Sound Waves• Sound waves are longitudinal waves traveling

through a medium• Vibrations are the source of all sound waves• Sound detectors (such as the ear drum) also

depend on vibrations induced from sound waves• Sound waves result from pressure differences in the

medium in which they travel• A vibrating tuning fork produces

sound waves when a vibrating tine creates alternating regions of high (compression) and low (rarefaction) molecular density and pressure

Sound Waves• The sound wave propagates when a succession of

compressions and rarefactions form and spread out from the source– Alternating regions of force due to air pressure differences– Remember longitudinal waves on a Slinky

(Tuning fork)

(Slinky) Creating Sound Waves

Frequencies of Sound Waves• Audible range lies within sensitivity of the human ear

– About 20 Hz to 20 kHz– Realistically, the sensitivity of the ear declines rapidly below

100 Hz and above 10 kHz

• Infrasonic range lies below audible range (< 20 Hz)– Seismic P waves are an example

• Ultrasonic range lies above audible range (> 20 kHz)• Audible ranges for animals can be quite different

– Dogs can hear frequencies as high as 50 kHz– Dolphins make use of frequencies as high as 250 kHz– Elephants communicate over long distances (up to 4 km)

using frequencies as low as 14 Hz– A rhinoceros uses frequencies down to 10 Hz

Applications of Sound Waves• Echolocation

– Bats, dolphins, whales, and some birds emit sound waves to locate prey and to “see” their environment

– Oilbirds of northern South America and cave swiftlets of Borneo and East Asia use sound waves to find their way around dark caves (echo return time yields information about where obstacles are)

– Sonar (sound navigation and ranging) used to determine depth of water below a ship (some autofocus cameras use same concept)

– Low-frequency active (LFA) sonar used for long-range detection of submarines

– Note that radar is very similar to sonar except electromagnetic waves (rather than sound) are used

Applications of Sound Waves• Medical applications of ultrasound

– Benefits of ultrasound: not damaging to tissue like x rays; images captured in real time; can resolve details at different depths; some organs (liver, spleen) detected which cannot be viewed with x rays

– Higher frequencies mean smaller wavelengths, so more detail can be seen in the images (wavelength ~ smallest detail that can be resolved)

– Prenatal care– Examination of organs– Diagnosis of heart conditions– Assessment of heart valve function and monitoring of

blood flow in large blood vessels– Guide to biopsies and other procedures (due to real-time

imaging)

Sound Intensity• As the sound wave moves, it carries energy

– Remember movie of object bobbing up and down in wave tank

• The intensity of a wave = rate energy flows per unit area perpendicular to direction of travel:– Faintest (loudest) sounds we can hear have

I = 1 10–12 W/m2 (1 W/m2)– Correspond to threshold of hearing and threshold of pain

• The ear is a very sensitive detector of sound waves– Can detect pressure fluctuations as small as about 3

parts in 1010

– Also covers enormous range of intensities

A

PI

Area

Power

Intensity Level in Decibels• The sensation of loudness is logarithmic in our ears• Relative intensity of sound is called the intensity

level or decibel level:

– Threshold of hearing = 0 dB– Threshold of pain = 120 dB– 90 dB is the occupational safety limit– Two rules of thumb: (1) When intensity increases by

factor of 10, intensity level adds 10 dB; (2) Adding 3.0 dB to intensity level doubles intensity (since log102 = 0.30)

0

log10I

I

(Itot = I1 + I2 + …)

(tot …)

(I0 = 1.0 10–12 W/m2 = reference intensity at the threshold of hearing)

Sound Intensity• Sound intensity drops off rapidly with distance

(as r –2)

(see http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 for a sound intensity “calculator”) (graphics from website given on this page)

CQ 1: If the intensity of a sound is doubled, the decibel level will increase by:

A) less than 10 dB. B) exactly 10 dB.C) more than 10 dB.D) exactly 20 dB.

Example Problem #14.18

Solution (details given in class):

(a) 1.32×10–4 W/m2

(b) 81.2 dB

A family ice show is held at an enclosed arena. The skaters perform to music playing at a level of 80.0 dB. This intensity level is too loud for your baby, who yells at 75.0 dB. (a) What total sound intensity engulfs you? (b) What is the combined sound level?

The Doppler Effect• Shift in frequency detected from a (quickly) moving

car:

• The opposite shift in frequency would be heard if the car were moving toward the observer

v

Sound waves

Observer

S

From perspective of car:

Sound waves

The Doppler Effect• From perspective of observer:

OCar at pt. Av

Car at pt. Bv

Car at pt. C

vObserver

Doppler Effect Stationary Observers

Doppler Effect Moving Source and Observer

Video Demo

OO

vf

The Doppler Effect• The observed wavelength O is smaller (larger)

when the source is moving toward (away) from the observer– Since f = v / , the observed frequency is larger (smaller)

when the source is moving toward (away from) the observer

• The Doppler formula generalizes this result for a moving source and/or observer:– fO (fS) = frequency measured by

observer (source)– vO (vS) = velocity of observer (source)– Positive sense of direction is from observer to source– All velocities are measured relative to the medium (usually

air) through which the sound waves propagate• If either source (vS) or observer (vO) is stationary, insert “0” for that

value into the formula

S

OSO vv

vvff (v = speed

of sound)

CQ 2: Which of the following factors by itself will increase the frequency at which an observer hears a sound emanating from a source?

A) A wind blows from the source to the observer.

B) The source and the observer move away from each other at the same speed.

C) The source and the observer move in the same direction at the same speed.

D) The source moves away from the observer more slowly than the observer moves toward the source.

CQ 3: Interactive Example Problem: The Happy Whistling Jogger

(ActivPhysics Online Exercise #10.9, copyright Addison Wesley publishing)

A) 960 Hz B) 1000 HzC) 1042 HzD) 1057 Hz

What frequency sound do you hear as you and the jogger are moving toward each other?

Example Problem #14.73

Solution (details given in class):439 Hz and 441 Hz

A block with a speaker bolted to it is connected to a spring having spring constant k = 20.0 N/m, as shown below. The total mass of the block and speaker is 5.00 kg, and the amplitude of the unit’s motion is 0.500 m. If the speaker emits sound waves of frequency 440 Hz, determine the lowest and highest frequencies heard by the person to the right of the speaker.

Special Cases of Doppler Effect• What happens if observer moves away from source

at the speed of sound (vO = v)?– = 0

– Sound never reaches observer!

• If vS v (speed of sound), then wave crests pile up on top of each other and a sound barrier is formed

• If vS > v, wave crests form cone-shaped shock waves– Speed of supersonic craft typically given by Mach number

(ratio of speed of craft to speed of sound)– A V-shaped bow wave from canoe similar to shock wave

• The sound of a shock wave is a sonic boom– When Space Shuttle lands, two “crack” sounds can be

heard, due to 2 shock waves created by the nose and tail

Of

(graphic from College Physics, Giambattista et al.)

Interference of Sound Waves• Sound waves from the same source (or different

sources driven by same oscillator) can interfere with each other

• Sound from speaker traveling different paths:– Constructive interference occurs if

path difference r1 – r2 is zero or some integer multiple of wavelength

– Destructive interference occurs if path difference r1 – r2 is some half-integer multiple of wavelength

• Important to remember when connecting wires between speakers and amplifier

...),2,1,0(12 nnrr

...),2,1,0(2

112

nnrr

Video Demo

Example Problem #14.36

Solution (details given in class):

1.44 m

Two loudspeakers are placed above and below one another, as shown below, and are driven by the same source at a frequency of 450 Hz. An observer is in front of the speakers (to the right) at point O, at the same distance from each speaker. If the speed of sound is 345 m/s, what minimum vertical distance upward should the top speaker be moved to create destructive interference at point O?

Standing Waves• A wave pulse is reflected at a boundary point (fixed

or free end)• A sinusoidal wave will reflect so that traveling waves

will move in both directions• If the frequency is just right, the wave pattern

doesn’t appear to move– Standing waves

• For a string fixed at both ends:– (b) represents string plucked at

midpoint– (c) and (d) represent string

displaced at multiple locations– node = zero displacement (N)– antinode = maximum displacement (A)

(distance between adjacent nodes is ½ )

Standing Waves• (b) represents the lowest

frequency of vibration for a standing wave– fundamental frequency

– L = 1 / 2 (or 1 = 2L)

– f1 = v / 1 = v / 2L

• All other frequencies (like those in (c) and (d)) are based on integer multiples of the fundamental frequency:

• Frequencies form a harmonic series– f1 = first harmonic, f2 = second harmonic, etc.

F

Lf

2

11

F

L

nnffn 21 (“natural” or “characteristic”

frequencies)

(since v = (F / )1/2 )

CQ 4: If a guitar string is 0.5 m long, what is the wavelength of its third harmonic?

A) 0.25 m B) 0.33 mC) 0.5 mD) 1 m

CQ 5: Interactive Example Problem: Tuning a Stringed Instrument

(ActivPhysics Online Exercise #10.5, copyright Addison Wesley publishing)

A) 0.4 N B) 0.8 NC) 1.6 ND) 16 N

What string tension is needed to produce a second-harmonic vibration frequency of 4.0 Hz?

Resonance• Many mechanical systems oscillate at one or more

natural frequencies– Simple harmonic motion has 1 natural frequency– Guitar strings oscillate at many natural frequencies

• Forced vibration occurs when a time-varying force causes the system to vibrate at the same frequency as the force

• If applied force is “in phase” with one of the system’s natural frequencies, large amplitude vibrations occur– “Resonance”– Example: pushing child on swing at just the right times

• Resonance can sometimes have rather destructive consequences

Standing Waves in Air Columns

f1 = v / 1 = v / 2L

f2 = v / 2 = v / L = 2f1

f3 = v / 3 = 3v / 2L = 3f1 (graphic from College Physics, Giambattista et al.)

),3,2,1(2 1 nfn

L

vnfn

(Both ends open)

Standing Waves in Air Columns

),5,3,1(4 1 nfn

L

vnfn

f1 = v / 1 = v / 4L

f2 = v / 2 = 3v / 4L = 3f1

f3 = v / 3 = 5v / 4L = 5f1

(graphic from College Physics, Giambattista et al.)

(One end closed)

Example Problem #14.54

Solution (details given in class):

f1 = 50 Hz, L = 1.70 m

Pipe is open only on one end.

Two adjacent natural frequencies of an organ pipe are found to be 550 Hz and 650 Hz. Calculate the fundamental frequency and length of this pipe. (Use v = 340 m/s.) Determine whether the pipe is open at both ends or open at only one end.

Beats• Beats are formed by the combination of 2 waves of

slightly different frequencies traveling in the same direction

• Beat frequency = difference in frequency between the 2 sources

• Listening to beat pattern is a way of tuning musical instruments