Sound and Hearing c16

Chapter 16. Sound and Hearing

Dr. Yousef [email protected]

Syllabus

16.1 Sound Waves

16.7 Beats

16.8 The Doppler Effect

University Physics, Chapter 16Dr. Y. Abou-Ali, IUST

Goals for Chapter 16

• To present sound as a standing longitudinal wave.

• To study beats.

• To solve for frequency shifts (the Doppler Effect).


Introduction

• Of all the mechanical waves that occur in nature, the most

important in our everyday lives are longitudinal waves in a medium,

usually air, called sound waves.

• The human ear is tremendously sensitive and can detect sound

waves even of very low intensity.waves even of very low intensity.

• A mechanical wave is a disturbance that travels through some

material or substance called the medium for the wave.

• As the wave travels through the medium, the particles that make

up the medium undergo displacements of various kinds, depending

on the nature of wave.


Introduction

• Waves in a fluid are the

result of a mechanical

disturbance.


• At right, a stone disturbs

water and creates visually

observable traveling waves.

Introduction

• Mechanical waves described primarily in terms of displacement.

• A description of sound waves in terms of pressure fluctuations is

often appropriate, largely because the ear is primarily sensitive to

changes in pressure.

• When a source of sound or listener moves through the air, the• When a source of sound or listener moves through the air, the

listener may hear a different frequency than the one emitted by the

source. This is Doppler effect.


16.1 Sound Waves

• The most general definition of sound is that is a longitudinal wave

in a medium. Our main concern in this chapter is with sound waves

in air.

• But sound can travel through any gas, liquid, or solid.

• The simplest sound waves are sinusoidal waves, which have•definite frequency, amplitude and wavelength.

• Human ear is sensitive to waves in the frequency range from about

20 – 20000 Hz /audible range/.

• But we also use the term sound for similar waves with frequencies

above (ultrasonic) and below (infrasonic) the range of human

hearing.


16.1 Sound Waves

• Sound waves usually travel out in all directions from the source of

sound, with an amplitude that depends on the direction and distance

from the source.

• On the idealised case of a sound wave that propagates in the

positive x-direction only.

• Such a wave is described by a wave function y(x, t), which gives the

instantaneous displacement y of a particle in the medium at position

x at time t. If the wave is sinusoidal:

(sou n d w ave p ro pagatin g in th e + -d irec tio n )

16 1( ) cos( )x

y x t A k x tω= − ( . ),


� Where k is wave number, A is the amplitude and ω angular frequency.

16.1 Sound Waves

• In a longitudinal wave the displacements are parallel to the

direction of travel of the wave, so distances x and y are measured

parallel to each other.

• In a transverse wave the displacements are perpendicular to the

direction of travel of the wave.

• The amplitude A is the maximum displacement of a particle in the

medium from its equilibrium position. It is also called the

displacement amplitude.

• Sound waves may also be described in terms of variations of

pressure at various points.


16.1 Sound Waves

• Let p(x, t) be the instantaneous pressure fluctuation in a sound

wave at any point x at time t.

• p(x, t) is the amount by which the pressure differs from normal

atmospheric pressure pa.

• Think of p(x, t) as the gauge pressure; it can be either positive or• Think of p(x, t) as the gauge pressure; it can be either positive or

negative. The absolute pressure at a point is then pa + p(x, t).

• For the sinusoidal wave:

1 6 4( ) s in ( )p x t B k A k x tω= − ( . ),

� Where B is the bulk modulus.


16.1 Sound Waves

B u lk s tress ∆

B u lk s tra in ∆ / o

pB

V V= = − (1 1 .1 3 )

(a) Displacement amplitude and,

(b) Pressure amplitude versus

position for a sinusoidal

longitudinal wave.


16.1 Sound Waves

• Equation (16.4) shows that the quantity BkA represents the

maximum pressure fluctuation.

• We call this the pressure amplitude, denoted by pmax:

m axp B k A= (16.5)m axp B k A=

• The pressure amplitude is directly proportional to the

displacement amplitude A, and it also depends on wavelength.

• Waves of shorter wavelength λ (larger wave number k = 2π/λ) have

greater pressure variations for a given amplitude.


16.1 Sound Waves

Example 16.1 ( Amplitude of a sound wave in air):

In a sinusoidal sound wave of moderate loudness the maximum

pressure variations are of the order of 3.0 × 10-2 Pa above and below

atmospheric pressure pa (nominally 1.013 × 105 Pa at sea level). Find

the corresponding maximum displacement if the frequency is

1000 Hz. In air at normal atmospheric pressure and density, the1000 Hz. In air at normal atmospheric pressure and density, the

speed of sound is 344 m/s and the bulk modulus is 1.42 × 105 Pa.

Solution:

• Identify and Set Up: We are given the pressure amplitude pmax,

wave speed v, frequency f, and bulk modulus B. Our target variable

is the displacement A, which is related to pmax by Eq. (16.5). We also

use the relationship ω = v k to determine the wave number k.


16.1 Sound Waves

• Execute: From Eq. (16.5), the maximum displacement is

A = pmax/Bk. The wave number is:

2 (2 rad)(1000 Hz)18 3 rad m

344 m s

fk

v v

ω π π= = = = . /

/

� Then:

-28max

5

3.0 10 Pa1 2 10 m

(1 42 10 Pa 18 3 rad/m)

pA

Bk

−×= = = ×

×

.. )( .

• Read Example 16.2/530).


16.1 Sound Waves

Perception of Sound Waves

• The physical characteristics of a sound wave are directly related to

the perception of that sound by a listener.

• For a given frequency, the greater the pressure amplitude of a

sinusoidal sound wave, the greater the perceived loudness.

• The frequency of a sound wave is the primary factor in

determining the pitch of a sound, the quality that lets us classify the

sound as “high” or “low”. The higher the frequency of a sound, the

higher the pitch.


16.1 Sound Waves

Perception of Sound Waves

• Two tones have the same frequency (the same pitch) but sound

different because of the presence of different amounts of the various

harmonics. The difference is called tone color, quality, or timber and

often described in subjective terms such as reedy, golden, round,

mellow, and tinny.mellow, and tinny.


16.7 Beats

• Let’s look at what happens when we have two waves with equal

amplitude but slightly different frequencies.

� This occurs, for examples, when two tuning forks with slightly

different frequencies are sounded together.

�� When two organ pipes that are supposed to have exactly the same

frequency are slightly “out of tune”.

• Consider a particular point in space where the two waves overlap.

The displacements of the individual waves at this point are plotted as

functions of time in the figure below (a).


16.7 Beats

• The result is the graph of figure (b).

• Appling the principle of superposition, we add the two

displacements at each instant of time to find the total displacement

at that time.

• At certain times the two waves are in phase; their maxima coincide

and their amplitudes add.


16.7 Beats


16.7 Beats

• The resultant wave in the figure above looks like a single

sinusoidal wave with a varying amplitude that goes from a

maximum to zero and back.

• The amplitude variation causes variations of loudness called beats,

and the frequency with which the loudness varies is called the beat

frequency.frequency.

• In this example the beat frequency is the difference of the two

frequencies.

• We can prove that the beat frequency is always the difference of

the two frequencies fa and fb, where fa is larger than fb.

beat frequencybeat a bf f f= − ( ) (16.24)



• You have probably noticed that when a car approaches you with

its horn sounding, the pitch seems to drop as the car passes.

• This phenomena, first described by the 9th century Austrian

scientist Christian Doppler, is called the Doppler effect.

• When a source of sound and a listener are in motion relative to• When a source of sound and a listener are in motion relative to

each other, the frequency of the sound heard by the listener is not the

same as the source frequency.

• To analyse the Doppler effect for sound, we will work out a

relation between the frequency shift and the velocities of source and

listener relative to the medium (usually air) through which the sound

waves propagate.



• To keep things simple, we consider only the special case in which

the velocities of both source and listener lie along the line joining

them.

• Let vs is the velocity component of source and vL is the velocity

component of listener along this line.component of listener along this line.

• We choose the positive direction for both vs and vL to be the

direction from the listener L to the source S.

• The speed of sound relative to the medium, v, is always positive.



Moving Listener

• Let’s think first about a listener L moving with velocity vL toward

a stationary source S.

• The source emits a sound wave with frequency fs and wavelength λ

= v/f .= v/fs.


• The figure below shows several wave crests, separated equal

distance λ.

• The wave crests approaching the moving listener have a speed of

propagating relative to the listener of (v + vL).

• So the frequency fL (the frequency the listener hears) is:


Moving Listener



Moving Listener

L LL

s

v v v vf

v fλ

+ += = (16.25)

/� Or:

L LL s s1

v v vf f f

+ = = + (16.26)

L LL s s1

(moving listener, stationary source)

f f fv v

= = +

(16.26)


• So a listener moving toward a source (vL > 0) hears a higher

frequency (higher pitch) than does a stationary listener.

• A listener moving away from the source (vL < 0) hears a lower

frequency (lower pitch) than does a stationary listener.


Moving Source and Moving Listener

• Now suppose the source is also moving, with velocity vs, The wave

speed relative to the wave medium (air) is still v.

• The wavelength is:

s sin front

s s s

wavelength in front of a moving source

v v vv

f f fλ

−= − = (16.27)

( )


• In the region to the left of the source (behind the source), it is:


Moving Source and Moving Listener

w aveleng th beh ind a m oving source

sbeh ind

s

v v

fλ

+= (16 .28)

( )w aveleng th beh ind a m oving source( )


• The frequency heard by the listener behind the source:

D o p p le r e ffe c t , m o v in g s o u rc e a n d m o v in g lis te n e r

LL s

s

v vf f

v v

+=

+(1 6 .2 9 )

( )


Example 16.15 ( Doppler effect I: Wavelengths):

A police siren emits a sinusoidal wave with frequency fs = 300 Hz.

The speed of sound is 340 m/s. a) Find the wavelength of the waves if

the siren is at rest in the air. b) If the siren is moving at 30 m/s

(108 km/h, or 67 mi/h), find the wavelengths of the waves ahead of

and behind the source.and behind the source.


Solution:

• Identify: The Doppler effect is not involved in part (a), since

neither the source nor the listener is moving. In part (b), the source

is in motion and we must invoke the Doppler effect.

• Set Up: We use the relationship v = f λ to determine the wavelength

when the police siren is at rest. When it is in motion, we find the

wavelength on either side of the siren using Eqs. (16.27) and (16.28).


• Execute: a) When the source is at rest,

s

340 m s1 13 m

300 Hz

v

fλ = = =

/.

b) From Eq. (16.27), in front of the siren:


sin front

s

340 m s 30 m s1 03 m

300 Hz

v v

fλ

− −= = =

/ /.

� From Eq. (16.28), behind the siren:

sbehind

s

340 m s 30 m s1 23 m

300 Hz

v v

fλ

+ += = =

/ /.


Example 16.16 ( Doppler effect II: Frequencies):

If a listener L is at rest and the siren in Example 16.16 is moving away

from L at 30 m/s (Fig. 16.31), what frequency does the listener hear?

• Identify and Set Up: Our target variable is the listener’s frequency

f . We know f = 300 Hz from Example 16.16, and we have v = 0 and

Solution:

fL. We know fs = 300 Hz from Example 16.16, and we have vL= 0 and

vs= 30 m/s. (The source velocity vs is positive because the siren is

moving in the same direction as the direction from listener to

source.)

• Execute: From Eq. (16.29),

L ss

340 m s(300 Hz) 276 Hz

340 m s 30 m s

vf f

v v= = =

+ +

/

/ /



Example 16.17 ( Doppler effect III: A moving listener):

If the siren is at rest and the listener is moving toward the left at

30 m/s (Fig. 16.32), what frequency does the listener hear?

• Identify and Set Up: The source is at rest (vs= 0) and the listener is

Solution:

• Identify and Set Up: The source is at rest (vs= 0) and the listener is

in motion. The positive direction (from listener to source) is still from

left to right, so vL= - 30 m/s.


• Execute: From Eq. (16.27),

LL s

340 m s ( 30 m s)(300 Hz) 274 Hz

340 m s

v vf f

v

+ + −= = =

/ /

/


Example 16.18 ( Doppler effect IV: Moving source, moving listener):

If the siren is moving away from the listener with a speed of 45 m/s

relative to the air and the listener is moving toward the siren with a

speed of 15 m/s relative to the air (Fig. 16.33), what frequency does

the listener hear?

Solution:

• Identify and Set Up: Both the listener and the source are in

motion, with vL= 15 m/s and vs= 45 m/s.

• Execute: Once again using Eq. (16.27), we find:

LL s

s

340 m s 15 m s(300 Hz) 277 Hz

340 m s 45 m s

v vf f

v v

+ += = =

+ +

/ /

/ /


Solution:

Sound and Hearing c16

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