Lectures 27-28 Chapter 17 Fall 2010

8/13/2019 Lectures 27-28 Chapter 17 Fall 2010

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PH 221-3A Fall 2010

Waves - II

Lectures 27-28

Chapter 17(Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

1



Chapter 17 Waves II

concentrate on the following topics:

Speed of sound wavesRelation between dis lacement and ressure am litudeInterference of sound wavesSound intensity and sound level

The Doppler effect

2



Sound waves are mechanical longitudinalwaves that propagate in solids liquids andgases. e sm c waves use y o exp orers

propagate in the earth’s crust. Sound wavesgenerated by a sonar system propagate in the

sea. An orchestra creates sound waves that propagate in the air.

The locus of the points of a sound wave that has the same displacement

is called a “ wavefront ”. Lines perpendicular to the wavefronts are“ ”

propagates. An example of a point source of sound waves is given inthe figure. We assume that the surrounding medium is isotropic i.e.sound propagates with the same speed for all directions. In this case thesound wave spreads outwards uniformly and the wavefronts are spherescentered at the oint source. The sin le arrows indicate the ra s. Thedouble arrows indicate the motion of the molecules of the medium inwhich sound propagates. 3



If we apply an overpressure on an object p

Bulk modulus Bv

Δ p ,

in the figure. The bulk modulus of the compressed material

p

The negative sign denotes the in volume

when is positive

.

/V

p

V Note : decrease

Using the above definition of the bulk modulus and combining it with

Newton's second law one can show that the s eed of sound in a

The speed of sound

homogeneous isotropic medium with bulk modulus and B density

is given by the equation: v B

Bulk modulus is smaller for more compressible pV B

V

Note 1 :

media. Such media exhibit lower speed of sound.

Denser materials (higher ) Note 2 : have lower speed of sound 4



The Speed of Sound

5



6



Example: Thunder, lightning and a rule of thumb

There is a rule of thumb for estimating how far away a thunderstorm is. You can estimate yours ance rom a o o g n ng y coun ng e secon s e ween see ng e as an ear ng e

thunder and then dividing by 3 to obtain the distance in km. Why does this rule work?

7



The Speed of SoundProblem : Both krypton (Kr) and neon (Ne) can be approximated at monatomic ideal gases. The

atomic mass of krypton is 83.8 u, while that of neon is 20.2 u. A loudspeaker produces a sound whosewavelength in krypton is 1.25m. If the loudspeaker were used to produce sound of the samefrequency in neon at the same temperature, what would be the wavelength?

8



Consider the tube filled with air shown in the figure.

Traveling sound waves.

We generate a harmonic sound wave traveling to the

right along the axis of the tube. One simple method

drive it at a particular frequency. Consider an air element of thickness which is located at position x

before the sound wave is generated. This is known

as the "e

x

qulibrium position" of the element. Under

these conditions the ressure inside the tube is constant

In the presence of the sound wave the element

oscillates about the equlibrium position. At the same

t me t e pressure at t e ocat on o t e e ementoscillates about its static value. The sound wave

in the tube can be described using one of two

parameters:9



One such parameter is the distance , of thes x t

Traveling sound waves. m m p v s

e ement rom ts equ r um pos t on

. The constant is

the of the wave. The

, cosm mss x t s kx t

dis lacement am litude

angular wavenumber and the angular frequencyhave the same meaning as in the case of the transverse

k

waves s u e n c ap er .

The second possibility is to use the pressure variation

from the static value. p , sinm p x t p kx t The constant p is the wave's pressure

The two amplitudes are connected by the equation:m amplitude.

The displacement and the pressure variation

have a phase differ

m m p v s Note :

ence of 90 . As a result when

one parameter has a maximum the other has a

minimum and vice versa. 10



1 2Consider two point sources of sound waves S and S

Interference

2 L

s own n e gure. e wo sources are n p ase an

emit sound waves of the same frequency.

Waves from both sources arrive at point P whose

1 2 1 2distance from S and S is and respectively.The two waves interfere at point P.

L L

1 1 1

2 2 2

At time the phase of sound wave 1 arriving from S at point P is

At time the phase of sound wave 2 arriving from S at point P is

t kL t

t kL t

2 1 2 1 2 1 2 1

2kL t kL t k L L L L

2 1e quan y s nown as e

between the two waves. Thus2

L

pat engt erence

Here is the wavelength of the two wav s. e

11



The wave at P resulting from the interference of the

Constructive intereference.

1 2two waves t at arr ve rom an as a max mum

amplitude when the phase difference 2

2

m

, , ,... .

0, , 2 ,. .. L

1 2

The wave at P resulting from the interference of the two waves that arrive

from S and S has a miniimum amplitude when the phase difference

.

2 1 0,1, 2,... .m m

1

2 2 1 L m

, , ,... 2

Δ equal to an integral multiple of λ L constructive interference

Δ equal to a half-integral multiple of λ L destructive interference12



At an open-air concert on a hot day (T c = 25 ° C, V s = 346.5 m/s), a person sits at a location that 7.0m and 9.1 m respectively from speakers at each side of the stage. A musician, warming up, plays a

single 494 Hz tone. What does the spectator hear?

13



Constructive and Destructive Interference of Sound WavesAssume that two loudspeakers in the figure are vibrating out of phase instead of in phase. (see

exam le 2 from the text The s eed of sound is 343 m/s. What is the smallest fre uenc that willproduce destructive interference at point C?

14



Constructive and Destructive Interference of Sound WavesSpeakers A and B are vibrating in phase. They are directly facing each other, are 7.80 m apart, and

. . .there are three points where constructive interference occurs. What are the distances of these threepoints from speaker A?

15



Consider a wave that is incident normally on a surface

Intensity of a sound wave

of area . The wave transports energy. As a result

power (energy per unit time) passes through .

A

P A

2 SI units: /W m

P A

I

2

2 2

The intensity of a harmonic wave with displacement amplitude is given by:

In terms of the pressure amplitude1

.

m

m m

s

v I s I p

Consider a point source S emitting a power in tP he form of sound waves

of a particular frequency. The surrounding medium is isotropic so the waves

spread uniformly. The corresponding wavefronts are spheres that have S as

their center. The sound i 2ntensity at a distance from S is:r P

I

2

1

The intensity of a sound wave for a point sources is proportional to r 16



The decibel

sound intensity . This allows the ear to percieve a wide range of

sound intensities. The threshold of hearing o

I

I is defined as the lowest12 2

0sound intensity that can be detected by the human ear. 10 W/mThe sound level is defined in such a way as to mimic the response

I

of the human 10loear. go

I I

is expressed in decibels (dB)

/10

10

For we have: 10 log1 0o

o

I I

I I

Note 1:

increases by 10 decibels every time increa I Note2 :4

ses by a factor of 10

For example 40 dB corresponds to 10 o I I

17



Example 1: Express the threshold of hearing (2.5 x 10 -12 W/m 2) and the threshold of pain (1 W/m 2)in decibels.

Exam le 2: At a distance of 60m from a et airline the intensit is 1 W/m 2. I = ?

18



The intensity of sound near a loud rock band is 120dB. What is theintensity of sound near two such rock bands playing together?

19



Problem (Decibels): Two identical rifles are shot at the same time and the sound intensity level is800dB. What would be the sound intensity level if only one rifle were shot? (hint: the answer is not400db)

20



Problem : Two sound sources each emit sound power uniformly in all directions. There are noreflections. Both sources are located on the x axis, one at the origin and the other at x = +123m. Thesource of the origin emits four times more power than the other source. Where on the x axis is theintensity of each sound equal? Note there are 2 answers.

21



Sound standing waves in pipes

.Sound waves that have walengths that satisfy a particular

relation with the length L of the pipe setup standing waves

that have sustained amplitudes.

The simplest pattern can be set up in a pipe that is open at both ends as shown in fig.a.

In such a pipe standing waves have a antinode (maximum) in the dispacement amplitud

The amplitude of the standing wave is plotted as function of distance in fig.b.

by a node (minimum). The distance between two adjacent antinodes is /2.

v v

The standing wave of fig.b is known as the " "

or " " of the tube.

2 L fundamental mode

irst harmonic

Antinodes in the displacement amplitude c r oNote : respond to nodes in the

pressure amplitude. This is because and are 90 out of phase.m ms p 22



The next three standing wave patterns are

Standing waves in tubes open at both ends 2

n

Ln

shown in fig.a. The wavelength

where 1, 2, 3, ... The integer n is

2n

n

Ln

known as the

The corr poes

harmonic number

nding frequencies n

nv f

for n=1,3,5,74

, ...n

L

n

The first four standing wave patterns are

an ng waves n u es open a one en

and closed at the other

shown in fig.a. They have an antinode at theopen end and a node at the closed end.

The wavelengt h n n

23





If we listen to two sound waves of equal amplitude and frequencies

Beats.

1 2 1 2 1 2

1 2

and and we perceive them as a sound of frequency

. in addition we also perceive "beats" which aav

f f f f f f

f f f

re variations in the

1 2

1 1 2 2

intensity of the sound with frequency . The displacements of the

two sound waves are given by the equations: cos , and cos .

beat

m m

f f f

s s t s s t

These are plotted in fig.a and fig.b.

Using the principle of superposition we can determine the resultant displacement as:

1 2 1 2

1 2 1 2

1

cos cos 2 cos cos2 2

2 cos cos where

m ms s s s t t s t t

s s t t

2 1 2and

m

1 2

2 2

Since

25



T beat

1 2 beat f f f

T'

1 2 1 2 2 2

The displacement s is plotted as function of time in the figure. We can regard

m

The am

mplitude is time dependent but varies slowly with time. The amplitude

exhibits a maximum whenever cos is equal to either +1 or -1 which happenst

twice within one period of the cos function.

Thus the

t 1 2

1 2angual frequency of the beats 2 2beat

1 2 1 2The frequency of the beats 2 2 2beat beat f f f 26



Consider the source and the detector of sound waves

The Doppler effect

shown in the figure. We assume that the frequency

of the source is equal to . f

We ta e as t e re erence rame t at surroun ng a r t roug w c t e soun waves

propagate. If there is relative motion between the source and the detector then thedetector erceives the fre uenc of the sound as . If the source or the

detector move towards to each other . if on the other hand the source or

the detector move away from each other . This is known as the " "

f f

f f

Doppler

effect. The frequecy is given by the equation: . Here and

are the speeds of the source and detector with respect to air, respsctively.

D

S S D

v v f f

vv v

v f

When the motion of the detector or source is each other the sign of the speedmust give an shift in frequency. If on the other hand the motion is from

towardsupward away

The four possible combinantions are illustrated in the next p

.

age.27





The Doppler EffectA train approaching a siren. The train A train receding from the siren. Theencounters more wave fronts per unittime than when stationary.

train encounters fewer wave fronts perunit time than when stationary.

f = f + additional # of condensationsadditional # of condensations in a time t= (V r · t)/ λ

in 1 sec additional # = V r / λ-> f = f + V r / λ = f(1 + V r /f λ ) = f(1 + V r /V)

A receiver on the train will detecta higher frequency whenapproaching a siren, and a lowerfrequency when receding.

29



The source (train) is in motion, the receiver is stationary

A stationary receiver will detecta higher frequency when it isfront of the train and a lowerfrequency when behind thetrain.

The wavelength ahead of thetrain is shorter λ ’ = λ – V E /f andbehind the train is longer λ ’ = λ

+ V E /f when the train isstationary.

30



Problem: The trucks travel at the same speed. They are far apart on adjacent lanes and approacheach other essentially head-on. One driver hears the horn of the other truck at a frequency that is1.20 times the frequency he hears when the truck is stationary. The speed of sound is 343 m/s. At

w a spee s eac ruc mov ng

31

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Lectures 27-28 Chapter 17 Fall 2010

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