Raymond A. Serway
Chris Vuille
Chapter Thirteen
Vibrations and Waves
Waves and Optics
Wave motion (including sound) (1.5 weeks)
Learning Objectives:
1. Traveling waves
Students should understand the description of traveling waves, so they can:
a) Sketch or identify graphs that represent traveling waves and determine the
amplitude, wavelength and frequency of a wave from such a graph.
b) Apply the relation among wavelength, frequency and velocity for a wave.
c) Understand qualitatively the Doppler effect for sound in order to explain why there
is a frequency shift in both the moving-source and moving-observer case.
d) Describe reflection of a wave from the fixed or free end of a string.
e) Describe qualitatively what factors determine the speed of waves on a string and
the speed of sound.
2. Wave propagation
a) Students should understand the difference between transverse and longitudinal
waves, and be able to explain qualitatively why transverse waves can exhibit
polarization.
b) Students should understand the inverse-square law, so they can calculate the
intensity of waves at a given distance from a source of specified power and
compare the intensities at different distances from the source.
3. Standing waves
Students should understand the physics of standing waves, so they can:
a) Sketch possible standing wave modes for a stretched string that is fixed at both
ends, and determine the amplitude, wavelength and frequency of such standing
waves.
b) Describe possible standing sound waves in a pipe that has either open or closed
ends, and determine the wavelength and frequency of such standing waves.
4. Superposition
Students should understand the principle of superposition, so they can apply it to
traveling waves moving in opposite directions, and describe how a standing wave
may be formed by superposition.
Read and take notes on pages 372-376 in
Conceptual Physics Text
Or
Read and take note on Physics Classroom WAVES Lesson 1
B and C
Watch either
Link to Bright Storm on Wave Characteristics
Or
Khan Academy on Wave Characteristics
(start to Minute 6:12)
Read and take notes on
pages 443-445 in
College Physics Text
Wave Motion
• A wave is the motion of a disturbance
• Mechanical waves require
– Some source of disturbance
– A medium that can be disturbed
– Some physical connection or mechanism though which
adjacent portions of the medium influence each other
• All waves carry energy and momentum
Section 13.7
Types of Waves – Traveling Waves
• Flip one end of a long
rope that is under
tension and fixed at the
other end
• The pulse travels to the
right with a definite
speed
• A disturbance of this
type is called a traveling
wave
Section 13.7
Types of Waves – Transverse
• In a transverse wave, each element that is disturbed
moves in a direction perpendicular to the wave
motion
Section 13.7
Types of Waves – Longitudinal
• In a longitudinal wave, the elements of the medium
undergo displacements parallel to the motion of the
wave
• A longitudinal wave is also called a compression wave
Section 13.7
Other Types of Waves
• Waves may be a combination of transverse
and longitudinal
• A soliton consists of a solitary wave front that
propagates in isolation
– First studied by John Scott Russell in 1849
– Now used widely to model physical phenomena
Section 13.7
Khan Academy on Frequency, Period,
Wave Speed of a Periodic Wave
(Long but very informative)
Waveform – A Picture of a Wave
• The brown curve is a
“snapshot” of the wave
at some instant in time
• The blue curve is later
in time
• The high points are
crests of the wave
• The low points are
troughs of the wave
Section 13.7
Longitudinal Wave Represented as a Sine Curve
• A longitudinal wave can also be represented as a sine
curve
• Compressions correspond to crests and stretches
correspond to troughs
• Also called density waves or pressure waves
Section 13.7
Producing Waves
Section 13.8
Description of a Wave
• A steady stream of
pulses on a very long
string produces a
continuous wave
• The blade oscillates in
simple harmonic
motion
• Each small segment of
the string, such as P,
oscillates with simple
harmonic motion
Section 13.8
Amplitude and Wavelength
• Amplitude is the
maximum displacement
of string above the
equilibrium position
• Wavelength, λ, is the
distance between two
successive points that
behave identically
Section 13.8
EXAMPLE 13.8 A Traveling Wave
Goal Obtain information about a wave directly from its graph.
Problem A wave traveling in the positive x-direction is pictured in figure a. Find the
amplitude, wavelength, speed, and period of the wave if it has a frequency of 8.00
Hz. In figure a, Δx = 40.0 cm and Δy = 15.0 cm.
Strategy The amplitude and wavelength can be read directly from the figure: The
maximum vertical displacement is the amplitude, and the distance from one crest to
the next is the wavelength. Multiplying the wavelength by the frequency gives the
speed, whereas the period is the reciprocal of the frequency.
SOLUTION
The maximum wave displacement is the amplitude A:
A = Δy = 15.0 cm = 0.150 m
The distance from crest to crest is the wavelength:
λ = Δx = 40.0 cm = 0.400 m
Multiply the wavelength by the frequency to get the speed:
v = fλ = (8.00 Hz)(0.400 m) = 3.20 m/s
Take the reciprocal of the frequency to get the period:
1 1
= 0.125 s
T= =
f 8s
LEARN MORE
Remarks It's important not to confuse the wave with the medium it travels in. A
wave is energy transmitted through a medium; some waves, such as light waves,
don't require a medium.
Question Is the frequency of a wave affected by the wave's amplitude? (Select all
that apply.)
No. They are independent of each other.
increases frequency.
Yes. Increasing amplitude
Yes. Increasing amplitude decreases frequency.
Read and take notes on pages 376-383 in
Conceptual Physics Text
or
Read and take note on Physics Classroom WAVES Lesson 2
B, C and E
Link to Bright Storm on
Wave Speed
(Start to minute 2:10)
Speed of a Wave
• v = ƒλ
– Is derived from the basic speed equation of
distance/time
• This is a general equation that can be applied
to many types of waves
Section 13.8
Speed of a Wave on a String
• The speed on a wave stretched under some
tension, F
– m is called the linear density
• The speed depends only upon the properties
of the medium through which the disturbance
travels
Section 13.9
EXAMPLE 13.9 Sound and Light
Goal Perform elementary calculations using speed, wavelength, and frequency.
Problem A wave has a wavelength of 3.00 m. Calculate the frequency of the wave if
it is (a) a sound wave and (b) a light wave. Take the speed of sound as 343 m/s and
the speed of light as 3.00 108 m/s.
SOLUTION
(a) Find the frequency of a sound wave with λ = 3.00 m.
Solve the equation for the frequency and substitute.
(1)
v 343 m/s
= 114 Hz
f= =
λ 3.00 m
(b) Find the frequency of a light wave with λ = 3.00 m.
Substitute into Equation (1), using the speed of light for c.
c 3.00 108 m/s
= 1.00 108 Hz
f= =
3.00 m
λ
LEARN MORE
Remarks The same equation can be used to find the frequency in each case, despite
the great difference between the physical phenomena. Notice how much larger
frequencies of light waves are than frequencies of sound waves.
Question A wave in one medium encounters a new medium and enters it. Which of
the following wave properties will be affected in this process? (Select all that apply.)
wavelength
speed
frequency
Read and take notes on
pages 447in
College Physics Text
EXAMPLE 13.10 A Pulse Traveling on a String
The tension F in the string is maintained by the suspended block. The wave speed is given by the expression v = √ F/m .
Goal Calculate the speed of a wave on a string.
Problem A uniform string has a mass M of 0.0300 kg and a length L of 6.00 m.
Tension is maintained in the string by suspending a block of mass m = 2.00 kg from
one end (see figure). (a) Find the speed of a transverse wave pulse on this string. (b)
Find the time it takes the pulse to travel from the wall to the pulley. Neglect the mass
of the hanging part of the string.
Strategy The tension F can be obtained from Newton's second law for equilibrium
applied to the block, and the mass per unit length of the string is μ = M/L. With these
quantities, the speed of the transverse pulse can be found by substitution. Part (b)
requires the formula d = vt.
SOLUTION
(a) Find the speed of the wave pulse.
Apply the second law to the block: the tension F is equal and opposite to the force of
gravity.
ΣF = F - mg = 0 → F = mg
Substitute expressions for F and μ into the equation.
v=
√
F
μ
=
√
mg
M/ L
(2.00 kg)(9.80 m/s2)
=
(0.0300 kg)/(6.00 m) =
= 62.6 m/s
√
√
19.6 N
0.00500 kg/m
(b) Find the time it takes the pulse to travel from the wall to the pulley.
Solve the distance formula for time.
d 5.00 m
= 0.0799 s
t= =
v 62.6 m/s
LEARN MORE
Remarks Don't confuse the speed of the wave on the string with the speed of the
sound wave produced by the vibrating string.
Question If the mass of the block is quadrupled, by what factor is the speed of the
wave changed?
2
If you didn’t read in the Conceptual Physics text
Read and take note on Physics
Classroom WAVES Lesson 3 A, C and D
And
Lesson 4 A, B and C
Link to Bright Storm on
Interference
Read and take notes on
pages 448-450 in
College Physics Text
Interference of Waves
• Two traveling waves can meet and pass
through each other without being destroyed
or even altered
• Waves obey the Superposition Principle
– When two or more traveling waves encounter each
other while moving through a medium, the resulting
wave is found by adding together the displacements
of the individual waves point by point
– Actually only true for waves with small amplitudes
Section 13.10
Constructive Interference
• Two waves, a and b,
have the same
frequency and
amplitude
– Are in phase
• The combined wave, c,
has the same frequency
and a greater amplitude
Section 13.10
Constructive Interference in a String
• Two pulses are traveling in opposite directions
• The net displacement when they overlap is the sum of the
displacements of the pulses
• Note that the pulses are unchanged after the interference
Section 13.10
Destructive Interference
• Two waves, a and b,
have the same
amplitude and
frequency
• One wave is inverted
relative to the other
• They are 180° out of
phase
• When they combine,
the waveforms cancel
Section 13.10
Destructive Interference in a String
• Two pulses are traveling in opposite directions
• The net displacement when they overlap is decreased since the
displacements of the pulses subtract
• Note that the pulses are unchanged after the interference
Section 13.10
Reflection of Waves –
Fixed End
• Whenever a traveling
wave reaches a
boundary, some or all of
the wave is reflected
• When it is reflected
from a fixed end, the
wave is inverted
• The shape remains the
same
Section 13.11
Reflected Wave – Free End
• When a traveling wave
reaches a boundary, all
or part of it is reflected
• When reflected from a
free end, the pulse is
not inverted
Section 13.11
Link to Webassign complete
Unit 4 A Waves including Sound
Homework # 1-5
Raymond A. Serway
Chris Vuille
Chapter Fourteen
Sound
Read and take notes on
pages 390-400 in
Conceptual Physics Text
If you didn’t read in the Conceptual Physics text Chapter26
Read and take note on Physics Classroom
SOUNDWAVES
Lesson 1 A, B, C
And
Lesson 2 A, B, C
Either watch
Link to Bright Storm on Sound Waves
Or
Khan Academy on Sound Waves
(Minute 6:12 to the end)
Read and take notes on
pages 459-463 in
College Physics Text
Sound Waves
• Sound waves are longitudinal waves
• Characteristics of sound waves will help you
understand how we hear
Introduction
Producing a Sound Wave
• Any sound wave has its source in a vibrating object
• Sound waves are longitudinal waves traveling
through a medium
• A tuning fork can be used as an example of
producing a sound wave
Section 14.1
Using a Tuning Fork to Produce a Sound
Wave
• A tuning fork will produce a
pure musical note
• As the tines vibrate, they
disturb the air near them
• As the tine swings to the
right, it forces the air
molecules near it closer
together
• This produces a high density
area in the air
– This is an area of compression
Section 14.1
Using a Tuning Fork, cont.
• As the tine moves
toward the left, the air
molecules to the right
of the tine spread out
• This produces an area
of low density
– This area is called a
rarefaction
Section 14.1
Using a Tuning Fork, final
• As the tuning fork continues to vibrate, a succession of
compressions and rarefactions spread out from the fork
• A sinusoidal curve can be used to represent the longitudinal
wave
– Crests correspond to compressions and troughs to rarefactions
Section 14.1
Categories of Sound Waves
• Audible waves
– Lay within the normal range of hearing of the human ear
– Normally between 20 Hz to 20 000 Hz
• Infrasonic waves
– Frequencies are below the audible range
– Earthquakes are an example
• Ultrasonic waves
– Frequencies are above the audible range
– Dog whistles are an example
Section 14.2
Applications of Ultrasound
• Can be used to produce images of small objects
• Widely used as a diagnostic and treatment tool in
medicine
– Ultrasonic flow meter to measure blood flow
– May use piezoelectric devices that transform electrical energy into
mechanical energy
• Reversible: mechanical to electrical
– Ultrasounds to observe babies in the womb
– Cavitron Ultrasonic Surgical Aspirator (CUSA) used to surgically remove
brain tumors
– High-intensity Focused Ultrasound (HIFU) is also used for brain surgery
• Ultrasonic ranging unit for cameras
Section 14.2
Speed of Sound in a Liquid
• In a fluid, the speed depends on the fluid’s
compressibility and inertia
– B is the Bulk Modulus of the liquid
– ρ is the density of the liquid
– Compares with the equation for a transverse wave on a
string
Section 14.3
Speed of Sound, General
• The speed of sound is higher in solids than in gases
– The molecules in a solid interact more strongly
• The speed is slower in liquids than in solids
– Liquids are more compressible
Section 14.3
Speed of Sound in a Solid Rod
• The speed depends on the rod’s compressibility and
inertial properties
– Y is the Young’s Modulus of the material
– ρ is the density of the material
Section 14.3
Speed of Sound in Air
• 331 m/s is the speed of sound at 0° C
• T is the absolute temperature
Section 14.3
If you didn’t read in the Conceptual Physics text Chapter26
Read and take note on Physics Classroom
SOUNDWAVES
Lesson 3 A, B, C
Link to Bright Storm on
Wave Intensity
Intensity of Sound Waves
• The average intensity I of a wave on a given surface is defined
as the rate at which the energy flows through the surface, ΔE
/Δt divided by the surface area, A
• The direction of energy flow is perpendicular to the surface at
every point
• The rate of energy transfer is the power
• SI unit: W/m2
Section 14.4
Various Intensities of Sound
• Threshold of hearing
– Faintest sound most humans can hear
– About 1 x 10-12 W/m2
• Threshold of pain
– Loudest sound most humans can tolerate
– About 1 W/m2
• The ear is a very sensitive detector of sound waves
– It can detect pressure fluctuations as small as about 3
parts in 1010
Section 14.4
Link to Bright Storm on
Decibel Scale
Intensity Level of Sound Waves
• The sensation of loudness is logarithmic in the
human ear
• β is the intensity level or the decibel level of
the sound
• Io is the threshold of hearing
Section 14.4
Intensity vs. Intensity Level
• Intensity is a physical quantity
• Intensity level is a convenient mathematical
transformation of intensity to a logarithmic
scale
Section 14.4
Various Intensity Levels
•
•
•
•
Threshold of hearing is 0 dB
Threshold of pain is 120 dB
Jet airplanes are about 150 dB
Table 14.2 lists intensity levels of various
sounds
– Multiplying a given intensity by 10 adds 10 dB to
the intensity level
Section 14.4
Read and take notes on
pages 464-465 in
College Physics Text
EXAMPLE 14.2 A Noisy Grinding Machine
Goal Working with watts and decibels.
Problem A noisy grinding machine in a factory produces a sound intensity of 1.00 10-5 W/m2.
Calculate (a) the decibel level of this machine and (b) the new intensity level when a second,
identical machine is added to the factory. (c) A certain number of additional such machines are
put into operation alongside these two machines. When all the machines are running at the same
time the decibel level is 77.0 dB. Find the sound intensity.
Strategy Parts (a) and (b) require substituting into the decibel formula, with the intensity in part
(b) twice the intensity in part (a). In part (c), the intensity level in decibels is given, and it's
necessary to work backwards, using the inverse of the logarithm function, to get the intensity in
watts per meter squared.
SOLUTION
(a) Calculate the intensity level of the single grinder.
Substitute the intensity into the decibel formula.
1.00 10-5 W/m2
β = 10 log 1.00 10-12 W/m2 = 10 log (107) = 70.0 dB
(
)
(b) Calculate the new intensity level when an additional machine is added.
Substitute twice the intensity of part (a) into the decibel formula.
β = 10 log
(
2.00 10-5 W/m2
1.00 10-12 W/m2
) = 73.0 dB
(c) Find the intensity corresponding to an intensity level of 77.0 dB.
Substitute 77.0 dB into the decibel formula and divide both sides by 10:
β = 77.0 dB = 10 log (I/I0)
7.70 = log
( 10
I
)
W/m2
Make each side the exponent of 10. On the right-hand side, 10log(u) = u, by definition of
base 10 logarithms.
7.70
10
-12
7
= 5.01 10 =
I= 5.01 10-5 W/m2
I
1.00 10-12 W/m2
LEARN MORE
Remarks The answer is five times the intensity of the single grinder, so in part (c) there
are five such machines operating simultaneously. Because of the logarithmic definition
of intensity level, large changes in intensity correspond to small changes in intensity
level.
Question By how many decibels is the sound intensity level raised when the sound
intensity is doubled?
3 dB
Read and take notes on
pages 466-467 in
College Physics Text
Spherical Waves
• A spherical wave
propagates radially
outward from the
oscillating sphere
• The energy propagates
equally in all directions
• The intensity is
Section 14.5
Intensity of a Point Source
• Since the intensity varies as 1/r2, this is an inverse
square relationship
• The average power is the same through any spherical
surface centered on the source
• To compare intensities at two locations, the inverse
square relationship can be used
Section 14.5
Representations of Waves
• Wave fronts are the
concentric arcs
– The distance between
successive wave fronts is
the wavelength
• Rays are the radial lines
pointing out from the
source and
perpendicular to the
wave fronts
Section 14.5
Plane Wave
• Far away from the
source, the wave fronts
are nearly parallel
planes
• The rays are nearly
parallel lines
• A small segment of the
wave front is
approximately a plane
wave
Section 14.5
Plane Waves, cont
• Any small portion of a
spherical wave that is
far from the source can
be considered a plane
wave
• This shows a plane
wave moving in the
positive x direction
– The wave fronts are
parallel to the plane
containing the y- and zaxes
Section 14.5
EXAMPLE 14.3 Intensity Variations of a Point Source
Goal Relate sound intensities and their distances from a point source.
Problem A small source emits sound waves with a power output of 80.0 W. (a) Find the
intensity 3.00 m from the source. (b) At what distance would the intensity be one-fourth
as much as it is at r = 3.00 m? (c) Find the distance at which the sound level is 40.0 dB.
Strategy The source is small, so the emitted waves are spherical and the intensity in part
(a) can be found by substituting values. Part (b) involves solving for r, followed by
substitution. In part (c), convert from the sound intensity level to the intensity in W/m 2.
Then substitute and solve for r2.
SOLUTION
(a) Find the intensity 3.00 m from the source.
Substitute av = 80.0 W and r = 3.00 m.
80.0 W
av
=
= 0.707 W/m2
I=
2
2
4πr 4π(3.00 m)
(b) At what distance would the intensity be one-fourth as much as it is at r = 3.00 m?
Take I = (0.707 W/m2)/4, and solve for r.
1/2
( ) =[
av
r = 4πI
80.0 W
4π(0.707 W/m2)/4.0
1/2
]
= 6.00 m
(c) Find the distance at which the sound level is 40.0 dB.
Convert the intensity level of 40.0 dB to an intensity in W/m2 by solving for I.
40.0 = 10 log (I/I0) → 4.00 = log (I/I0)
104.00 = I/I0 → I = 104.00I0 = 1.00 10-8 W/m2
Solve for r22, substitute the intensity and the result of part (a), and take the square root:
I1
I2
=
r22
→ r22 = r12
r12
r2 = (3.00 m)
2
2
r2 = 2.52 104 m
(
I1
I2
0.707 W/m2
1.00 10-8 W/m2
)
LEARN MORE
Remarks Once the intensity is known at one position a certain distance away from the
source, it's easier to use the intensity ratio equation rather than the inverse square
equation to find the intensity at any other location. This is particularly true for part (b),
where, using intensity square equation, we can see right away that doubling the distance
reduces the intensity to one fourth its previous value.
Question The power output of a sound system is increased by a factor of 25. By what
factor should you adjust your distance from the speakers so the sound intensity is the
same?
5
Read and take notes on
pages 468-470 in
College Physics Text
Link to Bright Storm on
Doppler Effect
Link to Khan Academy on Doppler Effect (Khan Academy not required)
Doppler Effect
• A Doppler effect is experienced whenever
there is relative motion between a source of
waves and an observer.
– When the source and the observer are moving
toward each other, the observer hears a higher
frequency
– When the source and the observer are moving
away from each other, the observer hears a lower
frequency
Section 14.6
Doppler Effect, cont.
• Although the Doppler Effect is commonly
experienced with sound waves, it is a
phenomena common to all waves
• Assumptions:
– The air is stationary
– All speed measurements are made relative to the
stationary medium
Section 14.6
Doppler Effect, Case 1
(Observer Toward Source)
• An observer is moving
toward a stationary
source
• Due to his movement,
the observer detects an
additional number of
wave fronts
• The frequency heard is
increased
Section 14.6
Doppler Effect, Case 1
(Observer Away from Source)
• An observer is moving
away from a stationary
source
• The observer detects
fewer wave fronts per
second
• The frequency appears
lower
Section 14.6
Doppler Effect, Case 1– Equation
• When moving toward the stationary source, the
observed frequency is
• When moving away from the stationary source,
substitute –vo for vo in the above equation
Section 14.6
Doppler Effect, Case 2 (Source in Motion)
• As the source moves
toward the observer
(A), the wavelength
appears shorter and the
frequency increases
• As the source moves
away from the observer
(B), the wavelength
appears longer and the
frequency appears to be
lower
Section 14.6
Doppler Effect, Source Moving – Equation
• Use the –vs when the source is moving toward
the observer and +vs when the source is
moving away from the observer
Section 14.6
Doppler Effect, General Case
• Both the source and the observer could be moving
• Use positive values of vo and vs if the motion is
toward
– Frequency appears higher
• Use negative values of vo and vs if the motion is away
– Frequency appears lower
Section 14.6
Doppler Effect, Final Notes
• The Doppler Effect does not depend on
distance
– As you get closer, the intensity will increase
– The apparent frequency will not change
Section 14.6
EXAMPLE 14.4 Listen, but Don't Stand on the Track
Goal Solve a Doppler shift problem when only the source is moving.
Problem A train moving at a speed of 40.0 m/s sounds its whistle, which has a
frequency of 5.00 102 Hz. Determine the frequency heard by a stationary observer as
the train approaches the observer. The ambient temperature is 24.0°C.
Strategy Get the speed of sound at the ambient temperature using the relevant equation,
then substitute values into the Doppler shift equation. Because the train approaches the
observer, the observed frequency will be larger. Choose the sign of vS to reflect this fact.
SOLUTION
Calculate the speed of sound in air at T = 24.0°C.
v = (331 m/s) √ T/ 273 K
= (331 m/s) √(273+ 24.0) K/ 273 K = 345 m/s
The observer is stationary, so vO = 0. The train is moving toward the observer, so vS =
40.0 m/s (positive). Substitute these values and the speed of sound into the Doppler shift
equation.
345 m/s
v + vO
2
= (5.00 10 Hz)
fO= fS
345 m/s - 40.0 m/s
v - vS
= 566 Hz
(
)
(
)
LEARN MORE
Remarks If the train were going away from the observer, vS = -40.0 m/s would have
been chosen instead.
Question Does the Doppler shift change due to temperature variations? If so, why? For
typical daily variations in temperature in a moderate climate, would any change in the
Doppler shift be best characterized as nonexistent, small, or large?
No. Temperature does not enter into the calculation of the Doppler shift.
Yes. A change in temperature changes the speed of sound and the Doppler shift, causing
changes of a few percent.
Yes. A change in temperature changes the speed of sound
and the Doppler shift, easily doubling or tripling the size of the shift.
Yes. A change
in temperature changes the speed of sound and the Doppler shift, but only
insignificantly.
EXAMPLE 14.5 The Noisy Siren
Goal Solve a Doppler shift problem when both the source and observer are moving.
Problem An ambulance travels down a highway at a speed of 75.0 mi/h, its siren
emitting sound at a frequency of 4.00 102 Hz. What frequency is heard by a passenger
in a car traveling at 55.0 mi/h in the opposite direction as the car and ambulance (a)
approach each other and (b) pass and move away from each other? Take the speed of
sound in air to be v = 345 m/s.
Strategy Aside from converting mi/h to m/s, this problem only requires substitution into
the Doppler formula, but two signs must be chosen correctly in each part. In part (a) the
observer moves toward the source and the source moves toward the observer, so both vO
and vS should be chosen to be positive. Switch signs after they pass each other.
SOLUTION
Convert the speeds from mi/h to m/s.
0.447 m/s
= 33.5 m/s
vS = (75.0 mi/h)
1.00 mi/h
0.447 m/s
= 24.6 m/s
vO = (55.0 mi/h)
1.00 mi/h
(a) Compute the observed frequency as the ambulance and car approach each other.
Each vehicle goes toward the other, so substitute vO = +24.6 m/s and vS = +33.5 m/s into
the Doppler shift formula.
v + vO
fO = fS
v - vS
345 m/s + 24.6 m/s
= (4.00 102 Hz)
= 475 Hz
345 m/s - 33.5 m/s
(
)
(
(
)
(
)
)
(b) Compute the observed frequency as the ambulance and car recede from each other.
Each vehicle goes away from the other, so substitute vO = -24.6 m/s and vS = -33.5 m/s
into the Doppler shift formula.
fO = fS
(
= (4.00
v + vO
v - vS
)
345 m/s + (-24.6 m/s)
10 Hz) (
= 339 Hz
345 m/s - (-33.5 m/s) )
2
LEARN MORE
Remarks Notice how the signs were handled. In part (b) the negative signs were
required on the speeds because both observer and source were moving away from each
other. Sometimes, of course, one of the speeds is negative and the other is positive.
Question Is the Doppler shift affected by sound intensity level?
Yes, higher intensity and amplitude pushes the wave fronts closer together.
Yes,
higher intensity and amplitude pushes the wave fronts farther apart.
No, the
effect depends on how the motion of source or observer affects the number of wave
fronts passing the observer each second.
Sometimes. It depends on whether it is the
observer, the medium, or the source that is moving.
Read and take notes on
pages 472-474 in
College Physics Text
Shock Waves
• A shock wave results
when the source
velocity exceeds the
speed of the wave itself
• The circles represent
the wave fronts emitted
by the source
Section 14.6
Shock Waves, cont
• Tangent lines are drawn from Sn to the wave
front centered on So
• The angle between one of these tangent
lines and the direction of travel is given by
sin θ = v / vs
• The ratio vs /v is called the Mach Number
• The conical wave front is the shock wave
Section 14.6
Shock Waves, final
• Shock waves carry
energy concentrated on
the surface of the cone,
with correspondingly
great pressure
variations
• A jet produces a shock
wave seen as a fog of
water vapor
Section 14.6
Interference of Sound Waves
• Sound waves interfere
– Constructive interference occurs when the path
difference between two waves’ motion is zero or
some integer multiple of wavelengths
• Path difference = nλ (n = 0, 1, 2, … )
– Destructive interference occurs when the path
difference between two waves’ motion is an odd
half wavelength
• Path difference = (n + ½)λ (n = 0, 1, 2, … )
Section 14.7
EXAMPLE 14.6 Two Speakers Driven by the Same Source
Two loudspeakers driven by the same source can produce interference.
Goal Use the concept of interference to compute a frequency.
Problem Two speakers placed 3.00 m apart are driven by the same oscillator (see
figure). A listener is originally at point O, which is located 8.00 m from the center of the
line connecting the two speakers. The listener then walks to point P, which is a
perpendicular distance 0.350 m from O, before reaching the first minimum in sound
intensity. What is the frequency of the oscillator? Take the speed of sound in air to be vs
= 343 m/s.
Strategy The position of the first minimum in sound intensity is given, which is a point
of destructive interference. We can find the path lengths r1 and r2 with the Pythagorean
theorem and then use the equation for destructive interference to find the wavelength λ.
Using v = fλ then yields the frequency.
SOLUTION
Use the Pythagorean theorem to find the path lengths r1 and r2.
r1 = √(8.00 m)2 + (1.15 m)2 = 8.08 m
r2 = √(8.00 m)2 + (1.85 m)2 = 8.21 m
Substitute these values and n = 0 into the equation for destructive interference, solving
for the wavelength.
r2 - r1 = (n + ½)λ
8.21 m - 8.08 m = 0.13 m = λ/2 → λ = 0.26 m
Solve v = λf for the frequency f and substitute the speed of sound and the wavelength.
v 343 m/s
= 1.3kHz
f= =
0.26
m
λ
LEARN MORE
Remarks For problems involving constructive interference, the only difference is that
Equation r2 - r1 = nλ, would be used instead of the equation used above
Question True or False: In the same context, smaller wavelengths of sound would create
more interference maxima and minima than longer wavelengths.
False. The distance between minima increases with decreasing wavelength.
True. The distance between minima increases with decreasing wavelength.
The distance between minima decreases with decreasing wavelength.
distance between minima has no dependence on wavelength.
between minima decreases with decreasing wavelength.
False.
False. The
True. The distance
Link to Bright Storm on
Standing Waves
Read and take notes on
pages 475-477 in
College Physics Text
Standing Waves
• When a traveling wave reflects back on itself,
it creates traveling waves in both directions
• The wave and its reflection interfere according
to the superposition principle
• With exactly the right frequency, the wave will
appear to stand still
– This is called a standing wave
Section 14.8
Standing Waves, cont
• A node occurs where the two traveling waves have
the same magnitude of displacement, but the
displacements are in opposite directions
– Net displacement is zero at that point
– The distance between two nodes is ½λ
• An antinode occurs where the standing wave vibrates
at maximum amplitude
Section 14.8
Standing Waves on a String
• Nodes must occur at the ends of the string because
these points are fixed
Section 14.8
Standing Waves, cont.
• The pink arrows
indicate the direction of
motion of the parts of
the string
• All points on the string
oscillate together
vertically with the same
frequency, but different
points have different
amplitudes of motion
Section 14.8
Standing Waves on a String, final
• The lowest frequency of
vibration (b) is called
the fundamental
frequency (ƒ1)
• Higher harmonics are
positive integer
multiples of the
fundamental
Section 14.8
Standing Waves on a String – Frequencies
• ƒ1, ƒ2, ƒ3 form a harmonic series
– ƒ1 is the fundamental and also the first harmonic
– ƒ2 is the second harmonic or the first overtone
• Waves in the string that are not in the harmonic
series are quickly damped out
– In effect, when the string is disturbed, it “selects” the
standing wave frequencies
Section 14.8
If you didn’t read in the Conceptual Physics text Chapter26
Read and take note on Physics Classroom
SOUNDWAVES
Lesson 4 A, B, C
And
Lesson 5 A, B, C, D
Read and take notes on
pages 479-480 in
College Physics Text
Forced Vibrations
• A system with a driving force will cause a
vibration at the frequency of the driving force
• When the frequency of the driving force
equals the natural frequency of the system,
the system is said to be in resonance
Section 14.9
An Example of Resonance
• Pendulum A is set in
motion
• The others begin to
vibrate due to the
vibrations in the flexible
beam
• Pendulum C oscillates at
the greatest amplitude
since its length, and
therefore its natural
frequency, matches that
of A
Section 14.9
Other Examples of Resonance
• Child being pushed on a swing
• Shattering glasses
• Tacoma Narrows Bridge collapse due to
oscillations caused by the wind
• Upper deck of the Nimitz Freeway collapse
due to the Loma Prieta earthquake
Section 14.9
Link to Bright Storm on
Standing Waves in Air
Columns
Read and take notes on
pages 480-482 in
College Physics Text
Standing Waves in Air Columns
• If one end of the air column is closed, a node
must exist at this end since the movement of
the air is restricted
• If the end is open, the elements of the air
have complete freedom of movement and an
antinode exists
Section 14.10
Tube Open at Both Ends
Section 14.10
Resonance in Air Column Open at Both
Ends
• In a pipe open at both ends, the natural
frequency of vibration forms a series whose
harmonics are equal to integral multiples of
the fundamental frequency
Section 14.10
Tube Closed at One End
Section 14.10
Resonance in an Air Column Closed at One
End
• The closed end must be a node
• The open end is an antinode
• There are no even multiples of the
fundamental harmonic
Section 14.10
Read and take notes on
pages 484-485 in
College Physics Text
Beats
• Beats are alternations in loudness, due to
interference
• Waves have slightly different frequencies and
the time between constructive and
destructive interference alternates
• The beat frequency equals the difference in
frequency between the two sources:
Section 14.11
Beats, cont.
Section 14.11
Read and take notes on
pages 486-489 in
College Physics Text
Quality of Sound – Tuning Fork
• Tuning fork produces
only the fundamental
frequency
Section 14.12
Quality of Sound – Flute
• The same note played
on a flute sounds
differently
• The second harmonic is
very strong
• The fourth harmonic is
close in strength to the
first
Section 14.12
Quality of Sound – Clarinet
• The fifth harmonic is
very strong
• The first and fourth
harmonics are very
similar, with the third
being close to them
Section 14.12
Timbre
• In music, the characteristic sound of any
instrument is referred to as the quality of
sound, or the timbre, of the sound
• The quality depends on the mixture of
harmonics in the sound
Section 14.12
Pitch
• Pitch is related mainly, although not
completely, to the frequency of the sound
• Pitch is not a physical property of the sound
• Frequency is the stimulus and pitch is the
response
– It is a psychological reaction that allows humans
to place the sound on a scale
Section 14.12
The Ear
• The outer ear consists
of the ear canal that
terminates at the
eardrum
• Just behind the
eardrum is the middle
ear
• The bones in the middle
ear transmit sounds to
the inner ear
Section 14.13
Frequency Response Curves
• Bottom curve is the
threshold of hearing
– Threshold of hearing is
strongly dependent on
frequency
– Easiest frequency to hear is
about 3300 Hz
• When the sound is loud
(top curve, threshold of
pain) all frequencies can be
heard equally well
Section 14.13
Link to more advanced explanation
about waves with Walter Lewin
(Not Required)
Link to Webassign
Unit 4 A Waves including Sound
Discussion Questions
Link to Webassign complete
Unit 4 A Waves including Sound
Homework # 6-12
Grading Rubric for Unit 4 A Waves including Sound
Name: ______________________
Conceptual notes from Physics Classroom
Waves
Sound waves
Lesson
Lesson
Lesson
Lesson
1 b, c -------------------------------------------------____
2 b, c, e ----------------------------------------------____
3 a, c, d ----------------------------------------------____
4 a, b, c ----------------------------------------------____
Lesson 1 a, b, c ----------------------------------------------____
Lesson 2 a, b, c ----------------------------------------------____
Lesson 3 a, b, c ----------------------------------------------____
Lesson 4 a, b, c ----------------------------------------------____
Lesson 5 a, b, c, d -------------------------------------------____
Conceptual Physics Text
Pgs 372-376------------------------------------------------------_____
Pgs 376-383------------------------------------------------------_____
Pgs 390-400------------------------------------------------------_____
Advanced notes from text book:
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
Pgs
443-445 -----------------------------------------------------_____
447 ------------------------------------------------------------_____
448-450-------------------------------------------------------_____
459-463------------------------------------------------------_____
464-465-------------------------------------------------------_____
466-467-------------------------------------------------------_____
468-470------------------------------------------------------_____
472-474-------------------------------------------------------_____
475-477-------------------------------------------------------_____
479-480------------------------------------------------------_____
480-482-------------------------------------------------------_____
484-485-------------------------------------------------------_____
Example Problems and Questions:
13.8 (1-4) ----------------------------------------------------------_____
13.9 (a-b) ------ ---------------------------------------------------_____
13.10 (a-b) ----------- --------------------------------------------_____
14.2 (a-c) --- ------------------------------------------------------_____
14.3 (a-c) ----------------------------------------------------------_____
14.4 (1-2) ----------------------------------------------------------_____
14.5 (a-b) ----------------------------------------------------------_____
14.6 (a) -------------------------------------------------------------_____
Web Assign 1 (a-d), 2 (a-b), 3, 4 (a-b), 5 (a-b), 6, 7(a-b), 8(a-b), 9(a-b), 10(a-b), 11(a-c), 12(a-b) --___