SOUND
A vibrating object, such as
your voice box, stereo
speakers, guitar strings,
etc., creates longitudinal
waves in the medium
around it.
When these waves cause
our ear drums to
vibrate, we “hear”
sounds.
Sound is caused by
vibrations!
Sonic Spectrum


The frequency range over which longitudinal
waves occur.
That part of the sonic spectrum that the
human ear is sensitive to is called the
aural range (a.k.a. sounds)



20 Hz – 20,000 Hz (wavelength range: 17mm – 17m)
Ultrasonic: frequencies above 20,000 Hz
Infrasonic: frequencies below 20 Hz
Sonic Spectrum

Upper frequency limit is determined by
the medium.

If the wavelength of the sound is small
compared to the inter-particle spacing the
wave will not be transmitted.

Small wavelength means high frequency.
Gases – around 109 Hz at ordinary
temperature and pressure.
 Solids/liquids – higher than 109 Hz due to
closer spacing of particles.

Sound Waves

Wave energy is passed through the particles
of the medium as a periodic, longitudinal wave.


Remember: the wave travels, not the medium!
The particles alternately experience
compression and rarefaction.
Speed of Sound

Factors affecting the speed of sound

Temperature: the hotter the medium the
faster the speed of sound.


Air: vsound = 331.5 m/s + .6T
(T is in degrees C)
Density: the denser the medium the faster
the speed of sound.
Air @ 00C: 331.5 m/s
 Steel: 5200 m/s

Try one!

A ship sounds its fog horn to find out how far
away an iceberg is. If the captain hears the
echo 6 sec after sounding the horn, how many
meters away is the iceberg? (assume T = 60 F)




Given: t = 6.0 s, T = 60 F
60F to Celsius: C = (5/9)(F – 32) = -14.40 C
Find the speed of sound: vsound = 322.9 m/s
Calculate the distance. Remember, the sound has
to travel twice the distance between them!
2d = vt
so
d = ((322.9m/s)*(6s))/2 = 969m
Sound Barrier


Sound waves spread
out from their
source in spherical
shells, similar to
ripples in a pond.
If the source is
moving close to the
speed of sound, the
waves begin to pile
up in front of it.
Sound Barrier


When the source moves
at the speed of sound
each new crest is
created on top of the
last one causing
constructive
interference
This creates an area of
high pressure in front
of the source called the
“sound barrier.”
Super Sonic



Finally, when moving
faster than the speed
of sound, the source
outruns the wave crests
it creates.
The V pattern created
by the successive wave
crests is called a “shock
wave,” with the source
ahead of it.
This shock wave is the
“sonic boom” we hear
when something goes by
at supersonic speed.
Properties of Sound

Pitch (frequency)





High pitch (high frequency): shorter wavelength
sounds such as a siren or a flute.
Low pitch (low frequency): longer wavelength
sounds such as a sub-woofer or a fog horn.
Pitch can also be described with musical notes.
Pitch (frequency) does not change when a sound
wave passes from one medium to another.
Normal speech range; 1000 – 5000 Hz
Properties of Sound

Intensity (loudness)

A measure of the amount of sound energy that
passes through a given area over a given time



Intensity is inversely proportional to the square of
the distance from the source


I = P/A = Watts/cm2
Power is determined by the source, but the area
increases with the square of the distance from the
source.
A sound heard from 100m away is ¼ as intense when
heard from 200m away.
Intensity is also a measure of the amplitude of the
sound wave.

Intensity (continued)

The decibel scale relates sound intensity to
our perception of how loud sounds are.
Units: decibels (dB)
 β = 10log(I/Io)





I is the intensity of the sound heard
Io is the intensity of a sound at the threshold of
hearing Io = 1.00x10-12 W/m2
β = 0 dB is the threshold of hearing.
β = 110 dB is considered the threshold of pain.

However, sounds below 110 dB can still cause hearing
loss.
Try it!

The intensity of a sound is found to be
1x10-14 W/cm2. What is the sound level?



Given: I = 1x10-14 W/cm2
Convert Io to W/cm2, Io = 1x10-16 W/cm2
β = 10log[(1x10-14 W/cm2)/(1x10-16 W/cm2)]
= 20 dB

How much would the sound level change if the
intensity was doubled?
 β would increase 3 dB.
Doppler Effect

The change in frequency of a sound due
to the relative motion between the
source and listener
Doppler Effect

A decreasing distance between the source
and observer will cause a higher pitch to be
heard.

 v  vo 

fo  fs 
 v  vs 




fo = frequency heard by
observer
fs= frequency of the
source
v = speed of sound
vo = speed of the
observer
vs = speed of the source
Doppler Effect

An increasing distance between the
source and observer will cause a lower
pitch to be heard.

 v  vo 

fo  fs 
 v  vs 




fo = frequency heard by
observer
fs= frequency of the
source
v = speed of sound
vo = speed of the
observer
vs = speed of the source
Try it!

A driver travels northbound on a highway at a
speed of 25.0 m/s. A police car, traveling
southbound at a speed of 40.0 m/s, approaches
with its siren sounding at a frequency of 2,500 Hz.
What frequency does the driver hear as the police
car approaches?


Given: fs = 2,500 Hz, v = 343 m/s, vs = 40 m/s, vl = 25
m/s
The cars are getting closer so
 343m/s 25m/s 
fo  2,500Hz
  3036Hz
 343m/s 40m/s 
Doppler Effect (Light)

A similar effect, but the equation is slightly
different




fo = frequency seen by observer
fs = frequency of source
fo
c = speed of light
v = relative speed between
observer and source


v


 1   

c
 fs 

 1   v  


c
 

More Properties of Sound

Reflection

Refraction

Interference

Need 2 waves



same frequency
in phase
When the waves travel to the same point, the
difference in their path lengths determines what
type of interference occurs
1D Interference


If L2 – L1 = n(½) (for n = 1,2,3…) then there will be
total destructive interference
If L2 – L1 = m (for m =1,2,3…) then there will be total
constructive interference
L1
L2
2D Interference


As with 1D

L = n(½) results in

L = m results in
total destructive
interference
total constructive
interference
You have to rely
more on the path
lengths than visual
cues for 2D
Beats


When two tones are heard at the same
time they interfere with each other
causing a pulsing sound called beats.
The frequency of the beat pattern is
the difference between the frequencies
of the two tones.

fb = f1 - f2
Let’s try it!

Jane holds two slightly different tuning
forks next to her ear. What is the beat
frequency she hears if one tuning fork
vibrates at 440Hz and the other at 436Hz?
fb = 440Hz – 436Hz = 4Hz
Shake it Up

Forced Vibration



a vibrating object is touched to a second object
the second object begins to vibrate at the same
frequency
Resonance

a vibration caused in a medium due to a
disturbance that occurs at the medium’s natural
frequency


the natural frequency is the frequency of oscillation in an
object that will produce a standing wave
unrestricted, the amplitude of the vibrations will
continue to increase
The Sound of Music



Musical instruments are designed to resonate
at one or more natural frequencies
The strings on a string instrument have a
base frequency based on its length and
tension (remember last chapter?), but they
can produce other notes by putting pressure
on the fret board effectively shortening the
string
Wind and brass instruments create resonance
patterns in the “pipe-like” body
Resonance in Pipes


There are two kinds of pipes to consider
Open pipes – open at both ends


a standing wave is created in the pipe such that there is an
antinode at each end
the wavelength of the standing wave depends on the length of
the pipe
 v 
f = n

 2L 
for n = 1,2,3,…
Resonance in Pipes

Closed pipes – closed at one end open at the other


a standing wave is created in the pipe such that there is a
node at the closed end and an antinode at the open end
the wavelength of the standing wave depends on the length of
the pipe
 v 
f = n

 4L 
for n = 1,3,5
Wind and Brass Instruments

Wind instruments act like an open pipe

Brass instruments act like closed pipes
Let’s try it!

In an unheated 100C room, a 25cm pipe is
producing its 3rd harmonic. If the pipe is open at
both ends, what is the frequency of the tone
heard?
first you find the speed of sound in the room
v = 331.5 + .6(10) = 337.5m/s
then you use the open pipe equation for n = 3
f = 3(337.5/(2x0.25m)) = 2025Hz