chapter 20: mechanical waves -...

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 20 3-1

Chapter 20: Mechanical waves

! How do bats “see” in the dark?

! How can you transmit energy without transmitting matter?

! Why does the pitch of a train whistle change as it approaches or leaves a station?

Make sure you know how to:

1. Graph sinusoidal functions

2. Mathematically describe position-versus-time for simple harmonic motion.

3. Determine the period and the amplitude of vibrations using a graph.

CO: An article in the 2004 medical journal Rheumatology was titled “From bats and ships

to babies and hips” (Kane, Grassi, Sturrock and Balint, 2004). The article described the history of

the development of medical techniques used to detect birth defects, breast tumors, joint problems,

and other medical conditions. Interestingly, the article started by discussing bats. Lazzaro

Spallanzani (1729–1799), an Italian priest, studied them extensively and was ridiculed for

proposing that bats see with their ears. Spallanzani’s bat problem, as it was termed, remained a

scientific mystery until 1938, when finally two young Harvard students, Donald Griffin and

Robert Galambos, explained the mechanism of bat navigation. But even before that in 1912 a

Canadian, Reginald Fessenden, patented devices using a “bat mechanism” to locate sunken ships.

The same idea was used by Paul Langevin and Constantin Chilowsky to construct a “bat-like”

generator that on April 23, 1916 during World War 1 was used to detect and then sink a German

U-boat (UC-3). Finally, in the early 1950s Professor Ian Donald of Glasgow suggested using

“bat” signals to study soft abdominal tissues. He like Spallanzani was ridiculed for suggesting a

totally unnecessary procedure for something that one could investigate by a simple manual

examination. As it turned out the first observations done by Spallanzani, the explanations

provided by Griffin and Calambos, and testing and application experiments conducted by many

led to break though diagnostics not only in medicine, but also in metallurgy and astronomy. They

all relate to the same simple phenomenon, whose name and real explanation you will learn later

in this chapter.

Lead: Vibrational motion occurs when a system that is disturbed from its equilibrium

position moves back toward this position but overshoots. It stops beyond the equilibrium position

and returns only to overshoot again. This repetitive motion caused by a restoring mechanism

leads to vibrational motion. In the last chapter, we looked only at the vibrational motion of one

object but did not consider what happens in the environment contacting the vibrating object. In

this chapter we focus on the effect that the vibrating object has on the medium that surrounds it.


20.1 Observations: pulses and wave motion

Our everyday experience indicates that the medium surrounding a vibrating object is also

disturbed. For example, if you hit one prong of a tuning fork, it vibrates. You can see this

vibration if you shine light on a small mirror glued on a prong of the tuning fork (see Fig. 20.1a).

Simultaneously, you hear sound. If you dip the prongs of the vibrating tuning fork in a pan with

water, you see the water splashing (Fig. 20.1b). The vibrating fork disturbed the water. When you

touch the prongs and they stop vibrating, you stop hearing sound. The vibrating tuning fork must

be disturbing the air adjacent to the fork. The disturbance is transmitted to our ear causing the

sound we heard. It is difficult to follow vibrating prongs as they move very quickly.

Observational experiments described below help us slow down the process.

Figure 20.1 Vibrating tuning fork produces sound

Observational Experiment Table 20.1 Disturbances in different media.

Observational experiments Analysis

1 Tie a rope to a doorknob and then shake once

the end that you hold. A disturbance travels

along the rope.

The disturbance moves along the rope but the rope

fibers (see the little ribbon on the rope) remain in the

same position after the disturbance passes.

2. Place a beach ball in a swimming pool and

Styrofoam pieces on the surface. Push the ball

up and down once. A circle spreads outward

increasing in radius. The Styrofoam pieces move

up and down.

The Styrofoam pieces moves up and down but do not

travel across the pool. The circular disturbance

produced by the ball’s up and down motion spreads

but the water does not move across the pool.

Pattern

In the two experiments, we saw:

(1) that a disturbance propagated in the medium at an observable speed.

(2) the particles of the medium did not travel across space.


Waves and wave fronts

The moving circular disturbance in the last experiment is called a wave front (see the top

view in Fig. 20.2a). Every point on the wave front has the same displacement from equilibrium. If

instead of pushing the ball down and up once, you push it down and up in a regular pattern, you

see a continuous group of humps and dips (waves consisting of many wave fronts) of circular

shape moving outward on the surface of the water (Fig. 20.2b). You have seen such repetitive

waves many times—for example, waves on the ocean or on a large lake. Similar phenomena

happen if we continuously shake the end of a long rope up and down.

Figure 20.2 Wavefronts

Wave motion involves a disturbance produced by a vibrating object (a source). The

disturbance propagates though a medium and causes points in the medium to vibrate.

To explain this new type of motion, we repeat the beach ball experiment. When the

vibrating beach ball moves down, water under the ball’s bottom surface is pushed out to the side

of the ball. We have formed a hump in the water at the sides of the ball. This hump falls back

down towards its equilibrium level and overshoots as the ball lifts up out of the water. The falling

hump pushes neighboring water upward. The hump is starting to propagate away from the ball. A

disturbance moves (propagates) outward, but the water at a particular location moves up and

down and not in the direction the wave travels. Thus, the wave exists because of the interactions

of the neighboring sections of the water and their coordinated vibrations.

We can use the same idea to explain an experiment with a stretched slinky. According to

our understanding of the wave motion, if we push or pull one coil several times, it will pull the

slinky coil attached to it. This coil pulls the next coil, which in turn pulls the next coil, which so

on and so forth. We see a pattern of propagating compressions and rarefactions moving along the

slinky (Fig. 20.3). This is called a longitudinal wave; the disturbance propagates along the slinky

and each coil moves back and forth parallel to the direction of travel of that disturbance.

We can disturb a slinky in a different way. If you vibrate the coil you are holding up and

down perpendicular to the slinky, the coils moves up and down perpendicular to the orientation of


the slinky and to the direction that the disturbance travels (Fig. 20.4). This is called a transverse

wave.

Longitudinal and transverse waves In a longitudinal wave the vibrational disturbance in the

medium is parallel to the direction of propagation of the disturbance. In a transverse wave the

vibrational disturbance in the medium is perpendicular to the direction of propagation of the

disturbance.

The wave we observed on the surface of water is more complicated; the water layers move in an

elliptical path so the wave is part transverse and part longitudinal.

We observe another important property of waves when watching waves on the surface of

the water or on a slinky or on a rope. When a wave reaches the wall of the container or the end of

the slinky or rope, it reflects off the end and moves in the opposite direction (Figure 20.5 shows

this reflection for one pulse.) Thus, if we want to study simple wave motion and not worry about

reflected waves, we need to have a very long medium. Any boundary between different media

causes a reflected wave in the same medium.

Figure 20.3 Longitudinal Wave Figure 20.4 Transverse Wave


Figure 20.5 Pulse reflects off of fixed end

Review Question 20.1

Can we say that particles in the medium move parallel to the direction of wave propagation in

longitudinal waves and they move perpendicular to the direction of transverse wave propagation?

Explain your answer.

20.2 Physical quantities and wave equation

Waves are ubiquitous: water waves, string waves, sound waves, and waves in Earth

initiated by earthquakes. To understand waves and predict their behavior, we need to invent some

physical quantities that are useful for describing them and to find relationships between the

physical quantities. We then use these relationships for practical purposes and to predict new

phenomena.

Let us use a wave source that vibrates with simple harmonic motion—perhaps a motor

that can vibrate the end of the rope. We attach this vibration source to a very long rope (Fig.

20.6a). With a really long rope, we avoid the complications of wave reflections if we observe the

process for the time interval before the reflection occurs. The motor source moves the left end of

the rope up and down perpendicular to the rope and thus produces a transverse wave.

Figure 20.6 Wave produced by vibrating source

At time zero, the source starts at y A" # . The source vibrates sinusoidally; thus the

source’s displacement from equilibrium is described by a sinusoidal function of time (Fig. 20.6b):


2 cos ( )y A t

T

$" (20.1)

where y is the displacement of that end of the rope from its equilibrium position, A is the

amplitude of the vibration, and T is the period of vibration. The left end of the rope moves up

and down with the vibrator. This end pulls the next part of the rope, which pulls its neighboring

part, which pulls its neighboring part, and so on. The wave propagates along the rope at a

particular speed. We now have four quantities that describe the wave started by this vibrating

source.

Period T in seconds is the time interval for one complete vibration of a point in the medium at

any point along the wave’s path.

Frequency f in Hz (–1

s ) is the number of vibrations of a point in the medium as the wave

passes.

Amplitude A is the maximum displacement of a point of the medium from its equilibrium

position as the wave passes.

Speed v in m/s is the distance a disturbance travels during a time interval divided by that time

interval.

Wave Equation

We choose the positive x direction as the direction in which the disturbance travels at

speed v . The left end of the rope at 0x " is attached to the vibrating source. According to Eq.

(20.1), it oscillates up and down with a vertical displacement

2(0, ) cos( )y t A t

T

$" .

The shape of the rope at five different times is shown in Fig. 20.7. If you look at any other

position along the -axisx , the rope vibrates the same way as it does at 0x " only in a different

part of the vibration cycle. For example, at the first marker to the right of 0x " , you see that at

time zero the rope displacement is – A ; at time 4

Tt " , its y displacement is zero; then A# ;

then 0; and finally back to – A .


Figure 20.7 Shape of wave at five consecutive times

We can mathematically describe the disturbance ( , )y x t of the rope at some positive

position x to the right of 0x " at some arbitrary time t . Every point to the right of 0x "

vibrates with the same frequency, period, and amplitude as the rope at the 0x " . But there is a

time delay x

tv

% " for the disturbance at 0x " to reach the location x . The vibration

disturbance at position x at time t is the same as the disturbance at 0x " was at the earlier time

–x

tv

. Thus, the vibration of the rope at position x can be described as:

2( , ) cos[ ( )]

xy x t A t

T v

$" & . (20.2)

You see from the snapshot graphs of the disturbance (see Fig. 20.7e) that the disturbance

pattern repeats in a distance Tv . This distance separates neighboring locations on the rope that

have the same displacement y and the same slope or the locations of wavefronts. This distance is

called a wavelength and is given the Greek symbol ' (lambda):

vTv

f' " " ,

where v is the speed of travel of the wave, T is the period of vibration of each part of the rope,

and f is the frequency of vibration (1

T).

Wavelength ' in m equals the distance between two nearest points on a wave that have exactly

the same displacement and shape (slope) or the distance between the nearest wavefronts:

vTv

f' " " (20.3)

We can substitute this expression for wavelength into Eq. (20.2) and rearrange a little to get the

so-called the wave equation:


2( , ) cos[ ( )] cos[2 ( )] cos[2 ( )]

x t x t xy x t A t A A

T v T Tv T

$$ $

'" & " & " & .

Wave equation The wave equation indicates the displacement from equilibrium y at location x

at time t when a wave of wavelength ' travels at speed v toward the right through a medium.

cos 2 t x

y AT

$'

( )* +" &, -. /0 12 3 (20.4)

The vibration frequency f is the inverse of the vibration period (1

fT

" ).

Tip! Notice how Eq. (20.4) is mathematically symmetrical with respect to the period of the wave

and the wavelength. The wave has two repetitive processes. If the location x is fixed, the

equation shows that the point in the medium vibrates in time with period T . If the time t is

fixed, the space has a wavelike appearance with wavelength ' .

If we write equation 20.4 as cos 2 2 t x

y AT

$ $'

( )" &. /2 3, the part 2

x$'

shows how the

displacement of a point at a distance x from the origin of the wave is different from the

displacement of the point at the origin. The 2x

$'

is called a phase. Two points are said to “be in

phase” when at the same time their displacements are the same, or they have the same phase. By

definition of the wavelength, two points, separated by the distance equal to a multiple of

wavelengths are always in phase.

The wave equation describes the disturbance from equilibrium of a medium through

which a wave travels. When the medium is two dimensional, a line of points that have the same

disturbance at the same clock reading is the familiar wave front. As we already know, one

example of a wave front is a ripple in a water pond surrounding a vibrating object. All particles

forming a wave front on the ripple have the same displacement from their equilibrium position.

Conceptual Exercise 20.1 Make sense of the wave equation A wave on an infinitely long

slinky is described mathematically as (0.1 m)cos 2 0.7 s 1.8 m

t xy $

( )* +" &, -. /0 12 3. Say everything

you know about this wave and represent it graphically.

Sketch and Translate Compare the given equation with the wave equation [Eq. (20.4)]:

cos 2 t x

y AT

$'

( )* +" &, -. /0 12 3. Therefore 0.1 m, 0.7 s, and = 1.8 mA T '" " . We can also say

that at 0t " , the starting point with the coordinate 0x " had a displacement of


0 0(0,0) (0.1 m)cos 2 = 0.1 m

0.7 s 1.8 my $

( )* +" &, -. /0 12 3. We can write a mathematical equation

for the motion of the wave source at location 0x " : (0.1 m)cos 2 0.7 s

source

ty $ * +" , -

0 1. Notice

that at 0x " , y is a function of time only, as we are looking at the disturbance at only one

location. Rearranging the equation Tv' " , we get / (1.8 m)/(0.70 s) = 2.6 m/sv T'" " , a

reasonable number.

Simplify and Diagram This is an idealized situation where the amplitude does not change with

time or with the distance it propagates. Knowing the amplitude and the period, we can draw a

disturbance-versus-time graph for one particular point in the medium (we do it for 0x " in Fig.

20.8a). Knowing the amplitude and the wavelength, we can draw a graph that represents an

instantaneous picture of the whole slinky at a particular time (done for 0t " in Fig. 20.8b). The

latter graph is like a photograph of the slinky.

Figure 20.8 Wave disturbance at one (a) location and (b) one time

Try It Yourself: A slinky vibrates with amplitude 15 cm and period 0.50 s. The speed of the wave

on the slinky is 4.0 m/s. Write the wave equation for the slinky.

Answer: (0.15 m) cos 2 – 0.50 s 2.0 m

t xy

( )* +" $, -. /0 12 3.

Tip! As the wave equation is a function of two variables, when we represent waves graphically,

we need two graphs – one showing how one point of the medium changes its position with time

( -vs- ; y t x const" ) and the other one showing the displacements of multiple points of the

medium at the same time – a snapshot of the wave ( -vs- ; y x t const" ).

A wave has been characterized using five physical quantities: period, frequency, speed,

wavelength, and amplitude. The dependence of these quantities on characteristics of the medium

through which the wave travels is the subject of the next section.


One might say that there are two speeds to consider when talking about wave motion. What two

speeds might these be?


20.3 Dynamics of wave motion: speed and the medium

We defined the speed of a wave as the distance that a disturbance of the medium travels

in a time interval divided by that time interval. This is the operational definition of speed. It tells

us how to determine the speed but does not explain why a particular wave has a certain speed.

Our goal in this section is to investigate what determines the wave speed. We will conduct

experiments in which we record the changes in the speed of disturbances (pulses) on a slinky as

one property of the slinky is changed while other properties are held constant. We assume that the

speed of the pulse is the same as the speed of the wave for the same conditions and the pulse does

not slow down or speed up as it travels. The experiment is shown in Fig. 20.9. In the table we

only show the average results for multiple experiments and do not show the uncertainty for

simplicity, so you can focus on the patterns.

Figure 20.9 Measuring wave speed on a slinky

Observational Experiment Table 20.2 What affects wave speed?

Observational Experiment Analysis

Measure the distance between people holding the slinky

and the time interval for a pulse to travel that distance:

Distance (m) Time interval (s)

4.0 1.0

The speed of pulse is:

distance/time interval = /v d t" %

= (4.0 m)/(1.0 s)

= 4.0 m/s.

Effect of amplitude Change the amplitude of the pulses:

Amplitude (m) Distance (m) Time interval (s)

0.1 4.0 1.0

0.2 4.0 0.9

0.3 4.0 1.0

Effect of amplitude on speed:

Amplitude (m) Speed (m/s)

0.1 4.0

0.2 4.4

0.3 4.0

Effect of frequency Change the frequency of the pulses:

Frequency (Hz) Distance (m) Time interval (s)

2 4.0 1.1

4 4.0 1.0

6 4.0 1.0

Effect of frequency on speed.

Frequency (Hz) Speed (m/s)

2.0 3.6

4.0 4.0

6.0 4.0

Effect of stretching slinky Change force pulling on end of

slinky (length of slinky also changes):

Force (N) Distance (m) Time Interval (s)

2 4.0 1.0

4 8.0 1.1

6 12.0 1.0

Effect of force pulling on end of slinky (and

possibly on length of slinky):

Force (N) Speed (m/s)

2.0 4.0

4.0 7.3

6.0 12.0

Patterns

! The speed of a pulse does not depend on the amplitude or frequency.

! The speed does depend on how hard you pull the end.

The patterns that we found in the Observational Experiment Table 20.2 are surprising –

the speed of the pulse does not depend on the amplitude of the pulse or how often the pulses are


generated—on the frequency. In other words the source of the waves does not affect the speed of

the waves on the slinky!

However, pulling harder on the end of the slinky exerting a larger force leads to the

increase in the speed. When the slinky is pulled harder, the coils spread farther apart – thus not

only the tension changes but also what is called the “linear density” of the slinky changes – the

slinky now has less mass per unit length. Thus, these properties of the slinky affect how fast a

pulse or a wave propagates in it!

If we do similar experiments with stiff springs that stretch less than slinkies, we find that

the wave speed is proportional to the square root of the force pulling on the end of the spring.

This applies to other objects such as strings.

M on M v F4 .

The subscripts indicate the pulling force of one part of the medium on a neighboring part. This

result makes sense – pulling on a spring makes it “stiffer”. The stiffer the spring the more its parts

interact with each other and the faster they transfer the disturbance.

Other experiments indicate that the mass per unit length of vibrating particles (coils in the

experiments in Observational Experiment Table 20.2) affects the speed. The speed of a wave is

inversely proportional to the square root of the mass per unit length of the media (for example the

coils):

1/ 1/m

vL

54 " .

Here the Greek letter 5 is used for the linear density (not for the coefficient of friction). The

dependence on mass can be understood if we think of Newton’s second law. The greater its mass,

the less the acceleration of each particle (coil) of the vibrating medium, and the longer the time

needed to move the next coil. The pulse travels slowly if the mass of coils is large. We can

combine these two factors into one equation:

M on MF

v5

" .

Let us check whether the units are correct:

2

N kg•m m m

kg/m s kg sv

* +* +" " ", -, -0 10 1

.

The units are correct. To make sure that the speed does in fact depend on the above properties, we

can perform a testing experiment. Take a rope of a known mass and length, attach one end to a

wall and pull on the other end exerting a known force. Use the above expression for speed to

predict the time a pulse travels from one end to the wall and back again. As the pulse travels very

fast, to minimize experimental uncertainty we will count 10 trips.

Table 20.3 Testing the expression for wave speed.


Testing experiment Prediction Outcome

Predict the time interval for a pulse

to travel from one end of the rope to

the other end and back again. The

rope is pulled by a spring scale.

The wave speed is:

S on R

/

F

m Lv "

= 20 N

0.60 kg/5.0 m = 12 m/s.

The time interval should be:

2= 2(5.0 m)/(12 m/s) = 0.83 s

L

vt% " . We

predict that ten round trips will take 8.3 s.

We conduct the

experiment several

times and measure

the time to be

8.5 0.5 s6 .

The predicted

outcome lies within

this interval.

Conclusion

The outcome is consistent with the prediction. We did not disprove the relation describing how the speed of a

wave depends on the properties of the medium.

Wave speed The speed v of a wave on a string or in some other medium depends on how hard

the string or medium is pulled M on M

F (one part of the string or medium pulling on the adjacent

part) and the mass per unit length m

L5 " of the string or medium:

M on MF

v5

" (20.5)

Tip: We have two mathematical expressions for the speed of the pulse: d

vt

"%

and

M on MF

v5

" . The first expression is a definition of speed. The second is a cause effect

relationship—the speed depends on the elastic properties of the medium and on its mass per unit

length.

We just learned that for elastic waves the speed depends only on properties of the

medium. Period and frequency of a wave depend on the vibration source—the medium vibrates at

the same frequency as the source (there is an important exception discussed later in the chapter,

the Doppler effect). Thus the wavelength depends on both the vibration frequency of the source

and on the speed of the wave through the medium, which depends on properties of the medium. If

the vibration frequency is low and the wave speed high, a crest of the wave has more time to

travel before the next crest leaves the source. Thus the wavelength is large:

/ fv' " .

However if the speed is low and/or the frequency high, the wavelength is small:

/v f' " .


The amplitude depends on how big the source vibration is and on how big the vibrations in the

medium are—they are not necessarily the same as we learn next. The amplitude does not affect

the speed of wave propagation or the wavelength of the wave. Notice in the following table that

wave speed also depends on the type of wave. Transverse waves travel slower in the same

medium than longitudinal waves.

Table 20.4. The speeds of different types of waves in different media

Type of wave Medium Sample values of the speed

Sound in a gas Air 340 m/s

Sound in a liquid Water 1500 m/s

Compression (longitudinal) wave

in solids

Clay

Sandstone

Limestone

Granite

Salt

1000 m/s

2000 m/s

4000 m/s

5000 m/s

6000 m/s

Shear or transverse wave in solids About 0.6 times the speed of

compression waves given above

Compression wave in a thin rod

or bone

Bone 3000 m/s

Transverse wave on a string Violin A-string

Violin G-string

288 m/s

128 m/s

Example 20.2 Waves on a slinky You place a slinky on the floor. Your friend holds the far one

end with a spring scale and you hold the other end and stretch the slinky (Fig. 20.9). After

measuring with the spring scale the force needed to hold her end, she then holds that end securely

with her hand without changing the length of the stretched slinky. You then abruptly start a

longitudinal impulse down the slinky. What information do you need to collect to be able to

predict the time interval it takes the pulse to reach your friend? After you make the prediction,

what experiment can you conduct to check the correctness of the prediction?

Sketch and Translate The experimental setup is already sketched in Fig. 20.9. Your friend has

recorded the magnitude of the force she exerts on the end of the slinkyS on S

F . You can also

measure the mass of the slinky m and the distance d between you and the friend. The latter two

quantities are used to determine the slinky’s mass per unit length. The above quantities are used

in Eq. (20.5) to predict the speed. To determine the speed experimentally, you need to measure

the time interval t% it takes the pulse to travel between you and the friend, leading to an

independent measurement of the speed.

Simplify and Diagram The above reasoning assumes that there is no friction between the slinky

and the floor.


Represent Mathematically We know that S on S S on S

/

F Fv

m d5" " . This equation allows you to

predict the speed. To check if the prediction matches the speed, we measure using a more direct

method based on the equation d

vt

"%

.

Solve and Evaluate After we collect the measurements for force and mass per unit length and

make a prediction, we measure the speed using the time interval and distance. If the assumptions

are valid, the speed measured this way should be about the same as the predicted speed. To do the

experiment correctly, the uncertainty of the prediction and of the measurement should be

considered to be sure they overlap.

Try It Yourself: Large slinkys can stretch long distances—5 or 10 m. Try to estimate the percent

uncertainty in your ability to measure this distance and also the percent uncertainty in measuring

the time interval for one pulse to travel that distance.

Answer: The distance uncertainty depends on the instrument we use. If we use a tape measure

that is calibrated in centimeters, then the uncertainty in the measurement of the length will be half

of one cm, which is 0.005 m. This makes percent uncertainty about (0.005 m/5 m)(100 %) = 0.1

%. The time uncertainty is due to the use of the stopwatch or due to human reaction time which is

about 0.2 s. Depending on the stopwatch, we choose the biggest uncertainty for the estimate. For

example the smallest time measured by the stop watch is 0.01 s. Thus human reaction time

contributes more to the uncertainty of the time measurement. If the total time of travel of the

pulse is 2 s then percent time uncertainty of about (0.2 s/2 s)(100 %) = 10 %. Our measured value

will probably be uncertain by about 10 percent—maybe more.


Why do you think the speeds of the waves on the A-string and the lower frequency G-string of a

guitar are different?

20.4. Energy, power, and intensity of waves

So far the physical quantities that we used to describe waves were invented for a very

simple case: a slinky or rope with a simple harmonic motion vibrating source. If there is no

damping of the vibrations as the wave propagates along this one-dimensional medium, then the

amplitude of the vibration at any point in the medium is the same as that of the source. What

happens to the amplitude of vibration if we make the model a little more complicated—a two-

dimensional medium (like the surface of a small lake) or a three-dimensional medium (like the air

surrounding a balloon that pops and produces sound)? We will analyze situations in which the

source still undergoes simple harmonic motion and we neglect damping—conversion of

vibrational energy into thermal energy.


Wave amplitude and energy in a two-dimensional medium

To understand what happens to the wave amplitude in a two-dimensional medium,

consider Fig. 20.10. A beach ball bobs up and down in water in simple harmonic motion thus

producing circular waves that travel outward across the water surface in all directions. The

shaded circles in Fig. 20.10 represent the peaks of the waves and are called wave crests. We

observe that the amplitudes of the crests decrease as the waves move farther from the source.

Why?

Figure 20.10 Wave crests at one time on water surface

Suppose that the wave source uses 10 J of energy to produce a small number of waves

during a 1.0-s time interval and that all of that energy is transferred to these spreading waves. As

the wave propagates, more and more particles vibrate since the circular wave has a longer

circumference as it gets farther from the source. However, the total energy of all vibrating

particles cannot exceed 10 J. Since there are more vibrating particles as the wave moves outward,

the energy per particle will have to decrease.

Now, let us do a quantitative analysis. Surround the source with two pretend rings (the

dashed lines in Fig. 20.11), one at a distance R from the source and the other at distance 2 R . If

the energy of the source per unit time is 10 J/s, then this energy first moves through the

circumference 1

2C R$" and then through the second larger circumference

2 12 • 2 2C R C$" " . The circumference of the second ring is 2 times more than the first, but the

same energy per unit time moves past the second ring. Consequently the energy per unit

circumference length ( energy/ energy/2C r$" ) passing the second ring is one-half that passing

the first ring. For any arbitrary distance from the source, the energy per unit time and per unit

circumference length is inversely proportional to the distance from the source.

Two-dimensional waves produced by a point source The energy per circumference

length and per unit time crossing a line perpendicular to the direction the wave travels

decreases as 1/ r , where r is the distance from the point source of the wave. The power

per unit length at a distance 2 R from the source is half the power per unit length at a

distance R .


Figure 20.11 Wave crest amplitude decreases with distance from source

Three-dimensional waves

Consider now a three-dimensional space. Suppose, for example, that a balloon pops and

sound moves outward in all directions surrounding the popped balloon. Consider two imaginary

spheres that surround the source (Fig. 20.12), one at distance R from the source and the other at

distance 2 R . If the energy of the source per unit time is 10 J/s, then this energy first moves

through the surface area 2

14A R$" and then through the second larger area

2

2 14 (2 ) 4A R A$" " . The area of the second sphere is 4 times more than the first, but the same

energy per unit time moves through it. Consequently, the energy per unit area through the second

sphere is one-fourth that through the first sphere. For any arbitrary distance from the source, the

energy per unit time and per unit area will be inversely proportional to the distance squared.

Three-dimensional waves The energy per area and per unit time passing across a surface

perpendicular to the direction the wave travels decreases as 2

1/ r , where r is the

distance from the point source of the wave. The power per unit area at a distance 2 R

from the source is 1/4 of the power per unit area at a distance R .


Figure 20.12 Sound decreases as get farther from source

Tip! The letter symbol for the area and the amplitude of vibrations is the same - A . Make sure

that you think of the meaning of this symbol in the context of the situation to use the appropriate

quantity.

Wave power and wave intensity

We defined the power P of a process in Chapter 6 as energy change per unit time

interval ( /P U t" % % ). The intensity of a wave is defined as the energy per unit area and per

unit time interval that crosses perpendicular to an area in a medium in which a wave travels.

Energy Intensity

time•area •

U PI

t A A

%" " " "

% (20.6)

The units of intensity I are energy (joule) divided by the product area (2

m ) and time (s), or

2J/m •s . The unit J/s is called a watt; thus, the unit of intensity is watt/

2m (

2W/m ).

We found that the intensity of a wave propagating in a three-dimensional medium

decreases in proportion to the inverse of the distance squared from the source (2

1 /I r4 ). The

energy of a vibrating system depends on the amplitude of the vibrating medium. For example for

an object attached to a spring:

2

s max

1

2U kA"

Tip! In the equation above the letter A refers to the amplitude of vibrations.

Thus, if the energy (and intensity) on the left side of the above equation decreases by one-forth

(1/4) when the distance doubles, the amplitude on the right side decreases by 1/2 (since 1/2

squared is 1/4). This is true for waves in three-dimensional space (sound or light), whose

amplitudes of vibration decrease in proportion to 1/ r , where r is the distance from the point

source of the wave. What happens to the wave amplitude of two-dimensional waves?

Example 20.3 Water wave in pool You do a cannonball “dive” off the high board in a pool. The

wave amplitude is 50 cm at your friend’s location 3.0 m from where you entered the water. What

is the wave amplitude at a second friend’s location 5.0 m from where you entered the water?


Sketch and Translate Visualize the person jumping into the pool and the circular wave crests

moving outward. We consider the water wave to be a two-dimensional wave. The amplitude of

the impulse at 3.0 m from the source is 50 cm. We wish to find the amplitude of the impulse at a

distance r = 5.0 m from the source.

Simplify and Diagram Assume that there is no transfer of mechanical energy into internal energy

and the changing amplitude is only due to increased distance.

Represent Mathematically The energy per circumference length and per unit time passing

perpendicular to the direction of wave travel decreases as 1/ r , where r is the distance from the

source. Thus,

2 2 1

1 1 2

(1/ )

(1/ )

r

r

U r r

U r r" " .

We also know that the energy of a vibration is proportional to the square of the amplitude of the

vibration: 2U A4 . Thus:

2

2

2 2 1

1 21

r r

rr

AU r

U rA" " .

Therefore

1/2

1

2 1

2

rA A

r

* +" , -0 1

.

Solve and Evaluate Thus,

1/2 1/2

1

2 1

2

3.0 m(0.50 m) = (0.77)(0.50 m) = 0.39 m

5.0 m

rA A

r

* + * +" ", - , -0 10 1

.

This decreased amplitude farther from the source seems reasonable.

Try It Yourself: You set up longitudinal vibrations on a really long slinky. If there is no friction or

other form of energy transfer from mechanical energy, how does the amplitude at a distance of

2.0 m from the source compare to the amplitude 4.0 m from the source?

Answer: They are the same—this is a linear medium and the same vibrational energy passes

every position on the slinky.


What happens to the intensity of a wave as it propagates in a three-dimensional medium? Explain

your answer.

20.5 Reflection and impedance matching


Until now, we have considered waves moving through a homogeneous medium, a

medium with the same properties everywhere. However, when we first observed the pulses on a

string or a slinky, we saw that they came back after reaching the end attached to a door or being

held by a person. What happens to a wave when it reaches the end of a medium or when there is

an abrupt change from one medium to another?

Consider the following observation. You hold one end of a rope with the other end fixed

(Fig. 20.13). You shake the end once and observe a traveling pulse. When the pulse reaches the

fixed end, the pulse bounces back in the opposite direction (reflection). The amount of reflection

at an interface between different medium (the fixed end is a different medium for the wave

compared to the rope) and the nature of the reflected wave can provide considerable information

about differences of the two media. Geologists, for instance, have determined that the Antarctic

ice sheet rests on earth rather than water because radio waves reflected from the bottom of the ice

sheet are characteristic of an ice-earth interface rather than an ice-water interface. Consider more

experiments.

Figure 20.13 Snapshots of pulses on a string

A thin rope (small mass per unit length) on the left is woven to a thicker rope with a

larger mass per unit length on the right. Send an upward transverse pulse to the right in the thin

rope (Fig. 20.14). A partially reflected inverted pulse returns to our hand and a partially

transmitted upright pulse travels in the thicker rope.

Figure 20.14, 20.15 Snapshot of reflection and transmission at interface

With the ropes connected in the opposite order (the thick rope on the left and thin rope on

the right), send a transverse upright pulse along the thick rope toward the right (Fig. 20.15). Now,


a partially reflected upright pulse returns to our hand and a partially transmitted upright pulse

travels in the thin rope.

How do we explain these observations? In the first experiment, because of its large mass

per unit length, the second thick rope is harder to accelerate upward than if a thin rope was on the

right. A small-amplitude upward pulse is initiated in the thick rope. As the thick rope does not

"give," it exerts a downward force starting an inverted, reflected pulse back toward the left in the

thin rope.

In the second experiment when the upright pulse in the left thick rope reaches the

interface, the small rope restrains the thick rope less than usual. The thick rope initiates a large

upright transmitted pulse in the thin rope and because of the less resistance from the small rope

overshoots and starts a small upright reflected pulse back in the thick rope.

Why is it important to understand the patterns that occur when waves travel from one

medium to another? You saw for example that in the first experiment a significant amount of the

pulse energy travels back – thus less energy is transmitted forward. If an important wave signal is

traveling through the air and it all comes back when reaching an interface into another medium,

little information is transmitted into the second medium. For example, your hard skull reflects

most of the sound energy traveling through the air from another person’s voice or from a musical

event. However, your ear has been built so that much of the energy of sound waves in the air can

reach the inner ear and be detected. On the other hand, if we want to avoid detecting sound

waves, we can create an interface that reflects most of the energy back. For example, you may

wear earplugs while on a long airplane ride or if trying to sleep in a noisy environment.

Impedance

We can invent a physical quantity that characterizes the degree to which waves are

reflected and transmitted at the boundary between different media. In physics this quantity is

called impedance. The impedance Z of a medium is a measure of the difficulty a wave has in

distorting the medium and depends both on the elasticity and the density of the medium—the

medium’s elastic and inertial properties. Elastic property is related to the interactions between the

particles in the medium (the stronger the interactions, the greater the property). The inertial

property characterizes the density of the medium. Impedance is defined as of the square root of

the product of the elastic and inertial properties of the medium:

Impedance = (Elastic property)(Inertial property)Z " (20.7)

If two media connected together at an interface have different elastic and inertial properties, for

example the two different thickness (density) ropes woven together, a wave traveling in one

medium is partially reflected and partially transmitted at the boundary between the media. If both

the elastic property and the inertial property of two media are the same, a wave moves from one

medium to the other as if there is no change. If the impedances of two media are very different

(for example a wave traveling from a medium with small impedance to a medium with high

impedance (like a pulse on a rope attached to a wall), then most of the wave energy is reflected

back into the first medium and does not travel into the second medium. An example of this effect


is the use of ultrasound to detect anomalies in internal organs. Ultrasound reflects off organs

because of their different densities. The shape of a baby inside mother’s womb can be determined

by looking at reflected waves.

Matching impedances

Ultrasound is poorly transmitted from air outside the body to tissue inside the body. The

impedance of the body is much greater than that of air and most of the ultrasound energy is

reflected at the air-body interface and does not get inside to “look” at objects inside the body. To

avoid this problem, the ultrasound emitter is held directly against the body, which is covered with

a gel that "matches" the impedance between and the emitter and the body surface. This matching

allows the ultrasound wave to propagate inside the body.

Scientists and engineers often build special detectors that make a better impedance match

with a sound source—for example, when detecting seismic waves in Earth caused by

earthquakes. One part of the seismometer actually becomes part of Earth and vibrates with it

during an earthquake. They are so sensitive that they can detect a man jumping on the ground one

km from the seismometer.

Animals have also adapted to detect seismic signals. Elephants have stiff cartilage and

dense fat in their feet. This special tissue makes an impedance match with the earth’s surface.

Elephants at times lean forward on their front feet seemingly to detect vibrations in the earth, for

example prior to a new herd approaching a water hole. Reports describe Asian elephants as

vigorously responding to earthquakes, even trumpeting at the approach of an earthquake.

So far we focused on the pulses and waves that reflect at the boundary between two

media. In addition, some absorption of a wave's energy may take place at the boundary between

two media or in any part of a medium through which the wave travels. The coordinated vibrations

of the atoms in the medium is turned into random kinetic energy, that is, into thermal energy. The

rate of this conversion varies greatly. The slower the conversion, the longer the wave lives.


Why is it impossible to create only one traveling wave on a slinky?

20.6 Superposition principle and problem solving strategy

In past sections, we studied the behavior of a single wave traveling in a medium and

initiated by a simple harmonic oscillator. Most periodic or repetitive disturbances of a medium

are combinations of two or more waves of the same or different frequencies traveling through the

same medium at the same time. The sound coming from a violin, for instance, may be a

combination of almost 20 sinusoidal waves, each of different frequency and wavelength. To

understand what happens in a medium through which two or more waves simultaneously pass,

consider several simple experiments.

Observational Experiment Table 20.5 Adding two pulses passing through a medium.


Observational experiment Analysis

An upright pulse traveling right from the left

side of a rope meets an inverted pulse traveling

left from the right side of the rope. The rope

looks undisturbed as the pulses pass in the

middle. After passing, the pulses continue

moving in the same direction with the same

amplitude as if they had never met.

The part of a string

where the pulses met is

pulled up by the pulse

coming from the left,

and down by the pulse

coming from the right

causing a zero

disturbance.

Two upright pulses travel from each end of

the rope toward the center. When the pulses

pass the center, the disturbance is twice that if

only one pulse was present. After passing the

center of the string, the pulses continue

moving in the same direction with the same

amplitude as though they had never met.

The part of a string

where the pulses met is

pulled up by the pulse

coming from the left,

and up by the pulse

coming from the right

producing a double

amplitude pulse.

Pattern

When pulses meet, the resultant pulse is the sum of the two of them (direction is important).

It appears that each pulse pulls the string separately and the final displacement of the

string at a particular time and location is just a combination of the two pulses at that time and

location—adding them together:

net 1 2y y y" #

This looks like superposition we encountered before. For example, we added the forces that other

objects exerted on an object of interest to find the sum of the forces—each force produces an

effect on an object independently of other forces and the net effect is a result of adding the forces.

We encountered superposition when we studied electric and magnetic fields too.

Figure 20.16 Superposition of two waves

We can test if the superposition principle works for the waves. Imagine we have two

vibrating sources in water—for example, you push beach balls A and B up and down at the same

frequency, one in each hand (Fig. 20.16a). They send out sinusoidal waves, so that each particle

of the surface vibrates sinusoidally. If our understanding of the principle of superposition is

correct, then there will be some locations in water where the waves coming from two sources

cause the net displacement of the surface to be twice the displacement if there was only one wave

source. Consider point C in Fig. 20.16a. If the sources A and B vibrate with the same amplitude


and the same period, then they each send a wave toward point C. The source A wave will cause C

to vibrate as:

2cos[ ( )]A

ay A t

T v

$" &

The source B wave will cause C to vibrate as

2cos[ ( )]B

by A t

T v

$" &

If the distances a and b are equal, then the two vibrations arrive in the same phase at point

C ( 2 2a b

Tv Tv$ $" ). If one wave causes C to move up, the other one will also cause it to move

up. If one causes C to move down, the other one will also cause C to move down (Fig. 20.16b

shows the displacements from the A and B waves to C at one instant). If the idea of superposition

is correct and the point C is at the same distance from both sources, then the amplitude of

vibration at C will be twice the amplitude caused by one wave source. There are many points at

the same distance from both sources, and the vibrations at these points will have twice the

amplitude caused by one wave.

We should get the same effect for other points that are one wavelength farther from

source A than from B. This will also be true for points that are any integral multiple of

wavelengths farther from A than from B (if a b n'& " for n = 0, 1, 2, …). At any of these

points, the resultant vibration amplitude due to both waves will be about two times bigger than

the vibration amplitude if only one wave was present.

We can reason further. Suppose now that we choose a point D located so that the wave

from A arrives there at a maximum displacement and the wave from B arrives at a minimum

displacement (see Fig. 20.17a). Then the A wave pushes point D up while B wave pushes it down

the same distance. The net displacement is zero. This situation happens if one wave has to travel a

distance that is half a wavelength longer than the other – see Fig. 20.17b. We conclude that at

locations where 1

( )2

a b n '& " with n an odd integer ( n = 1, 3, 5, …), there will be no vibration

because the two waves will always cancel each other.

Figure 20.17 Waves cancel at D


If we know the source positions and the wavelengths of the waves, we can predict the

locations of these two types of places: where the waves either reinforce each other or cancel each

other. For this testing experiment we can use two speakers at different locations and predict

places where our ears will hear no sound and places that are extra loud. We try this experiment

next. A procedure for solving wave interference problems is described on the left side and applied

to the problem in Example 20.6 on the right side.

Example 20.4 Where is there no sound? Two sound speakers separated by 100 m face each

other and vibrate in unison at a frequency of 85 Hz. Determine three places between the speakers

where you cannot hear any sound.

Sketch and Translate

! Construct a sketch of the situation. Label

all known quantities.

Where do you hear no sound?

Simplify and Diagram

! Decide what simplifications you are

making.

! Draw displacement-versus-time or

displacement-versus-position graphs to

represent waves if necessary.

Assume that the amplitude of the sound does not change as

the wave travels away from the source (this is a reasonable

assumption if the two distances are not too different). The

person should hear no sound at places where the distance

from the left speaker to the person is some odd multiple of a

half wavelength farther than the distance from the right

speaker.

Represent Mathematically

! Represent the situation described in the

problem using the relationships between

physical quantities.

Should hear no sound if:

1( )2

a b n '& " .

Here a is x and b is 100 m - x :

3 5,

2 2 2– (100 m – ) , ....x x

' ' '"

where

–1

sound340 m/s 340 m/s

4.0 m85 Hz 85 s

v

f' " " " " !

Solve and Evaluate

! Solve the equations for an unknown

quantity.

! Evaluate the results to see if they are

reasonable (the magnitude of the answer,

its units, how the solution changes in

limiting cases, and so forth).

Solve the equation in the last step for places where we hear

no sound:

3 5,

2 2 2– (100 m – ) , ....x x

' ' '"

3 5,

2 2 22 100 m = , ...x

' ' '7 &

3 5,

2 2 22 100 m + , ...x

' ' '7 "

3 550 m 50 m + 50 m +

4 4 4, , ....x

' ' '( ) ( ) ( )#. / . / . /2 3 2 3 2 3

7 "

Using ' = 4.0 m, we expect no sound at:

51 m, 53 m, 55 m, ...x " .

Note that the first place is 51 m from the left speaker and 49


m from the right (one half wavelength different). The next

place is 53 m – 47 m = 6 m or 1.5 wavelengths. The answers

seem reasonable. When we conduct the experiment we find

that at the predicted locations we hear no sound.

Try it yourself: For the last situation, where should the person stand to hear sound with twice the

amplitude?

Answer: The condition for these places is that the ear should be the same distance or one or more

full wavelengths different distances from the sound sources:

[ – (100 m – ) 0, , 2 ,....x x ' '" ]!"This condition is satisfied at positions 50 m, 52 m, 54 m, ….

We can now assert the superposition principle with greater confidence.

Superposition principle When two waves pass through the same medium at the same time,

the net displacement at a particular time and location in the medium is the sum of the

displacements that would be caused by each wave if alone in the medium at that time.

Mathematically this statement can be written as:

net 1 2 n...y y y y" # # # (20.8)

The process when two or more waves of the same frequency travel in the same medium

and add together to form a smaller or larger wave is called interference. Locations where the

waves add to create a larger disturbance are called locations of constructive interference. Places

where two waves add to produce a smaller disturbance (to cancel each other) are called locations

of destructive interference.

Huygens Principle

The superposition principle allows us to explain another phenomenon that we observed

many times. Imagine a beach ball vibrating with simple harmonic motion in water. Circular

waves move outward from the ball. How can we explain their formation using the superposition

principle? C. Huygens (1629-1695) provided the answer. He was selected by King Louis XIV to

direct all scientific research in France, just as Aristotle was selected by Alexander the Great to be

his court scientist. During his tenure with the king, Huygens developed the best telescopes of his

time and invented a pendulum clock so accurate that it easily detected small differences in free

fall acceleration at different locations on the earth.

Huygens idea for how waves were created in an orderly way as they moved out from a

source is shown in Fig. 20.18. You see a wave front (like the crest of a wave) represented by line

AB moving at a speed v away from a point source S. To determine the location of the wave front

a short time t after it is at curved line AB, one needs to draw a large number of circular arcs,

each centered on a different point of wave front AB. Each arc represents a wavelet that moves

away from a point on the original wave front. The radius of a wavelet r depends on the speed v

of the wave front at the point where the wavelet originates and on the time t that the wavelet has

traveled away from the point. These three quantities are related by the equation r vt" . The

addition of these wavelets produces a new wave front whose shape can be determined if one adds


the disturbances due to all wavelets. The new wavefront is determined by drawing a tangent to

the front edges of the wavelets – line CD in Fig. 20.18. To test Huygens’ idea, consider two new

experiments.

Figure 20.18 Creating new wavefront

Testing Experiment Table 20.6 Testing Huygens’ principle

Testing experiment Prediction Outcome

Place a vibrating board with many closely placed

pointed pins in water so the pins move up and

down each creating a wavelet in the water.

The wavelets produced by each

pin when added should produce a

straight wave front.

We observe the

predicted

straight wave

front waves.

Place in a tray of water a barrier with a narrow slit

opening in the path of a plane wave.

The vibrations of water in the

narrow slit should produce a

circular wavelet beyond the slit.

We observe a

circular wavelet

spreading

behind the

narrow slit.

Conclusion

Predictions based on Huygens’ principle match the outcomes of the testing experiments. These

experiments do not disprove the principle and give us more confidence in the principle.


Your friend says that it is impossible for two waves to arrive at the same point and still have no

motion at this point. How can you convince him that this is possible?

20.7 Sound waves

We used sound in some of our examples in this chapter. However, we have not discussed

the nature of a sound wave. As we know, for a wave to propagate, there needs to be a vibrating


source and a medium of interacting particles. When some of the particles are disturbed from an

equilibrium position, they push and pull nearby particles and so on.

Early in this chapter, we attached a tiny mirror to one prong of a tuning fork (Fig. 20.1a).

When light from a laser pointer shined on the mirror, we observed the reflected spot of light on

the wall vibrating up and down making a line. If we touched the tuning fork so the sound stopped,

the vibration stopped and there was just a dot of light on the wall. If we hit the fork harder, the

sound was louder and the line on the wall was longer. If you touch the vibrating tuning fork, the

sound dies. If you hit the tuning fork and then put the ends of the prongs down on the surface of

water in a pan (Fig. 20.1b), we saw waves on the water surface produced by the vibrating prongs.

We conclude that sound sources are vibrating objects.

What happens so that our ears hear a vibrating tuning fork or any other sound? Let’s use

a tuning fork and a microphone to get a better feel for the nature of a sound wave in air.

Remember that a microphone has a magnet and a membrane with a coil attached. Varying air

pressure causes the membrane to vibrate and the vibrating coil in a magnetic field causes a

corresponding vibrating electric signal (electromagnetic induction from Chapter 18). Evidently,

the sound wave is a varying pressure in the air. The vibrating prongs of the tuning fork causes air

pressure changes and these pressure changes travel through the air from one place to another. We

also know that sound propagates in water and even in solid objects – you can easily hear a

vibrating tuning fork with the handle touching a table and your ear on the table. Does sound

propagate in a vacuum?

To answer this question we could theoretically perform the first experiment under a

vacuum jar, but it is difficult to evacuate the jar completely. In addition, the tuning fork sitting on

the bottom of the jar would vibrate the jar, which will produce sound. Thus we have to answer the

question by reasoning. If a tuning fork vibrates in a vacuum, there will be no medium for the

tuning fork prongs to vibrate against and consequently no way for a sound wave to originate and

no medium for it to travel from one place to another. Sound needs a medium in order to travel

from a source to a sound receiver, such as the ear.

The measurements of the speed of sound in different media show that sound speed

increases in solids compared to liquids and air, which supports the idea that the sound has the

same nature as waves on a slinky but instead of vibrating coils interacting with each other, there

must be the particles of matter doing the same thing. In summary, sound involves the

transmission of a disturbance between interacting particles that vibrate. The disturbance

propagates in all directions as a wave. The analysis of these compressions shows that sound is a

longitudinal wave.

How do humans hear sound? Your body is a different medium for the sound than the air

through which the sound travels. When a wave reaches a different medium, it is partially reflected

and partially transmitted into it. We perceive these vibrations as sound. We can detect loud

sounds with many parts of our body—even our hands placed on an inside car door when the radio

is playing. But we can’t distinguish with our hands much about the sound—is it a ballad, a


symphony, or a country western singer? To distinguish the details about the sound, we rely on our

ears.

The ear and hearing

A simplified sketch of the human ear is shown in Fig. 20.19. The part of the ear at the

side of our head, called the pinna, gathers and guides sound energy into the auditory canal.

Because of the pinna, the pressure variation at the eardrum due to a sound wave is about two

times the pressure variation that would occur without the pinna and auditory canal. The pressure

variation at the eardrum causes it to vibrate. A large fraction of the vibrational energy is

transmitted through three small bones in the middle ear known collectively as the ossicles and

individually as the hammer, anvil, and stirrup. These three bones constitute a lever system that

increases the pressure that can be exerted on a small membrane of the inner ear known as the oval

window. The pressure increase is possible for two reasons. (1) The hammer feels a small pressure

variation over the large area of its contact with the eardrum, whereas the stirrup exerts a large

pressure variation over the small area of its contact with the oval window. This difference in areas

causes the pressure to be increased by a factor of about 15 to 30. (2) The three bones act as a lever

that causes the force that the stirrup exerts on the oval window to increase, although the

displacement caused by the force is less than if the bones were not present. Because of all these

effects, pressure fluctuation of the stirrup against the oval window can be 180 times greater than

the pressure fluctuation of a sound wave in air before it reaches the ear (a factor of 2 due to the

pinna and auditory canal, a factor of up to 30 because of the area change from the eardrum to the

oval window, and a factor of 3 from the lever action of the ossicles).

Figure 20.19 Parts of the ear

The increased pressure fluctuation against the oval window causes a fluctuating pressure

in the fluid inside the cochlea of the inner ear. The fluctuating pressure is sensed by nerve cells

along the basilar membrane. Nerves nearest the oval window respond to high-frequency sounds

whereas nerves farther from the window respond to low-frequency sounds. Thus, our ability to

distinguish the frequency of a sound depends on the variation in sensitivity of different cells to

different frequencies along the basilar membrane. This ability is remarkable in that the basilar

membrane is only about three centimeter long, and yet a normal ear can distinguish sounds that

differ in frequency by about 0.3 percent (a frequency difference of 3 Hz can be distinguished for

two sounds near 1000 Hz). This ability to distinguish sounds of slightly different frequency spans


the human hearing range from about 20 Hz to 20,000 Hz. What about high frequency sounds—

greater than 20,000 Hz?

Ultrasound

Ultrasound waves are pressure waves at a frequency higher than the upper limit of human

hearing. We cannot hear such high frequencies because of limitations of the ear. On the other

hand, many animals such as bats, dolphins and whales, dogs, and some types of fish communicate

using ultrasound signals. Ultrasound waves are widely used for diagnostics in medicine. Pregnant

women have ultrasound scans to determine whether the unborn baby has certain health defects,

and sometimes to determine the sex of the baby. The ultrasound reflects

differently off various organs, thus producing an interpretable picture for

the doctor (Fig. 20.20). Many applications of ultrasound are based on the

short wavelength of the waves, which are smaller than internal body

parts and help distinguish the details of those parts. Sound waves of

longer wavelength do not allows us to obtain the same information.

However, some applications of ultrasound are not related to the human

body.

Figure 20.20 Ultrasound

photo of child in womb

Cavitation

When low-intensity high-frequency sound travels through non-elastic media such as

water and most liquids, there is a small uniform fluctuation of the pressure in the fluids about

their normal pressure (see Fig. 20.21). If the amplitude and intensity of the ultrasound wave

increases, the rarefaction part of the wave may have such low pressure that the liquid pulls apart

or fractures leaving a small cavity in the liquid, a phenomenon known as cavitation. When a

cavity is formed, the liquid evaporates into it forming a low-pressure bubble in the liquid, which

begins to collapse due to the high pressure surrounding liquid. As the bubble collapses, the

pressure and temperature of the vapor inside increases. The bubble eventually shrinks to a tiny

fraction of its original size, at which point the gas inside can have a rather violent explosion,

which releases a significant amount of energy in the form of a shockwave and as visible light.

The temperature of this vapor inside the collapsed bubble may be up to 5000 K and the pressure

up to 1000 atm.

Figure 20.21 Depiction of pressure variation due to sound wave


Ultrasonic devices use cavitation bubbles to clean surfaces. When used for cleaning

purposes, ultrasound can remove harmful chemicals. Cavitation is often used to homogenize or

mix and break down suspended particles in a colloidal liquid compound, such as paint mixtures

or milk. Many industrial mixing machines are based upon this design principle. Ultrasound and

cavitation is used to break kidney stones. Future applications are planned for joint and muscle

treatment, for corrections of blood clots, and for high intensity focused ultrasound non-invasive

treatment of cancer.


How can you decide if sound is indeed produced by vibrating objects?

20.8 Loudness and Sound Intensity

In every day life we describe different sounds using the words loudness, pitch, and timbre

(the quality of sound). In this and the next section we learn how these words correspond to

various physical quantities.

Loudness

We already know from previous experiments with a tuning fork and laser pointer that two

sounds of different loudness have different amplitudes. Is the loudness of sound determined only

by the amplitude of vibrations? The answer would be yes if our ears were equally sensitive to all

frequencies of sound. However, they are most sensitive to frequencies around 500 Hz; sounds of

less amplitude at this frequency are perceived as loud as sounds of greater amplitude at a

frequency of 2000 Hz. Sounds of very high amplitude at frequency of 20 Hz or 20,000 Hz are

perceived as not loud—our ears can barely hear them if at all. This leads to different measures of

the apparent loudness of a sound—the physical quantities intensity and intensity level.

Sound Intensity

As noted, our impression of the loudness of a sound depends partly on the frequency of

the sound and on the amplitude of the vibrations that the sound induces in our eardrums, which in

turn depends on the magnitude of the pressure variation caused by the sound wave. Normal

atmospheric pressure is 5 2

1.0 x 10 N/m . The pressure fluctuation above and below atmospheric

pressure of a barely audible sound is called the threshold of audibility. For the normal ear of a

young person, the threshold pressure is about –5 2

2 x 10 N/m , or less than one-billionth

atmospheric pressure. A sound that is harmful to the ear is about 6

10 times greater in pressure, or

about 2

20 N/m . Instruments built to measure the loudness of a sound do not measure the sound's

pressure amplitude but instead measure two other quantities called intensity and intensity level.

The intensity I of a wave of any type was defined before [see Eq. (20.6)] as the wave energy that

crosses an area A perpendicular to the wave's direction in a time interval t% divided by the area


and the time interval. Intensity can also be defined in terms of power P since power is the

energy that flows per unit time interval.

Quantitative Exercise 20.5 Sound energy absorbed at a construction site The intensity of

sound at a construction site is 2

0.10 W/m . If the area of the eardrum is 2

0.20 cm , how much

sound energy is absorbed by one ear in an 8-h workday?

Represent Mathematically The intensity I of the sound, the area A of the eardrum, and the time

interval t% of the exposure have been given. We wish to determine the sound energy U%

absorbed. Equation (20.7) relates all of these quantities:

Energy Intensity

time•area •

UI

t A

%" " "

%

To find the energy, multiply both sides by the time interval and the area:

• • U I t A% " %

Solve and Evaluate In using the above, we need to convert the time interval into seconds and the

area into 2

m .

-4 2

2 2

2

3600 s 10 m(0.10 W/m ) (8.0 h) (0.20 cm ) 0.060 J

1 h 1 cmU

( )* +( )* +% " ". /, -, -. /0 12 3 0 12 3.

This looks like a small number. However, our perception of sound energy depends on the power

per unit area, not the total energy. Our environment includes sounds that differ in intensity from

about –12 2

10 W/m (barely audible) to nearly 2

1 W/m (a painful sound). The noise in a

classroom is about –7 2

10 W/m .

Intensity level

Because of the wide variation in the range of sound intensities, a different quantity for

measuring sound intensities, called intensity level, has been developed. Intensity level is not a

measure of a sound's intensity but is a comparison of the intensity of one sound and the intensity

of a reference sound. Intensity level is defined on a logarithmic scale as follows:

10

0

10logI

I8 " (20.9)

where 0

I is a reference intensity to which other intensities I are compared (for sound,

–12 2

010 W/mI " ). The log in Eq. (20.10) is the logarithm to the base ten of the ratio

0/I I . The

logarithms to the base of 10 are often used in engineering and acoustics; they are called common

logarithms1. For simplicity common logarithms assume the base of 10 but it is not written. As I

and 0

I in Eq. (20.10) have the same units, intensity level is a unit-less quantity because the units

cancel. Nevertheless, intensity level has a dimensionless unit called the decibel (dB). The unit

1 If

yx b" , then log ( )by x" . For common logarithms 10b " , thus for log( )y x" ; 10yx " .


serves as a reminder of what we are calculating, much like the radian is a dimensionless reminder

of one way of specifying angles. The intensity levels 8 of several familiar sounds are listed in

Table 20. 6.

Table 20.6 Intensities and Intensity Levels of Common Sounds

Source of Sound Intensity

(2

W/m )

Intensity level

(dB)

Description

Large rocket engine (nearby) 610 180

Jet takeoff (nearby) 310

150

Pneumatic riveter; machine gun 10 130

Rock concert with amplifiers (2 m); jet

takeoff (60 m)

1 120 Pain threshold

Construction noise (3 m) –110

110

Subway train –210

100

Heavy truck (15 m)

Niagara Falls

–310

90 Constant exposure endangers

hearing

Noisy office with machines;

average factory

–410

80

Busy traffic –510

70

Normal conversation (1 m) –610

60

Quiet office –710

50 Quiet

Library –810

40

Soft whisper (5 m) –910

30

Rustling leaves –1010

20 Barely audible

Normal breathing –1110

10

–1210

0 Hearing threshold

Quantitative Exercise 20.6 Noisy classroom The sound in an average classroom has an intensity

of –7

102

W/m . (a) Determine the intensity level of that sound. (b) If the sound intensity is

doubled, what is the new sound intensity level?

Represent Mathematically To calculate the intensity level 8 , we need to compare the given

sound intensity I with the reference intensity for sound –12 2

010 W/mI " and then take 10 times

the log of that ratio:

0

10log I

I8

* +" , -

0 1.

Solve and Evaluate (a) The ratio of the sound intensity I and the reference intensity 0

I is:

7 2

5

12 2

0

10 10 .

10

W/m

W/m

I

I

&

&" "


Since 5

log 10 is 52, the sound intensity level is:

510log10 10(5) 50 dB8 " " " .

(b) If I is doubled, the ratio of I and 0

I becomes:

7 2

5

12 2

0

2 10 2 10

10

W/m

W/m

I

I

&

&

9" " 9 .

The sound intensity level is, then,

510log(2 10 ) 10(5.3) 53 dBL " 9 " " .

The classroom is pretty quiet.

Try It Yourself: The intensity at a concert is 2

0.10 J/s•m . What is the intensity level?

Answer: 110 dB.

Tip! We can calculate the intensity levels listed in Table 20.6 easily by considering the power of

10 by which the sound's intensity exceeds the reference intensity. For example, the sound

intensity in a noisy office is about –4 2

10 W/m which is 8

10 times more intense than the

reference intensity of –12 2

10 W/m . Since the noisy office is 8 powers of 10 greater than 0

I , the

intensity level is 10(8) = 80 dB.

The intensity scales are used in fields other than acoustics. An example is a familiar Richter

scale for seismic activity. The scale uses a single number to represent the amount of seismic

energy released by an earthquake. Sometimes the Richter scale is known as local magnitude

scale. To make Richter scale scientists calculate a base -10 logarithm of the combined horizontal

amplitude of the largest displacement from zero as recorded by a specific instrument (Wood-

Anderson Torsion seismometer). Thus an earthquake that is 6.0 on a Richter scale has a 10 times

bigger amplitude than the one that measures 5.0. The problem with the scale is that it saturates at

high amplitudes thus geologists use a different scale called moment magnitude scale.

Determining Intensity from Intensity Level

Sometimes in our problem solving we need to calculate the intensity of a sound wave (or

some other wave) from its known intensity level. To do this, we use another general property of

logarithms [see Eq. (A.4) in Appendix A.3]. If

0log( / )

10I I

8" ,

/10

0/ 10I I 87 "

/10

010I I87 " (20.11)

For example if a shout produces an intensity level of 75 dB, the intensity of the sound is:

2We need to find

5log 10y " . From the knowledge of logs:

510 10

y x" " , thus in this case 5y " .


75/10 7.5 0.5 7 7 7 12 5 2

0 0 0 010 10 (10 )(10 ) 3.2 10 3.2 10 (10 ) 3.2 10

2W/m W/mI I I I I & &" " " " 9 " 9 " 9


How is the loudness of sound related to the physical quantities characterizing wave motion?

20.9 Pitch, frequency, and complex sounds

If you take several tuning forks of different sizes and hit each fork with a mallet, they

produce sounds of approximately the same intensity but you hear them as different pitch sounds.

The smaller fork makes a higher pitch sound and the bigger fork a lower pitch sound. The smaller

the fork has a higher vibration frequency, and the bigger the fork has a lower vibration frequency.

Thus, pitch is related to the frequency of sound.

Pitch is not a physical quantity (there are no units for pitch) but is a physiological

characteristic of sound. It is used to indicate a person’s subjective impression about the frequency

of a sound. A normal ear is sensitive to sounds in the audio-frequency range from about 20 Hz to

20,000 Hz. As a person ages, the limits of this hearing range shrink.

Complex sounds, waveforms and frequency spectra

There is more to sound than the sensations of its loudness and pitch. For example, an

oboe and a violin playing concert A equally loud at 440 Hz sound very different. What other

property of sound causes this different sensation to our ears? Let’s use a microphone and a

computer to look at a pressure-versus-time graph of a sound wave produced by playing one key

on a piano (see Fig. 20.22 - notice that the graph has negative values, thus it must be representing

the gauge pressure). There is a characteristic frequency, but the wave is not a pure sinusoidal like

wave. What causes this shape?

Figure 20.22 Pressure-versus-time of sound from one piano note

To help answer this question, lets use two tuning forks with the second fork vibrating at

twice the frequency of the first. A microphone can detect the sound when either or both forks

vibrate. A computer records the electric signal produced by the microphone. When we strike each

fork separately, the computer shows sinusoidal waves (Fig. 20.23a and b). When we hit the two

forks simultaneously, the signal looks different (Fig. 20.23c)—it is more like that recorded when

playing the piano.


Figure 20.23 Pressure-versus-time graphs caused by one or two tuning forks

To understand the shape of the wave in Fig. 20.23c, recall the superposition principle.

Instead of one wave arriving at the microphone, the sound from two tuning forks simultaneously

arrives at the microphone. We can use the superposition principle to explain the shape of the

pressure-versus-time graph for the sound produced by both tuning forks (Fig. 20.23c). To

construct the graph you add the pressure variations caused by each wave at each instant in time.

The result is the signal shown in Fig 20.23c. The fluctuating pressure pattern is called a

waveform.

Sometimes we wish to determine the frequencies and amplitudes of the sinusoidal pure

sounds that produced a complex wave. This is how it is done. A microphone detects a sound

wave and converts the pressure variation into variable electric current which changes with time

exactly the same way as the pressure near the microphone. This electric signal passes through a

device called a spectrum analyzer. This device electronically analyzes the complex waveform and

plots on a screen the amplitude of waves at different frequencies that were added together to form

the complex wave. The amplitude-versus-frequency graph is called a frequency spectrum of the

complex wave. A frequency spectrum of the complex waveform in Fig. 20.23c is shown in Fig.

20.24. The height of the line at each frequency is proportional to the amplitude of the wave at that

frequency. Since the amplitude of wave 2 is twice that of wave 1, its vertical height is twice the

height of wave. Wave 2 is also twice the frequency of wave 1 (in this example only).

Figure 20.24 Frequency spectrum produced by two tuning forks


Most sounds we hear are complex waves consisting of sinusoidal waves of many

different frequencies. The sound we call noise includes waves of a broad range of frequencies,

and we seldom associate a pitch with the noise. However, if the complex sound is the result of the

addition of the wave of high amplitude at some low frequency (called fundamental) and waves at

smaller amplitudes but at higher frequencies that are frequencies are multiples of the fundamental

frequency (called harmonics), then we usually identify the pitch of the sound as being that of the

fundamental frequency. This is true even when the amplitudes of the higher harmonics are greater

than that of the fundamental. The waveforms and frequency spectra of complex waves from a

piano and from a violin playing concert A at 440 Hz are shown in Fig. 20.25. Notice that the

violin A has many frequency components, which are all multiples of 440 Hz (for example, 880

Hz, 1320 Hz, 1760 Hz, 2200 Hz, and even higher frequencies). It is a complex wave.

Figure 20.25 Piano and violin concert A waveforms and frequency spectra

Several physiological aspects of our hearing system can cause variations in the pitch we

hear even though the frequency of the sound is unchanged. For example, a loud, high-frequency

sound may seem to have a higher pitch than the same frequency sound played softly. Under some

conditions these variations may be considerable. Fortunately, these conditions do not often arise

in music, and the pitch change can usually be ignored.

Quality of Sound

A violin while playing concert A at 440 Hz sounds different from an oboe or French horn

playing the same note. We say that the sounds of the instruments have a different quality or

timbre, the characteristic that distinguishes one sound from others of equal pitch and loudness.

The quality of a sound at a particular pitch depends on the frequency spectrum of the sound—on

the number of higher harmonics that are part of the sound, the relative amplitudes of these

harmonics, and on other more complex characteristics of the sounds.

Beat frequencies

The quality of sound is also affected when several instruments all play the same music

(for example, the first violin section of an orchestra). Although they try to play identical

frequencies, there can be slight differences. One violinist may be playing concert A at 439 Hz and

another at 441 Hz. What happens when these two violinists play together?


Example 20.7 Beats Two sound sources are equidistant from a microphone that is connected to a

computer that records the varying sound pressure as a function of time due to the two sound

sources (Fig. 20.26a). Source 1 vibrates at 10 Hz and source 2 at 12 Hz. During 1.0 s, the

variation of pressure at the microphone due to wave 1 is shown in Fig. 20.26b and that due to

wave 2 is shown in c. These are pressure variations due to only one wave source playing at a

time. Determine the pressure-versus-time wave pattern that would be seen if both sound sources

played at the same time.

Figure 20.26(a)

Sketch and Translate We need to add the two waves shown in Fig. 20.26b and c to see the pattern

produced when both sounds play simultaneously (Fig. 20.26d). Note that the pressures shown in

the Fig. 20.26b and c graphs are the pressures above or below atmospheric pressure caused by

each sound if that was the only sound present.

Simplify and Diagram Note in Fig. 20.26d that at time 0.00 s the positive pressure caused by

wave 1 cancels the negative pressure caused by wave 2. The net disturbance at time 0.00 s is zero.

At 0.25 s, both waves have a maximum negative pressure variation and the net disturbance is

twice that negative pressure. At 0.50 s, the two wave pressure variations again cancel each other

causing a zero pressure disturbance.

Figure 20.26(b)(c)(d) Beats produced by two sounds

Represent Mathematically The pattern shown in Fig. 20.26d from 0.00 s to 0.50 s is called a beat.

It has a frequency of about 11 Hz (the average of the two wave frequencies) but its amplitude is

modulated from zero to twice the amplitude of the individual waves and back to zero again at

0.50s. A second beat is produced from 0.50 s to 1.00 s. The beat frequency is the number of beats


per second—in this case 2 beats/s. This is just the difference in the two wave frequencies (12 Hz

– 10 Hz) = 2 Hz. In general, the number of beats produced per unit time, the beat frequency beat

f ,

equals the magnitude of the difference in the frequencies of the two waves causing the beat

pattern:

beat 1 2

–f f f"

Solve and Evaluate As noted above, the beat frequency in this case is

beat10 Hz – 12 Hz 2 Hzf " " .

Note that we took the absolute value of the frequency difference.

Try It Yourself: What would be the period of the amplitude modulation on the graph in Fig.

20.36d if one source vibrates at 67 Hz and the other one at 69 Hz.

Answer: The beat frequency would be the same (2 Hz) and the period of the beat frequency

pattern would be 1 / 0.5 sT f" " .

We can now answer the question posed before the above example. When the two violins play

concert A simultaneously, the waves that they make add together and produce vibrations whose

amplitude changes. The amplitude changes with the frequency that equals the difference of the

two violin frequencies—2 Hz in that example.

beat 1 2–f f f" (20.10)

These vibrations of changing amplitude are beats and the frequency of the amplitude

change is called the beat frequency. With many instruments playing together, the amplitudes of

the sounds produced are modulated at beat frequencies. This produces a pleasing so-called chorus

effect.

Beats are useful in precise frequency measurements. For example, a piano tuner can

easily set the frequency of the middle C string on the piano to 262 Hz by observing the beat

frequency produced when the piano string and a 262-Hz tuning fork are sounded simultaneously.

If a beat frequency of 3 Hz is observed, then the piano string must be vibrating at either 259 Hz or

265 Hz. The string is then tightened or loosened until the beat frequency is reduced to zero.


What happens when two sinusoidal sounds, one at twice the frequency of the other, are

simultaneously produced? How do we know?

20.10 Applications: Standing waves on strings

The superposition of waves (adding waves) is the basis for much that is needed to

understand how musical instruments work. Consider string instruments first. Start by observing

the vibration of a rope with one end attached to a fixed support (a wall) while the other end is


held in your hand. You start shaking that end up and down and observe an interesting effect.

Depending on the frequency with which you shake your hand, the rope has either very small

amplitude and irregular vibrations or large amplitude sine-shaped vibrations. The latter are not

traveling waves but are instead coordinated vibrations with parts the rope vibrating up and down

with large amplitude and other parts not vibrating at all. Graphs of these large amplitude

vibrations are shown in Fig. 20.27 and are called standing waves. How is the first standing wave

vibration produced?

Figure 20.17 Standing waves on a string

Tip! Notice that graphs in Fig. 20.27 represent several snapshots of the same wave. The times at

for which the solid line and the dashed lines represent the positions of different points are

different.

Observational Experiment Table 20.7 Producing the first standing wave vibration.


You shake the rope at a relatively

low frequency and observe the

large amplitude vibration shown

below.

A series of sketches of the disturbances

caused by your shaking at frequency 1f

is shown at the right. A pulse starts

moving to the right (b-d). When it

reaches the fixed end, the upward pulse

is reflected and inverted to a downward

pulse traveling back toward your hand

(e-g). When it reaches your hand, it once


again reflects and inverts to an upward pulse (part h). To make this

pulse bigger, you give your hand another upward shake, making sure to

time the shake so that it adds to the previous upward pulse that has just

been reflected (part h).

The subscript 1 means the lowest standing wave frequency. The word vibration is not a unit and

can be removed from the above equation. Thus, this lowest frequency vibration of the rope is one

shake per time interval 2L

v: " or

1

1 1

2 2

vf

L L

v

:" " " (20.13)

This frequency is traditionally called the fundamental frequency. If the shaking frequency is less

than this, each new pulse will interfere destructively with previous pulses, and the rope will

vibrate little. Should we raise and lower our hand at a frequency 1

2

vf

L" , each new pulse adds

to the last and creates a large-amplitude vibration. The wavelength of this fundamental frequency

vibration is:

1

1

2

2

v vL

vf

L

' " " "

The length of the rope equals one half of the wavelength of the wave that has this fundamental

frequency. Let’s test this idea.

Testing Experiment Table 20.8 Fundamental vibration frequency

Testing Experiment Prediction Outcome

Predict the

fundamental

vibration frequency

of a banjo D string.

The mass of the

string is 0.178 g, its

length is 0.690 m,

and the wooden peg

pulls on its end

exerting an 80.0-N

The speed of a wave in the string is:

1/21/2

Peg on String

–3string string

(80.0 N)807 m/s

( / ) (0.179 x 10 kg)/(0.690 m)

F

m Lv

( )( ). /. / " ". /. /2 3 2 3

" .

Consequently, we predict that the fundamental frequency should

be:

–1

1

807 m/s585 s 585 Hz

2 2(0.690 m)

vf

L" " " "

When we pluck a

banjo D string

and compare it to

the frequency of

a tuning fork, we

find 1Df = 587

Hz.

Pattern

You must shake the rope upward each time a previous pulse returns. The amplitude of the disturbance

traveling along the rope grows. You need one shake during the time interval 2L

v that it takes for the pulse

to travel down the string and back:

1 2

1 vibration fundamental frequency =

2L

v

vf

L" "


force.

Conclusion

The agreement is excellent. We did not disprove the frequency expression, but instead built confidence in it.

Now consider the vibration that produces a standing wave on the rope at twice the

frequency of the fundamental vibration (see Fig. 20.27). You vibrate the rope so that your hand

has two up and down shakes during the time interval that is needed for a pulse to make a round

trip down the rope and back (Fig. 20.28).

Figure 20.28 Producing second standing wave vibration

The new pulse adds to a pulse started earlier and that has completed a round trip back to your

hand. This earlier pulse reflects from your hand and joins the new pulse becoming a larger

amplitude pulse. Thus, the frequency 2

f of this second standing wave vibration is:

2

2 vibrations second standing wave vibration = 2

2 2

vf

L L

v

" " .

We can see a pattern now.

Standing wave frequencies on a string The frequencies n

f that a string with fixed ends

can vibrate are:

nfor 1, 2,3,...

2

vf n n

L" " (20.11)

where v is the speed of the wave on the string and L is the string length.


Since v

f' " , we can write an expression for the wavelengths of these allowed standing wave

vibrations:

2for 1, 2,3, , ,

( )2

n

n

v v Ln

vf nn

L

' " " " " . (20.12)

Note that a rope or string will vibrate with large amplitude at any frequency for which a pulse

produced by the wave source (for example, the vibrating hand) is reinforced by a new pulse after

completing a round-trip up the string and back again. The wave patterns that are produced on the

rope for these different values of n do not appear to travel or move. They appear to be stationary

vibrations, called standing waves. Standing waves are caused by the superposition of waves

moving in opposite directions.

How are standing waves different from familiar to us traveling waves? When a traveling

wave moves along the string each point on the string vibrates between the maximum and

minimum disturbances (between y A" 6 ). However different points reach the maximum

displacements at different times, in other words different points have different phases at the same

time. In a standing wave different points on the string move in different patterns—in some places

they vibrate with large amplitude vibration (called antinodes) and in other places with no

vibration at all (called nodes). However, different points reach their maximum displacements at

the same time – they vibrate in phase.

When one plays a musical instrument, she excites more than one of its standing wave

vibrations (see complex waves in the previous section). For example, when you bow a violin

string, you excite simultaneously 10 or 20 standing-wave vibrations. The pitch of the sound is

determined by the standing wave of the lowest frequency present in the complex wave. The

quality of the sound depends on the number and relative amplitudes of the other standing-wave

frequencies that make up the complex sound wave and on the way that this spectrum of tones

changes with time.

Example 20.8 A Playing a banjo A banjo D-string is 0.69 m long and has a fundamental

frequency of 587 Hz. The string vibrates at a higher fundamental frequency if its length is shorter.

To shorten the string you can hold the string down over a fret. Where should you place the fret to

play the note F-sharp, which has a fundamental frequency of 734 Hz?

Sketch and Translate We can visualize the situation by picturing a string of different lengths. We

know that when the length of the string is 0.69 mL " , the fundamental frequency is

1587 Hzf " . What is the new length of the string 'L for which the fundamental frequency is

1' 734 Hzf " ?

Simplify and Diagram Assume that the speed of the wave on the string is independent of where

the finger pushes down on the string.


Represent Mathematically The fundamental frequency of vibration 1

f relates to the string length

L by the equation 1

2

vf

L" , or

12

vL

f" . We cannot solve immediately for the unknown length

when the string is vibrating at 734 Hz since we do not know the speed v of the wave on the

string. But we can use this idea to write an expression for the speed of the wave for both cases

(the speed is the same for both cases):

1 1(2 ) '(2 ')v f L f L" " .

Solve and Evaluate

Dividing both sides of the expression for the speed sides by 1'f and multiplying by 2, we get an

expression for 'L :

1 1

1 1

'' '

f L fL L

f f

* +" " , -

0 1.

The answer is reasonable as it is less than 0.69 m. To find the fret location, we need to find the

difference between the original length L and the new length 'L :

– ' 0.69 m –0.55 m = 0.14 mL L " .

The fret should be 0.14 m from the end of the string.

Try It Yourself: If tuned correctly, the fundamental frequency of the A string of a violin is 440

Hz. List the frequencies of several other standing-wave vibrations of the A-string.

Answer: The other standing wave frequencies are integral multiples of the fundamental and

include: 880 Hz, 1320 Hz, 1760 Hz, and so forth.


A horizontal string of length L has one end passing over a pulley with a block hanging at the

end. A motor is attached to the other end of the string and exerts a constant horizontal force on

the string. When the motor is turned on, it vibrates the end of the string up and down at

frequencies that change slowly from zero to some high frequency. What do you observe?

20.11 Application: Standing Waves in Air Columns

So far we discussed how standing waves are formed on strings that are fixed at both ends.

In this section we learn how musical instruments made of pipes or tubes such as organs, clarinets,

flutes, and trumpets work.

Standing waves in open pipes

1

1

587 Hz' (0.69 m) = 0.55 m

' 734 Hz

fL L

f

* + * +" ", - , -0 10 1


Suppose you have pipes of different length and blow across their ends. They produce

sounds of different pitch. Or you take a bottle, fill it with water, and blow across the top of the

bottle. You again hear sound whose pitch depends on the amount of water in the bottle. Your can

even play music by having several bottles filled with different carefully measured amounts of

water and different people blowing into them. Each bottle can be one note in a song. How can we

explain these phenomena?

When we blow in a pipe, or flute, or through a vibrating reed, such as in a clarinet, we

initiate pressure pulses at that end of the pipe (for example, by a vibrating reed). Alternating high

and low pressure pulses leave the reed and move in the air column squishing the air in the air

column in a repetitive fashion.

Consider the fate of one of these pressure pulses while

it is moving in an air column with both ends open.

Such pipe is called an open pipe. The pressure pulse

moves along the pipe as shown in Fig. 20.29a-c. At

the open end, a high-pressure pulse can suddenly

expand into open space causing a high-pressure pulse

outside of the tube and leaving behind a low pressure

reflected pulse that travels back into the tube. This

change from high to low pressure is similar to a phase

change of a pulse on a string reflected off a fixed end.

Unlike the phase of the pulse due to the displacement

of air molecules (density pulse) the pressure pulse

changes its phase when reaching an open end and

does not change phase when it reaches a closed end.

Figure 20.29 Producing standing wave in open pipe

This pulse of decreased pressure moves back toward the other end of the pipe where it started

(parts d-f) and is reflected at the open end on the left (part g). After reflection from both open

ends, the pulse again has increased pressure and is moving toward the right. If a new pressure

pulse is initiated at this time, it interferes constructively with the reflected pulse, and its amplitude

increases. A large-amplitude standing-wave vibration begins to form in the open pipe. If,

however, the new air pressure pulse is too early or too late, it interferes destructively with the

reflected pulse, and the amplitude decreases.

The frequency of standing-wave vibrations in pipes depends on the time interval needed

for a pulse to travel down the pipe and back again. If the sound pulse travels at a speed v along a

pipe of length L , then the time interval T needed to travel a distance 2L is 2L

Tv

" . To

reinforce the vibration of air in the pipe, pressure pulses should occur once each time period T or

at a frequency 1

2

vf

T L" " .


This is the same equation as for the fundamental frequency of vibration of a string. Just as we can

make a string vibrate at multiples of the fundamental frequency, the pipe can also vibrate with

large amplitude if excited by air pressure pulses at integral multiples of the fundamental

frequency. However, in for pipes open on both ends , v the speed of a pulse is independent of the

pipe as all pulses propagate through the air inside the pipe (about 340 m/s).

Standing wave vibration frequencies in open pipes The resonant frequencies n

f of air

in an open pipe of length L in which sound travels at speed v are:

( ) for 1, 2,3,...2

n

vf n n

L" " (20.13)

Musical instruments that function as open pipes include the flute, recorder, oboe, and bassoon.

Tip! Be sure that you use the speed of sound in the medium that fills the pipe in the above

expression for the standing wave frequencies.

Standing waves in one-end closed pipes

Next, consider a pipe, which is closed at one end and open at the other end—called a

closed pipe. Musical instruments that behave somewhat like closed pipes include the clarinet,

trumpet, trombone, and human voice. The closed end of these instruments is at the reed,

mouthpiece, and vocal cords, respectively.

It is more complicated to determine the frequencies of standing waves in closed pipes

than in open pipes. Suppose that you initiate a pulse of increased pressure by a reed near the open

end of the pipe (Fig. 20.30a). When the pulse is reflected at the closed end, its phase is unchanged

(part d). If the pressure in the pulse is above normal pressure, the pressure in the reflective pulse

is also above normal. When the reflected pulse travels back to the open end, it is once again

partially reflected. However, this time there is a change in phase; a high-pressure pulse travels

into open space leaving behind a low-pressure reflected pulse. There is now decreased pressure

pulse traveling back to the right (part g). If a new pulse of increased pressure is initiated, it

cancels the effect of the reflected pressure pulse. To add to the reflected pulse in a constructive

way, we must wait for it to make an additional trip down the pipe and back again. During each

round-trip, its phase is inverted once. After two round-trips it is again a pulse of increased

pressure (part m), and a new pulse adds constructively to increase its amplitude.

For a large-amplitude standing wave to be produced, then, the time interval T between

pulses must be 4L

v. The frequency of the fundamental is

1

1

4

vf

T L" " . The one-end closed pipe

is unique in that so-called overtone vibrations (higher frequency standing waves) do not occur at

all frequencies that are integral multiples of the fundamental frequency. For example, at twice the


frequency of the fundamental, a pulse makes only one trip up and down the pipe before another

pulse is started. The reflected pulse, which is inverted, interferes destructively with the new pulse.

The pipe cannot vibrate at twice the fundamental frequency. A closed pipe can, however, vibrate

at odd integral multiples of the fundamental frequency.

Figure 20.30 Producing standing waves in a “closed” pipe

Testing the frequency equation for closed pipes

If our reasoning that sound in pipe-like instruments is due to standing waves in them is

correct, then changing the medium inside a pipe should change the fundamental frequency

produced by the pipe without changing its size. Our vocal cord vibrations initiate sounds by our

voices. The mouth is like a chamber with one end closed at the vocal cord and the other end open

at the mouth—a closed pipe. If we have a different gas in the throat and mouth, the speed of a

wave in the throat and mouth will change, as should the standing wave frequencies of sounds

produced.

Conceptual exercise 20.9 Funny talk What should happen to sound from a voice after a person

inhales helium and sings a note?

Sketch and Translate Assume that a person’s vocal chamber is a resonant cavity like a one end-

closed pipe with the vocal cords at the closed end and the mouth at the open end. The speed v of

sound in helium gas is much greater than in air. Thus, the fundamental frequency (1

4

vf

L" ) of

sound from our voice should be greater when filled with helium than when filled with air.


Simplify and Diagram. The length of the chamber does not change during the experiment. The

speed of sound in air is 340 m/s and in helium it is almost three times higher – 930 m/s . Thus

the fundamental frequencies of the sound we produce should increase almost three times – a

difference easily heard. When we inhale helium and speak or sing with it in our mouth, a very

high pitch sound is heard—as predicted.

Try It Yourself: In our testing experiment with helium, we used gas with a greater speed of sound.

What happens if we inhale a gas that has a lower sound speed than in air?

Answer: An example of such gas is 2

CO . Inhale it (as this gas is dangerous, only trained

performers can do it) and say a few words. The pitch of your voice will be much lower.

Another testing experiment might involve the receiving end of sound – our ear. We know

already that the auditory canal of the outer ear is a 3.0 - cm -long pipe closed at the end by the

eardrum. If we fill it with water (being underwater for example), then the speed of sound will be

significantly different (1500 m/s as opposed to 340 m/s in air). Thus, we should hear high

frequency sounds better under water than when in air.

Quantitative Exercise 20.10 Listening under water Calculate the fundamental frequency of the

auditory canal when (a) in air and (b) in water. How might this resonance of the auditory canal

affect our ability to hear sounds at these frequencies?

Represent Mathematically Visualize the canal as a one end-closed pipe—open to the air and

closed at the eardrum. The length of the canal does not change but the speed of sound v in the

canal increases significantly if filled with water compared to air. Thus the fundamental standing

wave frequency 1

4

vf

L" should be at higher frequency in the water filled canal.

Solve and Evaluate (a) The fundamental frequency 1

f for a closed pipe of length

3.0 cm = 0.03 mL " filled with air through which sound travels at a speed 340 m/sv " is:

1

340 m/s1 1 2800 Hz

4 4(0.030 m)

vf

L

* +* +" " ", -, -0 1 0 1

.

The ear is, in fact, sensitive to sound in this frequency range.

(b) Sound travels at speed 1500 m/sv " in water (see Table 20.4). The fundamental frequency

of the closed pipe auditory canal filled with water is:

1

1500 m/s1 13,000 Hz

4(0.030 m)f

* +" ", -0 1

.

Our inner ears are not very sensitive at this very high frequency. However, the resonance of the

auditory canal when underwater helps us hear high-frequency sounds better than when the canal

is filled with air.


Try It Yourself: You sing middle C (256 Hz) above a graduated cylinder partially filled with

water. What should be the height of the air space at the top of the cylinder (water is at the bottom)

so you hear a resonance of your voice with the cylinder?

Answer: 33 cm.

We can now assert with confidence:

Standing wave vibration frequencies in one-end closed pipes The resonant frequencies

nf of air in a pipe of length L closed on one end in which sound travels at speed v are:

( ) for 1,3,5,7,...4

n

vf n n

L" " (20.14)

Standing-wave vibrations occur in many objects besides strings and pipes—the swaying

of tall buildings and the twisting vibrations of bridges are examples.


When we studied traveling waves, we decided that the wavelength of a wave depends on the

frequency of vibration and the speed of the wave in the medium. Is this statement true for

standing waves in pipes?

20.12 Putting it all together: Doppler Effect

Until now, we have been studying situations in which the source of a wave and the

observer were both stationary with respect to the earth and with respect to each other and the

medium in which the wave propagated was at rest with respect to both the source and the

observer. In this case, the frequency of a wave as it leaves its source is the same as the frequency

of the wave when detected by the observer. In real life objects emitting and detecting waves can

be moving. What happens then to the frequency of emitted or received waves? To answer this

question we start with some observational experiments.

Suppose you stand near a racetrack as fast-moving race cars pass. While a car

approaches, you hear a high-pitched, whirring sound; as the car passes, the frequency or pitch of

the sound you hear suddenly and noticeably drops. A similar phenomenon occurs when you listen

to the sound from the horn of a passing car or the whistle of a passing train. The change from a

higher pitch as the sound source approaches to a lower pitch as it moves away is an example of

the Doppler effect. The Doppler effect occurs when a source of sound and an observer move with

respect to each other and with respect to the medium in which the sound travels.

Is there a pattern in these observed frequency shifts?


Consider more phenomena to see if there is a pattern in the frequency change. We

connect a generator to an electronic whistle that makes a sound at a source frequency S

f and a

microphone and spectrum analyzer that records the frequency O

f of the sound detected by an

observer (the microphone). The source whistle is attached to a cart that can move in one direction

toward or away from another observer cart that can move independently toward or away from the

source cart. Consider what happens to the frequency of the observed sound when either the sound

source or sound observer is moving with respect to the ground. In all experiments the air in which

the sound propagates is at rest with respect to earth.

Observational Experiment Table 20.9 Effects of motion on observed frequency.


(a) Whistle source stationary with respect to

the ground.

Observer moving toward source.

The frequency spectrum for the signal emitted

by the source whistle (S

f ) and heard by the

observer (O

f ): S O

f f;

(b) Whistle moving toward observer.

Observer stationary with respect to the ground.

The frequency spectrum

S Of f;

(c) Whistle stationary with respect to the ground.

Observer moving away from source.

The frequency spectrum

S Of f<

(d) Whistle moving away from observer.

Observer stationary.

The frequency spectrum:

S Of f<

(e) Whistle moving with respect to the ground.

Observer moving at the same speed in the

same direction.

The frequency spectrum:

S Of f"

Pattern


(a) & (b) Source and/or observer are moving toward each other: the observed frequency is greater than

the emitted source frequency.

(c) & (d) Source and/or observer are moving away from each other: the observed frequency is less than

the source frequency.

(e) There is no relative motion: the observed and emitted frequencies are the same.

Note carefully the observed patterns in Table 20.9. We try next to develop a quantitative

explanation for these patterns.

Doppler effect for a moving source

To explain the Doppler effect, examine the waves created by a water beetle bobbing up

and down at a constant frequency on the water. If the beetle bobs up and down in the same place,

the pattern of wave crests appears as shown in Fig. 20.31. The crests move away from the source

at a constant speed. The large circle represents a crest produced by the beetle some time ago, and

the small circle is a crest produced more recently. The distance between adjacent crests is one

wavelength and is the same toward observer A as toward observer B. The frequency of waves

reaching both observers is the same.

Figure 20.31 Bobbin beetle creates wave crests Figure 20.32 Crests created as beetle hops to right

Now suppose the bobbing beetle moves to the right with respect to the water at a slower

speed than the speed at which the waves travel. Each new wave produced by the beetle originates

from a point farther to the right (Fig. 20.32). Wave crest 1, the first to be created, is a large circle

centered at position 1 where the beetle started bobbing. Wave crest 2 is a smaller circle centered a

step to the right at position 2, and so forth. Wave 1, which started an equal distance from

observers A and B, reaches them at the same time. However, wave 4 reaches B somewhat sooner

than it reaches A because the wave originated to the right of center, closer to B than to A. Thus, B

observes four waves in a shorter time interval than does A. In general, if a wave source moves

toward an observer, the observer detects a higher frequency than the frequency emitted by the

source (see Experiment (b) in Table 20.9). If the source moves away from the observer, the

observer detects a lower frequency than the frequency emitted by the source (see Experiment (d)

in Table 20.9).

To derive an equation for calculating this shift, first calculate the separation of crests in

front of the beetle. Consider the separation of wave crest 4 and the wave initiated by the beetle's

fifth step (wave crest 5 shown in Fig. 20.33). If the beetle bobs up and down at frequency S

f , the


time between the start of wave 4 and the start of wave 5 is

S

1T

f" . During this time interval, the

beetle moves a distance S

v T to the right, where S

v is the speed of the wave "source", in this case

the beetle. Wave 4 has moved a distance vT during that same time interval, where v is the speed

of the wave through the water.

Figure 20.33 Wavelength reduced in front of hopping beetle

We see in Fig. 20.33 that the crests of waves 4 and 5 arriving at observer B are separated by a

distance B

'' (a wavelength) given by

B S' –vT v T' "

Similar reasoning could be used for the wavelengths of waves reaching observer A. The beetle is

moving away from A (see Fig. 20.32) and the wavelengths of the waves reaching A are:

A S' vT v T' " #

The waves are incident on observer B at a frequency

OB

B

wave speed 1

wavelength ' ( – ) ( – )s s

v v vf

T v v T v v'" " " " .

and on A at frequency

OA

A

wave speed 1

wavelength ' ( + ) ( + )s s

v v vf

T v v T v v'" " " "

The inverse of the time interval T between the beetle’s bobs is the beetle bobbing frequency

S

1f

T" (the beetle is the wave source). Substituting for

1

T in the preceding equations, we have

an expression for determining the observed frequency either in front or in back of the source:

O S( )s

vf f

v v"

!. (20.15)

The minus sign applies for an observer B toward which the source is moving and the plus sign to

an observer A for which the source is moving away. Note that with the minus sign and the source

moving toward the observer, the observed frequency will be greater than the source frequency in

agreement with the qualitative experiments. With the plus sign and the source moving away from


the observer, the observed frequency is less than the source frequency, also in agreement with the

qualitative experiments.

Doppler effect for a moving observer

The observed frequency also differs from the source frequency if the observer moves

with respect to the ground and the source is stationary. Assume, for instance, that a friend remains

stationary in the water of a swimming pool and creates waves by pushing a beach ball up and

down. If you swim toward the source, you encounter the waves more frequently than if you

remain stationary (consistent with Experiment (a) in Table 20.9). If you swim away from the

source, you encounter the waves less frequently than if you remain stationary (consistent with

Experiment (c) in Table 20.9).

If the observer moves toward the source at speed O

v , then the waves appear to move past

the observer at speed O

v v# . Since the wavelength ' of the waves from a stationary source is

uniform in all directions, the observer will encounter waves at a frequency

O

O

+ speed of observer relative to waves

wavelength

v vf

'" " .

If the observer moves away from the source, then the waves appear to move past the observer at

speed O

–v v and the observed frequency will be:

O

O

– speed of observer relative to waves

wavelength

v vf

'" "

The wavelength ' ""relative to the source frequency S

f and the speed v of the wave in the medium

is S

/v f' " . Thus,

O O O

O S

S

/

v v v v v vf f

v f v'6 6 6* +" " " , -

0 1 . (20.16)

This equation can be used to calculate the frequency O

f detected by an observer moving at

speed O

v toward (plus sign) or away from (negative sign) a stationary source emitting waves at

frequency S

f .

General equation for Doppler effect for sound

We can get a general equation for the Doppler effect by combining Eqs. (20.15) and

(20.16):

Doppler effect for sound When a sound source or sound observer moves relative to each

other, the frequency O

f of the observed (detected) sound differs from the frequency S

f

of the sound source:

O

O S

S

v vf f

v v

* +6" , -

0 1!, (20.17)


where v is the speed of the wave through the medium in which it travels; O

v is the speed

of the observer relative to the medium (use the plus sign in Eq. (20.17) if the observer

moves toward the source and a negative sign if the observer moves away from the

source); and S

v is the speed of the source relative to the medium (use the negative sign in

Eq. (20.17) if the source moves toward the observer and a positive sign if the source

moves away from the observer).

Tip! If the signs look confusing, use the sign that makes a higher frequency if the source is

moving toward the observer or the observer is moving toward the source; similarly, choose signs

the make the observed frequency lower if the source or observer is moving away from the other.

Example 20.11 Listen to train whistle A train traveling at 40 m/s with respect to the ground has

a horn that vibrates at a frequency of 200 Hz. Determine the frequency of the horn’s sound heard

by a bicycle rider traveling at 10 m/s with respect to the ground in the same direction as the train

when the cyclist is (a) ahead of the train and (b) when behind the train after the train has passed.

Sketch and Translate A sketch of the situations is shown in Fig. 20.34a and b.

Figure 20.34 Bicycle rider hears train whistle

Simplify and Diagram Both speeds are assumed constant; the speed of sound is 340 m/s.

Represent Mathematically (a) The train source whistle is moving toward the observer, which

causes an increase in the observed frequency. This increase occurs if you use the minus sign in

Eq. (20.17) in front of the source speed S

v . The observer on bicycle is moving away from the

source causing a decrease in the observed frequency. This decrease occurs if you use a minus sign

in Eq. (20.17) in front of the observer’s speed O

v . (b) The train source whistle is now moving

away from the observer, which causes a decrease in the observed frequency. This decrease occurs

if you use the plus sign in Eq. (20.17) in front of the source speed S

v . The observer on bicycle is

now moving toward the source causing an increase in the observed frequency. This increase

occurs if you use a plus sign in Eq, (20.17) in front of the observer’s speed O

v .

Solve and Evaluate (a) The frequency heard by the observer (the cyclist) when moving ahead of

the train is:


0

O S

S

340 m/s – 10 m/s(200 Hz) 220 Hz

340 m/s – 40 m/s

v vf f

v v

* +6 ( )" " ", - . /2 30 1!.

(b) The frequency heard by the cyclist after the train has passed is:

O

340 m/s 10 m/s(200 Hz) 184 Hz

340 m/s + 40 m/sf

#( )" ". /2 3.

The frequencies of sound heard by the cyclist make sense. In (a), the train was moving toward the

cyclist faster than he was moving away; they were moving closer and the sound heard by the

cyclist was at a higher frequency than that emitted by the train. In (b) after the train had passed,

they were moving apart and the frequency of sound heard by the cyclist was less than that emitted

by the train whistle. It is good to do qualitative checks on your quantitative answers to see if they

make sense.

Try it yourself: What would the bicyclist hear if he were biking toward the on-coming train?

Answer: 233 Hz.

Measuring the speed of blood

The Doppler effect is the basis of a technique used to measure the speed of blood flow.

High-frequency sound waves ( 20,000 Hzf < ) called ultrasound are directed into an artery, as

shown in Fig. 20.35. The waves are reflected by red blood cells back to a receiver. The frequency

detected at the receiver R

f relative to that emitted by the source S

f indicates the cell’s speed and

the speed of the blood. (A similar arrangement is used to measure the speed of cars, but

microwaves are used instead of ultrasound). To derive an equation for the frequency shift, we

consider the process in two parts: (1) The waves leave the source at frequency S

f and strike the

cell at frequency 'f . (2) The waves reflected from the cell at frequency 'f return to the receiver

where they are detected at frequency R

f .

Figure 20.35 Use Doppler effect to measure blood flow speed

The frequency of the sound 'f striking a red blood cell is related to the frequency S

f

emitted from the source by the equation


S

b

0'

vf f

v v

* +#" , -&0 1

(20.18)

where v is the speed of sound in blood and b

v is the speed of the blood (a positive number as the

blood is moving toward the source). The blood red cells reflect this sound to a receiver, which

detects the sound. The cells act as a source at frequency 'f . The frequency detected at the

receiver R

f is related to 'f by the equation

b

R'

0

v vf f

v

#* +" , -&0 1

Substituting for 'f from Eq. (20.18), we find the frequency detected at the receiver R

f

relative to the frequency emitted by the sound source S

f :

R S S

0( )

0

b b

b b

v v v vvf f f

v v v v v

* + # ## * +" ", -, -& # &0 10 1. (20.19)

This result allows us to measure the speed of blood using the Doppler effect.

Example 20.12 Buzzer moves in circle A friend with a ball attached to a string stands on the

floor and swings it in a horizontal circle. The ball has a 400-Hz buzzer in it. When the ball moves

toward you on one side, you measure a frequency of 412 Hz. When the ball moves away on the

other side of the circle, you observe a 389 Hz frequency. Determine the speed of the ball.

Sketch and Translate Draw a sketch of the situation (Fig. 20.36). You detect a higher frequency

when the ball moves toward you and lower frequency when it moves away, consistent with our

understanding of the Doppler effect.

Simplify and Diagram The speed of sound in air is constant and equal to 340 m/s. The air is

stationary with respect to the ground.

Figure 20.36 Using Doppler effect to find speed during circular motion

Represent Mathematically We can use Eq. (20.17) to find an expression for the ratio of the

frequency when the source (the whistle) is moving toward you (the observer) and when moving


away. This ratio can then be rearranged to get an expression for the speed of the source (the

whistle):

S

S moving toward O S S

S moving away from O SS

S

0

–

0 –

vf

f v v v v

vf v vf

v v

##

" "##

,

where v is the speed of sound in air and S

v is the speed of the source (the whistle). .

Solve and Evaluate Rearranging the above, we get an expression that can be used to determine

the source speed:

S moving toward O S moving away from O

S

S moving toward O S moving away from O

–f fv v

f f

* +" , -, -#0 1

To solve for the speed of the source S

v in terms of the observed frequencies (412 Hz when source

is moving toward observer and 389 Hz when moving away), we get:

S

412 Hz – 389 Hz(340 m/s) = 9.8 m/s

412 Hz + 389 Hzv

* +" , -0 1

.

The ball was moving at 9.8 m/s—the correct unit and a reasonable magnitude.

Try It Yourself: What frequency do you hear when the ball is moving perpendicular to your line

of sight when in front of you?

Answer: 400 Hz. At that instant, the ball is not moving toward or away from you – it moves in the

perpendicular direction – no Doppler shift.

Note you can determine how fast the ball was moving in a circle by just comparing the

sound frequency when moving toward you and the frequency when moving away. The same

conceptual idea is used to determine the speed of a star moving in a circle in a galaxy (the

quantitative procedure is a little different as light is used instead of sound). The speed

measurement can be used to determine the mass pulling inward on the star by other objects in the

galaxy and is one source of a new mystery—there seems to be an excess mass pulling in on the

star. This mass is unseen and unaccounted for—you will learn more later about this interesting

dark matter problem.


The siren on an ambulance sounds continuously as the ambulance first approaches you and then

moves away. How does the sound from the siren change and why?

Summary

Word description Sketch and/or diagram Mathematical description


Speed of wave depends on properties

of the medium (for example, a force

like quantity FMonM (medium on

medium) and a mass/length quantity

5 ). The wavelength ' depends on

frequency f and wave speed v .

v

fvT' " "

MonM/v F 5"

Traveling Wave equation The

traveling wave equation allows us to

determine the displacement from

equilibrium y of a point in the

medium at location x at time t when

a wave of wavelength ' travels at

speed v .

cos 2 – t x

y AT

$'

"* +, -0 1

Intensity I of a wave is the energy

per unit area and time interval that

crosses perpendicular to an area in a

medium in which the wave travels.

Intensity level 8 ""of a sound is the

common log of the ratio of its

intensity I and a reference intensity

0I .

•

U P

t A AI

%"

%"

0

10 logI

I8

* +, -, -0 1

"

where –12 2

010 J/m •sI " .

Superposition principle When two or

more waves pass through the same

medium at the same time, the net

displacement of any point in the

medium is the sum of the

displacements that would be caused

by each wave if alone in the medium

at that time.

...1 2 3

y y y y# ##"

Huygens principle Every vibrating

point of a medium produces a wavelet

with a circular front surrounding it.

The new wave front of the wave is the

superposition of all wavelets due to

vibrations on all points of previous

wave front.

Standing waves on strings of length L

on which waves at speed v form in-

phase vibrations of frequencies nf .

The amplitudes of those vibrations are

different for different points of the

medium.

( )2

n

vf n

L"

for 1, 2, 3, ...n "

Standing waves in open pipes of

length L in which sound travels at

speed v are discrete large amplitude

vibrations of frequencies nf .

( )2

n

vf n

L"

for 1, 2, 3, ...n "


Standing waves in one end-closed

pipes of length L in which sound

travels at speed v are discrete large

amplitude vibrations of frequencies

nf .

( )4

n

vf n

L"

for 1, 3, 5, ...n "

Doppler effect for sound If a sound

source or sound observer move

relative to each other, the frequency

Of of observed sound differs from

frequency S

f of sound source. O

v and

Sv are the observer and source speeds

(magnitudes of the velocities) and v is

the speed of the wave through the

medium.

0

o s( ).

s

v v

v vf f

6"

!

Use + in front of O

v if

observer moves toward

source and – if away; use –

in front of S

v if source

moves toward observer and

+ if away.

Conceptual questions 1. What statements (one or more) are applicable to the propagation of a pulse:

(a) The particles in the pulse travel from one place to another.

(b) The particles in the pulse do not travel from one place to another.

(c) The particles in the medium interact with each other.

(d) The energy travels but the material does not.

2. When a bell is rung, what happens to the vibrational energy initially present in the bell?

(a) It is lost.

(b) It transfers to vibrational energy of nearby particles and then further away from the bell.

(c) It eventually transforms into internal energy of the surrounding medium and the bell. (d) Both b and c

are applicable.

3. What does it mean if a period of a wave is 0.20 s?

(a) The wave lasts for 0.20 s.

(b) The wavelength is 0.20 s.

(c) Each particle in the wave completes one vibration in 0.20 s.

4. What does it mean if a speed of a wave is 300 m/s?

(a) Each particle moves at a speed of 300 m/s.

(b) The wave disturbance travels 300 m during each second.

(c) The wave disturbance travels 300 m for each wavelength.

5. What does it mean if the wavelength of a wave is 4.0 m?

(a) The wave extends for 4.0 m.

(b) The period of the wave is 4.0 m.

(c) The distance between adjacent identical locations in the disturbance is 4.0 m.

6. If you wish to graphically represent one period of a wave on a string, which two variables should you

have on the axes?

(a) v and T .

(b) y and x .

(c) y and t .

(d) T and ' .

chapter 20: mechanical waves -...

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