Download Hearing Sound Intensity Sound Level

Lecture 1
Sound
Hearing
Sound Intensity
Sound Level
Assistant Prof. Matthias Möbius
[email protected]
Sound Waves
Gas, liquid or solid is mechanically disturbed
• Sound waves are produced
Speed of sound in a substance depends on
•physical properties
•e.g. (density, temperature)
When sound encounters a boundary between
substances,
some sound energy is
transmitted and some reflected
Reflection makes ultrasound imaging possible
Sound
Sound Waves (Longitudinal waves)
A plucked string will vibrate at its natural
frequency and alternately compresses and
rarefies the air alongside it.
direction
compression
Density of Air
rarefaction
Compressed air >>> increased pressure
Rarefied air
>>> reduced pressure
organised vibrations of air molecules>> sound
Sound
Sound waves-(variation in air pressure)
can cause objects to oscillate
Example: ear drum is forced to vibrate in
response to the air pressure variation
Depending on:
 intensity of the sound
 frequency of vibration
movement of the ear drum will
stimulate nerve cells and the sound will be
perceived.
Sound Waves
Speed of sound (v) in materials
Depends on
•Phase of the material
•Characteristics of the material
(elasticity, density & temperature)
In general
vsolids
vliquids
vgases
•Greater in solids because molecules interact
more strongly with each other
•Greater in rigid materials
Material Speed (ms-1)
Air
344
Helium
965
Water
1450
Blood
1570
Body
Tissue
1570
Copper
3750
Iron
5000
Glass
5000
Helium has a lower
density than air.
Resonant frequencies of
vocal cavity increase.
Spectral distribution of
sounds shift to higher
frequencies
-timbre of sound changes
Sound Waves
Speed of sound (v) Depends
on elasticity and density
Solid bar
v
Liquid
v
Gas
v
 kT
m
E

B

E Young’s Modulus
 density
B bulk modulus

cp
cv
Cp specific heat constant pressure
Cv specific heat constant volume
m molecular mass
k Boltzmann’s constant
T temperature (Kelvin)
Calculate the speed of sound in air at 20 oC
 =1.4. Boltzmann’s constant =1.38x10-23J/K
Avg. mass of “air molecule” = 47.97x10-27kg
V
 kT
m
23
1.4(1.38 10 J / K )[(20  273.15) K ]
V
47.97 1027 kg
V  343.6ms
1
Sound
Speed of sound
The speed of sound in water is 4.2 times
the speed of sound in air. A whistle on land
produces a sound wave with frequency f0. When
this sound wave enters water, its frequency is:
a) 4.2f0
b) f0
c) f0/4.2
d) Not enough information given
• Frequency (f) of a wave is independent
of the medium through which the wave
travels.
–It is determined by the frequency of the
oscillator that is the source of the waves.
Sound
Diffraction Longer the wavelength compared
to size of opening or object the
greater the diffraction
Light waves:
•Wavelengths « dimensions of everyday objects
•Little diffraction occurs
•Relatively sharp shadows occur
v 3 108 ms 1
 
 500nm
14
f
6 10 Hz
Sound waves:
• Wavelength ≥ size of everyday objects
v 344ms 1
•diffraction occurs
 
 34.4cm
f
Example
Sound
source
1KHz
Hearing
Hearing
Sound wave enters the ear.
Forces exerted on eardrum due to air pressure
variations cause it to vibrate.
three small bones (hammer, anvil, and stirrup) in the
middle ear amplify & transmit forces to fluid filled
inner ear through the oval window (very small
area compared with eardrum) result pressure x 30
Other amplification characteristics ??
The motion of the fluid disturbs hair cells within
the Cochlea, which transmit nerve impulses to
the brain corresponding to the sound heard.
Outer ear
hammer
Middle
anvil
Inner ear
Oval window
Cochlea
sound
Ear canal
stirrup
ear drum
Ear can detect very low intensity sounds
Hearing
All waves carry energy
Audible sound waves carries very little energy
Ear can detect extremely low intensity sounds
Power output: Talk ≈10-5 W
Talk 24 hours a day non stop for 114 years
≈106 hours
Total energy output is ≈10-5 w x106 hrs =10 Wh
Equivalent to quantity of energy consumed
by a 100W bulb in 6 minutes
Sound Waves
Intensity
Waves (energy) spread out from source
Intensity (I) of a wave is defined as
•Energy (E) carried per unit time per unit area (A)
E /t
I
A
therefore
Power (P)
E
P
t
P
I
A
Unit of intensity Watt per square metre (Wm-2)
Sunlight intensity at Earth ≈103 Wm-2
Hearing
Intensity
Human ear can detect extremely low intensities
≈10-12 Wm-2
Maximum intensity without ear damage
≈1 Wm-2
Large range 1012 logarithmic units useful
Human perception
If we listen to two sounds (I1 and I2)
and I2 seems twice as loud as I1
Measure intensities
I2 is approximately 6 to 10 times I1
Convenient scale to measure loudness is
the logarithm of the intensity
Hearing
Perceived loudness is roughly Logarithmic
Ear response to sound
• logarithmic
• not linear
Decibel scale for intensity
Sound (Intensity) level in decibels (b)
 I 
b  10 log10  
 I0 
12
2
I

10
Wm
where
(threshold of hearing
0
at 1000Hz)
decibel (b) is a relative sound level measurement
Threshold of discomfort = 1 Wm-2
Above this, pain is experienced, and there is
potential for long term damage
Logarithm
Logarithm is the inverse of exponentiation:
10x =120
log10 (10x) = log10 (120)
x log10 (10) = log10 (120)
x=log10 (120)
Note that logarithms can have different bases.
The most common ones are:
log10, log2, ln (natural logarithm with base e)
log(a b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(ab) = b log(a)
log(1) = 0 for all bases
Convert between different bases:
logx(A) = logy(a) / logy(x)
Hearing
Sensitivity of ear
Can detect sound intensity of ≈10-12Wm-2
Corresponds to pressure variation of ≈ 3x10-5 Pa
(Atm. Pressure ≈ 101,325 Pa)
Random fluctuation due to thermal motion
of molecules ≈ 5x10-6 Pa
Sensitivity:
essentially due to mechanical layout
•Area ratio: ear drum to oval window ≈ 30
•hammer, anvil and stirrup amplification ≈2
•canal resonance at 3kHz pressure increase ≈2
•Total pressure amplification ≈ 30x2x2 = 120
Intensity  ( pressure)2
Intensity increases by factor of 1202=14,400
Brain: discriminatory role
Filters unwanted noise
Suppression: non-awareness of background noise
ear is not equally sensitive at all frequencies
Sound levels and Intensities
Sound
level
(dB)
Intensity
(Wm-2)
Sounds
Vibration
amplitude.
air molecules
0
1x10-12
Threshold of hearing
1.1x10-11m
10
1x10-11
20
1x10-10
30
1x10-9
Quiet room
40
1x10-8
computer
50
1x10-7
60
1x10-6
Normal conversation
70
1x10-5
Busy traffic
80
1x10-4
Loud radio
90
1x10-3
100
1x10-2
110
1x10-1
120
1
Rock concert, Threshold
of pain
140
1x102
Jet airplane at 30m
160
1x104
Bursting eardrums
1mm
Computer 10 times louder than quiet room
Does not seem so because of the logarithmic
response of the ear
Sound levels and Intensities
Danger Hearing loss
Damage Threshold
5 hours/week at > 89dB
damage after 5 years
> 100dB deemed hazardous
10 minutes at 120dB
Temporarily changes your threshold of hearing
from 0dB to 30dB
Sound Waves
(a) Calculate the sound level in dB of a sound
intensity 10-8Wm-2
(b) Calculate the intensity in Wm-2 of a sound
level of 80 dB
 I 
b  10 log10  
(a)
 I0 
 108Wm2 
b  10log10  12
2 
10
Wm


b  10log10 104   10  4  40d b
(b)
 I 
80  10 log10  
 I0 
 I 
8
   10
 I0 
 I 
8  log10  
 I0 
I  108 1012Wm2   104Wm2
I  104Wm2
Hearing
Hearing ability
Loudness is a method of describing the acoustic
pressure (or the intensity) of a given sound
Intensity hearing range: 10-12Wm-2 →1Wm-2
Ability to hear is not only a function of
intensity but also frequency
Humans:
Frequency range: 20 Hz → 20 kHz
Infrasonic < 20 Hz
Elephants:
down to 1Hz
Pigeons:
down to 0.1 Hz
20 kHz < ultrasonic
Dogs:
up to 40 kHz
Dolphins:
up to 250 kHz.
Bats:
up to 120 kHz
Hearing
Human Hearing Ability
Sound
Intensity Level
dB
W/m2
100
120
10-2
100
10-4
80
10-6
60
10-8
40
10-10
20
10-12
0
20
Pain threshold
Hearing threshold
100
1k
10k
20k Hz
frequency
Hearing ability as a function of intensity and
frequency. The blue solid line is the pure tone
threshold curve, below which the subject does
not hear.
Ear most sensitive at 3000 Hz
Pain threshold almost frequency independent
Hearing
Why two ears
Main advantage
Sounds from different directions arrive at each
of our ears at slightly different times and with
slightly different intensities.
Time difference of sound arriving at both ears
used to locate the source of the sound
Example:
crossing a road
direction of the car
approximately how close it is
Other advantages
•easier to understand speech in noisy background
• help judge loudness
Sound intensity
Distance
Sound intensity is reduced by moving away
from source By how much?
Inverse Square Law
Consider imaginary spheres
Isotropic
source
power
Intensity 
area
P
I1 
4 r12
r1
r2
P
I2 
4 r22
I1 r22
 2
I 2 r1
As the person gets further away, the sphere that
intersects with them gets larger and larger
Fraction = Area of person  4 π r12
Fraction = Area of person  4 π r22
Sound intensity
Variation of Sound Intensity with
distance from a point source
Inverse square law
Intensity I1 at a distance r1 from source
Intensity I2 at a distance r2 from source
2
2
2
1
I1 r

I2 r
Intensity is inversely proportional to the
square of the distance from the source.
NOTE: Sound level (dB) is not inversely
proportional to distance squared !
Sound intensity
Examples
The intensity falls off as 1/r2 (where r is the
distance) so moving 4 times as far away will
decrease the exposure by a factor of 16.
A person near a source of loud noise wants to
decrease their exposure to it by a factor of 10.
How far away do they have to move?
2
2
2
1
I1 r

I2 r
r22
10  2
r1
I1
r22
 2
 I1  r1
 
 10 
r2
 10  3.16
r1
They have to move 3.16 times further away
Waves
Example
A bat can detect sound frequencies up to
120,000 Hz. What is the wavelength of sound
in the air at this frequency?
v f
v

f
v
344ms 1
 
 2.87 103 metres
f 120, 000 Hz
=.287cms
High frequency—short wavelength
Wave only disturbed by objects with dimensions
similar to or greater than the wavelength
Smaller objects have little effect
Bats use ultrasound for navigation
Can distinguish between insect and falling leaf
Waves
Resonance
Most objects have a natural frequency:
Determined by
• size
• shape
•composition
1
l
Simple pendulum
T   2
Only one natural frequency
f
g
Resonance occurs if
 frequency of the driving force equals
 natural frequency of the system
Example: child being pushed on a swing.
Swing is kept in motion at its natural frequency
by a series of appropriately timed pushes.
Difficult to get it to swing at any other frequency
If an object is subjected to an intense wave
oscillating at object’s natural frequency
a large response (Resonance) occurs
Waves
Resonance: examples
Opera singers with powerful voices
can set glasses into audible vibration
If frequency of note is the same as the natural
frequency of the glass, the glass may vibrate
with a large amplitude and may break
Roman foot soldiers were instructed to break step
when marching over a bridge
•Prevented possible resonance response and
bridge damage
Air passages of the
 mouth,
 larynx
 Nasal cavity
together form an acoustic resonator.
Voiced sound depend on
•resonant frequencies of the total system
------depends on system’s volume and shape
Resonance: examples
Half-closed pipe Resonance (e.g. ear canal):
f Re sonance   sound / 
Fundamental mode: f1   sound /( 4L)
Electrical Resonance:
Example: Tuning in radio station
Adjust resonant frequency of the electrical circuit
to the broadcast frequency of the radio station
To “pick up” signal
Sound Waves
Travel distance is a function of frequency
Traveling waves transfer energy from one
place to another
Sound energy dissipates to thermal energy
when sound travels in air.
Higher frequency sounds dissipate more quickly,
because more energy transferred to the medium;
so lower frequency sounds travel further.
Examples
• foghorns have a low frequency
•Elephants communicate over long distances
(up to 4 km), frequencies as low as 14 Hz
Lecture 2
Sound
Beats
Doppler Effect
Ultrasound
Applications
Waves
Superposition
Simple case: Addition of two waves with
same frequency and amplitude
Wave 1
Wave 2
resultant
Beats
If the two waves interfering have slightly
different frequencies (wavelengths), beats occur.
In step (in phase)
In step (in phase)
Out of step (out of phase)
Waves
Beats
If the two waves interfering have slightly
different frequencies (wavelengths), beats occur.
Wave 1
Wave 2
Resultant
envelope
Waves get in and out of step as time progresses
Result• constructive and destructive interference occurs
alternately
•Amplitude changes periodically at the beat
frequency
Beat frequency fb = f1-f2
Absolute value: beat frequency always positive
Waves
Beats
fb = f1-f2
If frequency difference = zero
No beats occur
Wave 1
Wave 2
resultant
Waves
Beats
Beats can happen with any type of waves
Sound waves
Beats perceived as a modulated sound:
loudness varies periodically at the beat frequency
Application
Accurate determination of frequency
Example
Piano tuning
Adjust tension in wire and listen for beats
between it and a tuning fork of known frequency
The two frequencies are equal when the beats
cease.
Easier to determine than when listening to
individual sounds of nearly equal frequencies
f1 = 264Hz
f2 = 266 Hz
Beat frequency 2Hz
Sound Waves
Doppler Effect
Change in perceived frequency depending on
the relative motion of the source and listener.
Occurs with all types of waves – most notable
•sound waves,
•light waves.
Christian Doppler 1803-1853
Austrian Physicist, Mathematician
Example:
Perceived pitch (or frequency) of a moving
source such as a fire engine siren changes as it
goes past
Frequency of sound emitted does not change
Longer 
Lower f
stationary
moving→
Shorter 
higher f
Waves
Doppler effect is observed because the distance
between the source of sound and the observer
is changing.
source always emits the same frequency.
Source moving towards the observer
•sound waves reaching observer perceived to
be at a more frequent rate (higher frequency)
sound waves compressed into shorter distance
Source moving away from the observer,
•sound waves reaching observer perceived to
be at a less frequent rate (lower frequency)
Sound waves expanded into longer distance
Waves
Observed frequency for a moving source
f observer


vwave

 f source
 vwave  vsource 
+ sign: source moving away from observer
- sign: source moving towards observer
Stationary source, moving observer
f observer
 vwave  vobserver 

 f source
vwave


+sign: observer moving towards source
- sign: observer moving away from source
f = Frequency
v = Speed
Waves
Example moving observer
A stationary siren has a frequency of 1000 Hz. What
frequency will be heard by drivers of cars moving at 15ms1?
a) away from the siren?
b) toward the siren?
f observer
(a)
 vwave  vobserver 

 f source
vwave


 vw  vo 
fo  
 fs
 vw 
 344ms 1  15ms 1 
fo  
1000 Hz  956 Hz

1
344ms


(b)
 vw  vo 
fo  
 fs
 vw 
 344ms 1  15ms 1 
fo  
1000 Hz  1044 Hz

1
344ms


Example: Moving Source
A Garda car with a 1000 Hz siren is moving at
20 ms-1. What frequency is heard by a
stationary listener when the police car is:
a) Moving away from
b) approaching the listener
If you were to replace the Garda car with 2 stationary
sirens emitting at the two frequencies as perceived in (a)
and (b), what would be the beat frequency between them?
(a)
f observer
(b)
f observer
f observer


vwave

 f source
 vwave  vsource 


344ms 1

1000 Hz  945Hz
1
1 
 344ms  20ms 
f observer


vwave

 f source
 vwave  vsource 


344ms 1

1000 Hz  1062 Hz
1
1 
 344ms  20ms 
Beat frequency
fbeat  f a  fb
fbeat  945Hz  1062Hz
fbeat  117 Hz
Waves
Doppler effect can be used to measure speed
of the source
Radar: RAdio Detecting And Ranging
Police radar uses radio waves:
measures Doppler shift to determine speed of car
•compares frequency of reflected wave from car
with that emitted from radar
Doppler RADAR
•Weather
•Rainstorms, tornadoes
•Wind sheer at airports
Swirling air & water droplets
RADAR
Wave source
Sound Waves
Reflection of waves (echoes)
•Caused by solid object
•Change in nature of medium
Sound waves applications
SONAR (sound navigation and ranging)
- Underwater navigation and observation
• Measuring the travel time of sound waves
in the ocean can help monitor sea
temperatures and global changes
Ultrasound
Frequency greater than range of human hearing
Sound with frequencies above 20 kHz
Normally 1 →20MHz
Applications
•Navigation
•Diagnostics
•Surgery
•Therapeutic
•Cleaning
Ultrasound
Bats can determine distance, speed and direction
of their prey
(using reflection time and Doppler effect)
Typical prey: moths (dimensions
cms)
Bats use ultrasonic echolocation methods to
detect their presence.
why do bats use ultrasound?
v 344ms 1
Audible   
 344 103 m  34.4cm
f
1kHz
v 344ms 1
Ultrasound   
 6.88 103 m  0.7cm
f
50kHz
Ultrasound- Shorter wavelength
•Reflection, not diffraction occurs at moth.
Submarines, dolphins and bats use ultrasound
for navigation 30-100kHz
Ultrasound
Medical applications
Ultrasound Imaging
Ultrasound probe passed over region of interest
Reflections of ultrasound pulses from patients
occur at interfaces between different tissues
of different density
Good contrast: reflection from boundaries
between materials of nearly the same density
Reflection time provides depth information
Image constructed from echo
and position information
Ultrasound
Medical applications
Ultrasound & Doppler effect
can be used to measure
• Blood flow speed in arteries and veins,
measure arterial occlusion
•Echocardiogram , examination of the heart
• measure blood flow in and out
•fetal heart beats
• pulsation of artery walls
Stroke: early warning
Monitor blood speed in carotid artery in neck
Ultrasonic Doppler flow meter
Transmitter
Receiver
Red blood cell
Ultrasound
Surgery
Ultrasonic scalpel (55kHz)
Precise cutting and coagulation
•Tumour removal
•Tonsillectomies
Medical ultrasound without harmful effects
•intensity kept low (≈10-2 Wm-2) to avoid tissue
damage
Ultrasound scanning during pregnancy
Ultrasound
Imaging
Why use ultrasound---not audible sound
Smallest detail observable ≈ one wavelength
Audible sound wavelength in tissue
v 1570ms 1
 
 0.5m
f
3000 Hz
Ultrasound wavelength in tissue
1570ms 1

 1cm
150kHz
In tissue, higher frequencies are attenuated more
Compromise between spatial resolution of image
and penetration depth
1MHz:
penetration depth ≈ 6cm
3MHz: superficial conditions (eg. Tennis elbow etc)
•Frequency is selected based on the depth
of the tissue to be treated.
Example: deep heat therapy (low frequency)
Ultrasound
Example
Ultrasound speed =1500m/s in tissue.
Using an ultrasound frequency of 2MHz,
calculate (a) smallest detail visible
(b) time for reflected wave to return to probe
from a depth of 5cm
v f
(a)
v 1500m / s
 
6
f
2 10 Hz
l = 0.75mm
(b)
time for reflected wave to return to probe
s 2  0.05m
5
t 

6.6

10
sec
1
v 1500ms
Ultrasound
Other uses in medicine
•Destructive effects
•Intense ultrasound produces large
density and pressure changes
• Results
− Large stresses
−Heat is produced in most materials
− microscopic vapour bubbles formed and
implode releasing energy (cavitation)
Non-invasive removal of kidney stones
Dental applications
ultrasonic scalar
Consists of a ultrasound probe with a small
tip. The ultrasound in combination with water
flow effective in plaque and tartar removal
Ultrasound
Component surface cleaning
Component placed in fluid in ultrasonic bath
Ultrasound creates a periodic compression
and expansion in the fluid.
Results in Acoustic cavitation
Bubbles formed, grow, and implosively collapse
localised heating (>1000K)
and high pressures (>100 atmospheres)
Result: effective surface cleaning
Auto-focus cameras
computes time taken (and hence distance
of subject) for the reflected ultrasonic sound
wave to reach the camera lens position and
then sets focus accordingly.
Sound
Supersonic speed
Moving source, approaching listener
When speed of source approaches the speed of
sound, waves ahead of source come close
together.
f observer


vwave

 f source
 vwave  vsource 
f observer approaches infinity
Nearly infinite number of wave crests reach
observer in very short time
At supersonic speeds the waves overlap and
there are many points of constructive interference,
shock wave results
Wave front produced when
vsource  vwave
is known as a shock wave
sonic boom
Sound
Supersonic speed
vs  0
Circles represent wave fronts
emitted from sound source
Stationary source v = 0
Speed of sound in air =vs
v
vs
Waves ahead of source
come closer together
subsonic
v  vs
Waves pile up at front
Mach 1
v
vs
supersonic
Waves overlap:
Shock wave,
Sonic boom.
Sound
Supersonic speed
Circles represent wave fronts
emitted from sound source
vs t

In time interval t
vt
Sound wave travels a distance vst
Source travels distance vt
Tangent lines lie on surface of cone
vs t
vt
v
Ratio
is called Mach number M
vs
object  speed
v
1
M
M 
speed  of  sound
vs sin 
sin  
since sin   1
No shock unless
M 1
v
vs
vwwt
Bow sin  
vt
Waves
Speed of boat v
> Water wave speed Vww
Question
What is the speed of ultrasound with a wavelength of
0.25 mm and a frequency of 6 MHz? How does this
compare with the speed of sound in air?
v f
v   6 106 Hz  0.25 103 m   1.5 103 ms 1
Compare with speed of sound in air
1.5 103 ms 1
 4.4
1
344ms
Question
Lightening strikes 10 km away.
(a) How long after the strike will you see the light?
(b) How long after the strike will you hear the sound?
(a)
c = 3*108 m/s, s = 10 km, t = ?
s = vt  t = s/v
t = (10,000 m)/(3*108 m/s) = 3.3*10-5 s
(b)
v = 344 m/s, s = 10 km, t = ?
s = vt  t = s/v = (10,000 m)/(344 m/s) = 29 s
If you hear the sound 3 seconds after you see
the lightening how far away is the strike?
s = vt =(344 m/s)(3 s) = 1002 m
Question
(a) What is the sound level in decibels of a sound with an
intensity of 0.0200W/m2?
(b) If you had 3 such sounds what would the sound level
be?
 2 102Wm2 
b  10log10 
12
2 
10
Wm


(a)
 I 
b  10 log10  
 I0 
b  10log10  2 1010 
b  10  log10 2   10  log10 10 
b  10  0.3  10
(b)
b  103dB
 3  2 102Wm2 
b  10log10 

12
2
10
Wm


b  10log10  6 1010 
b  10  log10 6   10  log10 10 
b  10  0.78  10
b  107.8dB
Not equal to 3x103 dB !
Sound levels are logarithms of intensity