Chapter 10: Sound
• Section 1: The Nature of Sound
• Section 2: Properties of Sound
• Section 3: Music
• Section 4: Using Sound
Section 1: The Nature of Sound
Sound waves
• All sound waves are created by something that vibrates
 A stereo speaker:
1) The vibrating diaphragm of the stereo speaker collides
with nearby air molecules, transferring some energy
to them
2) These molecule then collide with other molecules and
pass the energy to them
3) The process of molecular collision and energy transfer
form a sound wave
• A sound wave is a compressional wave
 Requires a medium through which to travel
 Sound can travel through solids, liquids, or gases
 The speed of sound is different in different media
 Solids  fast, liquids  slower, gases  slowest
• The speed of sound in air varies with temperature
 At 0oC, Vsound = 331.5 m/s
 As the temperature increases, speed increases
 As temperature falls below 0oC, speed decreases
Section 1: The Nature of Sound
• The equation for speed of sound:
V = 331.5
m
m
± 0.6 o T
s
sC
Where: V = speed of sound (m/s)
T = change in temperature (from 0oC to actual
temperature)
use “+” if the temp. is greater than 0oC
use “-“ if the temp. is less than 0oC
Example 1: If the temperature is 15oC, what is the speed of a
sound wave?
Solution
𝑚
𝑚
T = 15oC
𝑉 = 331.5 ± 0.6 𝑜 ∆𝑇
𝑠
𝑠 𝐶
V=?
𝑚
𝑚
∆T = 15o C − 0o C
𝑉 = 331.5 + 0.6 𝑜 (15𝑜 𝐶)
𝑠
𝑠 𝐶
∆T = 15o C
𝑚
𝑚
𝑉 = 331.5 + 9.0
𝑠
𝑠
𝒎
𝑽 = 𝟑𝟒𝟎. 𝟓
𝒔
Section 1: The Nature of Sound
Example 2: What is the speed of sound when the temperature
is –10oC?
Solution
𝑚
𝑚
𝑇 = −100 𝐶
𝑉 = 331.5 ± 0.6 𝑜 ∆𝑇
𝑠
𝑠 𝐶
𝑉 =?
𝑚
𝑚
∆𝑇 = 0𝑜 𝐶 − (−10𝑜 𝐶)
𝑉 = 331.5 − 0.6
10℃
𝑠
𝑠℃
∆𝑇 = 10℃
𝑚
𝑚
𝑉 = 331.5 − 6.0
𝑠
𝑠
𝒎
𝑽 = 𝟑𝟐𝟓. 𝟓
𝒔
Example 3: The temperature is 10oC, how far will a sound
travel in 10.0-s?
Solution
𝑇 = 10℃
𝑡 = 10.0𝑠
𝑑 =?
∆𝑇 = 10℃ − 0℃
∆𝑇 = 10℃
𝑚
𝑚
± 0.6 𝑜 ∆𝑇
𝑠
𝑠 𝐶
𝑚
𝑚
𝑉 = 331.5 + 0.6
10℃
𝑠
𝑠℃
𝑚
𝑚
𝑉 = 331.5 + 6.0
𝑠
𝑠
𝑚
𝑉 = 337.5
𝑠
𝑉 = 331.5
𝑑
𝑡
𝑑
𝑡𝑉 =
𝑡
𝑡
𝑑 = 𝑡𝑉
𝑉=
𝑑 = 10.0𝑠(337.5
𝒅 = 𝟑, 𝟑𝟕𝟓. 𝟎𝒎
𝑚
)
𝑠
Section 2: Properties of Sound
Intensity and Loudness
• The amount of energy carried by a wave corresponds to the
wave’s amplitude
• In a compressional wave, amplitude is related to the density
of the particles in the compressions and rarefactions
 Low energy = low density
 High energy = high density
• Intensity – the amount of energy that flows through a certain
area in a specific amount of time
 Intensity influences how far away a sound can be heard
 Intensity influences how far a wave will travel because
some of the wave’s energy is converted to other forms of
energy when it is passed from particle to particle
• Loudness is the human perception of sound intensity
 The scale for measuring intensity (or loudness) of a sound
is the decibel scale
Pitch and Frequency
• Pitch – how high or low a sound seems to be
• Pitch and wave frequency are related
 The higher the frequency of a sound wave, the higher the
pitch
• The human hearing range: 20-Hz to 20,000-Hz
 Ultrasonic waves –waves with frequencies greater than
20,000Hz
 Infrasonic waves –waves with frequencies less than 20-Hz
Section 2: Properties of Sound
The Doppler Effect
• Doppler Effect – the change in pitch or frequency due to a
moving sound source or a moving observer
Example: a police car siren:
1) The siren from a
stationary police
car will sound the
same regardless
of where you are
in relation to the
car.
2) When the car is in
motion, a person standing in front on the car will hear the
siren at a frequency higher than it actually is while a
person behind the car will hear the siren at a frequency
lower than it actually is.
3) This occurs because the sound waves are compressed in
front of the car (higher frequency) , and stretched behind
the car (lower frequency)
That same compression and stretching occur if an observer is
moving in relation to a stationary source: moving towards the
source seems to compress the sound waves while moving away
from the source seems to stretch the sound waves.
Section 2: Properties of Sound
The Doppler Effect (continued)
• There are four (4) equations used to describe the Doppler
Effect:
V + Vs
1. Source moving toward observer:
fo = fs(
)
V
2. Source moving away from observer:
V - Vs
fo = fs(
)
V
3. Observer moving toward source:
V + Vo
fo = fs(
)
V
4. Observer moving away from source:
V - Vo
fo = fs(
)
V
Where:
fo = frequency observer hears
fs = frequency of sound source
V = speed of sound
Vs = speed of source
Vo = speed of observer
Two questions must be asked and answered before you can
solve a Doppler Effect problem. The questions:
1. What is moving, the sound source or the observer?
If the source is moving use Eqs. 1 or 2. If the observer is
moving use Eqs. 3 or 4.
2. What is the direction of motion, toward or away from the
observer or source?
If the direction of motion is toward the observer or source
use Eqs. 1 or 3. If the direction is away from observer or
source use Eqs. 2 or 4
Section 2: Properties of Sound
Example: Beth is standing on the corner of Main and Jackson
when she hears the sound of an ambulance approaching her at
35 m/s. The temperature is 25oC, and ambulance siren sounds
a steady 900 HZ. At what frequency does Beth hear the siren?
Solution
Vs = 35.0m/s
T = 25oC
fs = 900.0 Hz
fo = ?
m
m
V = 331.5
± 0.6 o T
s
sC
m
m
V = 331.5
+ 0.6 o (25 oC )
s
s C
m
m
V = 331.5
+ 15
s
s
m
V = 346.5
s
fo = fs(
V + Vs
)
V
m
m
+ 35
s
s )
fo = 900Hz(
m
346.5
2
m
381.5
s
fo = 900Hz(
)
m
346.5
s
fo = 900Hz(1.1)
346.5
fo = 990Hz
Solution Steps:
1. List the variables you know and the variable you are solving
for
2. Solve the equation for the speed of sound at that
temperature (V)
3. Ask and answer the two questions (What is moving?
Direction of motion?) to determine which Doppler Effect
equation to use.
In the example, the source is moving toward the observer
(use Equation 1)
4. Solve the problem showing all work and dimensional analysis.