The document discusses the decibel scale used to measure sound intensity levels. It explains that the decibel scale is logarithmic since sound intensities detected by the human ear can range from 10-12 W/m2 to 10 W/m2, or 1013 W/m2. The reference point for the decibel scale is set at the lowest intensity the human ear can detect, 0 decibels (10-12 W/m2). Any other sound intensity (I) can be expressed in decibels using the formula: β(I) = 0dB + 10 log10 (I/I0). Two examples are provided showing how decibel increases correspond to multiplicative increases in sound intensity.

1. The Decibel Scale
Brandon Morton PHYS 101 (L2I)

2. The Ear
 The quietest sound the ear can detect has an intensity
of 10-12 W/m2
 The loudest sound the ear can detect has an intensity
of 10 W/m2
 Due to this large range (1013 W/m2), it is convenient to
use a logarithmic scale
 The decibel is a common unit for logarithmic
comparison of power or intensity

3. Reference Point
 When using a logarithmic scale we need to define a
reference point.
 Since the lowest intensity the human ear can detect is
10-12 W/m2, we will set that to 0 decibels (I0 = 10-12
W/m2)
 To express any sound intensity (β) in decibels we use:
β(I) = 0dB + 10 log10 (I/I0)

4. Sample Question
 A mosquito's buzz is often rated with a decibel rating of
40 dB. Normal conversation is often rated at 60 dB.
How many times more intense is normal conversation
compared to a mosquito's buzz?
 a. 2
 b. 20
 c. 100
 d. 200

5. Answer
 Since the decibel scale is logarithmic each time you
increase by 10dB the intensity increases by 10 times.
 In this example, the sound increases by 20 dB which
means the intensity will be 100 times more intense.
 This is because it is 102 = 10 x 10 = 100
 Giving an answer of C. 100

6. Another Question
 Given that prolonged exposure to sounds greater
than 85 decibels can cause permanent hearing
damage, do you need to wear ear protection when
relaxing at the gun range?
 Given that a gunshot from a 0.22 rim-fire rifle has an
intensity of approximately I = (2.5 ×1013)I0

7. Answer
 We need to put the intensity of I = (2.50 ×1013)I0 into
the equation: β(I) = 0dB + 10 log10 (I/I0)
 β(I) = 10 log10 ((2.5 ×1013)I0 /I0)
 = 10 log10 (2.5 ×1013)
 = 133.979 dB
 = 1.34 x 102 dB

8. Answer Continued
 This means that at the rifle range I would be subjected
to a noise level of approximately 134 dB.
 Since this is significantly above the 85 dB level,
Hearing loss will occur.
 Therefore, I should wear the ear protection.

9. Bibliography
 http://www.cardinalspellman.org/ourpages/auto/2014/2/
13/40259160/Answers%2010%20Sound.pdf (accessed
February 21st 2015)
 http://logsanddecibles.blogspot.ca/ (Accessed February
21st 2015)