Raymond A. SerwayChris Vuille

Chapter ThirteenVibrations 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 C

lassroom 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:

T = 1

= 1

= 0.125 s f 8 s

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. Yes. Increasing amplitude

increases frequency. Yes. Increasing amplitude decreases frequency.

Read and take notes on pages 376-383 in

Conceptual Physics Text

orRead 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)

f = v

= 343 m/s

= 114 Hz 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.

f = c

= 3.00 108 m/s

= 1.00 108 Hz λ 3.00 m

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

= √ mg

μ M/L

= √ (2.00 kg)(9.80 m/s2)

= √ 19.6 N

(0.0300 kg)/(6.00 m) 0.00500 kg/m = 62.6 m/s

(b) Find the time it takes the pulse to travel from the wall to the pulley.

Solve the distance formula for time.

t = d

= 5.00 m

= 0.0799 s 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 DAnd

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 SoundHomework # 1-5

Raymond A. SerwayChris Vuille

Chapter FourteenSound

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

AndLesson 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.

β = 10 log ( 1.00 10-5 W/m2 ) = 10 log (107) = 70.0 dB 1.00 10-12 W/m2

(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 ) = 73.0 dB 1.00 10-12 W/m2

(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 ( I ) 10-12 W/m2

Make each side the exponent of 10. On the right-hand side, 10log(u) = u, by definition of base 10 logarithms.

107.70 = 5.01 107 = I

1.00 10-12 W/m2

I= 5.01 10-5 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 z-axes

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/m2. 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.

I = av

= 80.0 W

= 0.707 W/m2 4πr2 4π(3.00 m)2

(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.

r = ( av )

1/2 = [

80.0 W ]

1/2 = 6.00 m

4πI 4π(0.707 W/m2)/4.0

(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 =

r22 → r2

2 = r12

I1

I2 r12 I2

r22 = (3.00 m)2 (

0.707 W/m2 ) 1.00 10-8 W/m2

r2 = 2.52 104 m

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 EffectLink 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.

fO= fS ( v + vO ) = (5.00 102 Hz) (

345 m/s ) v - vS 345 m/s - 40.0 m/s = 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.

vS = (75.0 mi/h) ( 0.447 m/s ) = 33.5 m/s 1.00 mi/h

vO = (55.0 mi/h) ( 0.447 m/s ) = 24.6 m/s 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.

fO = fS ( v + vO

) v - vS

= (4.00 102 Hz) ( 345 m/s + 24.6 m/s ) = 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 ( v + vO

) v - vS

= (4.00 102 Hz) ( 345 m/s + (-24.6 m/s) ) = 339 Hz 345 m/s - (-33.5 m/s)

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.

f = v

= 343 m/s

= 1.3kHz λ 0.26 m

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. False.

The distance between minima decreases with decreasing wavelength. False. The

distance between minima has no dependence on wavelength. True. The distance between minima decreases with decreasing wavelength.

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 SoundDiscussion 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 Lesson 1 b, c -------------------------------------------------____ Lesson 2 b, c, e ----------------------------------------------____ Lesson 3 a, c, d ----------------------------------------------____ Lesson 4 a, b, c ----------------------------------------------____ Sound waves 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 443-445 -----------------------------------------------------_____ Pgs 447 ------------------------------------------------------------_____ Pgs 448-450-------------------------------------------------------_____ Pgs 459-463------------------------------------------------------_____ Pgs 464-465-------------------------------------------------------_____ Pgs 466-467-------------------------------------------------------_____ Pgs 468-470------------------------------------------------------_____ Pgs 472-474-------------------------------------------------------_____ Pgs 475-477-------------------------------------------------------_____ Pgs 479-480------------------------------------------------------_____ Pgs 480-482-------------------------------------------------------_____ Pgs 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) --___