Ch11 Physics

8/12/2019 Ch11 Physics

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11.1

WAVES AND SOUND

CONCEPTUAL QUESTIONS

1. SOLVE If the frequency doubles in a medium where the speed of the wave is independant of the frequency then

the relationship between speed frequency and wavelength determines the change in wavelength.

w o ov f=

where wv is the velocity of waves in water.

If the frequency 1 2 of f= is doubled the wavelength 1 is given by:

1

1

12 2

w wo

o

v vf f

= = = The wavelength is halved if the frequency is doubled.

REFLECT The relationship between frequency, velocity, and wavelength places each of these quantities on the

same level of mathematical importance. Physically, however, in most cases the velocity is fixed in homogeneous

media so the equation reflects an inverse relationship between frequency and wavelength.

1

f

While this relationship is generic for all wave phenomena, it cannot always be used in the way it is used above.

Consider the example of light of different frequencies traveling in glass. The speed of the light through the glass is

dependent upon frequency, which is the reason we have rainbows. For each color (frequency) the fundamental

relationship above still applies but because the velocity is dependant upon frequency, doubling the frequency will

not correspond to light with half the wavelength.

2. SOLVE Based on experience, the speed of sound in air is frequency independent. If velocity were frequency

dependant, music heard from a long distance away would sound very strange. The bass notes on the downbeat, for

example, would not arrive at your ears at the same as the treble notes on the downbeat.

REFLECT Explosive sounds (like the crack of a gun) are composed of high-frequency components and

low-frequency components. If velocity were frequency dependent in air you would not hear a crack, you would

hear the low/high frequency followed by the high/low frequencies, and this effect would be increased the farther

the source was away from the listener.

3. SOLVE The frequency and wavelength do not depend upon the amplitude of the wave, therefore the frequencyand wavelength will be unaffected by an increase in amplitude.

REFLECT The frequency is simply dependent upon how quickly the source in the medium is oscillating, and the

wavelength is dependent upon the frequency and the speed that the disturbance travels through the medium. The

amplitude may determine how far the wave travels before being dampened out by dispersive mechanisms, but it

does not affect the speed of travel of disturbances in the medium.

4. SOLVE Imagine a transfer station at the interface between the air and water consisting of a little bug that jiggles

in response to the wave in the air. Then, the jiggling bug causes the water to jiggle. This disturbance then radiates

into the bulk of the water as a wave. The frequency of the disturbance is unchanged as it goes from water to air.

11


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11.2 Chapter 11What is different between the two mediums is the speed of travel of the wave. The wave travels slower through the

air than it does through the water. From the fundamental relationship between wavelength, frequency, and speed

of travel v f= the fixed frequency and difference in speed of travel between the air and water produced longer

wavelengths in the water than in the air.

REFLECT This result would be true if the wave originated in the air and penetrated into the water as well as if the

wave originated in the water and was passed into the air. For the air water system the difference in speed is about

1.5 times so the wavelength is about 1.5 longer in water than it is in air. The invariance in the frequency means that

you will hear the same basic pitch of the voice above and below the water.

5. SOLVE The energy in a wave is not described by one point in space. The energy of a wave is distributed over a

volume containing the wave. If there is a point in space that destructive interference is observed there will be

points of constructive interference observed (i.e., where the local energy density is greater than each of the waves

individually). These points of constructive interference are where the energy goes. The energy of the wave is a

sum over all these points in space. Therefore, when considering all constructive and destructive interference, the

total energy of the wave is the sum of the energies radiating from the two sources of the wave.

REFLECT The principle of conservation of energy is one of the most important tools in our understanding of

nature. If you ever come up to a puzzle that appears to contradict this principle you can rest assured that if you look

hard enough you will find the missing energy. If you cannot find it you could be on your way to a Nobel prize.

6. SOLVE With two nodes there is one anti-node (think of a jump rope), three nodes correspond to 2 anti-nodes(imaging a figure eight), 4 nodes yield 3 anti-nodes. Extrapolating we discover for n nodes we will observe 1n

anti-nodes.

REFLECT Imagine a circular racetrack. One circuit around the track consists of two nodes (start and finish line)

and one anti-node (point at which you are as far away from the start/finish line as possible). Two laps introduces

one crossing of the starting line (one node) and one anti-node. Each successive lap added, adds one node and one

anti-node. We start out with one more node than anti-node so every lap will still leave a total of one more node

than anti-node.

7. SOLVE The velocity of a wave pulse on a rope increases as the tension in the rope increases. The tension in the

rope is maximum where it is fastened to the ceiling because at that point the rope must support the weight of the

entire rope below it. Therefore, assuming the rope is hanging under its own weight alone, the wave will increase in

speed as it travels up the rope because the tension in the rope increases as the position on the rope approaches theceiling. As the reflected wave travels back down the rope, the pulse will slow down because the tension on the rope

decreases as the rope has to support less of a percentage of the total weight of the rope.

REFLECT Pulse slowing down the rope is an easy effect to observe. Try it! If the rope supports a weight that is

much greater than the weight of the rope itself, this effect would diminish because the tension in the rope would be

due almost entirely to the weight hanging from the rope causing an effectively uniform wave velocity along the

entire rope.

8. SOLVE Tuning a brass instrument (as with other wind instruments) is achieved by changing the length of the

piping between the mouthpiece and the bell of the horn. Increase in temperature causes metals to expand hence

making the piping longer when it is warm than when it is cold. The goal of a tuned instrument is to produce

standardized frequencies during a performance. During a performance the instrument is being warmed by the

musician so it makes sense to have the temperature of the instrument match that of the performance temperaturebefore tuning the instrument.

REFLECT It does not take much change in length of the instrument to produce a noticeable detuning of the

instrument. The longer the pipe distance between the mouthpiece and the bell the more temperature can have an

effect on the tuning. Temperature is a very important consideration in many musical instruments from percussion

to the wind instruments.

9. SOLVE An octave difference between string 1 and string 2 ( 1S and 2S ) corresponds to a factor of 2 difference

between the fundamental frequencies of the two strings 2 12o of f= (we shall assume that string two has a

fundamental frequency an octave above 1S ). The fundamental frequency is proportional to the velocity of the wave


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Waves and Sound 11.3on the string. Since the velocity with fixed tension is inversely proportional to the square root of the mass per unitlength ( ), the frequency is inversely proportional to the mass per unit length.

1o

f

Assuming the volume density of the material of the two strings is the same, the mass per unit length is proportional

to the volume of the string which is proportional to 2D where D is the diameter of the string (the strings have the

same length).2D

Combining the above proportionalities yields:

1o

fD

If 1 22o of f= then =

1 2

1 2D Dwhich yields:

21

2

DD =

Consequently, a string whose fundamental frequency is twice that of another possesses, half the diameter.

REFLECT The high E and low E string of a guitar are under approximately the same tension and have

approximately the same volume mass density (larger steel strings are wound while the smaller strings are not). The

fundamental frequency of these two strings differs by two octaves (so differs by a factor of 4). The thicknesses (or

gauge) of the strings listed in Wikipedia also differs by a factor of 4 (approximately).

10. SOLVE Air is composed of roughly 80% N2and 20% O

2. Nitrogen is lighter than oxygen. Table 11.1 shows the

speed of sound in helium is approximately 3 times that of the speed of sound in oxygen, suggesting that the speed

of sound increases as the mass of the molecules of the gas decreases. Since air has oxygen in it, it has an average

molecular mass which is larger than nitrogen alone. Therefore, we infer from the table that the speed of sound in

Nitrogen alone is greater than it would be in a nitrogen/oxygen mixture.

REFLECT The speed of sound in nitrogen under standard temperature 27 C is 353 m/s, greater than the speed

of sound in air under similar temperature and pressure conditions. This is consistent with the fact that sound travels

faster in humid air than it does in dry air. Water has a smaller molecular mass (18 amu) while airs is 29 amu.

11. SOLVE Assuming a point source of the sound, the sound intensity will decrease as a function of 21r/ where r is

the distance of the observer from the source. Doubling the distance will decrease the intensity by a factor of 4.

The sound intensity level is measured in units of decibels on a logarithmic scale given by:

( )10 log oSIL I I = / Let the SIL at one distance be ( )1 110 log .oSIL I I = / Doubling the distance produces

( ) ( )2 1 1 110 log 4 10 log 10 log4 6 dBo oSIL I I I I SIL= / = / = Therefore, every time the distance is doubled the sound intensity level decreases by 6 decibels.

REFLECT The SIL is constructed to produce a value of 0 dB when a sound cannot be heard anymore by the

typical human auditory system. Normal talking dB levels at 1 m distant range from 4060 dB. By our calculation,at 2 m normal talking would have a range of 3454 dB and so on until at about 100 meters the above range is 020

dB. Which means that with a soft talker in a very quiet environment you could just barely tell that they were

talking if you were 100 meters away, and the loud talker would sound like a whisper.

12. SOLVE The standing wave mode frequencies in a tube are dependant upon the length of the tube. Longer tubes

produce lower frequencies. When all the holes are covered in a flute the tube is as long as the flute. To a first

approximation, the tube will be as long as the distance between the mouthpiece and the first open hole, so opening

holes shortens that length of the flute, increasing the fundamental pitch played.


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11.4 Chapter 11REFLECT This basic idea is generally true for all wind instruments. Of course the detailed physics involved are

more complicated. The above is true for the so-called open tone holes (they are bigger). However, so-called

register holes (smaller) are a bit more complicated and have to do with shaping the harmonic composition of the

note much like light fingering of a guitar string can omit harmonics from a tone.

13. SOLVE Sliding a trombone changes the length of the air column supporting the standing wave mode which

ultimately produces the sound emanating from the instrument. The longer the air column, the lower the pitch of the

fundamental mode. The same effect is seen in a slide whistle.REFLECT Directly changing the length of the tube is the principle behind all brass wind instruments. However,

while the trombone allows for a continuous change in the lengths (producing a distinctive glissando capability),

other instrumentslike the trumpetproduce discrete changes in length by rerouting through paths of different

length.

14. SOLVE No, if the relative velocity of source and observer is zero (which is the case in this problem) then there is

no Doppler shift.

REFLECT If the Doppler shift depended on a velocity relative to the ground or to the average position of the gas

then you would notice many strange effects that do not occur; for example, the pitch of the music in your car

would sound differently if you were traveling on the highway (in a convertible) than if you were parked.

MULTIPLE-CHOICE PROBLEMS

15. ORGANIZE AND PLAN Given values: frequency f and wavelength .

We are asked to find speed .v The fundamental relationship connecting these three values is:

v f= SOLVE Plugging in values yields:

0 40 Hz 2 0 m 0 80 m/sv= . . = . Choice (d)

REFLECT A straightforward application of the fundamental relationship between , ,v f and .

16. ORGANIZE AND PLAN Given values: frequency v and wavelength .

We are asked to find frequency .f The fundamental relationship connecting these three values is:

v f= Solving for f yields:

vf =

SOLVE Plugging in values yields:

360 m/s240 Hz

1 5 mf = =

.

Choice (c)

REFLECT Another straightforward application of the fundamental relationship between , ,v f and .

17. ORGANIZE AND PLAN The wavelength is independent from the amplitude of the wave so changes in amplitude

will produce no change in wavelength.

SOLVE Doubling the amplitude does not affect the original wavelength of 2 67 m = .

Choice (b)


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Waves and Sound 11.5REFLECT If amplitude had an affect upon wavelength musical instruments would be even more challenging to

play than they are. Pianos would change pitch with loudness so playing a specific pitch would require both a

dynamic level and key. Music would change in pitch on your stereo depending on volume. Experience tells us this

simply is not the case.

18. ORGANIZE AND PLAN We are given the frequency f and the velocity .c

The fundamental relationship is:

v f= Solving for lambda : v

f =

Additional notes: Frequency is given in units of 6MHz 10 Hz=

SOLVE Plugging in values:

8

6

3 0 10 m/s3 1 m

98 1 10 Hz

. = = .

.

REFLECT Another straightforward application of the fundamental relationship with a slight wrinkle of units. We

learn that the wavelength of an FM radio station is around 3 meters. This wavelength gives an idea of the size of

objects that the waves can diffract (or bend) around and why you cannot receive FM radio under a bridge.

19. ORGANIZE AND PLAN The first harmonic is the fundamental frequency which has 2 nodes and 1 anti-nodes.

From Conceptual Question 6 we deduced that for the thn harmonic there are 1n + nodes and n anti-nodes.

SOLVE For the third harmonic there are 3 1+ nodes and 3 anti-nodes.

Choice (b)

REFLECT A straightforward application of previous result and a little vocabulary.

20. ORGANIZE AND PLAN Sound intensity is inversely proportional to the square of the distance from the source:

21I d / We are given a sound intensity

oI at a known distance

od and asked to find the intensity at 1 2 :od d= The

proportionality yields:

2 21 1o oI d I d=

Solving for 1I :

2

1 21

o

o

dI I

d=

Since, in this problem, 1 2 :od d=

14

oI

I = SOLVE Plugging in values:

5 25 2

1

4 0 10 W/m1 0 10 W/m

4I

. = = .

Choice (c)

REFLECT In open space, doubling the distance decreases the intensity by a factor of 1/4. The 21d/ dependence

does not hold in all spaces however. In a long, narrow tunnel where the sound can bounce freely off the walls

without being absorbed the sound does not diminish in intensity with distance.


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11.6 Chapter 1121. ORGANIZE AND PLAN The sound intensity level is given by:

10 log( )oSIL I I = / Where the intensity 12 210 W/moI

= is chosen to be the threshold intensity that can be heard by typical human

ears.

If we are given the intensity of a sound as we are in this problem, the determination of the SIL is a one-step

calculation.


7 1210 log(2 0 10 10 ) 10 (log2 0 5) 53 dBSIL = . / = . + = Choice (b)

REFLECT As a student, you may ask why not just use the intensity. One reason is the operational intensity range

for human hearing spans approximately 12 orders of magnitude from 20 000000000001 W/m. to 21 W/m . It is

common to use a logarithmic scale when dealing with such widely ranging values.

22. ORGANIZE AND PLAN We are given the intensity level 1SIL a distance 1d from a source then asked to find 2SIL

if the distance is doubled. Recalling the result from CQ 11 we note that doubling the distance produces a decrease

of 6 dB in the SIL value (reestablished below).

The sound intensity level is measured in units of decibels on a logarithmic scale given by:

( )10 log oSIL I I = / Let the SIL at one distance be ( )1 110 log oSIL I I = / . Doubling the distance produces

( ) ( )2 1 1 110 log 4 10 log 10 log4 6 dBo oSIL I I I I SIL= / = / = SOLVE So if the SIL is 72 dB at 25 m it must be 72 dB 6 dB 66 dB = at 50 m

Choice (a)

REFLECT Building intuition about SIL values takes some time. However, even though most of you will never

use the decibel scale in your chosen profession it is not uncommon to require some skill in manipulating

logarithms. The 6 dB rule is a handy thing to have, and if you know how to derive such rules you can produce

them for yourself in other areas.

23. ORGANIZE AND PLAN We are given an approach velocity of the train and the source frequency of a sound and

asked to find the frequency perceived by a stationary observer. We call upon Equation 11.7:

1 s

ff

v v =

/

where f is the frequency as heard from the train, sv is the velocity of the train approaching the stationary

observer, v is the speed of the wave (in this case the speed of sound 343 m/s).


1 13 kHz

1 2 kHz1 (20 m/s) (343 m/s)f

.

= = . /

Choice (b)

REFLECT This is a significant difference in pitch and corresponds to roughly the pitch difference between any

two adjacent keys on a piano.


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Waves and Sound 11.724. ORGANIZE AND PLAN The fractional value of the observed frequency with respect to the emitted frequency from

a retreating source is obtained by manipulations of Equation 11.8:

1s

ff

v v =

+ /

It follows directly that:

11 s

ff v v

=+ /

In this problem we are givenf

f

and asked to solve for .

sv For cleanliness let

f

f

= and isolate .sv

(1 1)sv v= / The value 0 8 = . in this problem or 80% and we shall use 343 m/s for the speed of sound .v


343 m/s 0 25 86 m/ssv = . = Choice (d)

REFLECT It is interesting to note that if the retreating velocity is the speed of sound the observed frequency is a

factor of 2 (or an octave) below the emitted frequency. Though it may seem like an object moving away at the

speed of sound would not be heard, it is not consistent with experience. You can hear a jet travel away from you

after you have heard the sonic boom. The situation is different if you are moving relative to the air away from the

source greater than the speed of sound. In this case, you cannot hear sound from the source. Study conceptual

Example 11.13 to understand this more deeply.

25. ORGANIZE AND PLAN From Figure 11.15b the fundamental and first overtone (third harmonic) is depicted. The

first overtone in the half-opened pipe is 3 times the frequency of the first harmonic since the first harmonic allows

for 1/4 of a wavelength in the tube while the first overtone allows for 3/4 of a wavelength.

We are given the fundamental frequency .of To find the frequency of the first overtone we simply multiply by

three.SOLVE The frequency of the first overtone is 3 3 220 Hz 660 Hzof = =

Choice (d)

REFLECT One end closed pipe produces only odd harmonics because the standing wave modes require a node at

one end and an anti-node at the other. The one end closed pipe corresponds to instruments such as trumpets,

clarinets, oboes. If both ends are open (in a flute, for example) then all harmonics are produced. This distinct

difference in harmonic content is responsible for the distinctive sounds of the two types of instruments.

26. ORGANIZE AND PLAN From Figure 11.15c the fundamental and first overtone (second harmonic) is depicted. The

first overtone in the open pipe is 2 times the frequency of the first harmonic since the first harmonic allows for 1/2

of a wavelength in the tube while the first overtone allows for a full wavelength.

We are given the fundamental frequency .of To find the frequency of the first overtone we simply multiply by

two.

SOLVE The frequency of the first overtone is 2 2 440 Hz 880 Hzof = =

Choice (b)

REFLECT The first overtone in open pipes is twice the frequency (or a full octave above) the fundamental

frequency.


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11.8 Chapter 11PROBLEMS

27. ORGANIZE AND PLAN We are given values for wavelength and frequency and asked to determine wave speed.

We will employ the fundamental relationship: v f=

SOLVE Plugging in values: Part (a): 1 55 m 0 365 Hz 0 566 m/sv= . . = .

Part (b): 1 55 m 0 730 Hz 1 13 m/sv= . . = .

REFLECT Proportionality between frequency and velocity for fixed wavelength yields a doubling of the velocity

with a doubling of the frequency.

28. ORGANIZE AND PLAN We are given values for wavelength and speed and asked to determine frequency. We will

employ the fundamental relationship: vf =

SOLVE Plugging in values: Part (a): = =.

343 m/s 312 Hz1 10 m

f

Part (b): If the wavelength is halved the frequency is doubled:

343 m/s624 Hz

0 55 mf = =

.

REFLECT These frequencies correspond roughly to D sharp in the middle of the piano keyboard and one octave

above. So now we know that the frequencies near middle C on the piano correspond to wavelengths of roughly 1

meter.

29. ORGANIZE AND PLAN We assume the waves originated from the same spot at 0.t= The time of travel between

source and observer is t d v= / where d is the distance between source and observer and v is the velocity of the

wave.

The time of travel for the p waves is p pt d v= / while the time of travel for the s waves is .s st d v= / Since the p

waves are faster than the s waves .p st t<

We are given the difference in arrival times 24 .t s = We can derive the relationship between d and t as

follows:

1 1( )s p

s p

t t t d v v

= = Solving for d yields:

p s

p s

v vt d

v v =


2

2km6 4s24 s 288 km

km km6 4s s

d

= =

REFLECT P waves (primary waves) and s waves (secondary waves) are used to deduce many things including

location of earthquakes and the structure of the interior of the earth. Fundamental in making the connections

between the phenomena and the deduced information is an understanding of waves.

30. ORGANIZE AND PLAN We are given the wavelength and frequency of waves in order to deduce the velocity of

the waves. This is achieved via the fundamental relationship .v f=


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Waves and Sound 11.9From this and the relationship between the velocity and the depth v gd= we can determine d:

2 2 2v fd

g g

= =

where 29 8 m/sg= . (this will necessitate changing the units of the wavelength from cm to m).

As the depth doubles, the the velocity increases by a factor of 2. The frequency of oscillation will remain

unchanged at an interface so from the fundamental relationship the wavelength will increase by the same factor as

the velocity.

SOLVE The shallow depth:

2 2

2

(0 0110 m) (23 5 Hz)0 0068 m

9 8 m/sd

. .= = .

.

If the depth is doubled the wavelength is

2 2 1 10 cm 1 56 cmo = = . = . REFLECT The shallow water equation can be applied only when the wavelength is larger than the depth. In this

case, this relationship holds. Beyond this regime, the surface waves do not depend upon depth. An interesting

consequence of dependence upon depth is a kind of piling up of waves as water waves approach a sloping shore.

31. ORGANIZE AND PLAN Establishing the requested relationships require study of the arguments and understanding

the periodicity requirements of these arguments.

SOLVE The wave form ( ) cos( )y x y A kx t, = describes the displacement as a function of x for all time. At

0t= the wave completes one full cycle when the argument equals 2 (shown in figure below). We shall use this

fact to define the value of k as follows:

2k = Solving for k yields: 2k =

A

A

0

y =Acos(kx)

y

x =l

If we fix the position 0x= and watch the point oscillate in time. One full oscillation occurs over one period .t T=

As above, one full oscillation corresponds to the argument equaling 2 . We use this fact to define the value of

as follows:

2T = Solving for yields:

22 fT = =

We shall determine the wave speed by looking at how the position of the peak changes with time. The location of

one peak is obtained by setting the argument of the sinusoid equal to zero and finding x as a function of time.

0kx t =


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11.10 Chapter 11Solving for x yields:

x tk

=

The slope of the position as a function of time is the velocity. Consequently the velocity is:

vk

=

REFLECT The wave number and the angular frequency are the most natural numbers to describe waves, and in

more advanced physics these numbers dominate the theory. It is useful to be able to derive these relationships. For

quick recall these relationships can also be derived using unit analysis.

32. SOLVE See graph below.

0

y =Acos(kx p/2)

y =Acos(kx p)

y =Acos(kx 3p/2)

y =Acos(kx 2p)y

x

REFLECT

The wave form moves to the right as time increases. After one period the wave has moved to the right one full

wavelength.

33. ORGANIZE AND PLAN The direction of the velocity of the wave is determined by the relative signs of the spatial

and temporal arguments in the expression for the waveform.

Recall from Problem 36 that the velocity was determined by setting the argument to zero:

0kx t = Isolating for x yields:

x tk

=

SOLVE If the slope of the position as a function of time is negative then the wave travels in the negative x

direction. The sign in the argument that corresponds to a negative velocity is positive. Consequently, the sinusoidal

form is:

( ) cos( ) cos 2 x t

y x t A kx t AT

, = + = +

REFLECT Knowing the functional form for a wave traveling in both directions is required to describe waves

arising out of any disturbance. The preceding three problems are important prerequisites for more advanced

mathematical treatments of waves.

34. ORGANIZE AND PLAN We are given frequency and the speed of waves in the two mediums. We use the

fundamental relationship to derive the wavelength as follows .v f = /


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Waves and Sound 11.11SOLVE Plugging in values:

Part (a): Wavelength in air is6

343 m/s 71 m.4 8 10 Hz

= =.

Part (b): Wavelength in muscle is6

1580 m/s 330 m.4 8 10 Hz

= =.

Part (c): The resolution is approximately 330 m 0 33 mm. = . The dimensions of typical muscle cells are

approximately 3 m. Therefore the image will resolve details of hundreds cells but no better.

REFLECT Wave phenomena are widespread in many technologies. Ultrasound itself has saved countless lives.Study your physics, kids!

35 ORGANIZE AND PLAN Our goal is to determine the units of the quantity representing the velocity of a wave on a

string given by / ,T where T is the tension in a string (units of force) and is the linear mass density (units of

mass per unit length). In mks, the units of force are2

kg m[ ]

sF

= and the linear mass density is

kg[ ] .m =

SOLVE Inserting the constituent units yields

kg[ ]v =

2m

kg 2m

ss=

REFLECT Unit analysis can be a very powerful tool. Students should get in the habit of always checking units tocheck solutions.

36. ORGANIZE AND PLAN The velocity of a wave on a string in terms of the tension T and the linear mass density

is given by .v T = /


6250 10 N250 m/s

4100 kg/mv

= =

REFLECT The George Washington bridges highest cables are about 150 m long. If a car struck one cable it

would take about 1.2 s for the pulse to travel up and back down the cable.

37. ORGANIZE AND PLAN Recall from Problem 40 the condition for total destructive interference (the dead spot)occurs when

+ = = , , ,...1 2

(2 1)0 1 2

2

nd d n

where 1d and 2d are the distances between the observer and the two speakers.

At the midpoint the difference is 1 2 0.d d = The location of the first dead spot is when 1 2 .2d d =

The amount that you have to move away from the midpoint 1 2 .2

d dx

= In terms of the given

x =4


You must move

86 0 cm21 5 cm

4x

. = = .

to reach the first dead spot.


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11.12 Chapter 11REFLECT Another hot spot occurs at 43 cm, dead spot at 64.5, etc. For a tone that is an octave above the given

wavelength, the first dead spot will occur at 10.8 cm, hot spot at 21.5 cm. Different frequencies will have different

hot spots and dead spots.

38. ORGANIZE AND PLAN We are given the linear mass density and tension in a string. This will allow us to deduce

the velocity v of the wave on the string using the relationship .Tv =

The wavelengths of the first three harmonics (standing wave modes) are 1 2 322 .3LL L = = = Using the

fundamental relationship f v= /we find the frequencies of the first three harmonics.


Part (a): The wave velocity is .= =.

36 5 M 417 m/s42 1 10 kg/m

v

Part (b): The frequencies of the first three harmonics are: 1417 m/s 278 Hz,

2 0 75 mf = =

. 2 12 556 Hz,f f= = and

3 13 834 Hzf f= =

REFLECT The fundamental frequency of the string corresponds to C sharp just above middle C. The second

harmonic is one octave above the fundamental frequency.

39. ORGANIZE AND PLAN We are given the length and the fundamental frequency of a violin string and asked to find

the velocity of the wave on the string. We can use the expression for the fundamental frequency, 12

vf

L

= and

isolate for v to find:

12v Lf= Given the velocity determined above and the tension in the string we can use the relationship v T = / and isolate

for to find:

2

2 214

TT v

L f= / =


Part (a): The velocity is 2 0 60 m 196 Hz 235 m/sv= . =

Part b: The linear mass density is2 4

49 N (235 m/s) 8 9 10 kg/m

= / = . REFLECT Tuning, or changing the fundamental frequency of the a string, consists of changing the tension of the

string. The design of the violin has roughly similar tensions on each of the four strings so the different open string

fundamental frequencies must come primarily from differing linear mass densities. See the next problem for a

quantitative description of this.

40. ORGANIZE AND PLAN From the previous problem we derived the equation for the linear mass density as a

function of tension, length of the string, and the fundamental frequency:

2 21

4

T

L f=

Each string has the same length, we simply need to calculate for each string.


G String: = = . .

4

2 2

49 8 8 10 kg/m4(0 60 m) (196 Hz)

D String: = = . .

4

2 2

53 4 3 10 kg/m4(0 60 m) (294 Hz)

A String: = = . .

4

2 2

60 2 2 10 kg/m4(0 60 m) (440 Hz)


13/34

Waves and Sound 11.13E String: = = .

.4

2 2

83 1 3 10 kg/m4(0 60 m) (659 Hz)

REFLECT Assuming each string is made of the same material, the thickness of the strings decreases as the

fundamental frequency increases. From a structural standpoint, the long-term wear on the instrument and feel of

the string on the bow, an approximately uniform tension for each string necessitates changing gauge to obtain the

range in frequencies instead of changing the tension.

41. ORGANIZE AND PLAN If the fundamental frequency is given by 1f , the first two overtones of a vibrating string

are given by 12f and 13f .


G String: Second harmonic is: 2 2 196 Hz 392 Hz.f = = Third harmonic is: 3 13 588 Hzf f= =

D String: Second harmonic is: 2 2 294 Hz 588 Hz.f = = Third harmonic is: 3 13 882 Hzf f= =

A String: Second harmonic is: 2 2 440 Hz 880 Hz.f = = Third harmonic is: 3 13 1320 Hzf f= =

E String: Second harmonic is: 2 2 659 Hz 1318 Hz.f = = Third harmonic is: 3 13 1977 Hzf f= =

REFLECT The harmonic content of musical instruments produces the tone of a note beyond just the note on the

music (which corresponds only to the fundamental frequency). Skilled violinist can shape the harmonic content by

positioning the bow or using light pressure on the strings to dampen certain harmonics.

42. ORGANIZE AND PLAN The standing wave modes on a string occur when 1 12 ,L= 2

22L= and 3 32L=

corresponding to 1, 2, and 3 anti-nodes, respectively. The wavelength associated with a standing wave mode with 3anti-nodes is given by 3 32 .L=

Given the tension and the linear mass density we can determine the velocity of the wave on the string using

.v T = / The derived velocity and wavelength we can determine the frequency of the third harmonic using the

fundamental relationship 3 3f v= / .

SOLVE Part (a): The wavelength of the third harmonic is = = . = .32 2 0 76 m 0 51 m3 3L

Part (b): The velocity of the wave on the string is5

10 2 N 514 m/s3 86 10 kg/m

v

.= =.

The corresponding frequency of the third harmonic is 3514 m/s 1 0 kHz0 51 m

f = = ..

REFLECT The fundamental frequency is 31 3f

f = which reveals this string to be a high E string on a guitar. The

guitar player can reduce the amplitude of the third harmonic by placing the finger lightly on one of the anti-nodes1/3 of the way down the length of the string.

43. ORGANIZE AND PLAN The velocity as a function of tension and linear mass density is .Tv = The tension is

the total force applied to the wire. The total force per unit area is the pressure P T A= / where A is the cross-sectional area of the cable. The equation of the velocity in terms of the pressure is then:

P A P Pv

A

= = =

/

where is the volume mass density.

The function of the velocity is now in the form given by the problem.

Note on units: The MPa is 6 210 N/m

SOLVE The velocity of the wave on the copper wire right at the threshold pressure is

6 2

3

331 10 N/m193 m/s

8890 kg/mv

= =


14/34

11.14 Chapter 11The answer is independent of the diameter because as the diameter increases, more force would have to be applied

to break the wire. This would increase the tension and therefore increase the velocity. However, as the diameter is

increased, the linear mass density increases by the same factor as the tension. This would decrease the velocity.

These two effects cancel out.

REFLECT This velocity is the maximum velocity possible on a copper wire. Previous problems have shown wave

velocities on wires greater than 193 m/s. We can then conclude that guitar and violin wires cannot be made of

copper.

44. ORGANIZE AND PLAN The previous problem is similar to part a. Derivation shown above is used as a solution in

this part.

Finding the fundamental frequency is done here using the wavelength 1 2L = and the fundamental relationship:

1 1f v= / SOLVE Part (a): The velocity as a function of tension and linear mass density is .Tv = The tension is the total

force applied to the wire. The total force per unit area is the pressure T A = / where A is the cross-sectional areaof the cable. The equation of the velocity in terms of the pressure is then:

A P Pv

A

= = =

/

where is the volume mass density.

Part (b): The string model predicts a fundamental frequency of

3

1 3

1 17 5 10 Pa144 Hz

2 0 0142 m 1040 kg/mf

. = =

.

REFLECT If the vocal ligament is the structure to produce high-pitched sounds then it does not do so as a

vibrating string with a dominant first harmonic as a guitar or violin string. The note on a piano that corresponds to

this frequency is about an octave below middle C, which does not seem to correspond to the high-pitched

sounds. Instead, the harmonic shaping of the resonant cavity formed by the mouth and throat shapes the harmonic

content produced by the vocal ligaments. In short, as is often the case in real evolved systems, the simple physics

provides just enough understanding to get started in a study.

45. ORGANIZE AND PLAN As is discussed in the text the beat frequency is half the difference in the individual waves

frequencies (proved in Problem 102). Therefore, the beat frequency will be

= 1 22b

f ff


440 Hz 439 6 Hz0 2 Hz

2bf

.= = .

REFLECT This frequency will be heard as an oscillation with about 5 periods per second. As the pitches get

closer, the period increases. Musicians get around this high tolerance required for a unnoticeable period by

employing vibrato, which is a manual modulation of the pitch frequency. The interested student can investigate

what sort of resolution is required in adjusting the tension of the string to actually be able to tune the instruments

(the A string of the violin) to even this precision.

46. ORGANIZE AND PLAN We are given the frequency, we know the speed of sound at 20 , and we are asked to find

the wavelength. The fundamental relationship is .v f = /



15/34

Waves and Sound 11.15The wavelength associated with the highest frequency a dog can hear is

3

3

343 m/s6 86 10 m 6 86 mm

50 10 Hz = = . = .

REFLECT

The shortest wavelength humans can hear, roughly 17 mm corresponds to 20 kHz. The wavelength determines

the length scale of objects around which sound can diffract. Such considerations aid in developing theories ofsound localization and mechanisms for hearing.

47. ORGANIZE AND PLAN The distance traveled over the time it took the signal to reflect off the bottom and come

back is twice the distance the submarine is from the bottom. Recalling that rate times time equals distance we

obtain the required distance as follows:

2 h vt= The depth of the ocean D at that point is simply the depth of the submarine d plus the distance between the sub

and the sea floor.

2

vtD d h d= + = +

Note: The velocity of sound in water is 1480 m/sv=


The depth of the ocean is

1480 m/s 0 86 s65 m 700 m

2D

.= + =

REFLECT The Abyssal Plain (the deepest flat part of the ocean) is around 4000 m so this part of the ocean is one

of the shallower parts. For example, about 30 miles west of the San Francisco coast has an ocean depth of about

700 m.

48. ORGANIZE AND PLAN As in Problem 54, the distance traveled of the sound from source, off cliffs, back to source

is two times the distance x between the source and the cliffs. Using rate times time equals distance yields:

2 xt

v

=

where v is the velocity of sound in air.


The time it takes to hear the echo is

2 175 m1 02 s

343 m/st

= = .

REFLECT You have produced a useful rule of thumb: Every second of time you wait for the echo corresponds to

175 m.

49. ORGANIZE AND PLAN Again, we recall .vt d= At 20 C the speed of sound is 343 m/s while at 0 C the speed

is 331 m/s. Therefore, it will take longer for the sound to travel a fixed distance in colder air.

SOLVE The distance traveled in 35 s is

3343 m/s 35 s 12 10 m 12 km = =


16/34

11.16 Chapter 11It will take

312 10 m36 s

331 m/st

= =

REFLECT As expected, it takes longer for sound to travel a fixed distance d when it is colder but only a second

difference over 12 km.

50. ORGANIZE AND PLAN We shall treat the sound source as a point source in open space. The sound intensity is

inversely proportional to the distance squared, 21 1 1( )I P d= / where 1P is a property of the source only and

1 25 m.d = The intensity at 50 md= is1

2 4

II = while the intensity at 250 md= is 12 100 .

II =

The sound intensity level is defined to be

10 logo

ISIL

I=

Manipulating the SIL into a form more appropriate for the given information we note:

1 110 log 10 log 10 logo o

I ISIL

I I= =

We are given 155 dB 10 logo

I

I= so the SIL for this problem is:

55 dB 10 logSIL= SOLVE Part (a): The SIL for 50 md= (corresponding to 4 = ) is: 49 dB

Part (b): The SIL for 250 md= (corresponding to 100 = ) is: 35 dB

REFLECT The SIL of this source at around 2 5 m. corresponds to roughly 80 dB (a busy city street). At

approximately 1.5 blocks away the SIL diminishes to below that of a soft conversation heard at a distance. A

useful thing to know if you are a real estate agent or home buyer on the hunt.

51. ORGANIZE AND PLAN Total power tP will be obtained from the impinging intensity 1I since intensity is power

per unit area.

1tP I A=

Determination of 1I is achieved by inverting the SIL for intensity:

101 10

SIL

oI I= Preliminary calculations:

12 2 8 5 3 5 2 21 10 W/m 10 10 W/m 0 000316 W/mI

. .= = = . SOLVE The power incident on the eardrum for an 85 dB sound is

2 2 90 000316 W/m (0 005 m) 25 10 W 25 nWt

P = . . = = REFLECT There is a very small amount of power incident upon the eardrum. It is about the same power that

would be required to lift a flea in the Earths gravitational field at a rate of about 5/10 of an inch per minute.

52. ORGANIZE AND PLAN The wavelength is obtained using the fundamental relationship vf =

The speed of sound in air is 343 m/sav = while in glass it is much faster 5200 m/s.gv = We expect to find much

longer wavelength in glass.


17/34

Waves and Sound 11.17SOLVE The wavelength of the 550 Hz sound in air:

343 m/s0 62 m

550 Hza = = .

The wavelength of the 550 Hz sound in glass:

5200 m/s9 5 m

550 Hzg

= = . REFLECT This result brings up an interesting point. It is not common to have a piece of glass that is 9.5 m deep.

At no point will a full wavelength be present in the glass so describing the wavelength of the wave in the glass is

not very illuminating for 550 Hz sound waves.

53. ORGANIZE AND PLAN As we derived in Problem 57, the SIL can be given as

110 log 10 logro

ISIL SIL

I= =

when there is a known reference .rSIL The decrease in SIL levels when the intensity diminishes by a factor of

1/is 10 log .

We are given a drop of dBx and asked to find the corresponding value of . This can be found as follows: The

desired drop is

dB 10 logx = Isolating for

1010x/ = SOLVE If 3x= then 2. In words, if the intensity decreases by a factor of 1/2 then the SIL diminishes by

3 dB

REFLECT These types of rules of thumb are very handy when dealing with measurement scales that are not

immediately intuitive.

54. ORGANIZE AND PLAN The intensity of sound for a given dBx is achieved by inverting the SIL definition forintensity :I

( 10)10 xoI I /=

where = 12 210 W/moI (the threshold of hearing for the typical human)

SOLVE If x = 95, then 12 2 9 5 2 5 2 210 W/m 10 10 W/m 0 0032 W/mI . .= = = .

REFLECT The SIL of 95 dB corresponds to a fairly loud sound. This is an intuition building calculation.

55. ORGANIZE AND PLAN The power output from the band is .oP The intensity level a distance daway is

determined by remembering that the total power is spread over the surface of a sphere with the radius given by .d

Intensity is power per unit area or

24

oP

I P A d= / = If we want to decrease the SIL by 15 dB we can recall the derivation in Problem 60:

We are given a drop of dBx and asked to find the corresponding value of (where 1/is the fraction by which

the intensity is diminished). This can be found as follows: The desired drop is

dB 10 logx =


18/34

11.18 Chapter 11Isolating for

1010x/ = SOLVE Part (a): The intensity level at 25 md= is

2

2

6 5 W0 00083 W/m

4 (25 m)I

.= = .

which corresponds to the sound intensity level

4 2

12 20

8.3 10 W/mlog log 89.2 dB

10 W/m

ISIL

I

= = =

Part (b): The fractional change in intensity required to diminish the SIL level by 15 dB is

1 51 10 0 0316 ./ = = . REFLECT A difference of 15 dB can also be seen as a decrease of 3 dB five times. In Problem 60 we discovered

that the intensity decreases by a factor of roughly 1/2 for every 3 db of diminished .SIL So, a decrease of 15 dB

corresponds to a decrease by a factor of5

1

2which is approximately equal to the value obtained above.

56. ORGANIZE AND PLAN We are given a drop of dBx and asked to find the corresponding value of (where 1/

is the fraction by which the intensity is diminished). This can be found as follows: The desired drop is


1010x/ = The value of when (105 92) dB 13 dBx= = is

1 310 20. = = The intensity at 1 15 md = is 2

11 .

oP

dI = The new diminished intensity will be 12 .

II

= In terms of the distance, the

intensity is

2 2 21 2

o oP P

Id d

= =

The new distance is 2 1d d=


The distance required to diminish the SIL from 105 dB to 92 dB is 2 15 m 20 67 m.d = =

REFLECT As an approximate guess at the answer using our calculated rules of thumb we note that the intensity is

about 4 times 3 dB, which means a decrease by a factor of 1/16. From the squared value of the distance this factor

corresponds to approximately a quadrupled distance, which is close to the more precisely derived value. The rules

of thumb are good things to have, especially if you dont have a calculator available.

57. ORGANIZE AND PLAN Because the intensity is inversely proportional to the distance squared a factor of p

increase in the distance results in a factor of 21p/ change in the intensity (derived in problem 57 where 2p = ).

With a factor change of 21p/ in the intensity the original sound intensity level before the change in distance oSIL

is changed as follows:

20 log dBoSIL SIL p= The value 20 log dBSIL p = is the intensity level change.


Part (a): If 2p= 20 log 2 dB 6 dBSIL = =


19/34

Waves and Sound 11.19Part (a): If 10p= 20 log10 dB 20 dBSIL = =

Part (a): If 100p= 20 log100 dB 40 dBSIL = =

REFLECT Because of the logarithmic scale notice that changes in orders of magnitude of the intensity produce

linear changes in the .SIL In the previous problem at the rock concert if we moved 100 times farther away from

the original distance of 15 m (which is 1.5 km away), the very loud 105 dB sound diminishes to 65 dB, one could

still hear the sound as if watching the television.

58. ORGANIZE AND PLANAdding instruments causes the intensity levels to add due to the superposition principle of

waves. So, if we have Nidentical sources each contributing an intensity of 1,I the combined intensity is simply

1.tI NI= The SIL of the combined sources is

1 1110 log 10 log 10 log 10 logN

o o

NI ISIL N SIL N

I I= = + = +

where 1SIL is the sound intensity level of an individual source.

The change in the sound intensity level is 10 log .N For a given change SIL we can determine the required N

as follows:

( 10)10 SILN /= SOLVE An increase from 60 dB to 70 dB is an increase of 10 dB. The required number of clarinets is

110 10N= =

REFLECT The first 10 decibels make it appear that you get one decibel increase for every clarinet. However,

observe what happens if we have 100 clarinets: 100 clarinets only produces a 20 dB increase in the .SIL So if one

clarinet produces a SIL of 60 dB it would take 10000000000 clarinets to burst your ear drums (more than the

number of people living on the Earth now).

59. ORGANIZE AND PLAN The intensity of sound is the power per unit area. If you are a distance d away from a

speaker, the total power P from the speaker is spread over the surface of a sphere with the radius ,d so the

intensity at d is

2( )

4

PI d

d=

The threshold of hearing of the human ear depends upon the frequency of the sound (this is obvious and implicit in

the fact that a frequency range of human hearing exists). In Figure 11.14 we see the response curve for the human

ear. We read of the curve that the threshold of hearing at 1000-Hz is 12 21000 Hz 10 W/moI

, = and interpolate the

threshold of hearing at 100-Hz to be 8 8 2100 Hz 10 W/m .oI .

, =

Given the intensity and power we can isolate for the required distance:

4

Pd

I=


Distance required to reach threshold of hearing for 1000 Hz frequencies:

12 2

0 190 000 m 90 km

4 10 W/m

Wd

.= = , =

Distance required to reach threshold of hearing for 100 Hz frequencies:

8 8 2

0 1 W2 000 m 2 km

4 10 W/md

.

.= = , =


20/34

11.20 Chapter 11REFLECT This is an interesting result, because of the response curve of the human ear we will hear frequencies

around 1000-Hz for much greater distances than other frequencies. It turns out that the human vocal range lies

between about 300-Hz to 3000-Hz. The apparent coincidence between these two facts is not surprising.

60. ORGANIZE AND PLAN This problem is similar to Problem 63.

We are given a drop of dBx and asked to find the corresponding value of (where 1/is the fraction by which

the intensity is diminished). This can be found as follows: The desired drop is


1010x/ = If we let 1I and 2I be the two intensities. We note from above that 2 1I I= /. So 1 2I I = / (the requested quantity

in the problem).

SOLVE The value of for 1x= is 1 1010 1 25/ = = .

REFLECT Therefore, the human ear can detect a volume change between a sound with an intensity of 1I and a

sound with intensity 11 25 .I. These considerations are important in the digitization and subsequent storage of

music. If humans cannot hear a volume change less than a 1.25 times change in intensity then there is no benefit of

storing more information than that.

61. ORGANIZE AND PLAN The difference in dB between person A and person B is 9 4 dB 2 4 dB 7 dB.. . = As in

Problem 67 we find the value of

1010B x

A

I

I/= =

where x is the change in dB level.

SOLVE The ratio of sound intensities is 0 710 5.. = = In other words, the threshold intensity for person B is five

times that of person A.

REFLECT Converting between changes in dB levels and changes in intensity is good practice for other log

relationships such as pH values in chemistry. There is almost always another reason to do your physics homework

than simply answering the question.

62. ORGANIZE AND PLAN The fundamental frequency of a flute is = /1 2 ,f v L where v is the velocity of sound in air

and L is the length of the pipe. As shown in the text, playing low B of frequency 247 Hz requires a length of

0 694 m..

If the length is changed by 0 01 m . the frequency will be determined by:

12( 0 01 m)

o

vf

L =

.

SOLVE If the flute is made 1 cm longer, then the fundamental frequency is:

1

343 m/s244 Hz

2(0 704 m)f = =

.

If the flute is made 1 cm shorter, then the fundamental frequency is:

1

343 m/s251 Hz

2(0 684 m)f = =

.


21/34

Waves and Sound 11.21REFLECT These frequency shifts are not enough to produce a half-step difference in pitch required to reach

another note on the scale. For example, a half-step lower than low B (B flat) has a frequency of 233 Hz. A change

of length of 1 cm is a tuning change of the instrument. In fact, this is how a flute is tuned. There is a movable

junction called the headjoint between the mouthpiece and the body of the flute. By sliding this, the length of the

flute is changed, tuning the instrument.

63. ORGANIZE AND PLAN The fundamental frequency of a flute is = /1 2 ,f v L where v is the velocity of sound in air

andL

is the length of the pipe. As shown in the text, playing a low B of frequency 247 Hz at temperature

20 Crequires a length of 0 694 m.. If the length is fixed and the temperature suddenly becomes chilly the speed of

sound will decrease which will decrease the fundamental frequency and thereby the pitch of the instrument.

The speed of sound at 0 C is 0 331 m/s.v =

SOLVE The fundamental frequency of the flute in the B position tuned for 20 C taken to 0 C temperatures is:

1

331 m/s238 Hz

2(0 694 m)f = =

.

REFLECT As expected, the instrument becomes flat.

64. ORGANIZE AND PLAN A pipe open at both ends has a fundamental frequency of 1 2 .f v L= / The first three

overtones have the frequencies 2 1 3 12 3f f f f= , = , and 4 14f f=

SOLVE The fundamental frequency is 1 343 m/s (2 (4 3 m)) 40 Hzf = / . = with overtones 2 80 Hz,f =

3 120 Hz,f = and 4 160 Hzf =

REFLECT The fundamental frequency corresponds most closely to the lowest note on a stand-up bass and has the

same harmonic content as a vibrating string fixed at both ends.

65. ORGANIZE AND PLAN If the pipe has one end open and the other closed the fundamental frequency is 1 4 .f v L= /

In the open-closed pipe only the odd harmonics are present so the first 3 overtones are 3 1 5 13 5f f f f= , = , and

7 17f f= .

SOLVE The fundamental frequency is 1 343 m/s (4 (4 3 m)) 20 Hzf = / . = with overtones 3 60 Hz,f =

5 100 Hz,f = and =7 140 Hzf

REFLECT The flute and the clarinet are approximately the same length but the flute has a higher pitch and

distinctly different tonal flavor. The root of these fundamental differences is simply that fact that a flute is open atboth ends and the clarinet is closed at one end and open at the other. (The reed in the clarinet acts as a driver to the

oscillating cavity and is technically open sometimes but the pressure at the mouth piece is not the ambient

pressure. It is effectively a closed end).

66. ORGANIZE AND PLAN With a fundamental wavelength of ,o the frequency is .o of v= / The second harmonic

has twice the frequency and half the wavelength. The distance between nodes for the second harmonic is 1/2 the

distance between nodes of the fundamental, which is /2.o Therefore the distance between nodes in the second

harmonic is /4.o

SOLVE The fundamental frequency is 343 m/s 2 16 m 159 Hzof = / . =

The distance between nodes for the second harmonic is 2 16 m 4 0 54 m. / = .

REFLECT The fundamental frequency of this pipe corresponds to a flat E or sharp E flat. You could place a small

hole in the pipe (like a register hole in a flute) at the node of the second harmonic and diminish the fundamentals

contribution to the tone but not affect the second harmonic.


22/34

11.22 Chapter 1167. ORGANIZE AND PLAN The wavelength required for a particular frequency is given by the fundamental

relationship:

v f = / For open-closed pipes the wavelength of the fundamental is 4 times the length of the pipe. In other words, the

length of the pipe4.L = Combining the relationships:

4vLf

= SOLVE Plugging in values for the different frequencies:

Part (a): L for 56 Hz: 343 m/s 1.53 m4 56 Hz

L= =

Part (b): L for 262 Hz: 343 m/s 0 33 m4 262 Hz

L= = .

Part (c): L for 523 Hz: 343 m/s 0 16 m4 523 Hz

L= = .

Part (d): L for 1200 Hz: 343 m/s 0 07 m4 1200 Hz

L= = .

REFLECT This is a very wide range of sizes. The details of how pipe organs are constructed and how they

produce their sound is beyond the scope of this book but you know enough now to ask some of the right questions

if you are interested in knowing more.

68. ORGANIZE AND PLAN The frequencies of successive keys on the piano are simple multiplicative factors of 1 122 ./

So if we sit at one key with a frequency 1f the next key up will have frequency1 12

2 12 .f f/= It follows that the

frequency of a note n keys away from a key with frequency 1f will be12

12 .n

nf f/= There are 12 steps per octave

(which is where the 12 comes from). The difference in frequency between octaves is a factor of 2.

SOLVE The frequency of the key above concert A (B flat) is 1 12440 Hz2 466 Hz/ =

The frequency of the third key above concert A (octave above middle C) is 3 12440 Hz2 523 Hz/ =

The frequency of the 12th key above concert A (octave above concert A) is 440 Hz2 880 Hz=

REFLECT The structure of music is beautifully mathematical. It is notable that such beauty and emotional content

can be expressed within such a numerically rational framework. Math and physics are not separate from humanbeauty, they are an integral part of it.

69. ORGANIZE AND PLAN The fundamental frequency of a closed-open pipe is 1/2 the fundamental frequency of an

open pipe:

2 co of f= The first overtone of an open pipe is twice the frequency of the fundamental, so the frequency of the first overtone

in the now opened pipe is

2 2 4o o cof f f= = SOLVE Plugging in values:

The frequency of the first overtone of the now-opened pipe is 2 4 512 Hz 2048 Hzof = =

REFLECT The first overtone of the now-opened pipe is two octaves above the fundamental of the closed-open

pipe.


23/34

Waves and Sound 11.2370. ORGANIZE AND PLAN The expression for the observed frequency f when the source is in relative motion to the

medium (where the observer is stationary with respect to the medium) is:

1 s

ff

v v =

/

where f is the emitted frequency of sound, v is the speed of sound in the medium, sv corresponds to the

velocity of the source approaching the observer, and sv+ corresponds to the velocity of the source receding awayfrom the observer.

We are given f and sv in this problem.

SOLVE Observed frequency of sound when the jet approaches at 232 m/s :

850 Hz2630 Hz

1 (232 m/s) (343 m/s)f = =

/

Observed frequency of sound when the jet recedes at 232 m/s :

850 Hz507 Hz

1 (232 m/s) (343 m/s)f = =

+ /

REFLECT The jet is traveling at roughly 70% the speed of sound. The sound when approaching differs from thereceding sound by roughly 2.5 octaves. Remember this the next time you hear a fighter plane fly toward you.

71. ORGANIZE AND PLAN In this problem the observer is moving relative to the air and the emitter is stationary.

Following conceptual Example 11.13 the observed frequency f is related to the velocity and emitted frequency as

follows:

( )1 of f v v = + / We are given ,f f and we know the speed of sound in the medium 343 m/s.v= We are asked to deduce the

velocity. Isolating the above equation for ov yields:

( 1)ov v f f = / SOLVE Plugging in values:

The velocity of the runner is

359 Hz343 m/s( 1) 6 8 m/s

352 Hzo

v = = . REFLECT The result is a reasonable speed for a fast runner (100 m dash record of just under 10 seconds

corresponds to about 10 ).unitm s/ /

72. ORGANIZE AND PLAN The expression for the observed frequency f when the source is in relative motion to the


1 s

ff

v v =

/


velocity of the source approaching the observer, and sv+ corresponds to the velocity of the source receding away

from the observer.

We are given f and sv in this problem.


24/34

11.24 Chapter 11In part (b) we are given the velocity in km/hr. We shall have to convert this to standard units of m/s.

1 km/hr 0 278 m/s= .

Preliminary calculation:

Speed of bullet train in m/s is 250 km/hr 250 0 278 m/s 69 m/s.= . =

SOLVE Part (a):

Observed frequency of sound when the slow train approaches at 27 0 m/s :.

1200 Hz 1300 Hz1 (27 0 m/s) (343 m/s)

f = = . /

Observed frequency of sound when the slow train recedes at 27 0 m/s :.

1200 Hz1100 Hz

1 (27 0 m/s) (343 m/s)f = =

+ . /

Part (b):

Observed frequency of sound when the bullet train approaches at 69 m/s :

1200 Hz1500 Hz

1 (69 m/s) (343 m/s)f = =

/

Observed frequency of sound when the bullet train recedes at 69 m/s :

1200 Hz1000 Hz

1 (69 m/s) (343 m/s)f = =

+ /

REFLECT As expected, the bullet train approaches with a higher pitched sound and recedes with a lower pitched

sound than the freight train. The difference is noticeable. The difference between receding and approaching in the

freight train is about 3 keys on the piano while the same difference for the bullet train in 6 keys on the piano.

73. ORGANIZE AND PLAN As in the previous problem, we use the relationship:

1 s

ff

v v =

/


velocity of the source approaching the observer, and sv+ corresponds to the velocity of the source receding away

from the observer.

We are given sv and f in this problem. Isolating to find f if the source is approaching the observer:

(1 )sf f v v= / In the second part of the problem, we use the result from the first part and apply

1 s

ff

v v =

+ /

SOLVE The emitted frequency of the squawk:

257 Hz (1 (13 m/s) (343 m/s)) 247 Hzf = / = Observed frequency as goose is receding:

247 Hz238 Hz

1 (13 m/s) (343 m/s)f = =

+ /


25/34

Waves and Sound 11.25REFLECT These differences are noticeable. The difference in frequency between approaching goose and receding

goose corresponds approximately to adjacent keys on the piano.

74. ORGANIZE AND PLAN For Problem 80 the frequency of the approaching goose is 257 Hz while when receding

you hear a sound of frequency 238 Hz. The wavelengths associated with these frequencies are obtained using the

fundamental relationship:

v f = / SOLVE Wavelength of squawk of approaching goose. 343 m/s 257 Hz 1 33 m = / = .

Wavelength of squawk of receding goose. 343 m/s 238 Hz 1 44 m = / = .

REFLECT The difference in wavelength is roughly 10%, which may be surprising given the relatively slow

velocity of geese.

75. ORGANIZE AND PLAN To determine the observed frequency from the stationary helicopter we must know the

velocity of the source (the parachutist). Recall kinematic equations for free-fall. After time t the velocity is v gt=

for constant acceleration. With the deduced velocity we employ the Doppler effect relation for a receding source:

1 /s

ff

v v =

+

In this problem:

=+1 /v

ff

gt


The observed frequency of the shout is

2

425 Hz381 Hz

9 8 m/s 4 s1

343 m/s

f = =.

+

REFLECT The frequency is diminished as expected and the effect is noticeable. The glissando of the falling man

is embedded in our cultural experience. What you hear in movies (or mimic with your voice when you arepretending you are falling) is a mixture between the Doppler effect and free-fall under constant acceleration.

76. ORGANIZE AND PLAN The difference between the receding and approaching frequencies is obtained as follows:

2

1 12

1 1 1a r

s s

f f f fv v v v

= = / + /

where sv v /

This produces the quadratic equation:

22

1 0a r

f

f f + =

We approximate the emitted frequency +2

a bf ff and define +=

a r

a r

f fBf f

which yields the quadratic equation:

2 1 0B + =


26/34

11.26 Chapter 11The possible mathematical solutions are

2 4

2

B B + =

Only positive values of make sense in this problem so the only physically viable solution is:

2 4

2

B B + + =

Backing out the value of the velocity yields:

2 4

2s

B Bv v

+ +=

Preliminary calculation:

628 Hz 686 Hz22 66

686 Hz 628 HzB

+= = .

SOLVE The velocity of the firetruck is

2

22 66 (22 66) 4343 m/s 15 1 m/s2

sv . + . += = . REFLECT We can check the answer by placing the approximate emitted frequency of 657 Hz and the derived

velocity into the receding equation:

629 Hz1 s

ff

v v = =

+ /

as well as the approaching equation:

687 Hz1 s

ff

v v = =

/

This is consistent with the problem parameters (although the emitted frequency is probably closer to 656 Hz).

A speed of 15.1 m/s corresponds to 34 mph. One can only guess the firetruck is in traffic or not in a hurry.



1 s

ff

v v =

/

In this case we are given the combined value 0 99.sv v/ = .

SOLVE Plugging in appropriate number and signs:

When the jet is flying toward us the observed frequency is

1200 Hz120 000 Hz

1 0 99f = = ,

.

When the jet is flying away from us the observed frequency is

1200 Hz603 Hz

1 0 99f = =

+ .


27/34

Waves and Sound 11.27REFLECT The 1200 Hz sound is not audible to humans when the plane is approaching (human range of hearing

is 2020,000 Hz). After the jet passes the sound is approximately lowered a full octave from the emitting

frequency. Of course, at and past the speed of sound, the sound of the jet approaches us only when the jet hits us in

the same way that if you hear a supersonic bullet that has been successfully aimed at you, hearing it means you are

already bleeding.



1 s

ff

v v =

/

For the observed frequency to be twice that of the emitted frequency the object must be approaching and the

following condition must hold:

1 0 5sv v / = . Solving for :sv

2sv v= / SOLVE The source would have to move half of the velocity of the wave in the medium. In air, this velocity is172 m/s.

REFLECT Double frequency corresponds to an octave above the emitted frequency.

79. ORGANIZE AND PLAN From conceptual Exercise 11.13, the expression for the observed frequency f when the

source is stationary with respect to the medium and the observer is moving with respect to the source is:

(1 )of f v v = / where ov is the velocity of the observer and v is the velocity of the wave in the medium. The sign is positive for

approaching and negative for receding.

To double the observed frequency the observer must be approaching the source and the following condition must

be met:

(1 ) 2ov v+ / = SOLVE Solving for ov yields: ov v= To double the observed frequency when moving with respect to a

stationary source you must travel the speed of sound in the medium.

REFLECT The difference between moving source and moving observer in this and the preceding problems

highlights the importance of knowing who is moving and who is not with respect to the medium.

80. ORGANIZE AND PLAN In this case we have both the source and the observer moving towards each other. We

shall first determine the frequency that would be observed due to the moving source moving towards a stationary

point in the medium:

1 s

ff

v v

=

/

Now, from the perspective of the medium in this region there is no difference between a stationary source emitting

a frequency of f and the moving source emitting a frequency of .f Consequently, we may now treat the

observer as if he is moving toward a stationary source emitting at frequency .f The observed frequency in this

case is:

(1 )of f v v = + /


28/34

11.28 Chapter 11SOLVE Subbing in the value of f in terms of the actual emitted frequency fyields:

1

1

o

s

v vf f

v v

+ / =

/

REFLECT Notice that nothing pathological happens when the observer is traveling at the speed of sound but the

observed frequency diverges when the source reaches the speed of sound.

81. ORGANIZE AND PLAN In this case we have both the source and the observer moving away from each other. Weshall first determine the frequency that would be observed due to the moving source moving away from a

stationary point in the medium:

1s

ff

v v =

+ /

Now, from the perspective of the medium in this region there is no difference between a stationary source emitting

a frequency of f and the moving source emitting a frequency of .f Consequently, we may now treat the

observer as if he is moving away from a stationary source emitting at frequency .f The observed frequency in

this case is:

(1 )of f v v = / SOLVE Subbing in the value of f in terms of the actual emitted frequency f yields:

1

1

o

s

v vf f

v v

/ =

+ /

REFLECT In contrast to the previous problem, the unphysical results occur here when the observer is moving

away at and beyond the speed of sound (negative frequency does not make sense). Of course, this is a reflection of

the fact that the observer can actually outrun the sound if she is going away from it.

82. ORGANIZE AND PLAN Let the reference intensity be 1.I The sound intensity level of this sound is:

11 10 log

o

ISIL

I=

If we consider another sound with intensity 12I the corresponding 2SIL is:

1 11 1

210 log 10 log 10 log2 10 log2

o o

I ISIL SIL

I I= = + = +

SOLVE The value 10 log2 3 01 dB= .

REFLECT Since we never specified a specific value for the reference intensity, the result that a doubling of

intensity always produces an increase of 3 dB in the SIL is generic.

83. ORGANIZE AND PLAN Sound intensity is inversely proportional to the distance squared. Consequently, if you

double your distance from a sound source the intensity decreases by a factor of 1/4.

As in Problem 89: Let the reference intensity be 1I . The sound intensity level of this sound is:

11 10 log

o

ISIL

I=


29/34

Waves and Sound 11.29If we consider another sound with intensity 1 4I/ the corresponding 2SIL is:

1 11 110 log 10 log 10 log4 10 log4

4 o o

I ISIL SIL

I I= = =

SOLVE Doubling the distance decreases the intensity by a factor of 1/4.

Since the value 10 log4 6 02 dB,= . doubling the distance reduces the SIL by 6 dB

REFLECT Going the other way we find that halving the distance doubles the intensity twice (quadruples) theintensity. Either way you look at it you get the same answer using the two rules of thumb found in Problems 89

and 90. Doubling twice increases SIL by 3 3 dB.+ Quadrupling intensity increases the SIL by 6 dB.

84. ORGANIZE AND PLAN The difference between the receding and approaching frequencies is obtained as follows:

( )(1 ) (1 ) 2 ( )a r o o of f f v v v v f v v = + / / = / Solving for :ov

2

a ro

f fv v

f

=


The velocity required to observe a difference in frequency of 2 Hz for source frequency of 720 is

2 Hz343 m/s 0 5 m/s

2 720 Hzov = = .

REFLECT This is a walking velocity, but this change in frequency cannot likely be distinguished by a typical

human (the typical minimum change in pitch humans can distinguish corresponds to a frequency change of about5 12002 / ).

85. ORGANIZE AND PLAN We have a measured value of intensity at a reference point:

21 1

I P d= / and the corresponding

1

80 dBSIL =

If we change d from 10 m to 2600 m we increase the distance by a factor of 260. The new intensity level 2I is

then

22 1 (260)I I= /

The corresponding SIL is then:

2 1 20 log260SIL SIL= SOLVE The SIL level at the removed location is

2 80 dB 48 dB 32 dBSIL = = This SIL is about 10 dB louder than a quiet whisper.REFLECT Considering that absorption effects were neglected and there is ambient noise outside even in the

quietest of places, this noise level is quite good. In urban environments, 40 dB is the low limit of ambient sound, so

32 dB will likely be drowned out by the noise.


30/34

11.30 Chapter 1186. ORGANIZE AND PLAN The overtone sequence of a pipe with one end open is as follows:

, , , ... , , , ...1 3 5 1 1 1 3 5f f f f f f The separation between successive overtones is 1 1(2 1 2( 1) 1) 2n n f f + + =

We are given the frequencies of two successive overtones nf and 2.nf + The difference between these frequencies

must equal 12f yielding the following relationship for the fundamental frequency:

21

2

n nf ff + = With the fundamental frequency in hand we note that the fundamental frequency for a open-closed pipe is =1 4 .

v

Lf

The velocity of sound is then

14v Lf= SOLVE Plugging in values:

The fundamental frequency is 1375 Hz 225 Hz 75 Hz

2f = =

The speed of sound is 4 1 12 m 75 Hz 336 m/sv= . =

REFLECT The pipe produces a fundamental note corresponding to lowest note of an upright bass. The overtones

we were given were the second and third overtones (3rd and 5th harmonics). The speed of sound is closer to the

speed of sound at a temperature of 0 C than at 20 . It must be chilly where the pipe is.

87. ORGANIZE AND PLAN The bat emits a frequency that is Doppler shifted due to its velocity directed toward the

wall. The wall sees a frequency of

1s

ff

v v =

/

where sv is the velocity of the bat and f is the emitted frequency of the sound from the bat.

Once the sound has left the bat, the bat is now an observer and the wall is the source. The wave emitted by the

wall is Doppler shifted with respect to the bat. Since the bat is moving and the source is stationary the frequency

that the bat receives is

(1 )of f v v = + / In terms of the primary frequency:

1

1

o

s

v vf f

v v

+ / =

/


Ultimately the bat receives the frequency

7 5 m/s1343 m/s52 0 kHz 54 3 kHz7 5 m/s1343 m/s

f

.+ = . = .

.

REFLECT That speed is a respectable clip (it would hurt if you hit a wall going that fast). It is interesting to

contrast how we determine relative velocity with our eyes. We have to use some sort of geometrical calculation

built into our brains. Bats map pitch into a dynamic map of what is going on around them.

88. ORGANIZE AND PLAN The fundamental frequency of a pianos A string is 440 Hz. We are given the length of

this string .L Given these bits of information we can deduce the velocity of the waves on this string:


31/34

Waves and Sound 11.31The wavelength associated with the fundamental frequency of a string fixed at both ends is = 2 .L

The fundamental relationship for waves is ,v f= which in terms of given quantities is equivalent to:

= 2v L f The material properties of the string that determine the velocity are the tension T and the linear mass density .

v T = / Isolating for and combining with the velocity in terms of L and f yields:

2 24

T

L f=

However, we are asked to find the mass of the string. The mass M is just the linear mass density times the length

24

TM

Lf=

SOLVE Plugging in the values:

The mass of the string is2

667 N 0 0022 kg 2 2 g4 0 389 m (440 Hz)

M= = . = . .

REFLECT Checking units is always a good idea. Make it a habit. Considering that a typical paper clip has a mass

of about 1 g, our piano wire has a mass equivalent to roughly two paper clips.

89. ORGANIZE AND PLAN A closed-open pipe has a fundamental frequency of v .4

fL

= The velocity of sound at

body temperature (37 C) can be obtained by Equation 11.4:

m( ) 331 m/s 0 60 37C 353 m/s

sCv T = + . =

Given the fundamental frequency and velocity the length of the pipe is:

4

vL

f=


The predicted length of the pipe from the crude model is 353 m/s 0 142 m4 620 Hz

L= = .

REFLECT An effective length of 14 cm does not seem unreasonable. The physics of speech is very interesting

and this problem is but the tip of a significant iceberg.

90. ORGANIZE AND PLAN If the pipes fundamental frequency is to be 22 Hz it will need to be longer if it is open at

both ends than if it is open at one end only.

If it is open at both ends the fundamental frequency is = /1 2f v L while for an open-closed pipe = /1 4 .f v L

Isolating for the lengths we find a length of 12L v f= / required for the open pipe and a length of 14L v f= / .


An open pipe must be 343 m/s 7 8 m2 22 Hz

L= = .

to provide the lowest frequency that can be heard as a pitch.

An closed-open pipe must be343 m/s

3 9 m4 22 Hz

L= = .

to provide the lowest frequency that can be heard as a pitch.

REFLECT While both pipes will provide the same fundamental note, they will sound different because of the

differing harmonic content. In other words, if you want it all, you will need both types of pipe. Where would be a

good place to house an 8 m long tube? You got it, a cathedral!


32/34

11.32 Chapter 1191. ORGANIZE AND PLAN Given the SIL at a given point, the intensity of the sound 1I can be directly deduced:

101 10

SILoI I

/= Since the intensity is the total power per unit area (I P A= / ) we can extract the total power by multiplying the

intensity by the surface area of the sphere with a radius d where dis the distance from the source.

214P d I=

Combining the relationships:

2 104 10SILoP d I /=

SOLVE The total power emitted from the speakers

2 12 2 90 104 (8 0 m) 10 W/m 10 0 80 WP /= . = . REFLECT What this demonstrates is that it does not take much power to hurt your eardrums. It is important to

note that the power rating of a speaker is not the power that comes out in the form of sound. Compare this output

(less than a watt) to the output of a lightbulb (60 watts).

92. ORGANIZE AND PLAN Similar to Problem 75:

The separation in frequencies between notes in western music have been constructed such that octaves (noteswhose frequencies differ by a factor of 2) are broken into 12 steps. Two notes separated by one step differ in

frequency by a factor of 1 122 ./ Notes that differ by N steps differ in frequency by a factor of 122 .N/ If middle C

corresponds to a frequency Cf and we know the reference frequency of concert A (440 Hz) is nine steps above

middle C then the following relation must hold:

9 122 440 HzCf / =

Solving for cF yields:

3 42 440 HzCf /=

The C note two octaves above middle C has a frequency of 4 Cf f= . The length of a closed-open pipe with that

fundamental frequency will be

4 16 CL v f v f= / = / SOLVE The frequency of middle C is 261 6 Hz.Cf = .

The length of a closed-open pipe with a fundamental frequency two octaves above middle C is

343 m/s0 082 m

16 261 6 HzL= = .

.

REFLECT A pipe length of 8 cm seems a little small especially if you look at a clarinet and know that two

octaves above middle C are well within the range of the standard clarinet. !There are no holes within 8 cm of the

mouthpiece! This is when you realize that things are a bit more complicated than presented here. Higher registers

are achieved by shaping the harmonic content of lower notes, in effect, removing lower harmonics to producehigher pitched notes. Again, we present here only the tip of a very interesting subject.

93. ORGANIZE AND PLAN The moving heart wall acts as a moving observer of a stationary source as well as a

moving source of sound with respect to a stationary observer.

The moving observer observes a frequency of

(1 )of f v v = /


33/34

Waves and Sound 11.33where f is the frequency of the ultrasound machine emitter and vis the speed of sound in water.

The observed frequency is now the frequency emitted by a moving source. The reflected frequency is:

1s

ff

v v

=

/

In this case, both ov and sv are the same, say .wv Finding the reflected frequency in terms of the primary

frequency of the ultrasound emitter f :

1

1

w

w

v vf f

v v

/ =

/

Solving for wv :

w

f fv v

f f

=

+

The frequency shift is small compared to the source frequency so 2 .f f f +

2w

f fv v

f

=

(n.b.: We are only concerned with the speed of the heart wall (not the velocity) so the sign is not needed.)

SOLVE The speed of the heart wall is

6

100 Hz1480 m/s 0 015 m/s

2 5 0 10 Hzw

v = = . .

REFLECT If you model the oscillation of the heart wall as a harmonic oscillator, the maximum velocity is the

amplitude of the oscillation times the angular frequency. The angular frequency is approximately 15 rad/s (typical

heart rate is 140 bpm). If you assume the amplitude of the oscillation is on the order of 1 mm, the maximum

velocity is 1.5 cm/s, exactly consistent with the obtained result.

94. ORGANIZE AND PLAN When the relative velocity of source and observer are small compared to the speed of the

wave in the medium, the distinction between a moving source and a moving object is not important. As a result ofthis approximation, we will treat the double Doppler shift as if each shift was the same.

So, as shown in Problem 100, the frequency received and emitted from the car is:

2(1 )cf f v c = + / where cv is the speed of the car approaching the police car and

83 0 10 m/s.c= .

We shall make one more approximation by carrying out the squaring in the above equation:

22 ( )c cf f fv c f v c = + / + / Since the velocity of the car is much, much less than the speed of light the squared term is negligible leaving:

2 cf f fv c = / Preliminary calculation: 1 h130 km/h 130 000 m/h 36 m/s

3600 s= , =

SOLVE The frequency shift is then

9

8

2 60 10 Hz 36 m/s14 4 kHz

3 0 10 m/sf f

= = .

.


34/34

11.34 Chapter 11REFLECT The frequency shift is fractionally small to the original frequency. Detecting such a small shift is

achieved using beat phenomena combing the source wave with the reflected wave.

95. ORGANIZE AND PLAN Take a wave formed by the superposition of two waves with frequencies 1f and 2.f The

wave at a given point is described

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Ch11 Physics