P
LP ( in decibels ) 20 log
P0
Example 2
The sound pressure level of a sound wave is 6.3 105 micropascal. The decibel value
of this sound wave relative to 20 micropascal is
a) 63.
b) 72.
c) 90.
d) 85.
Select (c)
Solution
P
L p 20 log
P0
6.3 105
L p 20 log
20
L P 89.96 dB
1. When two decibel levels are equal or within 1 dB of each other, their sum is 3 dB
higher than the higher individual level. For example, 80 dBA + 80 dBA = 83 dBA;
100 dBA + 101 dBA = 104 dBA.
2. When two decibel levels are 2 or 3 dBA apart, their sum is 2 dBA higher than the
higher individual level. For example, 87 dBA + 85 dBA = 89 dBA; 75 dBA + 77
dBA = 79 dBA.
3. When two decibel levels are 4 to 9 dBA apart, their sum is 1 dBA higher than the
higher individual level. For example, 80 dBA + 86 dBA = 87 dBA; 32 dBA + 36
dBA = 37 dBA.
4. When two decibel levels are 10 or more dBA apart, their sum is the same as the
higher individual level. For example, 80 dBA + 95 dBA = 95 dBA.
Note: for every doubling of the distance from the source, the sound intensity is reduced by
approximately 6 decibels. For example, if the sound intensity at 10 ft. from a source is 96
dB, the intensity at 20 ft. is approximately 90 dB.
Difference between Two Amount to be Added to
Decibel Levels to be Larger Level to Obtain
Added (dB) Decibel Sum (dB)
0 3.0
1 2.6
2 2.1
3 1.8
4 1.4
5 1.2
6 1.0
7 0.8
8 0.6
9 0.5
10 0.4
11 0.3
12 0.2
Table 9. Obtaining Decibel Sum of Two Decibel Levels. This table is, in
essence, a summary of adding logarithms.
In situations where several sound levels have to be added to obtain the combined effect,
the following procedure can be used:
Example 3
Assume that five machines are operating in a machine shop, and each is generating a
sound level as summarized below:
Machine 1: 79 dBA
Machine 2: 85 dBA
Machine 3: 87 dBA
Machine 4: 95 dBA
Machine 5: 93 dBA
In order to calculate the combined sound level, we proceed as follows:
Therefore, the combined effect from all five machines operating simultaneously is
equivalent to a sound level of 98 dBA.
The term “noise reduction coefficient” (NRC) refers to the arithmetic average of the
absorption coefficient at frequencies 250, 500, 1,000 and 2,000 Hz.
Ei
STL = 10 log
Et
Where:
Ei: is the incident energy
Et: is the transmitted energy.
The STL increases with increasing frequency at a rate of about 5 dB for each doubling of
frequency.
OSHA Allowable Times for Noise Exposure
The allowable time that a worker can be exposed to a continuous noise can be obtained
from:
8
t L 90 /5
2
Where:
t is the allowable exposure time in hours
L is the sound intensity expressed in decibels.
Example 4
The allowable time that a worker can be exposed to a continuous noise level of 95
decibels in hours is:
a) 5.
b) 3.
c) 4.
d) 2.
Select (c)
Solution
8
t 95 90 /5 4 hours
2
Sound Level Meters A, B, C Response
An SLM can measure sound on either A, B, or C scales depending on the frequency of
sound. OSHA regulations require A-scale noise measurement for assessment of noise
exposure to employees because this scale most closely resembles the response of the
human ear. Other scales such as B or C may be utilized for evaluation of noise as part of
engineering control or design along with spectrum analysis.
Physiology of Hearing
Sound vibrations are transferred through ear canal to the eardrum and into the middle ear.
The middle ear contains three bones called the “ossicles” which transmit sound waves to the
inner ear. The hearing part of the inner ear is called “cochlea.”
a) Conductive impairment
Any condition which interferes with the transmission of sound waves to the cochlea.
b) Sensorineural impairment
This is an irreversible hearing loss caused by damage to the organ of “corti” and
degeneration of the neural elements of the auditory nerve.
c) Mixed impairment
Excessive noise can cause both conductive and sensorineural (mixed) impairment.
d) Central impairment
Refers to lack of ability of a person to interpret what is heard.
A high frequency noise is more harmful to human ear than a low frequency noise having
the same amount of sound energy.
Hearing Protection Equipment
Hearing protective devices can be classified into three general categories:
1. Ear plugs, or insert-type devices, which fit within the ear canal.
2. Ear muffs, which are devices that fit around the ears and are supported either from a hard
hat or from a head band that connects the individual muffs.
3. Canal Caps, which cover the entrance to the ear canal and are supported similarly to ear
muffs.
The noise reduction capability of hearing protective devices is a very important factor which
should always be considered.
Example 5
The root mean square sound pressure for a sound wave is 0.4 pascals. The sound pressure
level in decibels is:
a) 80.
b) 90.
c) 75.
d) 86.
Select (d)
The following relationship can be used to calculate the decibel value of a sound wave
from root mean square values P.
P
(decibels) = 20 log
20
0.4 106
dB = 20 log = 86
20
Relationship between noise dose and OSHA allowable times
In many workplace situations, an individual may be exposed to different levels of noise
during an 8-hour workday. The concept of noise dose is used to determine whether
OSHA allowable exposure times have been exceeded for a time varying noise level
exposure. Noise dose can be calculated from
n
ci
D 100
i 1 ti
Where:
D: is the noise dose
Ci: actual exposure times
ti: allowable times.
When noise dose is calculated using the above relationship, we can say that allowable
time has been exceeded if noise dose is greater than 1. Otherwise allowable time is not
exceeded. The following relationship can be used to calculate allowable times:
8
t (L 90) / 5
2
Where:
t is the allowable time in hours.
L is the sound pressure level in decibels.
We demonstrate how to use the above relationships with an example.
Example 6
What is the noise dose for a worker who is exposed to a 95 dB source for a period of 5 hours?
a) 165%.
b) 125%.
c) 93%.
d) 315%.
Select (b)
where Ci is the exposure time and ti the allowable time. The allowable time can be
8
calculated from t
2 ( L 90)/5
where L is the exposure in decibels.
8
t ( 95 90 )/ 5
4 hours
2
5
D 100
4
D 125%.
Relationship between noise reduction, perimeter and cross sectional area of a duct.
In many buildings, noise generated by a given source can be transmitted to other locations
through ventilation or heating ducts. In many instances ducts are lined with a sound
absorbing material. The amount of Noise Reduction (NR) expressed in decibels per foot
length of the duct can be obtained from the following relationship:
12.6P
1. 4
NR
A
Where:
NR is the sound absorption coefficient in decibels per foot.
P is the perimeter of the duct in inches
A is the cross sectional area of the duct in square inches
is is the absorption coefficient of the lining material.
Example 7
A square ventilation duct 2 feet on each side connects two rooms through a 20 feet length of
the duct. The sound absorption coefficient of the lining material is 0.2. The Noise
Reduction in dB/foot is:
a) 0.9.
b) 0.6.
c) 1.1.
d) 0.2.
Select (d)
12.6 8 12 0.21.4
NR
4 144
NR 0.2 dB / foot
Noise Induced Hearing Loss (HIHL) results from over-exposure to high levels of noise
.NIHL depends on sound intensity and t he exposure time. Which of the following
statements is generally true regarding NIHL ? (a) High frequency noise is more
damaging. (b) Low frequency noise is more damaging (c) NIHL depends only on the
sound intensity and time of exposure . (d) Only (b) and (c) above
Ans (a) Although there are some differences among individuals, exposure to high
frequency noise is more damaging to the ears than a low frequency noise having the
same sound intensity and for the same exposure time.
Ans. (c) TTS is caused as a result of over-exposure to loud noise which damages the
organ of Corti and its hair cells.
Sound absorption coefficient for a given material. (a) is more sensitive to frequency
changes than Noise Reduction Coefficient (NRC) (b) is less sensitive to frequency
changes than NRC. (c) is a synonym for NRC. (d) is a synonym for NRC and is
frequency independent.
Ans. (a) Sound absorption coefficient is given at a specific frequency. Noise reduction
coefficient is the average of sound absorption coefficient at frequencies of 250, 500,
1,000 and 2,000 HZ and is less frequency dependent than the sound absorption
coefficient.
Sound absorption coefficient for a given material. (a) is more sensitive to frequency
changes than Noise Reduction Coefficient (NRC) (b) is less sensitive to frequency
changes than NRC. (c) is a synonym for NRC. (d) is a synonym for NRC and is
frequency independent.
Ans. (a) Sound absorption coefficient is given at a specific frequency. Noise reduction
coefficient is the average of sound absorption coefficient at frequencies of 250, 500,
1,000 and 2,000 HZ and is less frequency dependent than the sound absorption
coefficient.
What is noise dose for a worker who is exposed to 95 dBA source for a period of 5 hours ?
a) 165% b) 125% c) 93% d) 31.5%
A worker is exposed to noise level of 110 dBA for one hour, 95 dBA for 3 hr., 90 dBA fro
3 hours. The noise dose for this worker is : a) 500% b) 275% c) 425% d) 312%
18 Two machines each generate noise level of 80 dBA. When both machines work
simultaneously, the noise level is approx. a) 85dBA, b) 83 dBA, c) 86 dBA, d)
160 dBA
Ans. (b)
When in a free hold and the difference between two sounds levels is zero, we add 3 dB
to obtain the combined effect.
20 A square ventilation duct 2 feet on each side connects two rooms through a 20 feet
length of the duct. A noise reduction of 0.2 dB/ft of duct is required. What should the
sound absorption coefficient of lining material be? A) 0.9176, B) 0.612, c) 0.526,
d) 0.186
Ans (d)
We use following equation - NR = 12.6 P a1.4
A
Where NR is noise reduction in decibels per foot, P is perimeter of deuct in inches, A
is cross section area of duct in square inches.
From this equation we can calculate a = 0.186
21 What is noise reduction in decibels per foot of a 2ft. diameter circular duct lined with a
material having a sound absorption coefficient of 0.75?
a) 1.4, b) 2.3, c) 3.8 or d) 2.9
Or we can say P/20 is equal to antilog of 4.75 (your calculator should have antilog
function)
From this equation we can calculate P = 1.12 pascal
24 The cochlea is main element of: a) inner ear b) middle ear c) outer ear d) tympanic
membrane
Ans(a)
The cochlea is main element of inner ear. It is filled with a fluid and shaped like a coil.
Sound waves travels through cochlea before reaching the hair like nerve cells for
transmission to brain.
25 Noise Induced hearing Loss (NIHL) results from over-exposure to high levels of noise.
NHIL depends on sound intensity and the exposure time. Which of the following
statements is generally true regarding NIHL ?
a) High frequency noise is more damaging.
b) Low frequency noise is more damaging.
c) NIHL depends only on sound intensity and time of exposure
d) Only b and c above
Ans.(a) Although there are some differences among individuals, exposure to high frequency
noise is more damaging to the ears than a low frequency noise having the same sound
intensity and for the same exposure time.
Ans(c) TTS is caused as a result of over exposure to loud noise which damages the organ of
Corti and its hair Cells.
The sound pressure level at a distance of 4 ft. from a sound is 96 decibels. The sound pressure
level at 12 ft. away from this sound source is - a) 90, b) 82 c 86 d) 94 decibel
If distance from a sound source is doubled, the reduction in sound pressure level is :-
a) 4 decibel, b) 8 decibel, c) 12 decibel d) 6 decibel
dB1 = dB0 + 20 log(d0/d1),, in this case d0/d1 = 0.5 distance doubled
we have to calculate dB1 - dB0
dB1 - dB0 = 20 log0.5 = - 6
In industrial noise control, every attempt must be made to reduce noise level by means of
engineering controls, one way to accomplish this is treatment of ceilings with sound absorbing
materials such as fiberglass. In a machine shop with a ceiling area of 6000 ft2 and sound
absorption coefficient of 0.04, the ceiling was treated with a sound absorbing material which has a
coefficient of absorption of 0.8. The amount of noise reduction in this machine shop is:-
a) 13, b) 48, c) 24 or d) 8 decibel.
For calculation of noise reduction is : dB = 10 log(A2/A1)
The time weighted average noise exposure for a worker is 95 decibels in a shift. The noise is
a) 120% b) 199% c) 50% d) 190%
Relationship between noise dose. A TWA is
TWA = 16.61 log(D/100)+90
TWA is time weighted noise exposure in dB, D is noise dose in percent.
From above equation we can calculate D = 199%
75 Which of following is an example of engineering noise control ?
a) Use of earmuffs, b) reducing number of hours or exposure, c) use of sound
absorbing material d) audiometric testing
Ans(c) Engineering controls are used to permanently eliminate or lessen severity of hazardous
conditions. In this case use of sound absorbing materials for control of over exposure to
noise is good example of engineering control.